Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.2

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 9 Probability Ex 9.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Probability Ex 9.2

Question 1.
First, 6 faced die which is numbered 1 to 6 is thrown, then a 5 faced die which is numbered 1 to 5 is thrown. What is the probability that sum of the numbers on the upper faces of the dice is divisible by 2 or 3?
Solution:
When a 6 faced die and a 5 faced die are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4,4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
∴ n(S) = 30
Let event A: The sum of the numbers on the upper faces of the dice is divisible by 2.
A = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4)}
∴ n(A) = 15
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{15}{30}\)
Let event B: Sum of the numbers on the upper faces of the dice is divisible by 3.
B = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3)}
∴ n(B) = 10
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{10}{30}\)
Now,
A ∩ B = {(1, 5), (2,4), (3, 3), (4, 2), (5, 1)}
∴ n(A ∩ B) = 5
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{5}{30}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{15}{30}+\frac{10}{30}-\frac{5}{30}\)
= \(\frac{20}{30}\)
= \(\frac{2}{3}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.2

Question 2.
A card is drawn from a pack of 52 cards. What is the probability that,
(i) card is either red or black?
(ii) card is either black or a face card?
Solution:
One card can be drawn from the pack of 52 cards in \({ }^{52} \mathrm{C}_{1}\) = 52 ways.
∴ n(S) = 52
The pack of 52 cards consists of 26 red and 26 black cards.
(i) Let event A: A red card is drawn.
∴ Red card can be drawn in \({ }^{26} \mathrm{C}_{1}\) = 26ways
∴ n(A) = 26
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{26}{52}\)
Let event B: A black card is drawn.
∴ Black card can be drawn in \({ }^{26} \mathrm{C}_{1}\) = 26 ways.
∴ n(B) = 26
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{26}{52}\)
Since A and B are mutually exclusive events,
P(A ∩ B) = 0
∴ Required probability
P(A ∪ B) = P(A) + P(B)
= \(\frac{26}{52}+\frac{26}{52}\)
= 1

(ii) Let event A: A black card is drawn.
∴ Black card can be drawn in \({ }^{26} \mathrm{C}_{1}\) = 26 ways.
n(A) = 26
n(A) 26 n(S) ~ 52
Let event B: A face card is drawn.
There are 12 face cards in the pack of 52 cards.
∴ 1 face card can be drawn in \({ }^{12} \mathrm{C}_{1}\) = 12 ways.
∴ n(B) = 12
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{12}{52}\)
There are 6 black face cards.
∴ n(A ∩ B) = 6
∴ P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{6}{52}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{26}{52}+\frac{12}{52}-\frac{6}{52}\)
= \(\frac{32}{52}\)
= \(\frac{8}{13}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.2

Question 3.
A girl is preparing for National Level Entrance exam and State Level Entrance exam for professional courses. The chances of her cracking National Level exam is 0.42 and that of State Level exam is 0.54. The probability that she clears both the exams is 0.11. Find the probability that
(i) she cracks at least one of the two exams.
(ii) she cracks only one of the two.
(iii) she cracks none.
Solution:
Let event A: The girl cracks the National Level exam.
∴ P(A) = 0.42
Let event B: The girl cracks the State Level exam.
∴ P(B) = 0.54
Also, P(A ∩ B) = 0.11
(i) P(the girl cracks at least one of the two exams)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.42 + 0.54 – 0.11
= 0.85

(ii) P(the girl cracks only one of the two exams)
= P(A) – P(B) – 2P(A ∩ B)
= 0.42 + 0.54 – 2(0.11)
= 0.74

(iii) P(the girl cracks none of the exams)
= P(A’ ∩ B’)
= P(A ∪ B)’
= 1 – P(A ∪ B)
= 1 – 0.85
= 0.15

Question 4.
A bag contains 75 tickets numbered from 1 to 75. One ticket is drawn at random. Find the probability that,
(i) number on the ticket is a perfect square or divisible by 4.
(ii) number on the ticket is a prime number or greater than 40.
Solution:
Out of the 75 tickets, one ticket can be drawn in \({ }^{75} \mathrm{C}_{1}\) = 75 ways.
∴ n(S) = 75
(i) Let event A: The number on the ticket is a perfect square.
∴ A = {1, 4, 9, 16, 25, 36, 49, 64}
∴ n(A) = 8
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{8}{75}\)
Let event B: The number on the ticket is divisible by 4.
∴ B = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72}
∴ n(B) = 18
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{18}{75}\)
Now, A ∩ B = {4, 16, 36, 64}
∴ n(A ∩ B) = 4
∴ P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{4}{75}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{8}{75}+\frac{18}{75}-\frac{4}{75}\)
= \(\frac{22}{75}\)

(ii) Let event A: The number on the ticket is a prime number.
∴ A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73}
∴ n(A) = 21
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{21}{75}\)
Let event B: The number is greater than 40.
∴ B = {41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75}
∴ n(B) = 35
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{35}{75}\)
Now,
A ∩ B = {41, 43, 47, 53, 59, 61, 67, 71, 73}
∴ n(A ∩ B) = 9
∴ n(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{9}{75}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{21}{75}+\frac{35}{75}-\frac{9}{75}\)
= \(\frac{47}{75}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.2

Question 5.
The probability that a student will pass in French is 0.64, will pass in Sociology is 0.45 and will pass in both is 0.40. What is the probability that the student will pass in at least one of the two subjects?
Solution:
Let event A: The student will pass in French.
∴ P(A) = 0.64
Let event B: The student will pass in Sociology.
∴ P(B) = 0.45
Also, P(A ∩ B) = 0.40
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.64 + 0.45 – 0.40
= 0.69

Question 6.
Two fair dice are thrown. Find the probability that the number on the upper face of the first die is 3 or sum of the numbers on their upper faces is 6.
Solution:
When two dice are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let event A: The number on the upper face of the first die is 3.
∴ A = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}
∴ n(A) = 6
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{6}{36}\)
Let event B: Sum of the numbers on their upper faces is 6.
∴ B = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
∴ n(B) = 5
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{5}{36}\)
Now, A ∩ B = {(3, 3)}
∴ n(A ∩ B) = 1
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{1}{36}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{6}{36}+\frac{5}{36}-\frac{1}{36}\)
= \(\frac{10}{36}\)
= \(\frac{5}{18}\)

Question 7.
For two events A and B of a sample space S, if P(A) = \(\frac{3}{8}\), P(B) = \(\frac{1}{2}\) and P(A ∪ B) = \(\frac{5}{8}\). Find the value of the following.
(a) P(A ∩ B)
(b) P(A’ ∩ B’)
(c) P(A’ ∪ B’)
Solution:
Here, P(A) = \(\frac{3}{8}\), P(B) = \(\frac{1}{2}\), P(A ∪ B) = \(\frac{5}{8}\)
(a) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
∴ P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= \(\frac{3}{8}+\frac{1}{2}-\frac{5}{8}\)
= \(\frac{1}{4}\)

(b) P(A’ ∩ B’) = P(A ∪ B)’
= 1 – P(A ∪ B)
= 1 – \(\frac{5}{8}\)
= \(\frac{3}{8}\)

(c) P(A’ ∪ B’) = P(A ∩ B)’
= 1 – P(A ∩ B)
= 1 – \(\frac{1}{4}\)
= \(\frac{3}{4}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.2

Question 8.
For two events A and B of a sample space S, if P(A ∪ B) = \(\frac{5}{6}\), P(A ∩ B) = \(\frac{1}{3}\) and P(B’) = \(\frac{1}{3}\), then find P(A).
Solution:
Here, P(A ∪ B) = \(\frac{5}{6}\), P(A ∩ B) = \(\frac{1}{3}\), P(B’) = \(\frac{1}{3}\)
P(B) = 1 – P(B’)
= 1 – \(\frac{1}{3}\)
= \(\frac{2}{3}\)
Since P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
\(\frac{5}{6}\) = P(A) + \(\frac{2}{3}-\frac{1}{3}\)
∴ \(\frac{5}{6}\) = P(A) + \(\frac{1}{3}\)
∴ P(A) = \(\frac{5}{6}-\frac{1}{3}\) = \(\frac{1}{2}\)

Question 9.
A bag contains 5 red, 4 blue and an unknown number m of green balls. If the probability of getting both the balls green, when two balls are selected at random is \(\frac{1}{7}\), find m.
Solution:
Total number of balls in the bag = 5 + 4 + m = 9 + m
Two balls are selected from (9 + m) balls in \({ }^{9+m} \mathrm{C}_{2}\) ways.
∴ n(S) = \({ }^{9+m} \mathrm{C}_{2}\)
Let event A: The two balls selected are green.
∴ 2 balls can be selected from m balls in \({ }^{\mathrm{m}} \mathrm{C}_{2}\) ways.
∴ n(A) = \({ }^{\mathrm{m}} \mathrm{C}_{2}\)
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.2 Q9
(9 + m)(8 + m) = 7m(m – 1)
72 + 9m + 8m + m2 = 7m2 – 7m
6m2 – 24m – 72 = 0
m2 – 4m – 12 = 0
(m – 6)(m + 2) = 0
m = 6 or m = -2
Since number of balls cannot be negative, m ≠ -2
∴ m = 6

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.2

Question 10.
From a group of 4 men, 4 women and 3 children, 4 persons are selected at random. Find the probability that,
(i) no child is selected.
(ii) exactly 2 men are selected.
Solution:
The group consists of 4 men, 4 women and 3 children, i.e., 4 + 4 + 3 = 11 persons.
4 persons are to be selected from this group.
∴ 4 persons can be selected from 11 persons in \({ }^{11} \mathrm{C}_{4}\) ways.
∴ n(S) = \({ }^{11} \mathrm{C}_{4}\)
(i) Let event A: No child is selected.
∴ 4 persons can be selected from 4 men and 4 women, i.e., from 8 persons in \({ }^{8} \mathrm{C}_{4}\) ways.
∴ n(A) = \({ }^{8} \mathrm{C}_{4}\)
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.2 Q10

(ii) Let event B: Exactly 2 men are selected.
∴ 2 men are selected from 4 men in \({ }^{4} \mathrm{C}_{2}\) ways, and remaining 2 persons are selected from 7 persons (i.e., 4 women and 3 children) in \({ }^{7} \mathrm{C}_{2}\) ways.
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.2 Q10.1

Question 11.
A number is drawn at random from the numbers 1 to 50. Find the probability that it is divisible by 2 or 3 or 10.
Solution:
One number can be drawn at random from the numbers 1 to 50 in \({ }^{50} \mathrm{C}_{1}\) = 50 ways.
∴ n(S) = 50
Let event A: The number drawn is divisible by 2.
∴ A = {2, 4, 6, 8, 10, …, 48, 50}
∴ n(A) = 25
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{25}{50}\)
Let event B: The number drawn is divisible by 3.
B = {3, 6, 9, 12, …, 48}
∴ n(B) = 16
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{16}{50}\)
Let event C: The number drawn is divisible by 10.
C = {10, 20, 30, 40, 50}
∴ n(C) = 5
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{5}{50}\)
Now, A ∩ B = {6, 12, 18, 24, 30, 36, 42, 48}
∴ n(A ∩ B) = 8
∴ P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{8}{50}\)
B ∩ C = {30}
∴ n(B ∩ C) = 1
∴ P(B ∩ C) = \(\frac{\mathrm{n}(\mathrm{B} \cap \mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{1}{50}\)
A ∩ C = {10, 20, 30, 40, 50}
∴ n(A ∩ C) = 5
∴ P(A ∩ C) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{5}{50}\)
A ∩ B ∩ C = {30}
∴ n(A ∩ B ∩ C) = 1
∴ P(A ∩ B ∩C) = \(\frac{n(A \cap B \cap C)}{n(S)}=\frac{1}{50}\)
∴ P(the number is divisible by 2 or 3 or 10)
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)
= \(\frac{25}{50}+\frac{16}{50}+\frac{5}{50}-\frac{8}{50}-\frac{1}{50}-\frac{5}{50}+\frac{1}{50}\)
= \(\frac{33}{50}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 9 Probability Ex 9.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Probability Ex 9.1

Question 1.
There are four pens: Red, Green, Blue, and Purple in a desk drawer of which two pens are selected at random one after the other with replacement. State the sample space and the following events.
(a) A : Select at least one red pen.
(b) B : Two pens of the same colour are not selected.
Solution:
The drawer contains 4 pens out of which one is red (R), one is green (G), one is blue (B) and the other one is purple (P).
From this drawer, two pens are selected one after the other with replacement.
∴ The sample space S is given by
S = {RR, RG, RB, RP, GR, GG, GB, GP, BR, BG, BB, BP, PR, PG, PB, PP}
(a) A : Select at least one red pen.
At least one means one or more than one.
∴ A = {RR, RG, RB, RP, GR, BR, PR}

(b) B : Two pens of the same colour are not selected.
B = {RG, RB, RP, GR, GB, GP, BR, BG, BP, PR, PG, PB}

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.1

Question 2.
A coin and a die are tossed simultaneously. Enumerate the sample space and the following events.
(a) A : Getting a tail and an odd number.
(b) B : Getting a prime number.
(c) C : Getting head and a perfect square.
Solution:
When a coin and a die are tossed simultaneously, the sample space S is given by
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
(a) A : Getting a tail and an odd number.
∴ A = {(T, 1), (T, 3), (T, 5)}

(b) B : Getting a prime number.
∴ B = {(H, 2), (H, 3), (H, 5), (T, 2), (T, 3), (T, 5)}

(c) C : Getting a head and a perfect square.
∴ C = {(H, 1), (H, 4)}

Question 3.
Find n(S) for each of the following random experiments.
(a) From an urn containing 5 gold and 3 silver coins, 3 coins are drawn at random.
(b) 5 letters are to be placed into 5 envelopes such that no envelope is empty.
(c) 6 books of different subjects are arranged on a shelf.
(d) 3 tickets are drawn from a box containing 20 lottery tickets.
Solution:
(a) There are 5 gold and 3 silver coins, i.e., 8 coins.
3 coins can be drawn from these 8 coins in \({ }^{8} \mathrm{C}_{3}\) ways.
∴ n(s) = \({ }^{8} \mathrm{C}_{3}=\frac{8 !}{5 ! 3 !}=\frac{8 \times 7 \times 6 \times 5 !}{5 ! \times 3 \times 2 \times 1}=56\)

(b) 5 letters have to be placed in 5 envelopes in such a way that no envelope is empty.
∴ The first letter can be placed into 5 envelopes in 5 different ways, the second letter in 4 ways.
Similarly, the third, fourth and fifth letters can be placed in 3 ways, 2 ways and 1 way, respectively.
∴ Total number of ways = 5!
= 5 × 4 × 3 × 2 × 1
= 120
∴ n(S) = 120

(c) 6 books can be arranged on a shelf in \({ }^{6} \mathrm{P}_{6}\) = 6! ways.
∴ n(S) = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

(d) 3 tickets are drawn at random from 20 tickets.
∴ 3 tickets can be selected in \({ }^{20} \mathrm{C}_{3}\) ways.
∴ n(S) = \({ }^{20} \mathrm{C}_{3}=\frac{20 !}{17 ! 3 !}=\frac{20 \times 19 \times 18 \times 17 !}{17 ! \times 3 \times 2 \times 1}\) = 1140

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.1

Question 4.
Two fair dice are thrown. State the sample space and write the favourable outcomes for the following events.
(a) A : Sum of numbers on two dice is divisible by 3 or 4.
(b) B : The sum of numbers on two dice is 7.
(c) C : Odd number on the first die.
(d) D : Even number on the first die.
(e) Check whether events A and B are mutually exclusive and exhaustive.
(f) Check whether events C and D are mutually exclusive and exhaustive.
Solution:
When two dice are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6,4), (6, 5), (6, 6)}
∴ n(S) = 36
(a) A: Sum of the numbers on two dice is divisible by 3 or 4.
∴ A = {(1, 2), (1, 3), (1, 5), (2, 1), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (3, 6), (4, 2), (4, 4), (4, 5), (5, 1), (5, 3), (5, 4), (6, 2), (6, 3), (6, 6)}

(b) B: Sum of the numbers on two dice is 7.
∴ B = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

(c) C: Odd number on the first die.
∴ C = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

(d) D: Even number on the first die.
∴ D = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

(e) A and B are mutually exclusive events as A ∩ B = Φ.
A ∪ B = {(1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2), (2, 4), (2, 5), (2, 6), (3, 1), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 6)} ≠ S
∴ A and B are not exhaustive events as A ∪ B ≠ S.

(f) C and D are mutually exclusive events as C ∩ D = Φ.
C ∪ D = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} = S
∴ C and D are exhaustive events.

Question 5.
A bag contains four cards marked as 5, 6, 7, and 8. Find the sample space if two cards are drawn at random
(a) with replacement.
(b) without replacement.
Solution:
The bag contains 4 cards marked 5, 6, 7, and 8. Two cards are to be drawn from this bag.
(a) If the two cards are drawn with replacement, then the sample space is
S = {(5, 5), (5, 6), (5, 7), (5, 8), (6, 5), (6, 6), (6, 7), (6, 8), (7, 5), (7, 6), (7, 7), (7, 8), (8, 5), (8, 6), (8, 7), (8, 8)}

(b) If the two cards are drawn without replacement, then the sample space is
S = {(5, 6), (5, 7), (5, 8), (6, 5), (6, 7), (6, 8), (7, 5), (7, 6), (7, 8), (8, 5), (8, 6), (8, 7)}

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.1

Question 6.
A fair die is thrown two times. Find the probability that
(a) the sum of the numbers on them is 5.
(b) the sum of the numbers on them is at least 8.
(c) the first throw gives a multiple of 2 and the second throw gives a multiple of 3.
(d) product of numbers on them is 12.
Solution:
When two dice are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6,4), (6, 5), (6, 6)}
∴ n(S) = 36
(a) Let event A: Sum of the numbers on uppermost face is 5.
∴ A = {(1, 4), (2, 3), (3, 2), (4, 1)}
∴ n(A) = 4
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)

(b) Let event B: Sum of the numbers on uppermost face is at least 8 (i.e., 8 or more than 8)
∴ B = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(B) = 15
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{15}{36}=\frac{5}{12}\)

(c) Let event C: First throw gives a multiple of 2 and second throw gives a multiple of 3.
∴ C = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6)}
∴ n(C) = 6
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)

(d) Let event D: The product of the numbers on uppermost face is 12.
∴ D = {(2, 6), (3, 4), (4, 3), (6, 2)}
∴ n(D) = 4
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)

Question 7.
Two cards are drawn from a pack of 52 cards. Find the probability that
(a) one is a face card and the other is an ace card.
(b) one is a club and the other is a diamond.
(c) both are from the same suit.
(d) both are red cards.
(e) one is a heart card and the other is a non-heart card.
Solution:
Two cards can be drawn from a pack of 52 cards in \({ }^{52} \mathrm{C}_{2}\) ways.
∴ n(S) = \({ }^{52} \mathrm{C}_{2}\)

(a) Let event A: Out of the two cards drawn, one is a face card and the other is an ace card.
There are 12 face cards and 4 ace cards in a pack of 52 cards.
∴ One face card can be drawn from 12 face cards in \({ }^{12} \mathrm{C}_{1}\) ways and one ace card can be drawn from 4 ace cards in \({ }^{4} \mathrm{C}_{1}\) ways.
∴ n(A) = \({ }^{12} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}\)
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{12} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{2}}=\frac{12 \times 4}{\frac{52 \times 51}{2 \times 1}}=\frac{8}{221}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.1

(b) Let event B: Out of the two cards drawn, one is club and the other is a diamond card.
There are 13 club cards and 13 diamond cards.
∴ One club card can be drawn from 13 club cards in \({ }^{13} \mathrm{C}_{1}\) ways and one diamond card can be drawn from 13 diamond cards in \({ }^{13} \mathrm{C}_{1}\) ways.
∴ n(B) = \({ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1}\)
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{{ }^{13} C_{1} \times{ }^{13} C_{1}}{{ }^{52} C_{2}}=\frac{13 \times 13}{\frac{52 \times 51}{2 \times 1}}=\frac{13}{102}\)

(c) Let event C: Both the cards drawn are of the same suit.
A pack of 52 cards consists of 4 suits each containing 13 cards.
∴ 2 cards can be drawn from the same suit in \({ }^{13} \mathrm{C}_{2}\) ways.
∴ n(C) = \({ }^{13} \mathrm{C}_{2} \times 4\)
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{4 \times{ }^{13} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{4 \times 13 \times 12}{52 \times 51}=\frac{4}{17}\)

(d) Let event D: Both the cards drawn are red.
There are 26 red cards in the pack of 52 cards.
∴ 2 cards can be drawn from them in \({ }^{26} \mathrm{C}_{2}\) ways.
∴ n(D) = \({ }^{26} \mathrm{C}_{2}\)
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{{ }^{26} C_{2}}{{ }^{52} C_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}\)

(e) Let event E: Out of the two cards drawn, one is heart and other is non-heart.
There are 13 heart cards in a pack of 52 cards, i.e., 39 cards are non-heart.
∴ One heart card can be drawn from 13 hdart cards in \({ }^{13} \mathrm{C}_{1}\) ways and one non-heart card can be drawn from 39 cards in \({ }^{39} \mathrm{C}_{1}\) ways.
∴ n(E) = \({ }^{13} \mathrm{C}_{1} \times{ }^{39} \mathrm{C}_{1}\)
∴ P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1} \times{ }^{39} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{2}}=\frac{13 \times 39}{\frac{52 \times 51}{2 \times 1}}=\frac{13}{34}\)

Question 8.
Three cards are drawn from a pack of 52 cards. Find the chance that
(a) two are queen cards and one is an ace card.
(b) at least one is a diamond card.
(c) all are from the same suit.
(d) they are a king, a queen, and a jack.
Solution:
3 cards can be drawn from a pack of 52 cards in \({ }^{52} \mathrm{C}_{3}\) ways.
∴ n(S) = \({ }^{52} \mathrm{C}_{3}\)

(a) Let event A: Out of the three cards drawn, 2 are queens and 1 is an ace card.
There are 4 queens and 4 aces in a pack of 52 cards.
∴ 2 queens can be drawn from 4 queens in \({ }^{4} \mathrm{C}_{2}\) ways and 1 ace can be drawn out of 4 aces in \({ }^{4} \mathrm{C}_{1}\) ways.
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.1 Q8

(b) Let event B: Out of the three cards drawn, at least one is a diamond.
∴ B’ is the event that all 3 cards drawn are non-diamond cards.
In a pack of 52 cards, there are 39 non-diamond cards.
∴ 3 non-diamond cards can be drawn in \({ }^{39} \mathrm{C}_{3}\) ways.
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.1 Q8.1

(c) Let event C: All the cards drawn are from the same suit.
A pack of 52 cards consists of 4 suits each containing 13 cards.
∴ 3 cards can be drawn from the same suit in \({ }^{13} \mathrm{C}_{3}\) ways.
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.1 Q8.2

(d) Let event D: The cards drawn are a king, a queen, and a jack.
There are 4 kings, 4 queens and 4 jacks in a pack of 52 cards.
∴ 1 king can be drawn from 4 kings in \({ }^{4} \mathrm{C}_{1}\) ways,
1 queen can be drawn from 4 queens in \({ }^{4} \mathrm{C}_{1}\) ways and
1 jack can be drawn from 4 jacks in \({ }^{4} \mathrm{C}_{1}\) ways.
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.1 Q8.3

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.1

Question 9.
From a bag containing 10 red, 4 blue, and 6 black balls, a ball is drawn at random. Find the probability of drawing
(a) a red bail.
(b) a blue or black ball.
(c) not a black ball.
Solution:
The bag contains 10 red, 4 blue, and 6 black balls,
i.e., 10 + 4 + 6 = 20 balls.
One ball can be drawn from 20 balls in \({ }^{20} \mathrm{C}_{1}\) ways.
∴ n(S) = \({ }^{20} \mathrm{C}_{1}\) = 20

(a) Let event A: Ball drawn is red.
There are total 10 red balls.
∴ 1 red ball can be drawn from 10 red balls in \({ }^{10} \mathrm{C}_{1}\) ways.
∴ n(A) = \({ }^{10} \mathrm{C}_{1}\) = 10
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{10}{20}=\frac{1}{2}\)

(b) Let event B: The ball drawn is blue or black.
There are 4 blue and 6 black balls.
∴ 1 blue ball can be drawn from 4 blue balls in \({ }^{4} \mathrm{C}_{1}\) ways
or 1 black ball can be drawn from 6 black balls in \({ }^{6} \mathrm{C}_{1}\) ways.
∴ n(B) = \({ }^{4} \mathrm{C}_{1}\) + \({ }^{6} \mathrm{C}_{1}\) = 4 + 6 = 10
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{10}{20}=\frac{1}{2}\)

(c) Let event C: Ball drawn is not black,
i.e., ball drawn is red or blue.
There are total 14 red and blue balls.
∴ 1 ball can be drawn from 14 balls in \({ }^{14} \mathrm{C}_{1}\) ways.
∴ n(C) = \({ }^{14} \mathrm{C}_{1}\) = 14
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{14}{20}=\frac{7}{10}\)

Question 10.
A box contains 75 tickets numbered 1 to 75. A ticket is drawn at random from the box. Find the probability that,
(a) number on the ticket is divisible by 6.
(b) the number on the ticket is a perfect square.
(c) the number on the ticket is prime.
(d) the number on the ticket is divisible by 3 and 5.
Solution:
The box contains 75 tickets numbered 1 to 75.
∴ 1 ticket can be drawn from the box in \({ }^{75} \mathrm{C}_{1}\) = 75 ways.
∴ n(S) = 75

(a) Let event A: Number on the ticket is divisible by 6.
∴ A = {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72}
∴ n(A) = 12
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{12}{75}=\frac{4}{25}\)

(b) Let event B: Number on the ticket is a perfect square.
∴ B = (1, 4, 9, 16, 25, 36, 49, 64}
∴ n(B) = 8
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{8}{75}\)

(c) Let event C: Number on the ticket is a prime number.
∴C = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73}

(d) Let event D: Number on the ticket is divisible by 3 and 5,
i.e., divisible by L.C.M. of 3 and 5,
i.e., 15.
∴D = {15, 30, 45, 60, 75}
∴ n(D) = 5
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{5}{75}=\frac{1}{15}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.1

Question 11.
What is the chance that a leap year, selected at random, will contain 53 Sundays?
Solution:
A leap year consists of 366 days.
It has 52 complete weeks and two more days.
These two days can be {(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thur), (Thur, Fri), (Fri, Sat), (Sat, Sun)}.
∴ n(S) = 7
Let event E : There are 53 Sundays.
∴ E = {(Sun, Mon), (Sat, Sun)}
∴ n(E) = 2
∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{2}{7}\)

Question 12.
Find the probability of getting both red balls, when from a bag containing 5 red and 4 black balls, two balls are drawn,
(i) with replacement
(ii) without replacement
Solution:
The bag contains 5 red and 4 black balls,
i.e., 5 + 4 = 9 balls.
(i) 2 balls can be drawn from 9 balls with replacement in \({ }^{9} \mathrm{C}_{1} \times{ }^{9} \mathrm{C}_{1}\) ways.
∴ n(S) = \({ }^{9} \mathrm{C}_{1} \times{ }^{9} \mathrm{C}_{1}\) = 9 × 9 = 81
Let event A: Balls drawn are red.
2 red balls can be drawn from 5 red balls with replacement in \({ }^{5} \mathrm{C}_{1} \times{ }^{5} \mathrm{C}_{1}\) ways.
∴ n(A) = \({ }^{5} \mathrm{C}_{1} \times{ }^{5} \mathrm{C}_{1}\) = 5 × 5 = 25
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{25}{81}\)

(ii) 2 balls can be drawn from 9 balls without replacement in \({ }^{9} C_{1} \times{ }^{8} C_{1}\) ways.
∴ n(S) = \({ }^{9} C_{1} \times{ }^{8} C_{1}\) = 9 × 8 = 72
2 red balls can be drawn from 5 red balls without replacement in \({ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}\) ways.
∴ n(B) = \({ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}\) = 5 × 4 = 20
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{20}{72}=\frac{5}{18}\)

Question 13.
A room has three sockets for lamps. From a collection of 10 bulbs of which 6 are defective. At night a person selects 3 bulbs, at random and puts them in sockets. What is the probability that
(i) room is still dark.
(ii) the room is lit.
Solution:
Total number of bulbs = 10
Number of defective bulbs = 6
∴ Number of non-defective bulbs = 4
3 bulbs can be selected out of 10 bulbs in \({ }^{10} \mathrm{C}_{3}\) ways.
∴ n(S) = \({ }^{10} \mathrm{C}_{3}\)

(i) Let event A: The room is dark.
For event A to happen the bulbs should be selected from the 6 defective bulbs. This can be done in \({ }^{6} \mathrm{C}_{3}\) ways.
∴ n(A) = \({ }^{6} \mathrm{C}_{3}\)
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{{ }^{6} C_{3}}{{ }^{10} C_{3}}=\frac{6 \times 5 \times 4}{10 \times 9 \times 8}=\frac{1}{6}\)

(ii) Let event A’: The room is lit.
∴ P(Room is lit) = 1 – P(Room is not lit)
∴ P(A’) = 1 – P(A) = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.1

Question 14.
Letters of the word MOTHER are arranged at random. Find the probability that in the arrangement
(a) vowels are always together.
(b) vowels are never together.
(c) O is at the beginning and end with T.
(d) starting with a vowel and ending with a consonant.
Solution:
There are 6 letters in the word MOTHER.
These letters can be arranged among themselves in \({ }^{6} \mathrm{P}_{6}\) = 6! ways.
∴ n(S) = 6!
(a) Let event A: Vowels are always together.
The word MOTHER consists of 2 vowels (O, E) and 4 consonants (M, T, H, R).
2 vowels can be arranged among themselves in \({ }^{2} \mathbf{P}_{2}\) = 2! ways.
Let us consider 2 vowels as one group.
This one group with 4 consonants can be arranged in \({ }^{5} \mathrm{P}_{5}\) = 5! ways.
∴ n(A) = 2! × 5!
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{2 ! \times 5 !}{6 !}=\frac{1}{3}\)

(b) Let event B: Vowels are never together.
4 consonants create 5 gaps, in which vowels are arranged.
Consider the following arrangement of consonants
_C_C_C_C_
2 vowels can be arranged in 5 gaps in \({ }^{5} \mathbf{P}_{2}\) ways.
Also 4 consonants can be arranged among themselves in \({ }^{4} \mathbf{P}_{4}\) = 4! ways.
∴ n(B) = 4! × \({ }^{5} \mathbf{P}_{2}\)
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{4 ! \times{ }^{5} P_{2}}{6 !}=\frac{4 ! \times 5 \times 4}{6 \times 5 \times 4 !}=\frac{2}{3}\)

(iii) Let event C: Word begin with O and end with T.
Thus first and last letters can be arranged in one way each and the remaining 4 letters can be arranged in \({ }^{4} \mathrm{P}_{4}\) = 4! ways
∴ n(C) = 4! × 1 × 1 = 4!
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{4 !}{6 !}=\frac{1}{30}\)

(d) Let event D: Word starts with a vowel and ends with a consonant.
There are 2 vowels and 4 consonants in the word MOTHER.
∴ The first place can be arranged in 2 different ways and the last place can be arranged in 4 different ways.
Now, the remaining 4 letters (3 consonants and 1 vowel) can be arranged in \({ }^{4} \mathrm{P}_{4}\) = 4! ways.
∴ n(D) = 2 × 4 × 4!
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{2 \times 4 \times 4 !}{6 !}=\frac{4}{15}\)

Question 15.
4 letters are to be posted in 4 post boxes. If any number of letters can be posted in any of the 4 post boxes, what is the probability that each box contains only one letter?
Solution:
There are 4 letters and 4 post boxes.
Since any number of letters can be posted in all 4 post boxes,
so each letter can be posted in different ways.
∴ n(S) = 4 × 4 × 4 × 4
Let event A: Each box contains only one letter.
∴ 1st letter can be posted in 4 different ways.
Since each box contains only one letter, 2nd letter can be posted in 3 different ways.
Similarly, 3rd and 4th letters can be posted in 2 different ways and 1 way respectively.
∴ n(A) = 4 × 3 × 2 × 1
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4 \times 3 \times 2 \times 1}{4 \times 4 \times 4 \times 4}=\frac{3}{32}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.1

Question 16.
15 professors have been invited for a round table conference by the Vice-chancellor of a university. What is the probability that two particular professors occupy the seats on either side of the Vice-chancellor during the conference?
Solution:
Since a Vice-chancellor invited 15 professors for a round table conference, there were all 16 persons in the conference.
These 16 persons can be arranged among themselves around a round table in (16 – 1)! = 15! ways.
∴ n(S) = 15!
Let event A: Two particular professors be seated on either side of the Vice-chancellor.
Those two particular persons sit on either side of a Vice chancellor in \({ }^{2} \mathrm{P}_{2}\) = 2! ways.
Thus the remaining 13 persons can be arranged in \({ }^{13} \mathrm{P}_{13}\) = 13! ways.
∴ n(A) = 13! 2!
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{13 ! \times 2 !}{15 !}=\frac{13 ! \times 2 \times 1}{15 \times 14 \times 13 !}=\frac{1}{105}\)

Question 17.
A bag contains 7 black and 4 red balls. If 3 balls are drawn at random, find the probability that
(i) all are black.
(ii) one is black and two are red.
Solution:
The bag contains 7 black and 4 red balls,
i.e., 7 + 4 = 11 balls.
∴ 3 balls can be drawn out of 11 balls in \({ }^{11} \mathrm{C}_{3}\) ways.
∴ n(S) = \({ }^{11} \mathrm{C}_{3}\)
(i) Let event A: All 3 balls drawn are black.
There are 7 black balls.
∴ 3 black balls can be drawn from 7 black balls in \({ }^{7} \mathrm{C}_{3}\) ways.
∴ n(A) = \({ }^{7} \mathrm{C}_{3}\)
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{7} \mathrm{C}_{3}}{{ }^{11} \mathrm{C}_{3}}=\frac{7 \times 6 \times 5}{11 \times 10 \times 9}=\frac{7}{33}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.1

(ii) Let event B: Out of 3 balls drawn, one is black and two are red.
There are 7 black and 4 red balls.
∴ One black ball can be drawn from 7 black balls in \({ }^{7} \mathrm{C}_{1}\) ways and 2 red balls can be drawn from 4 red balls in \({ }^{4} \mathrm{C}_{2}\) ways.
∴ n(A) = \({ }^{7} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{2}\)
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{{ }^{7} C_{1} \times{ }^{4} C_{2}}{{ }^{11} C_{3}}=\frac{7 \times \frac{4 \times 3}{2}}{\frac{11 \times 10 \times 9}{3 \times 2 \times 1}}=\frac{14}{55}\)

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8

(I) Select the correct option from the given alternatives:

Question 1.
If there are 10 values each equal to 10, then S.D. of these values is _________
(A) 100
(B) 20
(C) 0
(D) 6
Answer:
(C) 0
Hint:
Var (X) = \(\sigma_{x}^{2}=\frac{\sum x_{i}^{2}}{\mathrm{n}}-(\bar{x})^{2}\)
= \(\frac{1000}{10}\) – 100
= 0
∴ S.D. = 0

Question 2.
Number of patients who visited cardiologists are 13, 17, 11, 15 in four days, then variance (approximately) is
(A) 5 patients
(B) 4 patients
(C) 10 patients
(D) 15 patients
Answer:
(A) 5 patients

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8

Question 3.
If the observations of a variable X are, -4, -20, -30, -44 and -36, then the value of the range will be:
(A) -48
(B) 40
(C) -40
(D) 48
Answer:
(B) 40

Question 4.
The standard deviation of a distribution divided by the mean of the distribution and expressing in percentage is called:
(A) Coefficient of Standard deviation
(B) Coefficient of skewness
(C) Coefficient of quartile deviation
(D) Coefficient of variation
Answer:
(D) Coefficient of variation

Question 5.
If the S.D. of first n natural numbers is √2, then the value of n must be
(A) 5
(B) 4
(C) 7
(D) 6
Answer:
(A) 5

Question 6.
The positive square root of the mean of the squares of the deviations of observations from their mean is called:
(A) Variance
(B) Range
(C) S.D.
(D) C.V.
Answer:
(C) S.D.

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8

Question 7.
The variance of 19, 21, 23, 25 and 27 is 8. The variance of 14, 16, 18, 20 and 22 is:
(A) Greater than 8
(B) 8
(C) Less than 8
(D) 8 – 5 = 3
Answer:
(B) 8

Question 8.
For any two numbers, SD is always
(A) Twice the range
(B) Half of the range
(C) Square of the range
(D) None of these
Answer:
(B) Half of the range

Question 9.
Given the heights (in cm) of two groups of students:
Group A: 131 cm, 150 cm, 147 cm, 138 cm, 144 cm
Group B: 139 cm, 148 cm, 132 cm, 151 cm, 140 cm
Which of the following is/are true?
I. The ranges of the heights of the two groups of students are the same.
II. The means of the heights of the two groups of students are the same.
(A) I only
(B) II only
(C) Both I and II
(D) None
Answer:
(C) Both I and II

Question 10.
The standard deviation of data is 12 and mean is 72, then the coefficient of variation is
(A) 13.67%
(B) 16.67%
(C) 14.67%
(D) 15.67%
Answer:
(B) 16.67%

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8

(II) Answer the following:

Question 1.
Find the range for the following data.
76, 57, 80, 103, 61, 63, 89, 96, 105, 72
Solution:
Here, largest value (L) = 105, smallest value (S) = 57
∴ Range = L – S
= 105 – 57
= 48

Question 2.
Find the range for the following data.
116, 124, 164, 150, 149, 114, 195, 128, 138, 203, 144
Solution:
Here, largest value (L) = 203, smallest value (S) = 114
∴ Range = L – S
= 203 – 114
= 89

Question 3.
Given below the frequency distribution of weekly wages of 400 workers. Find the range.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q3
Solution:
Here, largest value (L) = 40, smallest value (S) = 10
∴ Range = L – S
= 40 – 10
= 30

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8

Question 4.
Find the range of the following data.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q4
Solution:
Here, upper limit of the highest class (L) = 175
lower limit of the lowest class (S) = 115
∴ Range = L – S
= 175 – 115
= 60

Question 5.
Find variance and S.D. for the following set of numbers.
25, 21, 23, 29, 27, 22, 28, 23, 21, 25
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q5

Question 6.
Find variance and S.D. for the following set of numbers.
125, 130, 150, 165, 190, 195, 210, 230, 245, 260
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q6

Question 7.
Following data gives no. of goals scored by a team in 90 matches. Compute the standard deviation.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q7
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q7.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q7.2

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8

Question 8.
Compute the variance and S.D. for the following data:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q8
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q8.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q8.2

Question 9.
Calculate S.D. from following data.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q9
Solution:
Since data is not continuous, we have to make it continuous.
let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-54.5}{10}\)
Calculation of variance of u:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q9.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q9.2

Question 10.
Given below is the frequency distribution of marks obtained by 100 students. Compute arithmetic mean and S.D.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q10
Solution:
Since data is not continuous, we have to make it continuous.
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-74.5}{10}\)
Calculation of variance of u:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q10.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q10.2
∴ Var (X) = h2 var (u)
= (10)2 × 1.4875
= 100 × 1.4875
= 148.75
∴ S.D. = σx = √Var(X)
= √148.75
= 12.2

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8

Question 11.
The arithmetic mean and standard deviation of a series of 20 items were calculated by a student as 20 cm and 5 cm respectively. But while calculating them, item 13 was misread as 30. Find the corrected mean and standard deviation.
Solution:
n = 20, \(\bar{x}\) = 20, σx = 5 …..(given)
\(\bar{x}=\frac{1}{\mathrm{n}} \sum x_{\mathrm{i}}\)
∴ \(\sum x_{\mathrm{i}}=\mathrm{n} \bar{x}\) = 20 × 20 = 400
Corrected Σxi = Σxi – (incorrect observation) + (correct observation)
= 400 – 30 + 13
= 383
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q11
Corrected \(\sum x_{\mathrm{i}}^{2}\) = \(\sum x_{\mathrm{i}}^{2}\) – (incorrect observation)2 + (correct observation)2
= 8500 – (30)2 + (13)2
= 8500 – 900 + 169
= 7769
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q11.1

Question 12.
The mean and S.D. of a group of 50 observations are 40 and 5 respectively. If two more observations 60 and 72 are added to the set, find the mean and S.D. of 52 items.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q12
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q12.1

Question 13.
The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. Find the number of boys and combined S.D.
Solution:
Let n1 and n2 be the number of boys and girls respectively.
Let n = 200, \(\bar{x}_{c}\) = 65, \(\bar{x}_{1}\) = 70, \(\bar{x}_{2}\) = 62, σ1 = 8, σ2 = 10
Here, n1 + n2 = n
n1 + n2 = 200 ……(i)
Combined mean (\(\bar{x}_{c}\)) = \(\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{i}+n_{2}}\)
65 = \(\frac{\mathrm{n}_{1}(70)+\mathrm{n}_{2}(62)}{200}\) …… [From (i)]
70n1 + 62n2 = 13000
35n1 + 31n2 = 6500 …..(ii)
Solving (i) and (ii), we get
n1 = 75, n2 = 125
Number of boys = 75
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q13

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8

Question 14.
From the following data available for 5 pairs of observations of two variables x and y, obtain combined S.D. for all 10 observations.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q14
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q14.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q14.2

Question 15.
Calculate the coefficient of variation of the following data.
23, 27, 25, 28, 21, 14, 16, 12, 18, 16
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q15
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q15.1

Question 16.
The following data relates to the distribution of weights of 100 boys and 80 girls in a school.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q16
Which of the two is more variable?
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q16.1
Here, C.V. of boys > C.V. of girls
∴ The Series of boys is more variable.

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8

Question 17.
The mean and standard deviations of the two brands of watches are given below:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q17
Calculate the coefficient of variation of the two brands and interpret the results.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q17.1
Here, C.V. (I) > C.V. (II)
∴ The brand I is more variable.

Question 18.
Calculate the coefficient of variation for the data given below:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q18
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q18.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q18.2

Question 19.
Calculate the coefficient of variation for the data given below:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q19
Solution:
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-6500}{1000}\)
Calculation of variance of u:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q19.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q19.2

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8

Question 20.
Compute the coefficient of variations for the following data to show whether the variation is greater in the field or in the area of the field.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q20
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q20.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q20.2
∴ the variation is greater in the area of the field.

Question 21.
There are two companies U and V which manufacture cars. A sample of 40 cars each from these companies is taken and the average running life (in years) is recorded.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q21
Which company shows greater consistency?
Solution:
Let f1 denote no. of cars of company U and f2 denote no. of cars of company V.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q21.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q21.2
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q21.3
Here, C.V. (U) < C.V. (V)
∴ Company U shows greater consistency in performance.

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8

Question 22.
The means and S.D. of weights and heights of 100 students of a school are as follows.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q22
Which shows more variability, weights, or heights?
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 II Q22.1
Here, C.V. for weight < C.V. for height
∴ Height shows more variability.

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 8 Measures of Dispersion Ex 8.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3

Question 1.
The means of two samples of sizes 60 and 120 respectively are 35.4 and 30.9 and the standard deviations are 4 and 5. Obtain the standard deviation of the sample of size 180 obtained by combining the two samples.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q1.1

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3

Question 2.
For certain data, the following information is available.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q2
Obtain the combined standard deviation.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q2.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q2.2

Question 3.
Calculate the coefficient of variation of marks secured by a student in the exam, where the marks are:
85, 91, 96, 88, 98, 82
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q3

Question 4.
Find the coefficient of variation of a sample that has a mean equal to 25 and a standard deviation of 5.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q4

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3

Question 5.
A group of 65 students of class XI has their average height as 150.4 cm with a coefficient of variance of 2.5%. What is the standard deviation of their heights?
Solution:
Given, n = 65, \(\bar{x}\) = 150.4, C.V. = 2.5%
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q5
∴ The standard deviation of students’ height is 3.76 cm.

Question 6.
Two workers on the same job show the following results:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q6
(i) Regarding the time required to complete the job, which worker is more consistent?
(ii) Which worker seems to be faster in completing the job?
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q6.1
(i) Since C.V. (P) < C.V.(Q), Worker P is more consistent regarding the time required to complete the job. (ii) Since \(\overline{\mathrm{p}}>\overline{\mathrm{q}}\),
i.e., the expected time for completing the job is less for worker Q.
∴ Worker Q seems to be faster in completing the job.

Question 7.
A company has two departments with 42 and 60 employees respectively. Their average weekly wages are ₹ 750 and ₹ 400. The standard deviations are 8 and 10 respectively.
(i) Which department has a larger bill?
(ii) Which department has larger variability in wages?
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q7
(i) Since \(\bar{x}_{1}>\bar{x}_{2}\),
i.e., average weekly wages are more for the first department.
∴ The first department has a larger bill.

(ii) Since C.V. (1) < C.V. (2),
second department is less consistent.
∴ The second department has larger variability in wages.

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3

Question 8.
The following table gives the weights of the students of two classes. Calculate the coefficient of variation of the two distributions. Which series is more variable?
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q8
Solution:
Let x denote the data of class A and y denote the data of class B.
Calculation of S.D. for class A:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q8.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q8.2
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q8.3
Since C.V. (Y) > C.V.(X),
C.V. (B) > C.V. (A)
∴ Series B is more variable.

Question 9.
Compute the coefficient of variation for team A and team B.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q9
Which team is more consistent?
Solution:
Let f1 denote no. of goals of team A and f2 denote no. of goals of team B.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q9.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q9.2
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q9.3
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q9.4
Since C.V. of team A > C.V. of team B,
Team B is more consistent.

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3

Question 10.
Given below is the information about marks obtained in Mathematics and Statistics by 100 students in a class. Which subject shows the highest variability in marks?
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q10
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.3 Q10.1
Since C.V. (S) > C.V. (M),
The subject statistics show higher variability in marks.

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 8 Measures of Dispersion Ex 8.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2

Question 1.
Find variance and S.D. for the following set of numbers.
7, 11, 2, 4, 9, 6, 3, 7, 11, 2, 5, 8, 3, 6, 8, 8, 2, 6
Solution:
Given data:
7, 11, 2, 4, 9, 6, 3, 7, 11, 2, 5, 8, 3, 6, 8, 8, 2, 6
The tabulated form of the above data is
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q1
Calculation of variance and S.D.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q1.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q1.2

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2

Question 2.
Find variance and S.D. for the following set of numbers.
65, 77, 81, 98, 100, 80, 129
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q2

Question 3.
Compute variance and standard deviation for the following data:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q3
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q3.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q3.2

Question 4.
Compute the variance and S.D.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q4
Solution:
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-34}{1}\)
Calculation of variance of u:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q4.1

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2

Question 5.
Following data gives ages of 100 students in a college. Calculate variance and S.D.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q5
Solution:
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-19}{1}\)
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q5.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q5.2

Question 6.
Find mean, variance and S.D. of the following data.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q6
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q6.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q6.2
Alternate Method:
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-55}{10}\)
Calculation of variance of u:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q6.3
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q6.4

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2

Question 7.
Find the variance and S.D. of the following frequency distribution which gives the distribution of 200 plants according to their height.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q7
Solution:
Since data is not continuous, we have to make it continuous.
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-31}{5}\)
Calculation of variance of u:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q7.1
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q7.2

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2

Question 8.
The mean of 5 observations is 4.8 and the variance is 6.56. If three of the five observations are 1, 3, and 8, find the other two observations.
Solution:
\(\bar{x}\) = 4.8, Var (X) = 6.56, n = 5, x1 = 1, x2 = 3, x3 = 8 ……(given)
Let the remaining two observations be x4 and x5.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q8
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2 Q8.1
From (i), we get
x5 = 5 or x5 = 7
∴ The two numbers are 5 and 7.

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 8 Measures of Dispersion Ex 8.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.1

Question 1.
Find a range of the following data:
19, 27, 15, 21, 33, 45, 7, 12, 20, 26
Solution:
Here, largest value (L) = 45, smallest value (S) = 7
∴ Range = L – S = 45 – 7 = 38

Question 2.
Find range of the following data:
575, 609, 335, 280, 729, 544, 852, 427, 967, 250
Solution:
Here, largest value (L) = 967, smallest value (S) = 250
∴ Range = L – S = 967 – 250 = 717

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.1

Question 3.
The following data gives a number of typing mistakes done by Radha during a week. Find the range of the data.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.1 Q3
Solution:
Here, largest value (L) = 21, smallest value (S) = 10
∴ Range = L – S = 21 – 10 = 11

Question 4.
The following results were obtained by rolling a die 25 times. Find the range of the data.
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.1 Q4
Solution:
Here, largest value (L) = 6, smallest value (S) = 1
∴ Range = L – S = 6 – 1 = 5

Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.1

Question 5.
Find range for the following data:
Maharashtra Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.1 Q5
Solution:
Here, upper limit of the highest class (L) = 72
lower limit of the lowest class (S) = 62
∴ Range = L – S = 72 – 62 = 10

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 7 Conic Sections Miscellaneous Exercise 7 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

(I) Select the correct option from the given alternatives.

Question 1.
The line y = mx + 1 is a tangent to the parabola y2 = 4x, if m is ________
(A) 1
(B) 2
(C) 3
(D) 4
Answer:
(A) 1
Hint:
y2 = 4x
Compare with y2 = 4ax
∴ a = 1
Equation of tangent is y = mx + 1
Compare with y = mx + \(\frac{a}{m}\)
\(\frac{a}{m}\) = 1
∴ a = m = 1

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

Question 2.
The length of latus rectum of the parabola x2 – 4x – 8y + 12 = 0 is ________
(A) 4
(B) 6
(C) 8
(D) 10
Answer:
(C) 8
Hint:
Given equation of parabola is
x2 – 4x – 8y + 12 = 0
⇒ x2 – 4x = 8y – 12
⇒ x2 – 4x + 4 = 8y – 12 + 4
⇒ (x – 2)2 = 8(y – 1)
Comparing this equation with (x – h)2 = 4b(y – k), we get
4b = 8
∴ Length of latus rectum = 4b = 8

Question 3.
If the focus of the parabola is (0, -3), its directrix is y = 3, then its equation is ________
(A) x2 = -12y
(B) x2 = 12y
(C) y2 = 12x
(D) y2 = -12x
Answer:
(A) x2 = -12y
Hint:
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7 I Q3
SP2 = PM2
⇒ (x – 0)2 + (y + 3)2 = \(\left|\frac{y-3}{\sqrt{1}}\right|^{2}\)
⇒ x2 + y2 + 6y + 9 = y2 – 6y + 9
⇒ x2 = -12y

Question 4.
The co-ordinates of a point on the parabola y2 = 8x whose focal distance is 4 are ________
(A) (\(\frac{1}{2}\), ±2)
(B) (1, ±2√2)
(C) (2, ±4)
(D) none of these
Answer:
(C) (2, ±4)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

Question 5.
The end points of latus rectum of the parabola y2 = 24x are ________
(A) (6, ±12)
(B) (12, ±6)
(C) (6, ±6)
(D) none of these
Answer:
(A) (6, ±12)

Question 6.
Equation of the parabola with vertex at the origin and directrix with equation x + 8 = 0 is ________
(A) y2 = 8x
(B) y2 = 32x
(C) y2 = 16x
(D) x2 = 32y
Answer:
(B) y2 = 32x
Hint:
Since directrix is parallel to Y-axis,
The X-axis is the axis of the parabola.
Let the equation of parabola be y2 = 4ax.
Equation of directrix is x + 8 = 0
∴ a = 8
∴ required equation of parabola is y2 = 32x

Question 7.
The area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the endpoints of its latus rectum is ________
(A) 22 sq. units
(B) 20 sq. units
(C) 18 sq. units
(D) 14 sq. units
Answer:
(C) 18 sq. units
Hint:
x2 = 12y
4b = 12
b = 3
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7 I Q7
Area of triangle = \(\frac{1}{2}\) × AB × OS
= \(\frac{1}{2}\) × 4a × a
= \(\frac{1}{2}\) × 12 × 3
= 18 sq. units

Question 8.
If P(\(\frac{\pi}{4}\)) is any point on the ellipse 9x2 + 25y2 = 225, S and S’ are its foci, then SP . S’P = ________
(A) 13
(B) 14
(C) 17
(D) 19
Answer:
(C) 17
Hint:
9x2 + 25y2 = 225
\(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)
Here, a = 5, b = 3
Eccentricity (e) = \(\frac{4}{5}\)
∴ \(\frac{\mathrm{a}}{\mathrm{e}}=\frac{5}{\left(\frac{4}{5}\right)}=\frac{25}{4}\)
Coordinates of foci are S(4, 0) and S'(-4, 0)
P(θ) = (a cos θ, b sin θ)
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7 I Q8

Question 9.
The equation of the parabola having (2, 4) and (2, -4) as end points of its latus rectum is ________
(A) y2 = 4x
(B) y2 = 8x
(C) y2 = -16x
(D) x2 = 8y
Answer:
(B) y2 = 8x
Hint:
The given points lie in the 1st and 4th quadrants.
∴ Equation of the parabola is y2 = 4ax
End points of latus rectum are (a, 2a) and (a, -2a)
∴ a = 2
∴ required equation of parabola is y = 8x

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

Question 10.
If the parabola y2 = 4ax passes through (3, 2), then the length of its latus rectum is ________
(A) \(\frac{2}{3}\)
(B) \(\frac{4}{3}\)
(C) \(\frac{1}{3}\)
(D) 4
Answer:
(B) \(\frac{4}{3}\)
Hint:
Length of latus rectum = 4a
The given parabola passes through (3, 2)
∴ (2)2 = 4a(3)
∴ 4a = \(\frac{4}{3}\)

Question 11.
The eccentricity of rectangular hyperbola is
(A) \(\frac{1}{2}\)
(B) \(\frac{1}{2^{\frac{1}{2}}}\)
(C) \(2^{\frac{1}{2}}\)
(D) \(\frac{1}{3^{\frac{1}{2}}}\)
Answer:
(C) \(2^{\frac{1}{2}}\)

Question 12.
The equation of the ellipse having one of the foci at (4, 0) and eccentricity \(\frac{1}{3}\) is
(A) 9x2 + 16y2 = 144
(B) 144x2 + 9y2 = 1296
(C) 128x2 + 144y2 = 18432
(D) 144x2 + 128y2 = 18432
Answer:
(C) 128x2 + 144y2 = 18432

Question 13.
The equation of the ellipse having eccentricity \(\frac{\sqrt{3}}{2}\) and passing through (-8, 3) is
(A) 4x2 + y2 = 4
(B) x2 + 4y2 = 100
(C) 4x2 + y2 = 100
(D) x2 + 4y2 = 4
Answer:
(B) x2 + 4y2 = 100

Question 14.
If the line 4x – 3y + k = 0 touches the ellipse 5x2 + 9y2 = 45, then the value of k is
(A) 21
(B) ±3√21
(C) 3
(D) 3(21)
Answer:
(B) ±3√21

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

Question 15.
The equation of the ellipse is 16x2 + 25y2 = 400. The equations of the tangents making an angle of 180° with the major axis are
(A) x = 4
(B) y = ±4
(C) x = -4
(D) x = ±5
Answer:
(B) y = ±4

Question 16.
The equation of the tangent to the ellipse 4x2 + 9y2 = 36 which is perpendicular to 3x + 4y = 17 is
(A) y = 4x + 6
(B) 3y + 4x = 6
(C) 3y = 4x + 6√5
(D) 3y = x + 25
Answer:
(C) 3y = 4x + 6√5

Question 17.
Eccentricity of the hyperbola 16x2 – 3y2 – 32x – 12y – 44 = 0 is
(A) \(\sqrt{\frac{17}{3}}\)
(B) \(\sqrt{\frac{19}{3}}\)
(C) \(\frac{\sqrt{19}}{3}\)
(D) \(\frac{\sqrt{17}}{3}\)
Answer:
(B) \(\sqrt{\frac{19}{3}}\)
Hint:
16x2 – 3y2 – 32x – 12y – 44 = 0
⇒ 16(x – 1)2 – 3(y + 2)2 = 48
⇒ \(\frac{(x-1)^{2}}{3}-\frac{(y+2)^{2}}{16}=1\)
Here, a2 = 3 and b2 = 16
\(e=\frac{\sqrt{a^{2}+b^{2}}}{a}=\frac{\sqrt{3+16}}{\sqrt{3}}=\sqrt{\frac{19}{3}}\)

Question 18.
Centre of the ellipse 9x2 + 5y2 – 36x – 50y – 164 = 0 is at
(A) (2, 5)
(B) (1, -2)
(C) (-2, 1)
(D) (0, 0)
Answer:
(A) (2, 5)
Hint:
9x2 + 5y2 – 36x – 50y – 164 = 0
⇒ 9(x – 2)2 + 5(y – 5)2 = 325
⇒ \(\frac{(x-2)^{2}}{\frac{325}{9}}+\frac{(y-5)^{2}}{65}=1\)
⇒ centre of the ellipse = (2, 5)

Question 19.
If the line 2x – y = 4 touches the hyperbola 4x2 – 3y2 = 24, the point of contact is
(A) (1, 2)
(B) (2, 3)
(C) (3, 2)
(D) (-2, -3)
Answer:
(C) (3, 2)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

Question 20.
The foci of hyperbola 4x2 – 9y2 – 36 = 0 are
(A) (±√13, 0)
(B) (±√11, 0)
(C) (±√12, 0)
(D) (0, ±√12)
Answer:
(A) (±√13, 0)

II. Answer the following.

Question 1.
For each of the following parabolas, find focus, equation of file directrix, length of the latus rectum and ends of the latus rectum.
(i) If 2y2 = 17x
(ii) 5x2 = 24y
Solution:
(i) Given equation of the parabola is 2y2 = 17x
y2 = \(\frac{17}{2}\)x
Comparing this equation with y2 = 4ax, we get
4a = \(\frac{17}{2}\)
a = \(\frac{17}{8}\)
Co-ordinates of focus are S(a, 0), i.e., S(\(\frac{17}{8}\), 0)
Equation of the directrix is x + a = 0
x + \(\frac{17}{8}\) = 0
8x + 17 = 0
Length of latus rectum = 4a = 4(\(\frac{17}{8}\)) = \(\frac{17}{2}\)
Co-ordinates of end points of latus rectum are (a, 2a) and (a, -2a)
i.e., \(\left(\frac{17}{8}, \frac{17}{4}\right)\) and \(\left(\frac{17}{8},-\frac{17}{4}\right)\)

(ii) Given equation of the parabola is 5x2 = 24y
x2 = \(\frac{24 y}{5}\)
Comparing this equation with x2 = 4by, we get
4b = \(\frac{24}{5}\)
b = \(\frac{6}{5}\)
Co-ordinates of focus are S(0, b), i.e., S(0, \(\frac{6}{5}\))
Equation of the directrix is y + b = 0
y + \(\frac{6}{5}\) = 0
5y + 6 = 0
Length of latus rectum = 4b = 4(\(\frac{6}{5}\)) = \(\frac{24}{5}\)
Co-ordinates of end points of latus rectum are (2b, b) and (-2b, b), i.e., \(\left(\frac{12}{5}, \frac{6}{5}\right)\) and \(\left(\frac{-12}{5}, \frac{6}{5}\right)\)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

Question 2.
Find the cartesian co-ordinates of the points on the parabola y2 = 12x whose parameters are
(i) 2
(ii) -3
Solution:
Given equation of the parabola is y2 = 12x
Comparing this equation with y2 = 4ax, we get
4a = 12
∴ a = 3
If t is the parameter of the point P on the parabola, then
P(t) = (at2, 2at)
i.e., x = at2 and y = 2at …..(i)
(i) Given, t = 2
Substituting a = 3 and t = 2 in (i), we get
x = 3(2)2 and y = 2(3)(2)
x = 12 and y = 12
∴ The cartesian co-ordinates of the point on the parabola are (12, 12).

(ii) Given, t = -3
Substitùting a = 3 and t = -3 in (i), we get
x = 3(-3)2 and y = 2(3)(-3)
∴ x = 27 and y = -18
∴ The cartesian co-ordinates of the point on the parabola are (27, -18).

Question 3.
Find the co-ordinates of a point of the parabola y2 = 8x having focal distance 10.
Solution:
Given equation of the parabola is y2 = 8x
Comparing this equation with y2 = 4ax, we get
4a = 8
∴ a = 2
Focal distance of a point = x + a
Given, focal distance = 10
x + 2 = 10
∴ x = 8
Substituting x = 8 in y2 = 8x, we get
y2 = 8(8)
∴ y = ±8
∴ The co-ordinates of the points on the parabola are (8, 8) and (8, -8).

Question 4.
Find the equation of the tangent to the parabola y2 = 9x at the point (4, -6) on it.
Solution:
Given equation of the parabola is y2 = 9x
Comparing this equation with y2 = 4ax, we get
4a = 9
∴ a = \(\frac{9}{4}\)
Equation of the tangent y2 = 4ax at (x1, y1) is yy1 = 2a(x + x1)
The equation of the tangent at (4, -6) is
y(-6) = 2(\(\frac{9}{4}\))(x + 4)
⇒ -6y = \(\frac{9}{2}\) (x + 4)
⇒ -12y = 9x + 36
⇒ 9x + 12y + 36 = 0
⇒ 3x + 4y + 12 = 0

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

Question 5.
Find the equation of the tangent to the parabola y2 = 8x at t = 1 on it.
Solution:
Given equation of the parabola is y2 = 8x
Comparing this equation with y2 = 4ax, we get
4a = 8
a = 2
t = 1
Equation of tangent with parameter t is yt = x + at2
∴ The equation of tangent with t = 1 is
y(1) = x + 2(1)2
y = x + 2
∴ x – y + 2 = 0

Question 6.
Find the equations of the tangents to the parabola y2 = 9x through the point (4, 10).
Solution:
Given equation of the parabola is y2 = 9x
Comparing this equation with y2 = 4ax, we get
4a = 9
∴ a = \(\frac{9}{4}\)
Equation of tangent to the parabola y2 = 4ax having slope m is
y = mx + \(\frac{a}{m}\)
y = mx + \(\frac{9}{4 m}\)
But, (4, 10) lies on the tangent.
10 = 4m + \(\frac{9}{4 m}\)
⇒ 40m = 16m2+ 9
⇒ 16m2 – 40m + 9 = 0
⇒ 16m2 – 36m – 4m + 9 = 0
⇒ 4m(4m – 9) – 1(4m – 9) = 0
⇒ (4m – 9) (4m – 1) = 0
⇒ 4m – 9 = 0 or 4m – 1 = 0
⇒ m = \(\frac{9}{4}\) or m = \(\frac{1}{4}\)
These are the slopes of the required tangents.
By slope point form, y – y1 = m(x – x1),
the equations of the tangents are
y – 10 = \(\frac{9}{4}\)(x – 4) or y – 10 = \(\frac{1}{4}\)(x – 4)
⇒ 4y – 40 = 9x – 36 or 4y – 40 = x – 4
⇒ 9x – 4y + 4 = 0 or x – 4y + 36 = 0

Question 7.
Show that the two tangents drawn to the parabola y2 = 24x from the point (-6, 9) are at the right angle.
Solution:
Given the equation of the parabola is y2 = 24x.
Comparing this equation with y2 = 4ax, we get
4a = 24
⇒ a = 6
Equation of tangent to the parabola y2 = 4ax having slope m is
y = mx + \(\frac{a}{m}\)
⇒ y = mx + \(\frac{6}{m}\)
But, (-6, 9) lies on the tangent
9 = -6m + \(\frac{6}{m}\)
⇒ 9m = -6m2 + 6
⇒ 6m2 + 9m – 6 = 0
The roots m1 and m2 of this quadratic equation are the slopes of the tangents.
m1m2 = -1
Tangents drawn to the parabola y2 = 24x from the point (-6, 9) are at a right angle.

Alternate method:
Comparing the given equation with y2 = 4ax, we get
4a = 24
⇒ a = 6
Equation of the directrix is x = -6.
The given point lies on the directrix.
Since tangents are drawn from a point on the directrix are perpendicular,
Tangents drawn to the parabola y2 = 24x from the point (-6, 9) are at the right angle.

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

Question 8.
Find the equation of the tangent to the parabola y2 = 8x which is parallel to the line 2x + 2y + 5 = 0. Find its point of contact.
Solution:
Given the equation of the parabola is y2 = 8x.
Comparing this equation with y2 = 4ax, we get
4a = 8
a = 2
Slope of the line 2x + 2y + 5 = 0 is -1
Since the tangent is parallel to the given line,
slope of the tangent line is m = -1
Equation of tangent to the parabola y2 = 4ax having slope m is y = mx + \(\frac{a}{m}\)
Equation of the tangent is
y = -x + \(\frac{2}{-1}\)
x + y + 2 = 0
Point of contact = \(\left(\frac{a}{m^{2}}, \frac{2 a}{m}\right)\)
= \(\left(\frac{2}{(-1)^{2}}, \frac{2(2)}{-1}\right)\)
= (2, -4)

Question 9.
A line touches the circle x2 + y2 = 2 and the parabola y2 = 8x. Show that its equation is y = ±(x + 2).
Solution:
Given equation of the parabola is y2 = 8x
Comparing this equation with y2 = 4ax, we get
4a = 8
a = 2
Equation of tangent to given parabola with slope m is
y = mx + \(\frac{2}{m}\)
m2x – my + 2 = 0 ….(i)
Equation of the circle is x2 + y2 = 2
Its centre = (0, 0) and Radius = √2
Line (i) touches the circle.
Length of perpendicular from the centre to the line (i) = radius
⇒ \(\left|\frac{m^{2}(0)-m(0)+2}{\sqrt{m^{4}+m^{2}}}\right|\) = √2
⇒ \(\frac{4}{m^{4}+m^{2}}\) = 2
⇒ m4 + m2 – 2 – 0
⇒ (m2 + 2)(m2 – 1) = 0
Since m2 ≠ -2,
m2 – 1 = 0
⇒ m = ±1
When m = 1, equation of the tangent is
y = (1)x + \(\frac{2}{(1)}\)
y = (x + 2) …..(i)
When m = -1, equation of the tangent is
y = (-1)x + \(\frac{2}{(-1)}\)
y = -x – 2
y = -(x + 2) …..(ii)
From (i) and (ii),
equation of the common tangents to the given parabola is y = ±(x + 2)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

Question 10.
Two tangents to the parabola y2 = 8x meet the tangents at the vertex in P and Q. If PQ = 4, prove that the locus of the point of intersection of the two tangents is y2 = 8(x + 2).
Solution:
Given parabola is y2 = 8x
Comparing with y2 = 4ax, we get,
4a = 8
⇒ a = 2
Let M(t1) and N(t2) be any two points on the parabola.
The equations of tangents at M and N are
yt1 = x + \(2 \mathrm{t}_{1}^{2}\) …..(1)
yt2 = x + \(2 \mathrm{t}_{2}^{2}\) …(2) ….[∵ a = 2]
Let tangent at M meet the tangent at the vertex in P.
But tangent at the vertex is Y-axis whose equation is x = 0.
⇒ to find P, put x = 0 in (1)
⇒ yt1 = \(2 \mathrm{t}_{1}^{2}\)
⇒ y = 2t1 …..(t1 ≠ 0 otherwise tangent at M will be x = 0)
⇒ P = (0, 2t1)
Similarly, Q = (0, 2t2)
It is given that PQ = 4
∴ |2t1 – 2t2| = 4
∴ |t1 – t2| = 2 …..(3)
Let R = (x1, y1) be any point on the required locus.
Then R is the point of intersection of tangents at M and N.
To find R, we solve (1) and (2).
Subtracting (2) from (1), we get
y(t1 – t2) = \(2 \mathrm{t}_{1}^{2}-2 \mathrm{t}_{2}^{2}\)
y(t1 – t2) = 2(t1 – t2)(t1 + t2)
∴ y = 2(t1 + t2) …..[∵ M, N are distinct ∴ t1 ≠ t2]
i.e., y1 = 2(t1 + t2) …..(4)
∴ from (1), we get
2t1(t1 + t2) = x + \(2 \mathrm{t}_{1}^{2}\)
∴ 2t1t2 = x i.e. x1 = 2t1t2 …..(5)
To find the equation of locus of R(x1, y1),
we eliminate t1 and t2 from the equations (3), (4) and (5).
We know that,
(t1 + t2)2 = (t1 + t2)2 + 4t1t2
⇒ \(\left(\frac{y_{1}}{2}\right)^{2}=4+4\left(\frac{x_{1}}{2}\right)\) …[By (3), (4) and (5)]
⇒ \(y_{1}^{2}\) = 16 + 8x1 = 8(x1 + 2)
Replacing x1 by x and y1 by y,
the equation of required locus is y2 = 8(x + 2).

Question 11.
The slopes of the tangents drawn from P to the parabola y2 = 4ax are m1 and m2, showing that
(i) m1 – m2 = k
(ii) \(\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)\) = k, where k is a constant.
Solution:
Let P(x1, y1) be any point on the parabola y2 = 4ax.
Equation of tangent to the parabola y2 = 4ax having slope m is y = mx + \(\frac{\mathrm{a}}{\mathrm{m}}\)
This tangent passes through P(x1, y1).
y1 = mx1 + \(\frac{\mathrm{a}}{\mathrm{m}}\)
my1 = m2x1 + a
m2x1 – my1 + a = 0
This is a quadratic equation in ‘m’.
The roots m1 and m2 of this quadratic equation are the slopes of the tangents drawn from P.
∴ m1 + m2 = \(\frac{y_{1}}{x_{1}}\), m1m2 = \(\frac{a}{x_{1}}\)
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7 II Q11
Since (x1, y1) and a are constants, m1 – m2 is a constant.
∴ m1 – m2 = k, where k is constant.

(ii) Since (x1, y1) and a are constants, m1m2 is a constant.
\(\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)\) = k, where k is a constant.

Question 12.
The tangent at point P on the parabola y2 = 4ax meets the Y-axis in Q. If S is the focus, show that SP subtends a right angle at Q.
Solution:
Let P(\(a t_{1}^{2}\), 2at1) be a point on the parabola and
S(a, 0) be the focus of parabola y2 = 4ax
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7 II Q12
Since the tangent passing through point P meet Y-axis at point Q,
equation of tangent at P(\(a t_{1}^{2}\), 2at1) is
yt1 = x + \(a t_{1}^{2}\) …..(i)
∴ Point Q lie on tangent
∴ put x = 0 in equation (i)
yt1 = \(a t_{1}^{2}\)
y = at1
∴ Co-ordinate of point Q(0, at1)
S = (a, 0), P(\(a t_{1}^{2}\), 2at1), Q(0, at1)
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7 II Q12.1
∴ SP subtends a right angle at Q.

Question 13.
Find the
(i) lengths of the principal axes
(ii) co-ordinates of the foci
(iii) equations of directrices
(iv) length of the latus rectum
(v) Distance between foci
(vi) distance between directrices of the curve
(a) \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)
(b) 16x2 + 25y2 = 400
(c) \(\frac{x^{2}}{144}-\frac{y^{2}}{25}=1\)
(d) x2 – y2 = 16
Solution:
(a) Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a2 = 25 and b2 = 9
∴ a = 5 and b = 3
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.
(i) Length of major axis = 2a = 2(5) = 10
Length of minor axis = 2b = 2(3) = 6
∴ Lengths of the principal axes are 10 and 6.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)
∴ e = \(\frac{\sqrt{25-9}}{5}\) = \(\frac{4}{5}\)
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0)
i.e., S(5(\(\frac{4}{5}\)), 0) and S'(-5(\(\frac{4}{5}\)), 0),
i.e., S(4, 0) and S'(-4, 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\)
i.e., x = ±\(\frac{5}{\frac{4}{5}}\)
i.e., x = ±\(\frac{25}{4}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(3)^{2}}{5}=\frac{18}{5}\)

(v) Distance between foci = 2ae = 2 (5) (\(\frac{4}{5}\)) = 8

(vi) Distance between directrices = \(\frac{2 a}{e}\) = \(\frac{2(5)}{\frac{4}{5}}\) = \(\frac{25}{2}\)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

(b) Given equation of the ellipse is 16x2 + 25y2 = 400
\(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a2 = 25 and b2 = 16
∴ a = 5 and b = 4
Since a > b,
X-axis is the major axis and Y-axis is the minor axis
(i) Length of major axis = 2a = 2(5) = 10
Length of minor axis = 2b = 2(4) = 8
Lengths of the principal axes are 10 and 8.

(ii) b2 = a2 (1 – e2)
16 = 25(1 – e2)
\(\frac{16}{25}\) = 1 – e2
e2 = 1 – \(\frac{16}{25}\)
e2 = \(\frac{9}{25}\)
e = \(\frac{3}{5}\) ……[∵ 0 < e < 1]
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S(5(\(\frac{3}{5}\)), 0) and S'(-5(\(\frac{3}{5}\)), 0),
i.e., S(3, 0) and S'(-3, 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\)
i.e., x = ±\(\frac{5}{\left(\frac{3}{5}\right)}\)
i.e., x = ±\(\frac{25}{3}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(16)}{5}=\frac{32}{5}\)

(v) Distance between foci = 2ae = 2(5)(\(\frac{3}{5}\)) = 6

(vi) Distance between directrices = \(\frac{2 a}{e}=\frac{2(5)}{\left(\frac{3}{5}\right)}=\frac{50}{3}\)

(c) Given equation of the hyperbola \(\frac{x^{2}}{144}-\frac{y^{2}}{25}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
a2 = 144 and b2 = 25
∵ a = 12 and b = 5
(i) Length of transverse axis = 2a = 2(12) = 24
Length of conjugate axis = 2b = 2(5) = 10
lengths of the principal axes are 24 and 10.

(ii) b2 = a2(e2 – 1)
25 = 144 (e2 – 1)
\(\frac{25}{144}\) = e2 – 1
e2 = 1 + \(\frac{25}{144}\)
e2 = \(\frac{169}{144}\)
e = \(\frac{13}{12}\) …….[∵ e > 1]
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0)
i.e., S(12(\(\frac{13}{12}\)), 0) and S'(-12(\(\frac{13}{12}\)), 0)
i.e., S(13, 0) and S'(-13, 0)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

(iii) Equations of the directrices are x = \(\pm \frac{a}{e}\)
i.e., x = \(\pm \frac{12}{\left(\frac{13}{12}\right)}\)
i.e., x = \(\pm \frac{144}{13}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2(25)}{12}=\frac{25}{6}\)

(v) Distance between foci = 2ae = 2(12)(\(\frac{13}{12}\)) = 26

(vi) Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}=\frac{2(12)}{\left(\frac{13}{12}\right)}\) = \(\frac{288}{13}\)

(d) Given equation of the hyperbola is x2 – y2 = 16
∴ \(\frac{x^{2}}{16}-\frac{y^{2}}{16}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a2 = 16 and b2 = 16
∴ a = 4 and b = 4
(i) Length of transverse axis = 2a = 2(4) = 8
Length of conjugate axis = 2b = 2(4) = 8

(ii) We know that
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7 II Q13(d)
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0),
i.e., S(4√2, 0) and S'(-4√2, 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\)
∴x = ± \(\frac{4}{\sqrt{2}}\)
∴ x = ±2√2

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2(16)}{4}\) = 8

(v) Distance between foci = 2ae = 2(4)(√2) = 8√2

(vi) Distance between directrices = \(\frac{2 a}{e}\) = \(\frac{2(4)}{\sqrt{2}}\) = 4√2.

Question 14.
Find the equation of the ellipse in standard form if
(i) eccentricity = \(\frac{3}{8}\) and distance between its foci = 6.
(ii) the length of the major axis is 10 and the distance between foci is 8.
(iii) passing through the points (-3, 1) and (2, -2).
Solution:
(i) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Given, eccentricity (e) = \(\frac{3}{8}\)
Distance between foci = 2ae
Given, distance between foci = 6
∴ 2ae = 6
∴ 2a(\(\frac{3}{8}\)) = 6
∴ \(\frac{3a}{4}\) = 6
∴ a = 8
∴ a2 = 64
Now, b2 = a2 (1 – e2)
= \(64\left[1-\left(\frac{3}{8}\right)^{2}\right]\)
= \(4\left(1-\frac{9}{64}\right)\)
= 64(\(\frac{55}{64}\))
= 55
∴ The required equation of the ellipse is \(\frac{x^{2}}{64}+\frac{y^{2}}{55}=1\)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

(ii) Let the equation of the ellipse be
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) ……(1)
Then length of major axis = 2a = 10
∴ a = 5
Also, distance between foci= 2ae = 8
∴ 2 × 5 × e = 8
∴ e = \(\frac{4}{5}\)
∴ b2 = a2(1 – e2)
= 25(1 – \(\frac{6}{25}\))
= 9
∴ from (1), the equation of the required ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)

(iii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
The ellipse passes through the points (-3, 1) and (2, -2).
∴ Substituting x = -3 and y = 1 in equation of ellipse, we get
\(\frac{(-3)^{2}}{a^{2}}+\frac{1^{2}}{b^{2}}=1\)
∴ \(\frac{9}{a^{2}}+\frac{1}{b^{2}}=1\) …..(i)
Substituting x = 2 and y = -2 in equation of ellipse, we get
\(\frac{2^{2}}{a^{2}}+\frac{(-2)^{2}}{b^{2}}=1\)
∴ \(\frac{4}{a^{2}}+\frac{4}{b^{2}}=1\) ……(ii)
Let \(\frac{1}{a^{2}}\) = A and \(\frac{1}{b^{2}}\) = B
∴ Equations (i) and (ii) become
9A + B = 1 ..…(iii)
4A + 4B = 1 …..(iv)
Multiplying (iii) by 4, we get
36A + 4B = 4 …..(v)
Subtracting (iv) from (v), we get
32A = 3
∴ A = \(\frac{3}{32}\)
Substituting A = \(\frac{3}{32}\) in (iv), we get
4(\(\frac{3}{32}\)) + 4B = 1
∴ \(\frac{3}{8}\) + 4B = 1
∴ 4B = 1 – \(\frac{3}{8}\)
∴ 4B = \(\frac{5}{8}\)
∴ B = \(\frac{5}{32}\)
Since \(\frac{1}{a^{2}}\) = A and \(\frac{1}{b^{2}}\) = B
\(\frac{1}{a^{2}}=\frac{3}{32}\) and \(\frac{1}{b^{2}}=\frac{5}{32}\)
∴ a2 = \(\frac{32}{3}\) and b2 = \(\frac{32}{5}\)
∴ The required equation of ellipse is
\(\frac{x^{2}}{\left(\frac{32}{3}\right)}+\frac{y^{2}}{\left(\frac{32}{5}\right)}\)
i.e., 3x2 + 5y2 = 32.

Question 15.
Find the eccentricity of an ellipse if the distance between its directrices is three times the distance between its foci.
Solution:
Let the equation of the ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
It is given that,
distance between directrices is three times the distance between the foci.
∴ \(\frac{2a}{e}\) = 3(2ae)
∴ 1 = 3e2
∴ e2 = \(\frac{1}{3}\)
∴ e = \(\frac{1}{\sqrt{3}}\) …..[∵ 0 < e < 1]

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

Question 16.
For the hyperbola \(\frac{x^{2}}{100}-\frac{y^{2}}{25}=1\), prove that SA . S’A = 25, where S and S’ are the foci and A is the vertex.
Solution:
Given equation of the hyperbola is \(\frac{x^{2}}{100}-\frac{y^{2}}{25}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 100 and b2 = 25
∴ a = 10 and b = 5
∴ Co-ordinates of vertex is A(a, 0), i.e., A(10, 0)
Eccentricity, e = \(\frac{\sqrt{a^{2}+b^{2}}}{a}\)
= \(\frac{\sqrt{100+25}}{10}\)
= \(\frac{\sqrt{125}}{10}\)
= \(\frac{5 \sqrt{5}}{10}\)
= \(\frac{\sqrt{5}}{2}\)
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0)
i.e., S(10(\(\frac{\sqrt{5}}{2}\)), 0) and S'(-10(\(\frac{\sqrt{5}}{2}\)), 0)
i.e., S(5√5, 0) and S'(-5√5, 0)
Since S, A and S’ lie on the X-axis,
SA = |5√5 – 10| and S’A = |-5√5 – 10|
= |-(5√5 + 10)|
= |5√5 + 10|
∴ SA . S’A = |5√5 – 10| |5√5 + 10|
= |(5√5)2 – (10)2|
= |125 – 100|
= |25|
SA . S’A = 25

Question 17.
Find the equation of the tangent to the ellipse \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\) passing through the point (2, -2).
Solution:
Given equation of the ellipse is \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a2 = 5 and b2 = 4
Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
Since (2, -2) lies on both the tangents,
-2 = 2m ± \(\sqrt{5 m^{2}+4}\)
∴ -2 – 2m = ±\(\sqrt{5 m^{2}+4}\)
Squaring both the sides, we get
4m2 + 8m + 4 = 5m2 + 4
∴ m2 – 8m = 0
∴ m(m – 8) = 0
∴ m = 0 or m = 8
These are the slopes of the required tangents.
∴ By slope point form y – y1 = m(x – x1),
the equations of the tangents are
y + 2 = 0(x – 2) and y + 2 = 8(x – 2)
∴ y + 2 = 0 and y + 2 = 8x – 16
∴ y + 2 = 0 and 8x – y – 18 = 0.

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

Question 18.
Find the equation of the tangent to the ellipse x2 + 4y2 = 100 at (8, 3).
Solution:
Given equation of ellipse is x2 + 4y2 = 100
∴ \(\frac{x^{2}}{100}+\frac{y^{2}}{25}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 100 and b2 = 25
Equation of tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) at (x1, y1) is \(\frac{x x_{1}}{a^{2}}+\frac{y y_{1}}{b^{2}}=1\)
Equation of tangent at (8, 3) is
\(\frac{8 x}{100}+\frac{3 y}{25}=1\)
\(\frac{2 x}{25}+\frac{3 y}{25}=1\)
2x + 3y = 25

Question 19.
Show that the line 8y + x = 17 touches the ellipse x2 + 4y2 = 17. Find the point of contact.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7 II Q19
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7 II Q19.1

Question 20.
Tangents are drawn through a point P to the ellipse 4x2 + 5y2 = 20 having inclinations θ1 and θ2 such that tan θ1 + tan θ2 = 2. Find the equation of the locus of P.
Solution:
Given equation of the ellipse is 4x2 + 5y2 = 20.
∴ \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a2 = 5 and b2 = 4
Since inclinations of tangents are θ1 and θ2,
m1 = tan θ1 and m2 = tan θ2
Equation of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{\mathrm{a}^{2} \mathrm{~m}^{2}+\mathrm{b}^{2}}\)
∴ y = mx ± \(\sqrt{5 m^{2}+4}\)
∴ y – mx = ±\(\sqrt{5 m^{2}+4}\)
Squaring both the sides, we get
y2 – 2mxy + m2x2 = 5m2 + 4
∴ (x2 – 5)m2 – 2xym + (y2 – 4) = 0
The roots m1 and m2 of this quadratic equation are the slopes of the tangents.
∴ m1 + m2 = \(\frac{-(-2 x y)}{x^{2}-5}=\frac{2 x y}{x^{2}-5}\)
Given, tan θ1 + tan θ2 = 2
∴ m1 + m2 = 2
∴ \(\frac{2 x y}{x^{2}-5}\)
∴ xy = x2 – 5
∴ x2 – xy – 5 = 0, which is the required equation of the locus of P.

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

Question 21.
Show that the product of the lengths of its perpendicular segments drawn from the foci to any tangent line to the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) is equal to 16.
Solution:
Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
∴ a2 = 25, b2 = 16
∴ a = 5, b = 4
We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)
∴ e = \(\frac{\sqrt{25-16}}{5}\) = \(\frac{3}{5}\)
ae = 5(\(\frac{3}{5}\)) = 3
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0),
i.e., S(3, 0) and S'(-3, 0)
Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{\mathrm{a}^{2} \mathrm{~m}^{2}+\mathrm{b}^{2}}\)
Equation of one of the tangents to the ellipse is
y = mx + \(\sqrt{25 \mathrm{~m}^{2}+16}\)
∴ mx – y + \(\sqrt{25 \mathrm{~m}^{2}+16}\) = 0 …..(i)
p1 = length of perpendicular segment from S(3, 0) to the tangent (i)
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7 II Q21
p2 = length of perpendicular segment from S'(-3, 0) to the tangent (i)
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7 II Q21.1

Question 22.
Find the equation of the hyperbola in the standard form if
(i) Length of conjugate axis is 5 and distance between foci is 13.
(ii) eccentricity is \(\frac{3}{2}\) and distance between foci is 12.
(iii) length of the conjugate axis is 3 and the distance between the foci is 5.
Solution:
(i) Let the required equation of hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Length of conjugate axis = 2b
Given, length of conjugate axis = 5
2b = 5
b = \(\frac{5}{2}\)
b2 = \(\frac{25}{4}\)
Distance between foci = 2ae
Given, distance between foci = 13
2ae = 13
ae = \(\frac{13}{2}\)
a2e2 = \(\frac{169}{4}\)
Now, b2 = a2(e2 – 1)
b2 = a2e2 – a2
\(\frac{25}{4}\) = \(\frac{169}{4}\) – a2
a2 = \(\frac{169}{4}-\frac{25}{4}\) = 36
∴ The required equation of hyperbola is \(\frac{x^{2}}{36}-\frac{y^{2}}{\frac{25}{4}}=1\)
i.e., \(\frac{x^{2}}{36}-\frac{4 y^{2}}{25}=1\)

(ii) Let the required equation of hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Given, eccentricity (e) = \(\frac{3}{2}\)
Distance between foci = 2ae
Given, distance between foci = 12
∴ 2ae = 12
∴ 2a(\(\frac{3}{2}\)) = 12
∴ 3a = 12
∴ a = 4
∴ a2 = 16
Now, b2 = a2(e2 – 1)
∴ b2 = \(\left[\left(\frac{3}{2}\right)^{2}-1\right]\)
∴ b2 = 16(\(\frac{9}{4}\) – 1)
∴ b2 = 16(\(\frac{5}{4}\))
∴ b2 = 20
∴ The required equation of hyperbola is \(\frac{x^{2}}{16}-\frac{y^{2}}{20}=1\)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

(iii) Let the required equation of hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Length of conjugate axis = 2b
Given, length of conjugate axis = 3
∴ 2b = 3
∴ b = \(\frac{3}{2}\)
∴ b2 = \(\frac{9}{4}\)
Distance between foci = 2ae
Given, distance between foci = 5
∴ 2ae = 5
∴ ae = \(\frac{5}{2}\)
∴ a2e2 = \(\frac{25}{4}\)
Now, b2 = a2(e2 – 1)
∴ b2 = a2e2 – a2
∴ \(\frac{9}{4}\) = \(\frac{25}{4}\) – a2
∴ a2 = \(\frac{25}{4}-\frac{9}{4}\)
∴ a2 = 4
∴ The required equation of hyperbola is \(\frac{x^{2}}{4}-\frac{y^{2}}{\left(\frac{9}{4}\right)}=1\)
i.e., \(\frac{x^{2}}{4}-\frac{4 y^{2}}{9}=1\)

Question 23.
Find the equation of the tangent to the hyperbola,
(i) 7x2 – 3y2 = 51 at (-3, -2)
(ii) x = 3 sec θ, y = 5 tan θ at θ = π/3
(iii) \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=1\) at P(30°).
Solution:
(i) Given equation of the hyperbola is 7x2 – 3y2 = 51
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7 II Q23(i)

(ii) Given, equation of the hyperbola is
x = 3 sec θ, y = 5 tan θ
Since sec2 θ – tan2 θ = 1,
\(\frac{x^{2}}{9}-\frac{y^{2}}{25}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 9 and b2 = 25
a = 3 and b = 5
Equation of tangent at P(θ) is
\(\frac{x \sec \theta}{\mathrm{a}}-\frac{y \tan \theta}{\mathrm{b}}=1\)
∴ Equation of tangent at P(π/3) is
\(\frac{x \sec \left(\frac{\pi}{3}\right)}{3}-\frac{y \tan \left(\frac{\pi}{3}\right)}{5}=1\)
\(\frac{2 x}{3}-\frac{\sqrt{3} y}{5}=1\)
10x – 3√3 y = 15

(iii) Given equation of hyperbola is \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a2 = 25 and b2 = 16
a = 5 and b = 4
Equation of tangent at P(θ) is
\(\frac{x \sec \theta}{\mathrm{a}}-\frac{y \tan \theta}{\mathrm{b}}=1\)
The equation of tangent at P(30°) is
\(\frac{x \sec 30^{\circ}}{5}-\frac{y \tan 30^{\circ}}{4}=1\)
\(\frac{2 x}{5 \sqrt{3}}-\frac{y}{4 \sqrt{3}}=1\)
8x – 5y = 20√3

Question 24.
Show that the line 2x – y = 4 touches the hyperbola 4x2 – 3y2 = 24. Find the point of contact.
Solution:
Given equation of die hyperbola is 4x2 – 3y2 = 24.
∴ \(\frac{x^{2}}{6}-\frac{y^{2}}{8}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 6 and b2 = 8
Given equation of line is 2x – y = 4
∴ y = 2x – 4
Comparing this equation with y = mx + c, we get
m = 2 and c = -4
For the line y = mx + c to be a tangent to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we must have
c2 = a2m2 – b2
c2 = (-4)2 = 16
a2m2 – b2 = 6(2)2 – 8 = 24 – 8 = 16
∴ The given line is a tangent to the given hyperbola and point of contact
= \(\left(-\frac{\mathrm{a}^{2} \mathrm{~m}}{\mathrm{c}},-\frac{\mathrm{b}^{2}}{\mathrm{c}}\right)\)
= \(\left(\frac{-6(2)}{-4}, \frac{-8}{-4}\right)\)
= (3, 2)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

Question 25.
Find the equations of the tangents to the hyperbola 3x2 – y2 = 48 which are perpendicular to the line x + 2y – 7 = 0.
Solution:
Given the equation of the hyperbola is 3x2 – y2 = 48.
∴ \(\frac{x^{2}}{16}-\frac{y^{2}}{48}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a2 = 16 and b2 = 48
Slope of the line x + 2y – 7 = 0 is \(-\frac{1}{2}\)
Since the given line is perpendicular to the tangents,
slope of the required tangent (m) = 2
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Equations of tangents to the ellipse having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
y = 2x ± \(\sqrt{16(2)^{2}-48}\)
y = 2x ± √16
∴ y = 2x ± 4

Question 26.
Two tangents to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) make angles θ1, θ2, with the transverse axis. Find the locus of their point of intersection if tan θ1 + tan θ2 = k.
Solution:
Given equation of the hyperbola is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Let θ1 and θ2 be the inclinations.
m1 = tan θ1, m2 = tan θ2
Let P(x1, y1) be a point on the hyperbola
Equation of a tangent with slope ‘m’ to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) is
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
This tangent passes through P(x1, y1).
y1 = mx1 ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
(y1 – mx1)2 = a2m2 – b2
\(\left(x_{1}{ }^{2}-\mathrm{a}^{2}\right) \mathrm{m}^{2}-2 x_{1} y_{1} \mathrm{~m}+\left(y_{1}{ }^{2}+\mathrm{b}^{2}\right)=0\) ……(i)
This is a quadratic equation in ‘m’.
It has two roots say m1 and m2, which are the slopes of two tangents drawn from P.
∴ m1 + m2 = \(\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}\)
Since tan θ1 + tan θ2 = k,
\(\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}=k\)
∴ P(x1, y1) moves on the curve whose equation is k(x2 – a2) = 2xy.

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 7 Conic Sections Ex 7.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3

Question 1.
Find the length of the transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices, and the length of the latus rectum of the hyperbolae.
(i) \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=1\)
(ii) \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=-1\)
(iii) 16x2 – 9y2 = 144
(iv) 21x2 – 4y2 = 84
(v) 3x2 – y2 = 4
(vi) x2 – y2 = 16
(vii) \(\frac{y^{2}}{25}-\frac{x^{2}}{9}=1\)
(viii) \(\frac{y^{2}}{25}-\frac{x^{2}}{144}=1\)
(ix) \(\frac{x^{2}}{100}-\frac{y^{2}}{25}=1\)
(x) x = 2 sec θ, y = 2√3 tan θ
Solution:
(i) Given equation of the hyperbola is \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a2 = 25 and b2 = 16
⇒ a = 5 and b = 4
Length of transverse axis = 2a = 2(5) = 10
Length of conjugate axis = 2b = 2(4) = 8
We know that
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q1 (i)

(ii) Given equation of the hyperbola is \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=-1\)
\(\frac{y^{2}}{16}-\frac{x^{2}}{25}=1\)
Comparing this equation with \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\), we get
b2 = 16 and a2 = 25
⇒ b = 4 and a = 5
Length of transverse axis = 2b = 2(4) = 8
Length of conjugate axis = 2a = 2(5) = 10
Co-ordinates of vertices are B(0, b) and B’ (0, -b)
i.e., B(0, 4) and B'(0, -4)
We know that
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q1 (ii)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3

(iii) Given equation of the hyperbola is 16x2 – 9y2 = 144.
\(\frac{x^{2}}{9}-\frac{y^{2}}{16}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a2 = 9 and b2 = 16
⇒ a = 3 and b = 4
Length of transverse axis = 2a = 2(3) = 6
Length of conjugate axis = 2b = 2(4) = 8
We know that
e = \(\frac{\sqrt{a^{2}+b^{2}}}{a}=\frac{\sqrt{9+16}}{3}=\frac{\sqrt{25}}{3}=\frac{5}{3}\)
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0),
i.e., S(3(\(\frac{5}{3}\)), 0) and S'(-3(\(\frac{5}{3}\)), 0)
i.e., S(5, 0) and S'(-5, 0)
Equations of the directrices are x = ±\(\frac{a}{e}\)
= \(\pm \frac{3}{\left(\frac{5}{3}\right)}\)
= \(\pm \frac{9}{5}\)
Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(16)}{3}=\frac{32}{3}\)

(iv) Given equation of the hyperbola is 21x2 – 4y2 = 84.
\(\frac{x^{2}}{4}-\frac{y^{2}}{21}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a2 = 4 and b2 = 21
⇒ a = 2 and b = √21
Length of transverse axis = 2a = 2(2) = 4
Length of conjugate axis = 2b = 2√21
We know that
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q1 (iv)

(v) Given equation of the hyperbola is 3x2 – y2 = 4.
\(\frac{x^{2}}{\left(\frac{4}{3}\right)}-\frac{y^{2}}{4}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a2 = \(\frac{4}{3}\) and b2 = 4
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q1 (v)

(vi) Given equation of the hyperbola is x2 – y2 = 16.
\(\frac{x^{2}}{16}-\frac{y^{2}}{16}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a2 = 16 and b2 = 16
⇒ a = 4 and b = 4
Length of transverse axis = 2a = 2(4) = 8
Length of conjugate axis = 2b = 2(4) = 8
We know that
e = \(\frac{\sqrt{a^{2}+b^{2}}}{a}=\frac{\sqrt{16+16}}{4}=\frac{\sqrt{32}}{4}=\frac{4 \sqrt{2}}{4}=\sqrt{2}\)
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0),
i.e., S (4√2, 0) and S’ (-4√2, 0)
Equations of the directrices are x = ±\(\frac{a}{e}\)
⇒ x = \(\pm \frac{4}{\sqrt{2}}\)
⇒ x = ± 2√2
Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(16)}{4}\) = 8

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3

(vii) Given equation of the hyperbola is \(\frac{y^{2}}{25}-\frac{x^{2}}{9}=1\).
Comparing this equation with \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\), we get
b2 = 25 and a2 = 9
⇒ b = 5 and a = 3
Length of transverse axis = 2b = 2(5) = 10
Length of conjugate axis = 2a = 2(3) = 6
Co-ordinates of vertices are B(0, b) and B’ (0, -b),
i.e., B(0, 5) and B’ (0, -5)
We know that
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q1 (vii)

(viii) Given equation of the hyperbola is \(\frac{y^{2}}{25}-\frac{x^{2}}{144}=1\).
Comparing this equation with \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\), we get
b2 = 25 and a2 = 144
⇒ b = 5 and a = 12
Length of transverse axis = 2b = 2(5) = 10
Length of conjugate axis = 2a = 2(12) = 24
Co-ordinates of vertices are B(0, b) and B’ (0, -b),
i.e., B(0, 5) and B’ (0, -5)
We know that
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q1 (viii)

(ix) Given equation of the hyperbola is \(\frac{x^{2}}{100}-\frac{y^{2}}{25}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a2 = 100 and b2 = 25
⇒ a = 10 and b = 5
Length of transverse axis = 2a = 2(10) = 20
Length of conjugate axis = 2b = 2(5) = 10
We know that
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q1 (ix)

(x) Given equation of the hyperbola is x = 2 sec θ, y = 2√3 tan θ.
Since sec2 θ – tan2 θ = 1,
\(\left(\frac{x}{2}\right)^{2}-\left(\frac{y}{2 \sqrt{3}}\right)^{2}=1\)
\(\frac{x^{2}}{4}-\frac{y^{2}}{12}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 4 and b2 = 12
⇒ a = 2 and b = 2√3
Length of transverse axis = 2a = 2(2) = 4
Length of conjugate axis = 2b = 2(2√3) = 4√3
We know that
e = \(\frac{\sqrt{a^{2}+b^{2}}}{a}\) = \(\frac{\sqrt{4+12}}{2}\) = 2
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0),
i.e., S(2(2), 0) and S'(-2(2), 0),
i.e., S(4, 0) and S'(-4, 0)
Equations of the directrices are x = ±\(\frac{a}{e}\).
⇒ x = ±\(\frac{2}{2}\)
⇒ x = ±1
Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(12)}{2}\) = 12

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3

Question 2.
Find the equation of the hyperbola with centre at the origin, length of the conjugate axis as 10, and one of the foci as (-7, 0).
Solution:
Given, one of the foci of the hyperbola is (-7, 0).
Since this focus lies on the X-axis, it is a standard hyperbola.
Let the required equation of hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Length of conjugate axis = 2b
Given, length of conjugate axis = 10
⇒ 2b = 10
⇒ b = 5
⇒ b2 = 25
Co-ordinates of focus are (-ae, 0)
ae = 7
⇒ a2e2 = 49
Now, b2 = a2(e2 – 1)
⇒ 25 = 49 – a2
⇒ a2 = 49 – 25 = 24
The required equation of hyperbola is \(\frac{x^{2}}{24}-\frac{y^{2}}{25}=1\)

Question 3.
Find the eccentricity of the hyperbola, which is conjugate to the hyperbola x2 – 3y2 = 3
Solution:
Given, equation of hyperbola is x2 – 3y2 = 3.
\(\frac{x^{2}}{3}-\frac{y^{2}}{1}=1\)
Equation of the hyperbola conjugate to the above hyperbola is \(\frac{y^{2}}{1}-\frac{x^{2}}{3}=1\)
Comparing this equation with \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\), we get
b2 = 1 and a2 = 3
Now, a2 = b2(e2 – 1)
⇒ 3 = 1(e2 – 1)
⇒ 3 = e – 1
⇒ e2 = 4
⇒ e = 2 …..[∵ e > 1]

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3

Question 4.
If e and e’ are the eccentricities of a hyperbola and its conjugate hyperbola respectively, prove that \(\frac{1}{e^{2}}+\frac{1}{\left(e^{\prime}\right)^{2}}=1\).
Solution:
Let e be the eccentricity of a hyperbola
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q4
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q4.1

Question 5.
Find the equation of the hyperbola referred to its principal axes:
(i) whose distance between foci is 10 and eccentricity is \(\frac{5}{2}\)
(ii) whose distance between foci is 10 and length of the conjugate axis is 6.
(iii) whose distance between directrices is \(\frac{8}{3}\) and eccentricity is \(\frac{3}{2}\).
(iv) whose length of conjugate axis = 12 and passing through (1, -2).
(v) which passes through the points (6, 9) and (3, 0).
(vi) whose vertices are (±7, 0) and endpoints of the conjugate axis are (0, ±3).
(vii) whose foci are at (±2, 0) and eccentricity is \(\frac{3}{2}\).
(viii) whose lengths of transverse and conjugate axes are 6 and 9 respectively.
(ix) whose length of transverse axis is 8 and distance between foci is 10.
Solution:
(i) Let the required equation of hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Given, eccentricity (e) = \(\frac{5}{2}\)
Distance between foci = 2ae
Given, distance between foci = 10
⇒ 2ae = 10
⇒ ae = 5
⇒ a(\(\frac{5}{2}\)) = 5
⇒ a = 2
⇒ a2 = 4
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q5 (i)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3

(ii) Let the required equation of hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Length of conjugate axis = 2b
Given, length of conjugate axis = 6
⇒ 2b = 6
⇒ b = 3
⇒ b2 = 9
Distance between foci = 2ae
Given, distance between foci = 10
⇒ 2ae = 10
⇒ ae = 5
⇒ a2e2 = 25
Now, b2 = a2 (e2 – 1)
⇒ b2 = a2 e2 – a2
⇒ 9 = 25 – a2
⇒ a2 = 25 – 9
⇒ a2 = 16
The required equation of hyperbola is \(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\)

(iii) Let the required equation of hyperbola be \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\)
Given, eccentricity (e) = \(\frac{3}{2}\)
Distance between directrices = \(\frac{2a}{e}\)
Given, distance between directrices = \(\frac{8}{3}\)
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q5 (iii)
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q5 (iii).1

(iv) Let the required equation of hyperbola be
\(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\) ……(i)
Length of conjugate axis = 2b
Given, length of conjugate axis = 12
⇒ 2b = 12
⇒ b = 6 …..(ii)
⇒ b2 = 36
The hyperbola passes through (1, -2)
Substituting x = 1 and y = -2 in (i), we get
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q5 (iv)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3

(v) Let the required equation of hyperbola be
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) ……(i)
The hyperbola passes through the points (6, 9) and (3, 0).
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q5 (v)

(vi) Let the required equation of hyperbola be
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Co-ordinates of vertices are (±a, 0).
Given that, co-ordinates of vertices are (±7, 0)
∴ a = 7
Endpoints of the conjugate axis are (0, b) and (0, -b).
Given, the endpoints of the conjugate axis are (0, ±3).
∴ b = 3
The required equation of hyperbola is \(\frac{x^{2}}{7^{2}}-\frac{y^{2}}{3^{2}}=1\)
i.e., \(\frac{x^{2}}{49}-\frac{y^{2}}{9}=1\)

(vii) Let the required equation of hyperbola be
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) ……(i)
Given, eccentricity (e) = \(\frac{3}{2}\)
Co-ordinates of foci are (±ae, 0).
Given co-ordinates of foci are (±2, 0)
ae = 2
⇒ a(\(\frac{3}{2}\)) = 2
⇒ a = \(\frac{4}{3}\)
⇒ a2 = \(\frac{16}{9}\)

(viii) Let the required equation of hyperbola be
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Length of transverse axis = 2a
Given, length of transverse axis = 6
⇒ 2a = 6
⇒ a = 3
⇒ a2 = 9
Length of conjugate axis = 2b
Given, length of conjugate axis = 9
⇒ 2b = 9
⇒ b = \(\frac{9}{2}\)
⇒ b2 = \(\frac{81}{4}\)
The required equation of hyperbola is
\(\frac{x^{2}}{9}-\frac{y^{2}}{\left(\frac{81}{4}\right)}=1\)
i.e., \(\frac{x^{2}}{9}-\frac{4 y^{2}}{81}=1\)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3

(ix) Let the required equation of hyperbola be
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Length of transverse axis = 2a
Given, length of transverse axis = 8
⇒ 2a = 8
⇒ a = 4
⇒ a2 = 16
Distance between foci = 2ae
Given, distance between foci = 10
⇒ 2ae = 10
⇒ ae = 5
⇒ a2e2 = 25
Now, b2 = a2 (e2 – 1)
⇒ b2 = a2 e2 – a2
⇒ b2 = 25 – 16 = 9
The required equation of hyperbola is \(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\)

Question 6.
Find the equation of the tangent to the hyperbola.
(i) 3x2 – y2 = 4 at the point (2, 2√2).
(ii) 3x2 – y2 = 12 at the point (4, 6)
(iii) \(\frac{x^{2}}{144}-\frac{y^{2}}{25}=1\) at the point whose eccentric angle is \(\frac{\pi}{3}\).
(iv) \(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\) at the point in a first quadrant whose ordinate is 3.
(v) 9x2 – 16y2 = 144 at the point L of the latus rectum in the first quadrant.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q6
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q6.1
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q6.2
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q6.3
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q6.4
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q6.5
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q6.6
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q6.7

Question 7.
Show that the line 3x – 4y + 10 = 0 is a tangent to the hyperbola x2 – 4y2 = 20. Also, find the point of contact.
Solution:
Given equation of the hyperbola is x2 – 4y2 = 20
\(\frac{x^{2}}{20}-\frac{y^{2}}{5}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 20 and b2 = 5
Given equation of line is 3x – 4y + 10 = 0.
y = \(\frac{3 x}{4}+\frac{5}{2}\)
Comparing this equation with y = mx + c, we get
m = \(\frac{3}{4}\) and c = \(\frac{5}{2}\)
For the line y = mx + c to be a tangent to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we must have
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q7

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3

Question 8.
If the line 3x – 4y = k touches the hyperbola \(\frac{x^{2}}{5}-\frac{4 y^{2}}{5}=1\), then find the value of k.
Solution:
Given equation of the hyperbola is
\(\frac{x^{2}}{5}-\frac{4 y^{2}}{5}=1\)
\(\frac{x^{2}}{5}-\frac{y^{2}}{\frac{5}{4}}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a2 = 5, b2 = \(\frac{5}{4}\)
Given equation of line is 3x – 4y = k
y = \(\frac{3}{4} x-\frac{\mathrm{k}}{4}\)
Comparing this equation with y = mx + c, we get
m = \(\frac{3}{4}\), c = \(-\frac{\mathrm{k}}{4}\)
For the line y = mx + c to be a tangent to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we must have
c2 = a2 m2 – b2
⇒ \(\left(\frac{-\mathrm{k}}{4}\right)^{2}=5\left(\frac{3}{4}\right)^{2}-\frac{5}{4}\)
⇒ \(\frac{\mathrm{k}^{2}}{16}=\frac{5}{16}(9-4)\)
⇒ \(\frac{\mathrm{k}^{2}}{16}=\frac{5}{16}(5)\)
⇒ k2 = 25
⇒ k = ±5

Alternate method:
Given equation of the hyperbola is
\(\frac{x^{2}}{5}-\frac{4 y^{2}}{5}=1\) …….(i)
Given equation of the line is 3x – 4y = k
y = \(\frac{3 x-\mathrm{k}}{4}\)
Substituting this value ofy in (i), we get
\(\frac{x^{2}}{5}-\frac{4}{5}\left(\frac{3 x-\mathrm{k}}{4}\right)^{2}=1\)
⇒ \(\frac{x^{2}}{5}-\frac{4}{5}\left(\frac{9 x^{2}-6 k x+k^{2}}{16}\right)=1\)
⇒ 4x2 – (9x2 – 6kx + k2) = 20
⇒ 4x2 – 9x2 + 6kx – k2 = 20
⇒ -5x2 + 6kx – k2 = 20
⇒ 5x2 – 6kx + (k2 + 20) = 0 …..(ii)
Since, the given line touches the given hyperbola.
The quadratic equation (ii) in x has equal roots.
(-6k)2 – 4(5)(k2 + 20) = 0
⇒ 36k2 – 20k2 – 400 = 0
⇒ 16k2 = 400
⇒ k2 = 25
⇒ k = ±5

Question 9.
Find the equations of the tangents to the hyperbola \(\frac{x^{2}}{25}-\frac{y^{2}}{9}=1\) making equal intercepts on the co-ordinate axes.
Solution:
Given equation of the hyperbola is \(\frac{x^{2}}{25}-\frac{y^{2}}{9}=1\).
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 25 and b2 = 9
Since the tangents make equal intercepts on the co-ordinate axes,
∴ m = -1
Equations of tangents to the hyperbola \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
⇒ y = -x ± \(\sqrt{25(-1)^{2}-9}\)
⇒ y = -x ± √16
⇒ x + y = ±4

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3

Question 10.
Find the equations of the tangents to the hyperbola 5x2 – 4y2 = 20 which are parallel to the line 3x + 2y + 12 = 0.
Solution:
Given equation of the hyperbola is 5x2 – 4y2 = 20
\(\frac{x^{2}}{4}-\frac{y^{2}}{5}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a2 = 4 and b2 = 5
Slope of the line 3x + 2y + 12 = 0 is \(-\frac{3}{2}\)
Since the given line is parallel to the tangents,
Slope of the required tangents (m) = \(-\frac{3}{2}\)
Equations of tangents to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3 Q10

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 7 Conic Sections Ex 7.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2

Question 1.
Find the
(i) lengths of the principal axes
(ii) co-ordinates of the foci
(iii) equations of directrices
(iv) length of the latus rectum
(v) distance between foci
(vi) distance between directrices of the ellipse:
(a) \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)
(b) 3x2 + 4y2 = 12
(c) 2x2 + 6y2 = 6
(d) 3x2 + 4y2 = 1
Solution:
(a) Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 25 and b2 = 9
a = 5 and b = 3
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(5) = 10
Length of minor axis = 2b = 2(3) = 6
Lengths of the principal axes are 10 and 6.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)
= \(\frac{\sqrt{25-9}}{5}\)
= \(\frac{\sqrt{16}}{5}\)
= \(\frac{4}{5}\)
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S(5(\(\frac{4}{5}\)), 0) and S'(-5(\(\frac{4}{5}\)), 0)
i.e., S(4, 0) and S'(-4, 0)

(iii) Equations of the directrices are x = ±\(\frac{\mathrm{a}}{\mathrm{e}}\)
= \(\pm \frac{5}{\frac{4}{5}}\)
= \(\pm \frac{25}{4}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(3)^{2}}{5}=\frac{18}{5}\)

(v) Distance between foci = 2ae
= 2(5)(\(\frac{4}{5}\))
= 8

(vi) Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)
= \(\frac{2(5)}{\frac{4}{5}}\)
= \(\frac{25}{2}\)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2

(b) Given equation of the ellipse is 3x2 + 4y2 = 12
\(\frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 4 and b2 = 3
a = 2 and b = √3
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(2) = 4
Length of minor axis = 2b = 2√3
Lengths of the principal axes are 4 and 2√3.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)
= \(\frac{\sqrt{4-3}}{2}\)
= \(\frac{1}{2}\)
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S(2(\(\frac{1}{2}\)), 0) and S'(-2(\(\frac{1}{2}\)), 0)
i.e., S(1, 0) and S'(-1, 0)

(iii) Equations of the directrices are x = ±\(\frac{\mathrm{a}}{\mathrm{e}}\)
= \(\pm \frac{2}{\frac{1}{2}}\)
= ±4

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(\sqrt{3})^{2}}{2}=3\)

(v) Distance between foci = 2ae = 2(2)(\(\frac{1}{2}\)) = 2

(vi) Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)
= \(\frac{2(2)}{\frac{1}{2}}\)
= 8

(c) Given equation of the ellipse is 2x2 + 6y2 = 6
\(\frac{x^{2}}{3}+\frac{y^{2}}{1}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 3 and b2 = 1
a = √3 and b = 1
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2√3
Length of minor axis = 2b = 2(1) = 2
Lengths of the principal axes are 2√3 and 2.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)
= \(\frac{\sqrt{3-1}}{\sqrt{3}}\)
= \(\frac{\sqrt{2}}{\sqrt{3}}\)
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S(√3(\(\frac{\sqrt{2}}{\sqrt{3}}\)), o) and S'(-√3(\(\frac{\sqrt{2}}{\sqrt{3}}\)), 0)
i.e., S(√2, 0) and S'(-√2, 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\),
= \(\pm \frac{\sqrt{3}}{\frac{\sqrt{2}}{\sqrt{3}}}\)
= \(\pm \frac{3}{\sqrt{2}}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(1)^{2}}{\sqrt{3}}=\frac{2}{\sqrt{3}}\)

(v) Distance between foci = 2ae
= \(2(\sqrt{3})\left(\frac{\sqrt{2}}{\sqrt{3}}\right)\)
= 2√2

(vi) Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)
= \(\frac{2 \sqrt{3}}{\frac{\sqrt{2}}{\sqrt{3}}}\)
= \(\frac{2 \times 3}{\sqrt{2}}\)
= 3√2

(d) Given equation of the ellipse is 3x2 + 4y = 1.
\(\frac{x^{2}}{\frac{1}{3}}+\frac{y^{2}}{\frac{1}{4}}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = \(\frac{1}{3}\) and b2 = \(\frac{1}{4}\)
a = \(\frac{1}{\sqrt{3}}\) and b = \(\frac{1}{2}\)
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(\(\frac{1}{\sqrt{3}}\)) = \(\frac{2}{\sqrt{3}}\)
Length of minor axis = 2b = 2(\(\frac{1}{2}\)) = 1
Lengths of the principal axes are \(\frac{2}{\sqrt{3}}\) and 1.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)
e = \(\frac{\sqrt{\frac{1}{3}-\frac{1}{4}}}{\frac{1}{\sqrt{3}}}=\frac{\sqrt{\frac{1}{12}}}{\frac{1}{\sqrt{3}}}=\sqrt{\frac{3}{12}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S\(\left(\frac{1}{\sqrt{3}}\left(\frac{1}{2}\right), 0\right)\) and S’\(\left(-\frac{1}{\sqrt{3}}\left(\frac{1}{2}\right), 0\right)\)
i.e., S(\(\frac{1}{2 \sqrt{3}}\), 0) and S'(-\(\frac{1}{2 \sqrt{3}}\), 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\),
= \(\pm \frac{\frac{1}{\sqrt{3}}}{\frac{1}{2}}\)
= \(\pm \frac{2}{\sqrt{3}}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}\)
= \(\frac{2\left(\frac{1}{2}\right)^{2}}{\frac{1}{\sqrt{3}}}\)
= \(\frac{\sqrt{3}}{2}\)

(v) Distance between foci = 2ae
= \(2\left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{2}\right)\)
= \(\frac{1}{\sqrt{3}}\)

(vi) Distance between directrices = \(\frac{2 a}{e}\)
= \(\frac{2\left(\frac{1}{\sqrt{3}}\right)}{\frac{1}{2}}\)
= \(\frac{4}{\sqrt{3}}\)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2

Question 2.
Find the equation of the ellipse in standard form if
(i) eccentricity = \(\frac{3}{8}\) and distance between its foci = 6.
(ii) the length of the major axis is 10 and the distance between foci is 8.
(iii) distance between directrices is 18 and eccentricity is \(\frac{1}{3}\).
(iv) minor axis is 16 and eccentricity is \(\frac{1}{3}\).
(v) the distance between foci is 6 and the distance between directrices is \(\frac{50}{3}\).
(vi) the latus rectum has length 6 and foci are (±2, 0).
(vii) passing through the points (-3, 1) and (2, -2).
(viii) the distance between its directrices is 10 and which passes through (-√5, 2).
(ix) eccentricity is \(\frac{2}{3}\) and passes through (2, \(\frac{-5}{3}\)).
Solution:
(i) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Given, eccentricity (e) = \(\frac{3}{8}\)
Distance between foci = 2ae
Given, distance between foci = 6
2ae = 6
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q2 (i)
The required equation of ellipse is \(\frac{x^{2}}{64}+\frac{y^{2}}{55}=1\).

(ii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Length of major axis = 2a
Given, length of major axis = 10
2a = 10
a = 5
a2 = 25
Distance between foci = 2ae
Given, distance between foci = 8
2ae = 8
2(5)e = 8
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q2 (ii)
The required equation of ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\).

(iii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Given, eccentricity (e) = \(\frac{1}{3}\)
Distance between directrices = \(\frac{2a}{e}\)
Given, distance between directrices = 18
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q2 (iii)
The required equation of ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{8}=1\)

(iv) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Length of minor axis = 2b
Given, length of minor axis = 16
2b = 16
b = 8
b2 = 64
Given, eccentricity (e) = \(\frac{1}{3}\)
Now, b2 = a2 (1 – e2)
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q2 (iv)
The required equation of ellipse is \(\frac{x^{2}}{72}+\frac{y^{2}}{64}=1\).

(v) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Distance between foci = 2ae
Given, distance between foci = 6
2ae = 6
ae = 3
a = \(\frac{3}{e}\) …….(i)
Distance between directrices = \(\frac{2a}{e}\)
Given, distance between directrices = \(\frac{50}{3}\)
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q2 (v)
The required equation of ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\).

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2

(vi) Given, the length of the latus rectum is 6, and co-ordinates of foci are (±2, 0).
The foci of the ellipse are on the X-axis.
Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Length of latus rectum = \(\frac{2 b^{2}}{a}\)
\(\frac{2 b^{2}}{a}\) = 6
b2 = 3a …..(i)
Co-ordinates of foci are (±ae, 0)
ae = 2
a2e2 = 4 …..(ii)
Now, b2 = a2 (1 – e2)
b2 = a2 – a2 e2
3a = a2 – 4 …..[From (i) and (ii)]
a2 – 3a – 4 = 0
a2 – 4a + a – 4 = 0
a(a – 4) + 1(a – 4) = 0
(a – 4) (a + 1) = 0
a – 4 = 0 or a + 1 = 0
a = 4 or a = -1
Since a = -1 is not possible,
a = 4
a2 = 16
Substituting a = 4 in (i), we get
b2 = 3(4) = 12
The required equation of ellipse is \(\frac{x^{2}}{16}+\frac{y^{2}}{12}=1\).

(vii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
The ellipse passes through the points (-3, 1) and (2, -2).
Substituting x = -3 and y = 1 in equation of ellipse, we get
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q2 (vii)
Equations (i) and (ii) become
9A + B = 1 …..(iii)
4A + 4B = 1 …..(iv)
Multiplying (iii) by 4, we get
36A + 4B = 4 …..(v)
Subtracting (iv) from (v), we get
32A = 3
A = \(\frac{3}{32}\)
Substituting A = \(\frac{3}{32}\) in (iv), we get
4(\(\frac{3}{32}\)) + 4B = 1
\(\frac{3}{8}\) + 4B = 1
4B = 1 – \(\frac{3}{8}\)
4B = \(\frac{5}{8}\)
B = \(\frac{5}{32}\)
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q2 (vii).1

(viii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Distance between directrices = \(\frac{2 a}{e}\)
Given, distance between directrices = 10
\(\frac{2 a}{e}\) = 10
a = 5e …..(i)
The ellipse passes through (-√5, 2).
Substituting x = -√5 and y = 2 in equation of ellipse, we get
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q2 (viii)
b2 = 15(\(\frac{2}{5}\))
b2 = 6
The required equation of ellipse is \(\frac{x^{2}}{15}+\frac{y^{2}}{6}=1\).

(ix) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Given, eccentricity (e) = \(\frac{2}{3}\)
The ellipse passes through (2, \(\frac{-5}{3}\)).
Substituting x = 2 and y = \(\frac{-5}{3}\) in equation of ellipse, we get
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q2 (ix)
The required equation of ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\).

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2

Question 3.
Find the eccentricity of an ellipse, if the length of its latus rectum is one-third of its minor axis.
Solution:
Let the equation of ellipse be \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), where a > b.
Length of latus rectum = \(\frac{2 b^{2}}{a}\)
Length of minor axis = 2b
According to the given condition,
Length of latus rectum = \(\frac{1}{3}\) (Minor axis)
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q3

Question 4.
Find the eccentricity of an ellipse, if the distance between its directrices is three times the distance between its foci.
Solution:
Let the required equation of ellipse be \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), where a > b.
Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)
Distance between foci = 2ae
According to the given condition,
distance between directrices = 3(distance between foci)
\(\frac{2 \mathrm{a}}{\mathrm{e}}\) = 3(2ae)
\(\frac{1}{\mathrm{e}}\) = 3e
\(\frac{1}{3}\) = e2
e = \(\frac{1}{\sqrt{3}}\) ……[∵ 0 < e < 1]
Eccentricity of the ellipse is \(\frac{1}{\sqrt{3}}\)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2

Question 5.
Show that the product of the lengths of the perpendicular segments drawn from the foci to any tangent line to the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) is equal to 16.
Solution:
Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\).
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 25, b2 = 16
a = 5, b = 4
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q5
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q5.1

Question 6.
A tangent having slope \(\left(-\frac{1}{2}\right)\) the ellipse 3x2 + 4y2 = 12 intersects the X and Y axes in the points A and B respectively. If O is the origin, find the area of the triangle AOB.
Solution:
Given equation of the ellipse is 3x2 + 4y2 = 12.
\(\frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a2 = 4, b2 = 3
Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
Here, m = \(-\frac{1}{2}\)
Equations of the tangents are
y = \(\frac{-1}{2} x \pm \sqrt{4\left(-\frac{1}{2}\right)^{2}+3}=\frac{-1}{2} x \pm 2\)
2y = -x ± 4
x + 2y ± 4 = 0
Consider the tangent x + 2y – 4 = 0
Let this tangent intersect the X-axis at A(x1, 0) and Y-axis at B(0, y1).
x1 + 0 – 4 = 0 and 0 + 2y1 – 4 = 0
x1 = 4 and y1 = 2
A = (4, 0) and B = (0, 2)
l(OA) = 4 and l(OB) = 2
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q6
Area of ∆AOB = \(\frac{1}{2}\) × l(OA) × l(OB)
= \(\frac{1}{2}\) × 4 × 2
= 4 sq.units

Question 7.
Show that the line x – y = 5 is a tangent to the ellipse 9x2 + 16y2 = 144. Find the point of contact.
Solution:
Given equation of the ellipse is 9x2 + 16y2 = 144
\(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a2 = 16 and b2 = 9
Given equation of line is x – y = 5, i.e., y = x – 5
c2 = a2 m2 + b2
Comparing this equation with y = mx + c, we get
m = 1 and c = -5
For the line y = mx + c to be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\)1, we must have
c2 = a2 m2 + b2
c2 = (-5)2 = 25
a2 m2 + b2 = 16(1)2 + 9 = 16 + 9 = 25 = c2
The given line is a tangent to the given ellipse and point of contact
= \(\left(\frac{-\mathrm{a}^{2} \mathrm{~m}}{\mathrm{c}}, \frac{\mathrm{b}^{2}}{\mathrm{c}}\right)\)
= \(\left(\frac{(-16)(1)}{-5}, \frac{9}{-5}\right)\)
= \(\left(\frac{16}{5}, \frac{-9}{5}\right)\)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2

Question 8.
Show that the line 8y + x = 17 touches the ellipse x2 + 4y2 = 17. Find the point of contact.
Solution:
Given equation of the ellipse is x2 + 4y2 = 17.
\(\frac{x^{2}}{17}+\frac{y^{2}}{\frac{17}{4}}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 17 and b2 = \(\frac{17}{4}\)
Given equation of line is 8y + x = 17,
y = \(\frac{-1}{8} x+\frac{17}{8}\)
Comparing this equation with y = mx + c, we get
m = \(\frac{-1}{8}\) and c = \(\frac{17}{8}\)
For the line y = mx + c to be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\)1, we must have
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q8

Question 9.
Determine whether the line x + 3y√2 = 9 is a tangent to the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\). If so, find the co-ordinates of the point of contact.
Solution:
Given equation of the ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 9 and b2 = 4
Given equation of line is x + 3y√2 = 9,
i.e., y = \(\frac{-1}{3 \sqrt{2}} x+\frac{3}{\sqrt{2}}\)
Comparing this equation with y = mx + c, we get
m = \(\frac{-1}{3 \sqrt{2}}\) and c = \(\frac{3}{\sqrt{2}}\)
For the line y = mx + c to be a tangent to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=\)1, we must have
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q9

Question 10.
Find k, if the line 3x + 4y + k = 0 touches 9x2 + 16y2 = 144.
Solution:
Given equation of the ellipse is 9x2 + 16y2 = 144.
\(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 16 and b2 = 9
Given equation of line is 3x + 4y + k = 0,
i.e., y = \(-\frac{3}{4} x-\frac{k}{4}\)
Comparing this equation with y = mx + c, we get
m = \(\frac{-3}{4}\) and c = \(\frac{-\mathrm{k}}{4}\)
For the line y = mx + c to be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\)1, we must have
c2 = a2 m2 + b2
\(\left(\frac{-k}{4}\right)^{2}=16\left(\frac{-3}{4}\right)^{2}+9\)
\(\frac{\mathrm{k}^{2}}{16}\) = 9 + 9
\(\frac{\mathrm{k}^{2}}{16}\) = 18
k2 = 288
k = ±12√2

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2

Question 11.
Find the equations of the tangents to the ellipse:
(i) \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\) passing through the point (2, -2).
(ii) 4x2 + 7y2 = 28 from the point (3, -2).
(iii) 2x2 + y2 = 6 from the point (2, 1).
(iv) x2 + 4y2 = 9 which are parallel to the line 2x + 3y – 5 = 0.
(v) \(\frac{x^{2}}{25}+\frac{y^{2}}{4}=1\) which are parallel to the line x + y + 1 = 0.
(vi) 5x2 + 9y2 = 45 which are ⊥ to the line 3x + 2y + 1 = 0.
(vii) x2 + 4y2 = 20 which are ⊥ to the line 4x + 3y = 7.
Solution:
(i) Given equation of the ellipse is \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\).
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 5 and b2 = 4
Equations of tangents to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
Since (2, -2) lies on both the tangents,
-2 = 2m ±\(\sqrt{5 m^{2}+4}\)
-2 – 2m = ±\(\sqrt{5 m^{2}+4}\)
Squaring both the sides, we get
4m2 + 8m + 4 = 5m2 + 4
m2 – 8m = 0
m(m – 8) = 0
m = 0 or m = 8
These are the slopes of the required tangents.
By slope point form y – y1 = m(x – x1),
the equations of the tangents are
y + 2 = 0(x – 2) and y + 2 = 8(x – 2)
y + 2 = 0 and y + 2 = 8x – 16
y + 2 = 0 and 8x – y – 18 = 0

(ii) Given equation of the ellipse is 4x2 + 7y2 = 28.
\(\frac{x^{2}}{7}+\frac{y^{2}}{4}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 7 and b2 = 4
Equations of tangents to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
Since (3, -2) lies on both the tangents,
-2 = 3m ± \(\sqrt{7 \mathrm{~m}^{2}+4}\)
-2 – 3m = ±\(\sqrt{7 \mathrm{~m}^{2}+4}\)
Squaring both the sides, we get
9m2 + 12m + 4 = 7m2 + 4
2m2 + 12m = 0
2m(m + 6) = 0
m = 0 or m = -6
These are the slopes of the required tangents.
By slope point form y – y1 = m(x – x1),
the equations of the tangents are
y + 2 = 0(x – 3) and y + 2 = -6(x – 3)
y + 2 = 0 and y + 2 = -6x + 18
y + 2 = 0 and 6x + y – 16 = 0

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2

(iii) Given equation of the ellipse is 2x2 + y2 = 6.
\(\frac{x^{2}}{3}+\frac{y^{2}}{6}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a2 = 3 and b2 = 6
Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
Since (2, 1) lies on both the tangents,
1 = 2m ± \(\sqrt{3 m^{2}+6}\)
1 – 2m = ±\(\sqrt{3 m^{2}+6}\)
Squaring both the sides, we get
1 – 4m + 4m2 = 3m2 + 6
m2 – 4m – 5 = 0
(m – 5)(m + 1) = 0
m = 5 or m = -1
These are the slopes of the required tangents.
By slope point form y – y1 = m(x – x1),
the equations of the tangents are
y – 1 = 5(x – 2) and y – 1 = -1(x – 2)
y – 1 = 5x – 10 and y – 1 = -x + 2
5x – y – 9 = 0 and x + y – 3 = 0

(iv) Given equation of the ellipse is x2 + 4y2 = 9.
\(\frac{x^{2}}{9}+\frac{y^{2}}{\frac{9}{4}}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a2 = 9 and b2 = \(\frac{9}{4}\)
Slope of the line 2x + 3y – 5 = 0 is \(\frac{-2}{3}\).
Since the given line is parallel to the required tangents, slope of the required tangents is
m = \(\frac{-2}{3}\)
Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q11 (iv)

(v) Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{4}=1\).
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a2 = 25 and b2 = 4
Slope of the given line x + y + 1 = 0 is -1.
Since the given line is parallel to the required tangents,
the slope of the required tangents is m = -1.
Equations of tangents to the ellipse
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q11 (v)

(vi) Given equation of the ellipse is 5x2 + 9y2 = 45.
\(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a2 = 9 and b2 = 5
Slope of the given line 3x + 2y + 1 = 0 is \(\frac{-3}{2}\)
Since the given line is perpendicular to the required tangents, slope of the required tangents is
m = \(\frac{2}{3}\)
Equations of tangents to the ellipse
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q11 (vi)

(vii) Given equation of the ellipse is x2 + 4y2 = 20.
\(\frac{x^{2}}{20}+\frac{y^{2}}{5}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 20 and b2 = 5
Slope of the given line 4x + 3y = 7 is \(\frac{-4}{3}\).
Since the given line is perpendicular to the required tangents,
slope of the required tangents is m = \(\frac{3}{4}\).
Equations of tangents to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
y = \(\frac{3}{4} x \pm \sqrt{20\left(\frac{3}{4}\right)^{2}+5}\)
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q11 (vii)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2

Question 12.
Find the equation of the locus of a point, the tangents from which to the ellipse 3x2 + 5y2 = 15 are at right angles.
Solution:
Given equation of the ellipse is 3x2 + 5y2 = 15.
\(\frac{x^{2}}{5}+\frac{y^{2}}{3}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a2 = 5 and b2 = 3
Equations of tangents to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
y = mx ± \(\sqrt{5 m^{2}+3}\)
y – mx =±\(\sqrt{5 m^{2}+3}\)
Squaring both the sides, we get
y2 – 2mxy + m2x2 = 5m2 + 3
(x2 – 5) m2 – 2xym + (y2 – 3) = 0
The roots m1 and m2 of this quadratic equation are the slopes of the tangents.
m1m2 = \(\frac{y^{2}-3}{x^{2}-5}\)
Since the tangents are at right angles,
m1m2 = -1
\(\frac{y^{2}-3}{x^{2}-5}=-1\)
y2 – 3 = -x2 + 5
x2 + y2 = 8, which is the required equation of the locus.

Alternate method:
The locus of the point of intersection of perpendicular tangents is the director circle of an ellipse.
The equation of the director circle of an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is x2 + y2 = a2 + b2
Here, a2 = 5 and b2 = 3
x2 + y2 = 5 + 3
x2 + y2 = 8, which is the required equation of the locus.

Question 13.
Tangents are drawn through a point P to the ellipse 4x2 + 5y2 = 20 having inclinations θ1 and θ2 such that tan θ1 + tan θ2 = 2. Find the equation of the locus of P.
Solution:
Given equation of the ellipse is 4x2 + 5y2 = 20.
\(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a2 = 5 and b2 = 4
Since inclinations of tangents are θ1 and θ2,
m1 = tan θ1 and m2 = tan θ2
Equation of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
y = mx ± \(\sqrt{5 \mathrm{~m}^{2}+4}\)
y – mx = ± \(\sqrt{5 \mathrm{~m}^{2}+4}\)
Squaring both the sides, we get
y2 – 2mxy + m2x2 = 5m2 + 4
(x2 – 5)m2 – 2xym + (y2 – 4) = 0
The roots m1 and m2 of this quadratic equation are the slopes of the tangents.
m1 + m2 = \(\frac{-(-2 x y)}{x^{2}-5}=\frac{2 x y}{x^{2}-5}\)
Given, tan θ1 + tan θ2 = 2
m1 + m2 = 2
\(\frac{2 x y}{x^{2}-5}\) = 2
xy = x2 – 5
x2 – xy – 5 = 0, which is the required equation of the locus of P.

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2

Question 14.
Show that the locus of the point of intersection of tangents at two points on an ellipse, whose eccentric angles differ by a constant, is an ellipse.
Solution:
Let P(θ1) and Q(θ2) be any two points on the given ellipse such that θ1 – θ2 = k, where k is a constant.
The equation of the tangent at point P(θ1) is
\(\frac{x \cos \theta_{1}}{\mathrm{a}}+\frac{y \sin \theta_{1}}{\mathrm{~b}}=1\) ……(i)
The equation of the tangent at point Q(θ2) is
\(\frac{x \cos \theta_{2}}{\mathrm{a}}+\frac{y \sin \theta_{2}}{\mathrm{~b}}=1\) ……(ii)
Multiplying equation (i) by cos θ2 and equation (ii) by cos θ1 and subtracting, we get
\(\frac{y}{b}\) (sin θ1 cos θ2 – sin θ2 cos θ1) = cos θ2 – cos θ1
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q14
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q14.1

Question 15.
P and Q are two points on the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) with eccentric angles θ1 and θ2. Find the equation of the locus of the point of intersection of the tangents at P and Q if θ1 + θ2 = \(\frac{\pi}{2}\).
Solution:
Given equation of the ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\).
θ1 and θ2 are the eccentric angles of a tangent.
Equation of tangent at point P is
\(\frac{x}{a} \cos \theta_{1}+\frac{y}{b} \sin \theta_{1}=1\) ……(i)
Equation of tangent at point Q is
\(\frac{x}{a} \cos \theta_{2}+\frac{y}{b} \sin \theta_{2}=1\) ………(ii)
θ1 + θ2 = \(\frac{\pi}{2}\) …..[Given]
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q15
i.e., points P and Q coincide, which is not possible, as P and Q are two different points.
cos θ1 – sin θ1 ≠ 0
Dividing equation (iii) by (cos θ1 – sin θ1), we get
\(\frac{x_{1}}{a}=\frac{y_{1}}{b}\)
bx1 – ay1 = 0
bx – ay = 0, which is the required equation of locus of point M.

Question 16.
The eccentric angles of two points P and Q of the ellipse 4x2 + y2 = 4 differ by \(\frac{2 \pi}{3}\). Show that the locus of the point of intersection of the tangents at P and Q is the ellipse 4x2 + y2 = 16.
Solution:
Given equation of the ellipse is 4x2 + y2 = 4.
\(\frac{x^{2}}{1}+\frac{y^{2}}{4}=1\)
Let P(θ1) and Q(θ2) be any two points on the given ellipse such that
θ1 – θ2 = \(\frac{2 \pi}{3}\)
The equation of the tangent at point P(θ1) is
\(\frac{x \cos \theta_{1}}{1}+\frac{y \sin \theta_{1}}{2}=1\) ……(i)
The equation of the tangent at point Q(θ2) is
\(\frac{x \cos \theta_{2}}{1}+\frac{y \sin \theta_{2}}{2}=1\)
Multiplying equation (i) by cos θ2 and equation (ii) by cos θ1 and subtracting, we get
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q16
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q16.1

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2

Question 17.
Find the equations of the tangents to the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\), making equal intercepts on co-ordinate axes.
Solution:
Given equation of the ellipse is \(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a2 = 16 and b2 = 9
Since the tangents make equal intercepts on the co-ordinate axes,
m = -1.
Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
y = -x ± \(\sqrt{16(-1)^{2}+9}\)
y = -x ± \(\sqrt{25}\)
x + y = ±5

Question 18.
A tangent having slope \(\left(-\frac{1}{2}\right)\) to the ellipse 3x2 + 4y2 = 12 intersects the X and Y axes in the points A and B respectively. If O is the origin, find the area of the triangle AOB.
Solution:
The equation of the ellipse is 3x2 + 4y2 = 12
\(\frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)
Comparing with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a2 = 4, b2 = 3
The equation of tangent with slope m is
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2 Q18
It meets X axis at A
∴ for A, put y = 0 in equation (1), we get,
x = ±4
∴ A = (±4, 0)
Similarly, B = (0, ±2)
∴ OA = 4, OB = 2
∴ Area of ΔOAB = \(\frac{1}{2}\) × OA × OB
= \(\frac{1}{2}\) × 4 × 2
= 4 sq. units

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 7 Conic Sections Ex 7.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1

Question 1.
Find co-ordinates of focus, equation of directrix, length of latus rectum and the co-ordinates of end points of latus rectum of the parabola:
(i) 5y2 = 24x
(ii) y2 = -20x
(iii) 3x2 = 8y
(iv) x2 = -8y
(v) 3y2 = -16x
Solution:
(i) Given equation of the parabola is 5y2 = 24x.
⇒ y2 = \(\frac{24}{5}\)x
Comparing this equation with y2 = 4ax, we get
⇒ 4a = \(\frac{24}{5}\)
⇒ a = \(\frac{6}{5}\)
Co-ordinates of focus are S(a, 0), i.e., S(\(\frac{6}{5}\), 0)
Equation of the directrix is x + a = 0.
⇒ x + \(\frac{6}{5}\) = 0
⇒ 5x + 6 = 0
Length of latus rectum = 4a
= 4(\(\frac{6}{5}\))
= \(\frac{24}{5}\)
Co-ordinates of end points of latus rectum are (a, 2a) and (a, -2a),
⇒ \(\left(\frac{6}{5}, \frac{12}{5}\right)\) and \(\left(\frac{6}{5}, \frac{-12}{5}\right)\)

(ii) Given equation of the parabola is y2 = -20x.
Comparing this equation with y2 = -4ax, we get
⇒ 4a = 20
⇒ a = 5
Co-ordinates of focus are S(-a, 0), i.e., S(-5, 0)
Equation of the directrix is x – a = 0
⇒ x – 5 = 0
Length of latus rectum = 4a = 4(5) = 20
Co-ordinates of end points of latus rectum are (-a, 2a) and (-a, -2a),
⇒ (-5, 10) and (-5, -10).

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1

(iii) Given equation of the parabola is 3x2 = 8y
⇒ x2 = \(\frac{8}{3}\) y
Comparing this equation with x2 = 4by, we get
⇒ 4b = \(\frac{8}{3}\)
⇒ b = \(\frac{2}{3}\)
Co-ordinates of focus are S(0, b), i.e., S(0, \(\frac{2}{3}\))
Equation of the directrix is y + b = 0,
⇒ y + \(\frac{2}{3}\) = 0
⇒ 3y + 2 = 0
Length of latus rectum = 4b = 4(\(\frac{2}{3}\)) = \(\frac{8}{3}\)
Co-ordinates of end points of latus rectum are (2b, b) and (-2b, b),
⇒ \(\left(\frac{4}{3}, \frac{2}{3}\right)\) and \(\left(-\frac{4}{3}, \frac{2}{3}\right)\).

(iv) Given equation of the parabola is x2 = -8y.
Comparing this equation with x2 = -4by, we get
⇒ 4b = 8
⇒ b = 2
Co-ordinates of focus are S(0, -b), i.e., S(0, – 2)
Equation of the directrix is y – b = 0, i.e., y – 2 = 0
Length of latus rectum = 4b = 4(2) = 8
∴ Co-ordinates of end points of latus rectum are (2b, -b) and (-2b, -b), i.e., (4, -2) and (-4, -2).

(v) Given equation of the parabola is 3y2 = -16x.
⇒ y2 = \(-\frac{16}{3}\)x
Comparing this equation withy = -4ax, we get
⇒ 4a = \(\frac{16}{3}\)
⇒ a = \(\frac{4}{3}\)
Co-ordinates of focus are S(-a, 0), i.e., (\(-\frac{4}{3}\), 0)
Equation of the directrix is x – a = 0,
⇒ x – \(-\frac{4}{3}\) = 0
⇒ 3x – 4 = 0
Length of latus rectum = 4a = 4(\(\frac{4}{3}\)) = \(\frac{16}{3}\)
Co-ordinates of end points of latus rectum are (-a, 2a) and (-a, -2a),
i.e., \(\left(-\frac{4}{3}, \frac{8}{3}\right)\) and \(\left(-\frac{4}{3},-\frac{8}{3}\right)\)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1

Question 2.
Find the equation of the parabola with vertex at the origin, the axis along the Y-axis, and passing through the point (-10, -5).
Solution:
Vertex of the parabola is at origin (0, 0) and its axis is along Y-axis.
Equation of the parabola can be either x2 = 4by or x2 = -4by
Since the parabola passes through (-10, -5), it lies in 3rd quadrant.
Required parabola is x2 = -4by.
Substituting x = -10 and y = -5 in x2 = -4by, we get
⇒ (-10)2 = -4b(-5)
⇒ b = \(\frac{100}{20}\) = 5
∴ The required equation of the parabola is x2 = -4(5)y, i.e., x2 = -20y.

Question 3.
Find the equation of the parabola with vertex at the origin, the axis along the X-axis, and passing through the point (3, 4).
Solution:
Vertex of the parabola is at the origin (0, 0) and its axis is along X-axis.
Equation of the parabola can be either y2 = 4ax or y2 = -4ax.
Since the parabola passes through (3, 4), it lies in the 1st quadrant.
Required parabola is y2 = 4ax.
Substituting x = 3 and y = 4 in y2 = 4ax, we get
⇒ (4)2 = 4a(3)
⇒ a = \(\frac{16}{12}=\frac{4}{3}\)
The required equation of the parabola is
y2 = 4(\(\frac{4}{3}\))x
⇒ 3y2 = 16x

Question 4.
Find the equation of the parabola whose vertex is O(0, 0) and focus at (-7, 0).
Solution:
Focus of the parabola is S(-7, 0) and vertex is O(0, 0).
Since focus lies on X-axis, it is the axis of the parabola.
Focus S(-7, 0) lies on the left-hand side of the origin.
It is a left-handed parabola.
Required parabola is y = -4ax.
Focus is S(-a, 0).
a = 7
∴ The required equation of the parabola is y2 =-4(7)x, i.e., y2 = -28x.

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1

Question 5.
Find the equation of the parabola with vertex at the origin, the axis along X-axis, and passing through the point
(i) (1, -6)
(ii) (2, 3)
Solution:
(i) Vertex of the parabola is at origin (0, 0) and its axis is along X-axis.
Equation of the parabola can be either y2 = 4ax or y2 = -4ax.
Since the parabola passes through (1, -6), it lies in the 4th quadrant.
Required parabola is y2 = 4ax.
Substituting x = 1 and y = -6 in y2 = 4ax, we get
⇒ (-6)2 = 4a(1)
⇒ 36 = 4a
⇒ a = 9
∴ The required equation of the parabola is y2 = 4(9)x, i.e., y2 = 36x.

(ii) Vertex of the parabola is at origin (0, 0) and its axis is along X-axis.
Equation of the parabola can be either y2 = 4ax or y2 = -4ax.
Since the parabola passes through (2, 3), it lies in 1st quadrant.
∴ Required parabola is y2 = 4ax.
Substituting x = 2 and y = 3 in y2 = 4ax, we get
⇒ (3)2 = 4a(2)
⇒ 9 = 8a
⇒ a = \(\frac{9}{8}\)
The required equation of the parabola is
y2 = 4(\(\frac{9}{8}\))x
⇒ y2 = \(\frac{9}{2}\) x
⇒ 2y2 = 9x.

Question 6.
For the parabola 3y2 = 16x, find the parameter of the point:
(i) (3, -4)
(ii) (27, -12)
Solution:
Given the equation of the parabola is 3y2 = 16x.
⇒ y2 = \(\frac{16}{3}\)x
Comparing this equation with y2 = 4ax, we get
⇒ 4a = \(\frac{16}{3}\)
⇒ a = \(\frac{4}{3}\)
If t is the parameter of the point P on the parabola, then
P(t) = (at2, 2at)
i.e., x = at2 and y = 2at ………(i)
(i) Given point is (3, -4)
Substituting x = 3, y = -4 and a = \(\frac{4}{3}\) in (i), we get
3 = \(\frac{4}{3}\) t2 and -4 = 2(\(\frac{4}{3}\)) t
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1 Q6
∴ The parameter of the given point is \(\frac{-3}{2}\)

(ii) Given point is (27, -12)
Substituting x = 27, y = -12 and a = \(\frac{4}{3}\) in (i), we get
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1 Q6.1
∴ The parameter of the given point is \(\frac{-9}{2}\)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1

Question 7.
Find the focal distance of a point on the parabola y2 = 16x whose ordinate is 2 times the abscissa.
Solution:
Given the equation of the parabola is y2 = 16x.
Comparing this equation with y2 = 4ax, we get
⇒ 4a = 16
⇒ a = 4
Since ordinate is 2 times the abscissa,
y = 2x
Substituting y = 2x in y2 = 16x, we get
⇒ (2x)2 = 16x
⇒ 4x2 = 16x
⇒ 4x2 – 16x = 0
⇒ 4x(x – 4) = 0
⇒ x = 0 or x = 4
When x = 4,
focal distance = x + a = 4 + 4 = 8
When x = 0,
focal distance = a = 4
∴ Focal distance is 4 or 8.

Question 8.
Find coordinates of the point on the parabola. Also, find focal distance.
(i) y2 = 12x whose parameter is \(\frac{1}{3}\)
(ii) 2y2 = 7x whose parameter is -2
Solution:
(i) Given equation of the parabola is y2 = 12x.
Comparing this equation with y2 = 4ax, we get
⇒ 4a = 12
⇒ a = 3
If t is the parameter of the point P on the parabola, then
P(t) = (at2, 2at)
i.e., x = at2 and y = 2at ……..(i)
Given, t = \(\frac{1}{3}\)
Substituting a = 3 and t = \(\frac{1}{3}\) in (i), we get
x = 3(\(\frac{1}{3}\))2 and y = 2(3)(\(\frac{1}{3}\))
x = \(\frac{1}{3}\) and y = 2
The co-ordinates of the point on the parabola are (\(\frac{1}{3}\), 2)
∴ Focal distance = x + a
= \(\frac{1}{3}\) + 3
= \(\frac{10}{3}\)

(ii) Given equation of the parabola is 2y2 = 7x.
⇒ y2 = \(\frac{7}{2}\)x
Comparing this equation with y2 = 4ax, we get
⇒ 4a = \(\frac{7}{2}\)
⇒ a = \(\frac{7}{8}\)
If t is the parameter of the point P on the parabola, then
P(t) = (at2, 2at)
i.e., x = at2 and y = 2at …..(i)
Given, t = -2
Substituting a = \(\frac{7}{8}\) and t = -2 in (i), we get
x = \(\frac{7}{8}\)(-2)2 and y = 2(\(\frac{7}{8}\))(-2)
x = \(\frac{7}{2}\) and y = \(\frac{-7}{2}\)
The co-ordinates of the point on the parabola are (\(\frac{7}{2}\), \(\frac{-7}{2}\))
∴ Focal distance = x + a
= \(\frac{7}{2}\) + \(\frac{7}{8}\)
= \(\frac{35}{8}\)

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1

Question 9.
For the parabola y2 = 4x, find the coordinates of the point whose focal distance is 17.
Solution:
Given the equation of the parabola is y2 = 4x.
Comparing this equation with y2 = 4ax, we get
⇒ 4a = 4
⇒ a = 1
Focal distance of a point = x + a
Given, focal distance = 17
⇒ x + 1 = 17
⇒ x = 16
Substituting x = 16 in y2 = 4x, we get
⇒ y2 = 4(16)
⇒ y2 = 64
⇒ y = ±8
∴ The co-ordinates of the point on the parabola are (16, 8) or (16, -8).

Question 10.
Find the length of the latus rectum of the parabola y2 = 4ax passing through the point (2, -6).
Solution:
Given equation of the parabola is y2 = 4ax and it passes through point (2, -6).
Substituting x = 2 and y = -6 in y2 = 4ax, we get
⇒ (-6)2 = 4a(2)
⇒ 4a = 18
∴ Length of latus rectum = 4a = 18 units

Question 11.
Find the area of the triangle formed by the line joining the vertex of the parabola x2 = 12y to the endpoints of the latus rectum.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1 Q11
Given the equation of the parabola is x2 = 12y.
Comparing this equation with x2 = 4by, we get
⇒ 4b = 12
⇒ b = 3
The co-ordinates of focus are S(0, b), i.e., S(0, 3)
End points of the latus-rectum are L(2b, b) and L'(-2b, b),
i.e., L(6, 3) and L'(-6, 3)
Also l(LL’) = length of latus-rectum = 4b = 12
l(OS) = b = 3
Area of ∆OLL’ = \(\frac{1}{2}\) × l(LL’) × l(OS)
= \(\frac{1}{2}\) × 12 × 3
Area of ∆OLL’ = 18 sq. units

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1

Question 12.
If a parabolic reflector is 20 cm in diameter and 5 cm deep, find its focus.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1 Q12
Let LOM be the parabolic reflector such that LM is the diameter and ON is its depth.
It is given that ON = 5 cm and LM = 20 cm.
LN = 10 cm
Taking O as the origin, ON along X-axis and a line through O ⊥ ON as Y-axis.
Let the equation of the reflector be y2 = 4ax ……(i)
The point L has the co-ordinates (5, 10) and lies on parabola given by (i).
Substituting x = 5 and y = 10 in (i), we get
⇒ 102 = 4a(5)
⇒ 100 = 20a
⇒ a = 5
Focus is at (a, 0), i.e., (5, 0)

Question 13.
Find co-ordinates of focus, vertex, and equation of directrix and the axis of the parabola y = x2 – 2x + 3.
Solution:
Given equation of the parabola is y = x2 – 2x + 3
⇒ y = x2 – 2x + 1 + 2
⇒ y – 2 = (x – 1)2
⇒ (x – 1)2 = y – 2
Comparing this equation with X2 = 4bY, we get
X = x – 1, Y = y – 2
⇒ 4b = 1
⇒ b = \(\frac{1}{4}\)
The co-ordinates of vertex are (X = 0, Y = 0)
⇒ x – 1 = 0 and y – 2 = 0
⇒ x = 1 and y = 2
The co-ordinates of vertex are (1, 2).
The co-ordinates of focus are S(X = 0, Y = b)
⇒ x – 1 = 0 and y – 2 = \(\frac{1}{4}\)
⇒ x = 1 and y = \(\frac{9}{4}\)
The co-ordinates of focus are (1, \(\frac{9}{4}\))
Equation of the axis is X = 0
x – 1 = 0, i.e., x = 1
Equation of directrix is Y + b = 0
⇒ y – 2 + \(\frac{1}{4}\) = 0
⇒ y – \(\frac{7}{4}\) = 0
⇒ 4y – 7 = 0

Question 14.
Find the equation of tangent to the parabola
(i) y2 = 12x from the point (2, 5)
(ii) y2 = 36x from the point (2, 9)
Solution:
(i) Given equation of the parabola is y2 = 12x.
Comparing this equation with y2 = 4ax, we get
⇒ 4a = 12
⇒ a = 3
Equation of tangent to the parabola y2 = 4ax having slope m is
y = mx + \(\frac{a}{m}\)
Since the tangent passes through the point (2, 5)
⇒ 5 = 2m + \(\frac{3}{m}\)
⇒ 5m = 2m2 + 3
⇒ 2m2 – 5m + 3 = 0
⇒ 2m2 – 2m – 3m + 3 = 0
⇒ 2m(m – 1) – 3(m – 1) = 0
⇒ (m- 1)(2m – 3) = 0
⇒ m = 1 or m = \(\frac{3}{2}\)
These are the slopes of the required tangents.
By slope point form, y – y1 = m(x – x1), the equations of the tangents are
⇒ y – 5 = 1(x – 2) and y – 5 = \(\frac{3}{2}\) (x – 2)
⇒ y – 5 = x – 2 and 2y – 10 = 3x – 6
⇒ x – y + 3 = 0 and 3x – 2y + 4 = 0

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1

(ii) Given equation of the parabola is y2 = 36x.
Comparing this equation with y2 = 4ax, we get
⇒ 4a = 36
⇒ a = 9
Equation of tangent to the parabola y2 = 4ax having slope m is
y = mx + \(\frac{a}{m}\)
Since the tangent passes through the point (2, 9),
⇒ 9 = 2m + \(\frac{9}{m}\)
⇒ 9m = 2m2 + 9
⇒ 2m2 – 9m + 9 = 0
⇒ 2m2 – 6m – 3m + 9 = 0
⇒ 2m(m – 3) – 3(m – 3) = 0
⇒ (m – 3)(2m – 3) = 0
⇒ m = 3 or m = \(\frac{3}{2}\)
These are the slopes of the required tangents.
By slope point form, y – y1 = m(x – x1), the equations of the tangents are
⇒ y – 9 = 3(x – 2) and y – 9 = \(\frac{3}{2}\) (x – 2)
⇒ y – 9 = 3x – 6 and 2y – 18 = 3x – 6
⇒ 3x – y + 3 = 0 and 3x – 2y + 12 = 0

Question 15.
If the tangents drawn from the point (-6, 9) to the parabola y2 = kx are perpendicular to each other, find k.
Solution:
Given equation of the parabola is y2 = kx
Comparing this equation with y2 = 4ax, we get
⇒ 4a = k
⇒ a = \(\frac{\mathrm{k}}{4}\)
Equation of tangent to the parabola y2 = 4ax having slope m is
y = mx + \(\frac{a}{m}\)
Since the tangent passes through the point (-6, 9),
⇒ 9 = -6m + \(\frac{k}{4m}\)
⇒ 36m = -24m2 + k
⇒ 24m2 + 36m – k = 0
The roots m1 and m2 of this quadratic equation are the slopes of the tangents.
m1m2 = \(\frac{-\mathrm{k}}{24}\)
Since the tangents are perpendicular to each other,
m1m2 = -1
⇒ \(\frac{-\mathrm{k}}{24}\) = -1
⇒ k = 24

Alternate method:
We know that, tangents drawn from a point on directrix are perpendicular.
(-6, 9) lies on the directrix x = -a.
⇒ -6 = -a
⇒ a = 6
Since 4a = k
⇒ k = 4(6) = 24

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1

Question 16.
Two tangents to the parabola y2 = 8x meet the tangents at the vertex in the points P and Q. If PQ = 4, prove that the equation of the locus of the point of intersection of two tangents is y2 = 8(x + 2).
Solution:
Given equation of the parabola is y2 = 8x
Comparing this equation with y2 = 4ax, we get
⇒ 4a = 8
⇒ a = 2
Equation of tangent to given parabola at A(t1) is y
t1 = x + 2\(\mathrm{t}_{1}^{2}\) …….(i)
Equation of tangent to given parabola at B(t2) is y
t2 = x + 2\(\mathrm{t}_{2}^{2}\) …..(ii)
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1 Q16
A tangent at the vertex is Y-axis whose equation is x = 0.
x-coordinate of points P and Q is 0.
Let P be(0, k1) and Q be (0, k2).
Then, from (i) and (ii), we get
Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1 Q16.1
∴ Equation of locus of R is y2 = 8(x + 2).

Question 17.
Find the equation of common tangent to the parabolas y2 = 4x and x2 = 32y.
Solution:
Given equation of the parabola is y2 = 4x
Comparing this equation with y2 = 4ax, we get
⇒ 4a = 4
⇒ a = 1
Let the equation of common tangent be
y = mx + \(\frac{1}{m}\) …..(i)
Substituting y = mx + \(\frac{1}{m}\) in x2 = 32y, we get
⇒ x2 = 32(mx + \(\frac{1}{m}\)) = 32 mx + \(\frac{32}{m}\)
⇒ mx2 = 32 m2x + 32
⇒ mx2 – 32 m2x – 32 = 0 ……..(ii)
Line (i) touches the parabola x2 = 32y.
The quadratic equation (ii) in x has equal roots.
Discriminant = 0
⇒ (-32m2)2 – 4(m)(-32) = 0
⇒ 1024 m4 + 128m = 0
⇒ 128m (8m3 + 1) = 0
⇒ 8m3 + 1 = 0 …..[∵ m ≠ 0]
⇒ m3 = \(-\frac{1}{8}\)
⇒ m = \(-\frac{1}{2}\)
Substituting m = \(-\frac{1}{2}\) in (i), we get
⇒ \(y=-\frac{1}{2} x+\frac{1}{\left(-\frac{1}{2}\right)}\)
⇒ \(y=-\frac{1}{2} x-2\)
⇒ x + 2y + 4 = 0, which is the equation of the common tangent.

Question 18.
Find the equation of the locus of a point, the tangents from which to the parabola y2 = 18x are such that sum of their slopes is -3.
Solution:
Given equation of the parabola is y2 = 18x
Comparing this equation with y2 = 4ax, we get
⇒ 4a = 18
⇒ a = \(\frac{9}{2}\)
Equation of tangent to the parabola y2 = 4ax having slope m is
⇒ y = mx + \(\frac{a}{m}\)
⇒ y = mx + \(\frac{9}{2m}\)
⇒ 2ym = 2xm2 + 9
⇒ 2xm2 – 2ym + 9 = 0
The roots m1 and m2 of this quadratic equation are the slopes of the tangents.
m1 + m2 = \(-\frac{(-2 y)}{2 x}=\frac{y}{x}\)
But, m1 + m2 = -3
\(\frac{y}{x}\) = -3
y = -3x, which is the required equation of locus.

Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1

Question 19.
The towers of a bridge, hung in the form of a parabola, have their tops 30 metres above the roadway and are 200 metres apart. If the cable is 5 metres above the roadway at the centre of the bridge, find the length of the vertical supporting cable 30 metres from the centre.
Solution:
Let CAB be the cable of the bridge and X’OX be the roadway.
Let A be the centre of the bridge.
From the figure, vertex of parabola is at A(0, 5).
Let the equation of parabola be
x2 = 4b(y – 5) …..(i)
Since the parabola passes through (100, 30).
Substituting x = 100 and y = 30 in (i), we get
⇒ 1002 = 4b (30 – 5)
⇒ 1002 = 4b(25)
⇒ 1002 = 100b
⇒ b = 100
Substituting the value of b in (i), we get
x2 = 400(y – 5) …..(ii)
Let l metres be the length of vertical supporting cable.
Then P(30, l) lies on (ii).
⇒ 302 = 400(l – 5)
⇒ 900 = 400(l – 5)
⇒ \(\frac{9}{4}\) = l – 5
⇒ l = \(\frac{9}{4}\) + 5
⇒ l = \(\frac{9}{4}\) m = 7.25 m
The length of the vertical supporting cable is 7.25 m.

Question 20.
A circle whose centre is (4, -1) passes through the focus of the parabola x2 + 16y = 0. Show that the circle touches the directrix of the parabola.
Solution:
Given equation of the parabola is x2 + 16y = 0.
⇒ x2 = -16y
Comparing this equation with x2 = -4by, we get
⇒ 4b = 16
⇒ b = 4
Focus = S(0, -b) = (0, -4)
Centre of the circle is C(4, -1) and it passes through focus S of the parabola.
Radius = CS
= \(\sqrt{(4-0)^{2}+(-1+4)^{2}}\)
= \(\sqrt{16+9}\)
= 5
Equation of the directrix is y – b = 0, i.e.,y – 4 = 0
Length of the perpendicular from centre C(4, -1) to the directrix
= \(\left|\frac{0(4)+1(-1)-4}{\sqrt{(0)^{2}+(1)^{2}}}\right|\)
= \(\left|\frac{-1-4}{1}\right|\)
= 5
= radius
∴ The circle touches the directrix of the parabola.