Class 11

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Miscellaneous Exercise 7 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7

I. Select the correct answer from the given alternatives.

Question 1.
\(\lim _{x \rightarrow 2}\left(\frac{x^{4}-16}{x^{2}-5 x+6}\right)=\)
(A) 23
(B) 32
(C) -32
(D) -16
Answer:
(C) -32
Hint:

Question 2.
\(\lim _{x \rightarrow-2}\left(\frac{x^{7}+128}{x^{3}+8}\right)=\)
(A) \(\frac{56}{3}\)
(B) \(\frac{112}{3}\)
(C) \(\frac{121}{3}\)
(D) \(\frac{28}{3}\)
Answer:
(B) \(\frac{112}{3}\)
Hint:

Question 3.
\(\lim _{x \rightarrow 3}\left(\frac{1}{x^{2}-11 x+24}+\frac{1}{x^{2}-x-6}\right)=\)
(A) \(-\frac{2}{25}\)
(B) \(\frac{2}{25}\)
(C) \(\frac{7}{25}\)
(D) \(-\frac{7}{25}\)
Answer:
(A) \(-\frac{2}{25}\)
Hint:

Question 4.
\(\lim _{x \rightarrow 5}\left(\frac{\sqrt{x+4}-3}{\sqrt{3 x-11-2}}\right)=\)
(A) \(\frac{-2}{9}\)
(B) \(\frac{2}{7}\)
(C) \(\frac{5}{9}\)
(D) \(\frac{2}{9}\)
Answer:
(D) \(\frac{2}{9}\)
Hint:

Question 5.
\(\lim _{x \rightarrow \frac{\pi}{3}}\left(\frac{\tan ^{2} x-3}{\sec ^{3} x-8}\right)=\)
(A) 1
(B) \(\frac{1}{2}\)
(C) \(\frac{1}{3}\)
(D) \(\frac{1}{4}\)
Answer:
(C) \(\frac{1}{3}\)
Hint:

Question 6.
\(\lim _{x \rightarrow 0}\left(\frac{5 \sin x-x \cos x}{2 \tan x-3 x^{2}}\right)=\)
(A) 0
(B) 1
(C) 2
(D) 3
Answer:
(C) 2
Hint:

Question 7.
\(\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{3 \cos x+\cos 3 x}{(2 x-\pi)^{3}}\right]=\)
(A) \(\frac{3}{2}\)
(B) \(\frac{1}{2}\)
(C) \(-\frac{1}{2}\)
(D) \(\frac{1}{4}\)
Answer:
(C) \(-\frac{1}{2}\)
Hint:

Question 8.
\(\lim _{x \rightarrow 0}\left(\frac{15^{x}-3^{x}-5^{x}+1}{\sin ^{2} x}\right)=\)
(A) log 15
(B) log 3 + log 5
(C) log 3 . log 5
(D) 3 log 5
Answer:
(C) log 3 . log 5
Hint:

Question 9.
\(\lim _{x \rightarrow 0}\left(\frac{3+5 x}{3-4 x}\right)^{\frac{1}{x}}=\)
(A) e³
(B) e⁶
(C) e⁹
(D) e^-3
Answer:
(A) e³
Hint:

Question 10.
\(\lim _{x \rightarrow 0}\left[\frac{\log (5+x)-\log (5-x)}{\sin x}\right]=\)
(A) \(\frac{3}{2}\)
(B) \(-\frac{5}{2}\)
(C) \(-\frac{1}{2}\)
(D) \(\frac{2}{5}\)
Answer:
(D) \(\frac{2}{5}\)
Hint:

Question 11.
\(\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{3^{\cos x}-1}{\frac{\pi}{2}-x}\right)=\)
(A) 1
(B) log 3
(C) \(3^{\frac{\pi}{2}}\)
(D) 3 log 3
Answer:
(B) log 3
Hint:

Question 12.
\(\lim _{x \rightarrow 0}\left[\frac{x \cdot \log (1+3 x)}{\left(e^{3 x}-1\right)^{2}}\right]=\)
(A) \(\frac{1}{\mathrm{e}^{9}}\)
(B) \(\frac{1}{\mathrm{e}^{3}}\)
(C) \(\frac{1}{9}\)
(D) \(\frac{1}{3}\)
Answer:
(D) \(\frac{1}{3}\)
Hint:

Question 13.
\(\lim _{x \rightarrow 0}\left[\frac{\left(3^{\sin x}-1\right)^{3}}{\left(3^{x}-1\right) \cdot \tan x \cdot \log (1+x)}\right]=\)
(A) 3 log 3
(B) 2 log 3
(C) (log 3)²
(D) (log 3)³
Answer:
(C) (log 3)²
Hint:

Question 14.
\(\lim _{x \rightarrow 3}\left[\frac{5^{x-3}-4^{x-3}}{\sin (x-3)}\right]=\)
(A) log 5 – 4
(B) log \(\frac{5}{4}\)
(C) \(\frac{\log 5}{\log 4}\)
(D) \(\frac{\log 5}{4}\)
Answer:
(B) log \(\frac{5}{4}\)
Hint:

Question 15.
\(\lim _{x \rightarrow \infty}\left[\frac{(2 x+3)^{7}(x-5)^{3}}{(2 x-5)^{10}}\right]=\)
(A) \(\frac{3}{8}\)
(B) \(\frac{1}{8}\)
(C) \(\frac{1}{6}\)
(D) \(\frac{1}{4}\)
Answer:
(B) \(\frac{1}{8}\)
Hint:

(II) Evaluate the following.

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{(1-x)^{5}-1}{(1-x)^{3}-1}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}[x]\) ([*] is a greatest integer function.)
Solution:

Question 3.
If f(r) = πr² then find \(\lim _{h \rightarrow 0}\left[\frac{f(r+h)-f(r)}{h}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 0}\left[\frac{x}{|x|+x^{2}}\right]\)
Solution:

Question 5.
Find the limit of the function, if it exists, at x = 1
\(f(x)=\left\{\begin{array}{lll}
7-4 x & \text { for } & x<1 \\
x^{2}+2 & \text { for } & x \geq 1
\end{array}\right.\)
Solution:

Question 6.
Given that 7x ≤ f(x) ≤ 3x² – 6 for all x. Determine the value of \(\lim _{x \rightarrow 3} f(x)\)
Solution:

Question 7.
\(\lim _{x \rightarrow 0}\left[\frac{\sec x^{2}-1}{x^{4}}\right]\)
Solution:

Question 8.
\(\lim _{x \rightarrow 0}\left[\frac{e^{x}+e^{-x}-2}{x \cdot \tan x}\right]\)
Solution:

Question 9.
\(\lim _{x \rightarrow 0}\left[\frac{x\left(6^{x}-3^{x}\right)}{\cos (6 x)-\cos (4 x)}\right]\)
Solution:

Question 10.
\(\lim _{x \rightarrow 0}\left[\frac{a^{3 x}-a^{2 x}-a^{x}+1}{x \cdot \tan x}\right]\)
Solution:

Question 11.
\(\lim _{x \rightarrow a}\left[\frac{\sin x-\sin a}{x-a}\right]\)
Solution:

Question 12.
\(\lim _{x \rightarrow 2}\left[\frac{\log x-\log 2}{x-2}\right]\)
Solution:

Question 13.
\(\lim _{x \rightarrow 1}\left[\frac{a b^{x}-a^{x} b}{x^{2}-1}\right]\)
Solution:

Question 14.
\(\lim _{x \rightarrow 0}\left[\frac{\left(5^{x}-1\right)^{2}}{\left(2^{x}-1\right) \log (1+x)}\right]\)
Solution:

Question 15.
\(\lim _{x \rightarrow \infty}\left[\frac{(2 x+1)^{2}(7 x-3)^{3}}{(5 x+2)^{5}}\right]\)
Solution:

Question 16.
\(\lim _{x \rightarrow a}\left[\frac{x \cos a-a \cos x}{x-a}\right]\)
Solution:

Question 17.
\(\lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{(\sin x-\cos x)^{2}}{\sqrt{2}-\sin x-\cos x}\right]\)
Solution:

Question 18.
\(\lim _{x \rightarrow 1}\left[\frac{2^{2 x-2}-2^{x}+1}{\sin ^{2}(x-1)}\right]\)
Solution:

Question 19.
\(\lim _{x \rightarrow 1}\left[\frac{4^{x-1}-2^{x}+1}{(x-1)^{2}}\right]\)
Solution:

Question 20.
\(\lim _{x \rightarrow 1}\left[\frac{\sqrt{x}-1}{\log x}\right]\)
Solution:

Question 21.
\(\lim _{x \rightarrow 0}\left(\frac{\sqrt{1-\cos x}}{x}\right)\)
Solution:

Question 22.
\(\lim _{x \rightarrow 1}\left(\frac{x+3 x^{2}+5 x^{3}+\cdots \cdots \cdots \cdots \cdots+(2 n-1) x^{n}-n^{2}}{x-1}\right)\)
Solution:

Question 23.
\(\lim _{x \rightarrow 0} \frac{1}{x^{12}}\left[1-\cos \left(\frac{x^{2}}{2}\right)-\cos \left(\frac{x^{4}}{4}\right)+\cos \left(\frac{x^{2}}{2}\right) \cdot \cos \left(\frac{x^{4}}{4}\right)\right]\)
Solution:

Question 24.
\(\lim _{x \rightarrow \infty}\left(\frac{8 x^{2}+5 x+3}{2 x^{2}-7 x-5}\right)^{\frac{4 x+3}{8 x-1}}\)
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Miscellaneous Exercise 2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2

(I) Select the correct answer from the given alternative:

Question 1.
The common ratio for the G.P. 0.12, 0.24, 0.48, is
(A) 0.12
(B) 0.2
(C) 0.02
(D) 2
Answer:
(D) 2

Question 2.
The tenth term of the geometric sequence is \(\frac{1}{4}, \frac{-1}{2}, 1,-2, \ldots\) is
(A) 1024
(B) \(\frac{1}{1024}\)
(C) -128
(D) \(\frac{-1}{128}\)
Answer:
(C) -128
Hint:

Question 3.
If for a G.P. \(\frac{t_{6}}{t_{3}}=\frac{1458}{54}\) then r = ?
(A) 3
(B) 2
(C) 1
(D) -1
Answer:
(A) 3
Hint:

Question 4.
Which term of the geometric progression 1, 2, 4, 8, ….. is 2048.
(A) 10th
(B) 11th
(C) 12th
(D) 13th
Answer:
(C) 12th
Hint:
Here, a = 1, r = 2
nth term of geometric progression = ar^n-1
∴ ar^n-1 = 2048
2^n-1 = 2¹¹
n – 1 = 11
∴ n = 12

Question 5.
If the common ratio of the G.P. is 5, the 5th term is 1875, the first term is
(A) 3
(B) 5
(C) 15
(D) -5
Answer:
(A) 3

Question 6.
The sum of 3 terms of a G.P. is \(\frac{21}{4}\) and their product is 1, then the common ratio is
(A) 1
(B) 2
(C) 4
(D) 8
Answer:
(C) 4
Hint:
Let three terms be \(\frac{a}{r}\), a, ar
According to the given conditions,
\(\frac{a}{r}\) + a + ar = \(\frac{21}{4}\) …..(i)
and \(\frac{a}{r}\) × a × ar = 1,
i.e., a³ = 1
∴ a = 1
∴ from equation (i), we get
\(\frac{1}{r}\) + 1 + r = \(\frac{21}{4}\)
By solving this, we get r = 4.

Question 7.
Sum to infinity of a G.P. 5, \(-\frac{5}{2}, \frac{5}{4},-\frac{5}{8}, \frac{5}{16}, \ldots\) is
(A) 5
(B) \(-\frac{1}{2}\)
(C) \(\frac{10}{3}\)
(D) \(\frac{3}{10}\)
Answer:
(C) \(\frac{10}{3}\)
Hint:
Here, a = 5, r = \(\frac{-1}{2}\), |r| < 1
∴ Sum to the infinity = \(\frac{a}{1-r}=\frac{5}{1+\frac{1}{2}}=\frac{10}{3}\)

Question 8.
The tenth term of H.P. \(\frac{2}{9}, \frac{1}{7}, \frac{2}{19}, \frac{1}{12}, \ldots\) is
(A) \(\frac{1}{27}\)
(B) \(\frac{9}{2}\)
(C) \(\frac{5}{2}\)
(D) 27
Answer:
(A) \(\frac{1}{27}\)
Hint:

Question 9.
Which of the following is not true, where A, G, H are the AM, GM, HM of a and b respectively, (a, b > 0)
(A) A = \(\frac{a+b}{2}\)
(B) G = \(\sqrt{a b}\)
(C) H = \(\frac{2 a b}{a+b}\)
(D) A = GH
Answer:
(D) A = GH

Question 10.
The G.M. of two numbers exceeds their H.M. by \(\frac{6}{5}\), the A.M. exceeds G.M. by \(\frac{3}{2}\) the two numbers are
(A) 6, \(\frac{15}{2}\)
(B) 15, 25
(C) 3, 12
(D) \(\frac{6}{5}\), \(\frac{3}{2}\)
Answer:
(C) 3, 12
Hint:

(II) Answer the following:

Question 1.
In a G.P., the fourth term is 48 and the eighth term is 768. Find the tenth term.
Solution:

Question 2.
Find the sum of the first 5 terms of the G.P. whose first term is 1 and the common ratio is \(\frac{2}{3}\).
Solution:

Question 3.
For a G.P. a = \(\frac{4}{3}\) and t₇ = \(\frac{243}{1024}\), find the value of r.
Solution:

Question 4.
For a sequence, if \(t_{n}=\frac{5^{n-2}}{7^{n-3}}\), verify whether the sequence is a G.P. If it is a G.P., find its first term and the common ratio.
Solution:
The sequence (t_n) is a G.P.
if \(\frac{\mathrm{t}_{\mathrm{n}+1}}{\mathrm{t}_{\mathrm{n}}}\) = constant for all n ∈ N.

Question 5.
Find three numbers in G.P. such that their sum is 35 and their product is 1000.
Solution:
Let the three numbers in G.P. be \(\frac{a}{r}\), a, ar.
According to the given conditions,
\(\frac{a}{r}\) + a + ar = 35
a(\(\frac{1}{r}\) + 1 + r) = 35 …..(i)
Also, (\(\frac{a}{r}\))(a)(ar) = 1000
a³ = 1000
∴ a = 10
Substituting the value of a in (i), we get

Hence, the three numbers in G.P. are 20, 10, 5, or 5, 10, 20.

Question 6.
Find five numbers in G.P. such that their product is 243 and the sum of the second and fourth numbers is 10.
Solution:
Let the five numbers in G.P. be \(\frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2}\).
According to the given condition,

Question 7.
For a sequence, S_n = 4(7ⁿ – 1), verify that the sequence is a G.P.
Solution:

∴ The given sequence is a G.P.

Question 8.
Find 2 + 22 + 222 + 2222 + … upto n terms.
Solution:
S_n = 2 + 22 + 222 +… upto n terms
= 2(1 + 11 + 111 + ….. upto n terms)
= \(\frac{2}{9}\) (9 + 99 + 999 + … upto n terms)
= \(\frac{2}{9}\) [(10 – 1) + (100 – 1) + (1000 – 1) + …… upto n terms]
= \(\frac{2}{9}\) [(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + ….. n times)]
Since 10, 100, 1000, ….. n terms are in G.P. with
a = 10, r = \(\frac{100}{10}\) = 10,

Question 9.
Find the nth term of the sequence 0.6, 0.66, 0.666, 0.6666,…
Solution:
0.6, 0.66, 0.666, 0.6666, …
∴ t₁ = 0.6
t₂ = 0.66 = 0.6 + 0.06
t₃ = 0.666 = 0.6 + 0.06 + 0.006
Hence, in general
t_n = 0.6 + 0.06 + 0.006 + …..upto n terms.
The terms are in G.P. with
a = 0.6, r = \(\frac{0.06}{0.6}\) = 0.1
∴ t_n = the sum of first n terms of the G.P.

Question 10.
Find \(\sum_{r=1}^{n}\left(5 r^{2}+4 r-3\right)\)
Solution:

Question 11.
Find \(\sum_{r=1}^{n} r(r-3)(r-2)\)
Solution:

Question 12.
Find \(\sum_{r=1}^{n} \frac{1^{2}+2^{2}+3^{2}+\ldots+r^{2}}{2 r+1}\)
Solution:
We know that

Question 13.
Find \(\sum_{r=1}^{n} \frac{1^{3}+2^{3}+3^{3}+\ldots r^{3}}{(r+1)^{2}}\)
Solution:

Question 14.
Find 2 × 6 + 4 × 9 + 6 × 12 + ….. upto n terms.
Solution:
2, 4, 6, ….. are in A.P.
∴ rth term = 2 + (r – 1) 2 = 2r
6, 9, 12, ….. are in A.P.
∴ rth term = 6 + (r – 1)(3) = (3r + 3)
∴ 2 × 6 + 4 × 9 + 6 × 12 + ….. to n terms

= n(n + 1) [2n + 1 + 3]
= 2n(n + 1)(n + 2)

Question 15.
Find 2 × 5 × 8 + 4 × 7 × 10 + 6 × 9 × 12 + …… upto n terms.
Solution:
2, 4, 6,… are in A.P.
∴ rth term = 2 + (r – 1) 2 = 2r
5, 7, 9, … are in A.P.
∴ rth term = 5 + (r – 1) (2) = (2r + 3)
8, 10, 12, … are in A.P.
∴ rth term = 8 + (r – 1) (2) = (2r + 6)
2 × 5 × 8 + 4 × 7 × 10 + 6 × 9 × 12 + ….. to n terms

= 2n (n + 1) [n(n + 1) + 3(2n + 1) + 9]
= 2n (n + 1)(n² + 7n + 12)
= 2n (n + 1) (n + 3) (n + 4)

Question 16.
Find \(\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{2}+\frac{1^{2}+2^{2}+3^{2}}{3}+\ldots\) upto n terms.
Solution:

Question 17.
Find 12² + 13² + 14² + 15² + ….. 20²
Solution:

Question 18.
If \(\frac{1+2+3+4+5+\ldots \text { upto } \mathrm{n} \text { terms }}{1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots \text { upto } \mathrm{n} \text { terms }}=\frac{3}{22}\), Find the value of n.
Solution:

Question 19.
Find (50² – 49²) + (48² – 47²) + (46² – 45²) +… + (2² – 1²).
Solution:

Question 20.
If \(\frac{1 \times 3+2 \times 5+3 \times 7+\ldots \text { upto } \mathrm{n} \text { terms }}{1^{3}+2^{3}+3^{3}+\ldots \text { upto } \mathrm{n} \text { terms }}=\frac{5}{9}\), find the value of n.
Solution:

Question 21.
For a G.P. if t₂ = 7, t₄ = 1575, find a.
Solution:

Question 22.
If for a G.P. t₃ = \(\frac{1}{3}\), t₆ = \(\frac{1}{81}\) find r.
Solution:

Question 23.
Find \(\sum_{r=1}^{n}\left(\frac{2}{3}\right)^{r}\).
Solution:

Question 24.
Find k so that k – 1, k, k + 2 are consecutive terms of a G.P.
Solution:
Since k – 1, k, k + 2 are consecutive terms of a G.P.,
\(\frac{k}{k-1}=\frac{k+2}{k}\)
k² = k² + k – 2
k – 2 = 0
∴ k = 2

Question 25.
If for a G.P. first term is (27)² and the seventh term is (8)², find S₈.
Solution:

Question 26.
If pth, qth and rth terms of a G.P. are x, y, z respectively. Find the value of \(x^{q-r} \cdot y^{r-p} \cdot z^{p-q}\).
Solution:
Let a be the first term and R be the common ratio of the G.P.

Question 27.
Which 2 terms are inserted between 5 and 40 so that the resulting sequence is G.P.
Solution:
Let the required numbers be G₁ and G₂.

∴ For the resulting sequence to be in G.P. we need to insert numbers 10 and 20.

Question 28.
If p, q, r are in G.P. and \(\mathrm{p}^{1 / \mathrm{x}}=\mathrm{q}^{1 / \mathrm{y}}=\mathrm{r}^{1 / \mathrm{z}}\), verify whether x, y, z are in A.P. or G.P. or neither.
Solution:

Question 29.
If a, b, c are in G.P. and ax² + 2bx + c = 0 and px² + 2qx + r = 0 have common roots, then verify that pb² – 2qba + ra² = 0.
Solution:
a, b, c are in G.P.
∴ b² = ac
ax² + 2bx + c = 0 becomes

Question 30.
If p, q, r, s are in G.P., show that (p² + q² + r²)(q² + r² + s²) = (pq + qr + rs)².
Solution:
p, q, r, s are in G.P.

Question 31.
If p, q, r, s are in G.P., show that (pⁿ + qⁿ), (qⁿ + rⁿ), (rⁿ + sⁿ) are also in G.P.
Solution:
p, q, r, s are in G.P.
Let the common ratio be R
∴ let p = \(\frac{a}{R^{3}}\), q = \(\frac{a}{R}\), r = aR and s = aR³
To show that (pⁿ + qⁿ), (qⁿ + rⁿ), (rⁿ + sⁿ) are in G.P,
i.e., we have to show

Question 32.
Find the coefficient x⁶ in the expression of e^2x using series expansion.
Solution:

Question 33.
Find the sum of infinite terms of \(1+\frac{4}{5}+\frac{7}{25}+\frac{10}{125}+\frac{13}{625}+\ldots\)
Solution:

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Ex 6.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Ex 6.3

Question 1.
Write the parametric equations of the circles:
(i) x² + y² = 9
(ii) x² + y² + 2x – 4y – 4 = 0
(iii) (x – 3)² + (y + 4)² = 25
Solution:
(i) Given equation of the circle is
x² + y² = 9
⇒ x² + y² = 3²
Comparing this equation with x² + y² = r², we get r = 3
The parametric equations of the circle in terms of θ are
x = r cos θ and y = r sin θ
⇒ x = 3 cos θ and y = 3 sin θ

(ii) Given equation of the circle is
x² + y² + 2x – 4y – 4 = 0
⇒ x² + 2x + y² – 4y – 4 = 0
⇒ x² + 2x + 1 – 1 + y² – 4y + 4 – 4 – 4 = 0
⇒ (x² + 2x + 1 ) + (y² – 4y + 4) – 9 = 0
⇒ (x + 1)² + (y – 2)² = 9
⇒ (x + 1)² + (y – 2)² = 3²
Comparing this equation with (x – h)² + (y – k)² = r², we get
h = -1, k = 2 and r = 3
The parametric equations of the circle in terms of θ are
x = h + r cos θ and y = k + r sin θ
⇒ x = -1 + 3 cos θ and y = 2 + 3 sin θ

(iii) Given equation of the circle is
(x – 3)² + (y + 4)² = 25
⇒ (x – 3)² + (y + 4)² = 5²
Comparing this equation with (x – h)² + (y – k)² = r², we get
h = 3, k = -4 and r = 5
The parametric equations of the circle in terms of θ are
x = h + r cos θ and y = k + r sin θ
⇒ x = 3 + 5 cos θ and y = -4 + 5 sin θ

Question 2.
Find the parametric representation of the circle 3x² + 3y² – 4x + 6y – 4 = 0.
Solution:
Given equation of the circle is 3x² + 3y² – 4x + 6y – 4 = 0
Dividing throughout by 3, we get

Comparing this equation with (x – h)² + (y – k)² = r², we get
h = \(\frac{2}{3}\), k = -1 and r = \(\frac{5}{3}\)
The parametric representation of the circle in terms of θ are
x = h + r cos θ and y = k + r sin θ
⇒ x = \(\frac{2}{3}\) + \(\frac{5}{3}\) cos θ and y = -1 + \(\frac{5}{3}\) sin θ

Question 3.
Find the equation of a tangent to the circle x² + y² – 3x + 2y = 0 at the origin.
Solution:
Given equation of the circle is x² + y² – 3x + 2y = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -3, 2f = 2, c = 0
⇒ g = \(-\frac{3}{2}\), f = 1, c = 0
The equation of a tangent to the circle
x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is xx₁ +yy₁ + g(x + x₁) + f(y + y₁) + c = 0
The equation of the tangent at (0, 0) is
x(0) + y(0) + (\(-\frac{3}{2}\)) (x + 0) + 1(y + 0) + 0 = 0
⇒ \(-\frac{3}{2}\)x + y = 0
⇒ 3x – 2y = 0

Question 4.
Show that the line 7x – 3y – 1 = 0 touches the circle x² + y² + 5x – 7y + 4 = 0 at point (1, 2).
Solution:
Given equation of the circle is x² + y² + 5x – 7y + 4 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = 5, 2f = -7, c = 4
⇒ g = \(\frac{5}{2}\), f = \(\frac{-7}{2}\), c = 4
The equation of a tangent to the circle x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
The equation of the tangent at (1, 2) is

7x – 3y – 1 = 0, which is same as the given line.
The line 7x – 3y – 1 = 0 touches the given circle at (1, 2).

Question 5.
Find the equation of tangent to the circle x² + y² – 4x + 3y + 2 = 0 at the point (4, -2).
Solution:
Given equation of the circle is x² + y² – 4x + 3y + 2 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -4, 2f = 3, c = 2
g = -2, f = \(\frac{3}{2}\), c = 2
The equation of a tangent to the circle x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
The equation of the tangent at (4, -2) is
x(4) + y(-2) – 2(x + 4) + \(\frac{3}{2}\)(y – 2) + 2 = 0
⇒ 4x – 2y – 2x – 8 + \(\frac{3}{2}\) y – 3 + 2 = 0
⇒ 2x – \(\frac{1}{2}\)y – 9 = 0
⇒ 4x – y – 18 = 0

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

Calculation of variance and S.D.

Question 2.
Find variance and S.D. for the following set of numbers.
65, 77, 81, 98, 100, 80, 129
Solution:

Question 3.
Compute variance and standard deviation for the following data:

Solution:

Question 4.
Compute the variance and S.D.

Solution:
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-34}{1}\)
Calculation of variance of u:

Question 5.
Following data gives ages of 100 students in a college. Calculate variance and S.D.

Solution:
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-19}{1}\)

Question 6.
Find mean, variance and S.D. of the following data.

Solution:


Alternate Method:
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-55}{10}\)
Calculation of variance of u:

Question 7.
Find the variance and S.D. of the following frequency distribution which gives the distribution of 200 plants according to their height.

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:

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.


From (i), we get
x₅ = 5 or x₅ = 7
∴ The two numbers are 5 and 7.

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}

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

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)}

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) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{12} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{2}}=\frac{12 \times 4}{\frac{52 \times 51}{2 \times 1}}=\frac{8}{221}\)

(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.

(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.

(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.

(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.

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) = \(\frac{n(D)}{n(S)}=\frac{5}{75}=\frac{1}{15}\)

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 – \(\frac{1}{6}\) = \(\frac{5}{6}\)

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) = \(\frac{n(A)}{n(S)}=\frac{4 \times 3 \times 2 \times 1}{4 \times 4 \times 4 \times 4}=\frac{3}{32}\)

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) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{7} \mathrm{C}_{3}}{{ }^{11} \mathrm{C}_{3}}=\frac{7 \times 6 \times 5}{11 \times 10 \times 9}=\frac{7}{33}\)

(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) = \(\frac{n(A)}{n(S)}=\frac{{ }^{7} C_{1} \times{ }^{4} C_{2}}{{ }^{11} C_{3}}=\frac{7 \times \frac{4 \times 3}{2}}{\frac{11 \times 10 \times 9}{3 \times 2 \times 1}}=\frac{14}{55}\)

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

Question 3.
The following data gives a number of typing mistakes done by Radha during a week. Find the range of the data.

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.

Solution:
Here, largest value (L) = 6, smallest value (S) = 1
∴ Range = L – S = 6 – 1 = 5

Question 5.
Find range for the following data:

Solution:
Here, upper limit of the highest class (L) = 72
lower limit of the lowest class (S) = 62
∴ Range = L – S = 72 – 62 = 10

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 = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)
= \(\frac{2(5)}{\frac{4}{5}}\)
= \(\frac{25}{2}\)

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

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

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

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

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)

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 = \(\frac{50}{3}\)

The required equation of ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\).

(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

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 = \(\frac{5}{32}\)

(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

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

The required equation of ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\).

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)

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 \(\frac{1}{\sqrt{3}}\)

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

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

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
= \(\left(\frac{-\mathrm{a}^{2} \mathrm{~m}}{\mathrm{c}}, \frac{\mathrm{b}^{2}}{\mathrm{c}}\right)\)
= \(\left(\frac{(-16)(1)}{-5}, \frac{9}{-5}\right)\)
= \(\left(\frac{16}{5}, \frac{-9}{5}\right)\)

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

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

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

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

(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

(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

(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

(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 = \(\frac{3}{4} x \pm \sqrt{20\left(\frac{3}{4}\right)^{2}+5}\)

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.

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

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]

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

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

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

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Miscellaneous Exercise 6 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Miscellaneous Exercise 6

(I) Choose the correct alternative.

Question 1.
Equation of a circle which passes through (3, 6) and touches the axes is
(A) x² + y² + 6x + 6y + 3 = 0
(B) x² + y² – 6x – 6y – 9 = 0
(C) x² + y² – 6x – 6y + 9 = 0
(D) x² + y² – 6x + 6y – 3 = 0
Answer:
(C) x² + y² – 6x – 6y + 9 = 0

Question 2.
If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 sq. units, then find the equation of the circle.
(A) x² + y² – 2x + 2y = 40
(B) x² + y² – 2x – 2y = 47
(C) x² + y² – 2x + 2y = 47
(D) x² + y² – 2x – 2y = 40
Answer:
(C) x² + y² – 2x + 2y = 47
Hint:
Centre of circle = Point of intersection of diameters.
Solving 2x – 3y = 5 and 3x – 4y = 7, we get
x = 1, y = -1
Centre of the circle C(h, k) = C(1, -1)
∴ Area = 154
πr² = 154
\(\frac{22}{7} \times r^{2}\) = 154
r² = 154 × \(\frac{22}{7}\) = 49
∴ r = 7
equation of the circle is
(x – 1)² + (y + 1)² = 72
x² + y² – 2x + 2y = 47

Question 3.
Find the equation of the circle which passes through the points (2, 3) and (4, 5), and the center lies on the straight line y – 4x + 3 = 0.
(A) x² + y² – 4x – 10y + 25 = 0
(B) x² + y² – 4x – 10y – 25 = 0
(C) x² + y² – 4x + 10y – 25 = 0
(D) x² + y² + 4x – 10y + 25 = 0
Answer:
(A) x² + y² – 4x – 10y + 25 = 0

Question 4.
The equation(s) of the tangent(s) to the circle x² + y² = 4 which are parallel to x + 2y + 3 = 0 are
(A) x – 2y = 2
(B) x + 2y = ±2√3
(C) x + 2y = ±2√5
(D) x – 2y = ±2√5
Answer:
(C) x + 2y = ±2√5

Question 5.
If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find the radius of the circle.
(A) \(\frac{3}{4}\)
(B) \(\frac{4}{3}\)
(C) \(\frac{1}{4}\)
(D) \(\frac{7}{4}\)
Answer:
(A) \(\frac{3}{4}\)
Hint:
Tangents are parallel to each other.
The perpendicular distance between tangents = diameter

Question 6.
The area of the circle having centre at (1, 2) and passing through (4, 6) is
(A) 5π
(B) 10π
(C) 25π
(D) 100π
Answer:
(C) 25π
Hint:

Question 7.
If a circle passes through the points (0, 0), (a, 0), and (0, b), then find the co-ordinates of its centre.
(A) \(\left(\frac{-a}{2}, \frac{-b}{2}\right)\)
(B) \(\left(\frac{a}{2}, \frac{-b}{2}\right)\)
(C) \(\left(\frac{-a}{2}, \frac{b}{2}\right)\)
(D) \(\left(\frac{a}{2}, \frac{b}{2}\right)\)
Answer:
(D) \(\left(\frac{a}{2}, \frac{b}{2}\right)\)

Question 8.
The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is
(A) x² + y² = 9a²
(B) x² + y² = 16a²
(C) x² + y² = 4a²
(D) x² + y² = a²
Answer:
(C) x² + y² = 4a²
Hint:
Since the triangle is equilateral.
The centroid of the triangle is same as the circumcentre
and radius of the circumcircle = \(\frac{2}{3}\) (median) = \(\frac{2}{3}\)(3a) = 2a
Hence, the equation of the circumcircle whose centre is at (0, 0) and radius 2a is x² + y² = 4a²

Question 9.
A pair of tangents are drawn to a unit circle with centre at the origin and these tangents intersect at A enclosing an angle of 60. The area enclosed by these tangents and the arc of the circle is
(A) \(\frac{2}{\sqrt{3}}-\frac{\pi}{6}\)
(B) \(\sqrt{3}-\frac{\pi}{3}\)
(C) \(\frac{\pi}{3}-\frac{\sqrt{3}}{6}\)
(D) \(\sqrt{3}\left(1-\frac{\pi}{6}\right)\)
Answer:
(B) \(\sqrt{3}-\frac{\pi}{3}\)
Hint:

Question 10.
The parametric equations of the circle x² + y² + mx + my = 0 are
(A) x = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \cos \theta\), y = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \sin \theta\)
(B) x = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \cos \theta\), y = \(\frac{+m}{2}+\frac{m}{\sqrt{2}} \sin \theta\)
(C) x = 0, y = 0
(D) x = m cos θ, y = m sin θ
Answer:
(A) x = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \cos \theta\), y = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \sin \theta\)

(II) Answer the following:

Question 1.
Find the centre and radius of the circle x² + y² – x + 2y – 3 = 0.
Solution:
Given equation of the circle is x² + y² – x + 2y – 3 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -1, 2f = 2 and c = -3
g = \(\frac{-1}{2}\), f = 1 and c = -3
Centre of the circle = (-g, -f) = (\(\frac{1}{2}\), -1)
and radius of the circle

Question 2.
Find the centre and radius of the circle x = 3 – 4 sin θ, y = 2 – 4 cos θ.
Solution:
Given, x = 3 – 4 sin θ, y = 2 – 4 cos θ
⇒ x – 3 = -4 sin θ, y – 2 = -4 cos θ
On squaring and adding, we get
⇒ (x – 3)² + (y – 2)² = (-4 sin θ)² + (-4 cos θ)²
⇒ (x – 3)² + (y – 2)² = 16 sin² θ + 16 cos² θ
⇒ (x – 3)² + (y – 2)² = 16(sin² θ + cos² θ)
⇒ (x – 3)² + (y – 2)² = 16(1)
⇒ (x – 3)² + (y – 2)² = 16
⇒ (x – 3)² + (y – 2)² = 4²
Comparing this equation with (x – h)² + (y – k)² = r², we get
h = 3, k = 2, r = 4
∴ Centre of the circle is (3, 2) and radius is 4.

Question 3.
Find the equation of circle passing through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0 and whose centre is the point of intersection of lines x + y + 1 = 0 and x – 2y + 4 = 0.
Solution:
Required circle passes through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0.
x + 3y = 0
⇒ x = -3y ……..(i)
2x – 7y = 0 ……(ii)
Substituting x = -3y in (ii), we get
⇒ 2(-3y) – 7y = 0
⇒ -6y – 7y = 0
⇒ -13y = 0
⇒ y = 0
Substituting y = 0 in (i), we get
x = -3(0) = 0
Point of intersection is O(0, 0).
This point O(0, 0) lies on the circle.
Let C(h, k) be the centre of the required circle.
Since, point of intersection of lines x + y = -1 and x – 2y = -4 is the centre of circle.
∴ x = h, y = k
∴ Equations of lines become
h + k = -1 ……(iii)
h – 2k = -4 …..(iv)
By (iii) – (iv), we get
3k = 3
⇒ k = 1
Substituting k = 1 in (iii), we get
h + 1 = -1
⇒ h = -2
∴ Centre of the circle is C(-2, 1) and it passes through point O(0, 0).
Radius(r) = OC
= \(\sqrt{(0+2)^{2}+(0-1)^{2}}\)
= \(\sqrt{4+1}\)
= √5
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = -2, k = 1
the required equation of the circle is
(x + 2)² + (y – 1)² = (√5)²
⇒ x² + 4x + 4 + y² – 2y + 1 = 5
⇒ x² + y² + 4x – 2y = 0

Question 4.
Find the equation of the circle which passes through the origin and cuts off chords of lengths 4 and 6 on the positive side of the X-axis and Y-axis respectively.
Solution:

Let the circle cut the chord of length 4 on X-axis at point A and the chord of length 6 on the Y-axis at point B.
∴ the co-ordinates of point A are (4, 0) and co-ordinates of point B are (0, 6).
Since ∠BOA is a right angle.
AB represents the diameter of the circle.
The equation of a circle having (x₁, y₁) and (x₂, y₂) as endpoints of diameter is given by
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
Here, x₁ = 4, y₁ = 0, x₂ = 0, y₂ = 6
∴ the required equation of the circle is
⇒ (x – 4) (x – 0) + (y – 0) (y – 6) = 0
⇒ x² – 4x + y² – 6y = 0
⇒ x² + y² – 4x – 6y = 0

Question 5.
Show that the points (9, 1), (7, 9), (-2, 12) and (6, 10) are concyclic.
Solution:
Let the equation of circle passing through the points (9, 1), (7, 9), (-2, 12) be
x² + y² + 2gx + 2fy + c = 0 …….(i)
For point (9, 1),
Substituting x = 9 andy = 1 in (i), we get
81 + 1 + 18g + 2f + c = 0
⇒ 18g + 2f + c = -82 …..(ii)
For point (7, 9),
Substituting x = 7 andy = 9 in (i), we get
49 + 81 + 14g + 18f + c = 0
⇒ 14g + 18f + c = -130 ……(iii)
For point (-2, 12),
Substituting x = -2 and y = 12 in (i), we get
4 + 144 – 4g + 24f + c = 0
⇒ -4g + 24f + c = -148 …..(iv)
By (ii) – (iii), we get
4g – 16f = 48
⇒ g – 4f = 12 …..(v)
By (iii) – (iv), we get
18g – 6f = 18
⇒ 3g – f = 3 ……(vi)
By 3 × (v) – (vi), we get
-11f = 33
⇒ f = -3
Substituting f = -3 in (vi), we get
3g – (-3) = 3
⇒ 3g + 3 = 3
⇒ g = 0
Substituting g = 0 and f = -3 in (ii), we get
18(0) + 2(-3) + c = – 82
⇒ -6 + c = -82
⇒ c = -76
Equation of the circle becomes
x² + y² + 2(0)x + 2(-3)y + (-76) = 0
⇒ x² + y² – 6y – 76 = 0 ……(vii)
Now for the point (6, 10),
Substituting x = 6 and y = 10 in L.H.S. of (vii), we get
L.H.S = 6² + 10² – 6(10) – 76
= 36 + 100 – 60 – 76
= 0
= R.H.S.
∴ Point (6,10) satisfies equation (vii).
∴ the given points are concyciic.

Question 6.
The line 2x – y + 6 = 0 meets the circle x² + y² + 10x + 9 = 0 at A and B. Find the equation of circle with AB as diameter. Solution:
2x – y + 6 = 0
⇒ y = 2x + 6
Substituting y = 2x + 6 in x² + y² + 10x + 9 = 0, we get
⇒ x² + (2x + 6)² + 10x + 9 = 0
⇒ x² + 4x² + 24x + 36 + 10x + 9 = 0
⇒ 5x² + 34x + 45 = 0
⇒ 5x² + 25x + 9x + 45 = 0
⇒ (5x + 9) (x + 5) = 0
⇒ 5x = -9 or x = -5
⇒ x = \(\frac{-9}{5}\) or x = -5
When x = \(\frac{-9}{5}\),
y = 2 × \(\frac{-9}{5}\) + 6
= \(\frac{-18}{5}\) + 6
= \(\frac{-18+30}{5}\)
= \(\frac{12}{5}\)
∴ Point of intersection is A\(\left(\frac{-9}{5}, \frac{12}{5}\right)\)
When x = -5,
y = -10 + 6 = -4
∴ Point of intersection in B (-5, -4).
By diameter form, equation of circle with AB as diameter is
(x + \(\frac{9}{5}\)) (x + 5) + (y – \(\frac{12}{5}\)) (y + 4) = 0
⇒ (5x + 9) (x + 5) + (5y – 12) (y + 4) = 0
⇒ 5x² + 25x + 9x + 45 + 5y² + 20y – 12y – 48 = 0
⇒ 5x² + 5y² + 34x + 8y – 3 = 0

Question 7.
Show that x = -1 is a tangent to circle x² + y² – 4x – 2y – 4 = 0 at (-1, 1).
Solution:
Given equation of circle is x² + y² – 4x – 2y – 4 = 0.
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -4, 2f = -2, c = -4
⇒ g = -2, f = -1, c = -4
The equation of a tangent to the circle
x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
the equation of the tangent at (-1, 1) is
⇒ x(-1) + y(1) – 2(x – 1) – 1(y + 1) – 4 = 0
⇒ -3x – 3 = 0
⇒ -x – 1 = 0
⇒ x = -1
∴ x = -1 is the tangent to the given circle at (-1, 1).

Question 8.
Find the equation of tangent to the circle x² + y² = 64 at the point P(\(\frac{2 \pi}{3}\)).
Solution:
Given equation of circle is x² + y² = 64
Comparing this equation with x² + y² = r², we get r = 8
The equation of a tangent to the circle x² + y² = r² at P(θ) is x cos θ + y sin θ = r
∴ the equation of the tangent at P(\(\frac{2 \pi}{3}\)) is
⇒ x cos \(\frac{2 \pi}{3}\) + y sin \(\frac{2 \pi}{3}\) = 9
⇒ \(x\left(\frac{-1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=8\)
⇒ -x + √3y = 16
⇒ x – √3y + 16 = 0

Question 9.
Find the equation of locus of the point of intersection of perpendicular tangents drawn to the circle x = 5 cos θ and y = 5 sin θ.
Solution:
The locus of the point of intersection of perpendicular tangents is the director circle of the given circle.
x = 5 cos θ and y = 5 sin θ
⇒ x² + y² = 25 cos² θ + 25 sin² θ
⇒ x² + y² = 25 (cos² θ + sin² θ)
⇒ x² + y² = 25(1) = 25
The equation of the director circle of the circle x² + y² = a² is x² + y² = 2a².
Here, a = 5
∴ the required equation is
x² + y² = 2(5)² = 2(25)
∴ x² + y² = 50

Question 10.
Find the equation of the circle concentric with x² + y² – 4x + 6y = 1 and having radius 4 units.
Solution:
Given equation of circle is
x² + y² – 4x + 6y = 1 i.e., x² + y² – 4x + 6y – 1 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -4, 2f = 6
⇒ g = -2, f = 3
Centre of the circle = (-g, -f) = (2, -3)
Given circle is concentric with the required circle.
∴ They have same centre.
∴ Centre of the required circle = (2, -3)
The equation of a circle with centre at (h, k) and radius r is (x – h)² + (y – k)² = r²
Here, h = 2, k = -3 and r = 4
∴ the required equation of the circle is
(x – 2)² + [y – (-3)]² = 4²
⇒ (x – 2)² + (y + 3)² = 16
⇒ x² – 4x + 4 + y² + 6y + 9 – 16 = 0
⇒ x² + y² – 4x + 6y – 3 = 0

Question 11.
Find the lengths of the intercepts made on the co-ordinate axes, by the circles.
(i) x² + y² – 8x + y – 20 = 0
(ii) x² + y² – 5x + 13y – 14 = 0
Solution:
To find x-intercept made by the circle x² + y² + 2gx + 2fy + c = 0,
substitute y = 0 and get a quadratic equation in x, whose roots are, say, x₁ and x₂.

These values represent the abscissae of ends A and B of the x-intercept.
Length of x-intercept = |AB| = |x₂ – x₁|
Similarly, substituting x = 0, we get a quadratic equation in y whose roots, say, y₁ and y₂ are ordinates of the ends C and D of the y-intercept.
Length of y-intercept = |CD| = |y₂ – y₁|
(i) Given equation of the circle is
x² + y² – 8x + y – 20 = 0 ……(i)
Substituting y = 0 in (i), we get
x² – 8x – 20 = 0 ……(ii)
Let AB represent the x-intercept, where
A = (x₁, 0), B = (x₂, 0)
Then from (ii),
x₁ + x₂ = 8 and x₁x₂ = -20
(x₁ – x₂)² = (x₁ + x₂)² – 4x₁x₂
= (8)² – 4(-20)
= 64 + 80
= 144
∴ |x₁ – x₂| = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}}\) = √144 = 12
∴ Length of x – intercept =12 units
Substituting x = 0 in (i), we get
y² + y – 20 = 0 …..(iii)
Let CD represent the y – intercept,
where C = (0, y₁) and D = (0, y₂)
Then from (iii),
y₁ + y₂ = -1 and y₁y₂ = -20
(y₁ – y₂)² = (y₁ + y₂)² – 4y₁y₂
= (-1)² – 4(-20)
= 1 + 80
= 81
∴ |y₁ – y₂| = \(\sqrt{\left(y_{1}-y_{2}\right)^{2}}\) = √81 = 9
∴ Length of y – intercept = 9 units.

Alternate Method:
Given equation of the circle is x² + y² – 8x + y – 20 = 0 ……(i)
x-intercept:
Substituting y = 0 in (i), we get
x² – 8x – 20 = 0
⇒ (x – 10)(x + 2) = 0
⇒ x = 10 or x = -2
length of x-intercept = |10 – (-2)| = 12 units
y-intercept:
Substituting x = 0 in (i), we get
y² + y – 20 = 0
⇒ (y + 5)(y – 4) = 0
⇒ y = -5 or y = 4
length of y-intercept = |-5 – 4| = 9 units

(ii) Given equation of the circle is
x² + y² – 5x + 13y – 14 = 0
Substituting y = 0 in (i), we get
x² – 5x – 14 = 0 ……(ii)
Let AB represent the x-intercept, where
A = (x₁, 0), B = (x₂, 0)
Then from (ii),
x₁ + x₂ = 5 and x₁x₂ = -14
(x₁ – x₂)² = (x₁ + x₂)² – 4x₁x₂
= (5)² – 4(-14)
= 25 + 56
= 81
∴ |x₁ – x₂| = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}}\) = √81 = 9
∴ Length of x-intercept = 9 units
Substituting x = 0 in (i), we get
y² + 13y – 14 = 0 ……(iii)
Let CD represent they-intercept,
where C = (0, y₁), D = (0, y₂).
Then from (iii),
y₁ + y₂ = -13 and y₁y₂ = -14
(y₁ – y₂)² = (y₁ + y₂)² – 4y₁y₂
= (-13)² – 4(-14)
= 169 + 56
= 225
∴ |y₁ – y₂| = \(\sqrt{\left(y_{1}-y_{2}\right)^{2}}\) = √225 = 15
∴ Length ofy-intercept = 15 units

Question 12.
Show that the circles touch each other externally. Find their point of contact and the equation of their common tangent.
(i) x² + y² – 4x + 10y + 20 = 0
x² + y² + 8x – 6y – 24 = 0
(ii) x² + y² – 4x – 10y + 19 = 0
x² + y² + 2x + 8y – 23 = 0
Solution:
(i) Given equation of the first circle is x² + y² – 4x + 10y + 20 = 0
Here, g = -2, f = 5, c = 20
Centre of the first circle is C₁ = (2, -5)
Radius of the first circle is
r₁ = \(\sqrt{(-2)^{2}+5^{2}-20}\)
= \(\sqrt{4+25-20}\)
= √9
= 3
Given equation of the second circle is x² + y² + 8x – 6y – 24 = 0
Here, g = 4, f = -3, c = -24
Centre of the second circle is C₂ = (-4, 3)
Radius of the second circle is
r₂ = \(\sqrt{4^{2}+(-3)^{2}+24}\)
= \(\sqrt{16+9+24}\)
= √49
= 7
By distance formula,
C₁C₂ = \(\sqrt{(-4-2)^{2}+[3-(-5)]^{2}}\)
= \(\sqrt{36+64}\)
= √1oo
= 10
r₁ + r₂ = 3 + 7 = 10
Since, C₁C₂ = r₁ + r₂
∴ the given circles touch each other externally.

Let P(x, y) be the point of contact.
∴ P divides C₁C₂ internally in the ratio r₁ : r₂ i.e. 3 : 7.
∴ By internal division,

Equation of common tangent is
(x² + y² – 4x + 10y + 20) – (x² + y² + 8x – 6y – 24) = 0
⇒ -4x + 10y + 20 – 8x + 6y + 24 = 0
⇒ -12x + 16y + 44 = 0
⇒ 3x – 4y – 11 = 0

(ii) Given equation of the first circle is x² + y² – 4x – 10y + 19 = 0
Here, g = -2, f = -5, c = 19
Centre of the first circle is C₁ = (2, 5)
Radius of the first circle is
r₁ = \(\sqrt{(-2)^{2}+(-5)^{2}-19}\)
= \(\sqrt{4+25-19}\)
= √10
Given equation of the second circle is x² + y² + 2x + 8y – 23 = 0
Here, g = 1, f = 4, c = -23
Centre of the second circle is C₂ = (-1, -4)
Radius of the second circle is
r₂ = \(\sqrt{(-1)^{2}+4^{2}+23}\)
= \(\sqrt{1+16+23}\)
= √40
= 2√10
By distance formula,
C₁C₂ = \(\sqrt{(-1-2)^{2}+(-4-5)^{2}}\)
= \(\sqrt{9+81}\)
= √90
= 3√10
r₁ + r₂ = √10 + 2√10 = 3√10
Since, C₁C₂ = r₁ + r₂
the given circles touch each other externally.
r₁ : r₂ = √10 : 2√10 = 1 : 2
Let P(x, y) be the point of contact.

∴ P divides C₁ C₂ internally in the ratio r₁ : r₂ i.e. 1 : 2
∴ By internal division,

Point of contact = (1, 2)
Equation of common tangent is
(x² + y² – 4x – 10y + 19) – (x² + y² + 2x + 8y – 23) = 0
⇒ -4x – 10y + 19 – 2x – 8y + 23 = 0
⇒ -6x – 18y + 42 = 0
⇒ x + 3y – 7 = 0

Question 13.
Show that the circles touch each other internally. Find their point of contact and the equation of their common tangent.
(i) x² + y² – 4x – 4y – 28 = 0,
x² + y² – 4x – 12 = 0
(ii) x² + y² + 4x – 12y + 4 = 0,
x² + y² – 2x – 4y + 4 = 0
Solution:
(i) Given equation of the first circle is x² + y² – 4x – 4y – 28 = 0
Here, g = -2, f = -2, c = -28
Centre of the first circle is C₁ = (2, 2)
Radius of the first circle is
r₁ = \(\sqrt{(-2)^{2}+(-2)^{2}+28}\)
= \(\sqrt{4+4+28}\)
= √36
= 6
Given equation of the second circle is x² + y² – 4x – 12 = 0
Here, g = -2, f = 0, c = -12
Centre of the second circle is C₂ = (2, 0)
Radius of the second circle is
r₂ = \(\sqrt{(-2)^{2}+0^{2}+12}\)
= \(\sqrt{4+12}\)
= √16
= 4
By distance formula,
C₁C₂ = \(\sqrt{(2-2)^{2}+(0-2)^{2}}\)
= √4
= 2
|r₁ – r₂| = 6 – 4 = 2
Since, C₁C₂ = |r₁ – r₂|
∴ the given circles touch each other internally.
Equation of common tangent is
(x² + y² – 4x – 4y – 28) – (x² + y² – 4x – 12) = 0
⇒ -4x – 4y – 28 + 4x + 12 = 0
⇒ -4y – 16 = 0
⇒ y + 4 = 0
⇒ y = -4
Substituting y = -4 in x² + y² – 4x – 12 = 0, we get
⇒ x² + (-4)² – 4x – 12 = 0
⇒ x² + 16 – 4x – 12 = 0
⇒ x² – 4x + 4 = 0 .
⇒ (x – 2)² = 0
⇒ x = 2
∴ Point of contact is (2, -4) and equation of common tangent is y + 4 = 0.

(ii) Given equation of the first circle is x² + y² + 4x – 12y + 4 = 0
Here, g = 2, f = -6, c = 4
Centre of the first circle is C₁ = (-2, 6)
Radius of the first circle is
r₁ = \(\sqrt{2^{2}+(-6)^{2}-4}\)
= \(\sqrt{4+36-4}\)
= √36
= 6
Given equation of the second circle is x² + y² – 2x – 4y + 4 = 0
Here, g = -1, f = -2, c = 4
Centre of the second circle is C₂ = (1, 2)
Radius of the second circle is
r₂ = \(\sqrt{(-1)^{2}+(-2)^{2}-4}\)
= \(\sqrt{1+4-4}\)
= √1
= 1
By distance formula,
C₁C₂ = \(\sqrt{[1-(-2)]^{2}+(2-6)^{2}}\)
= \(\sqrt{9+16}\)
= √25
= 5
|r₁ – r₂| = 6 – 1 = 5
Since, C₁C₂ = |r₁ – r₂|
the given circles touch each other internally.
Equation of common tangent is
(x² + y² + 4x – 12y + 4) – (x² + y² – 2x – 4y + 4) = 0
⇒ 4x – 12y + 4 + 2x + 4y – 4 = 0
⇒ 6x – 8y = 0
⇒ 3x – 4y = 0
⇒ y = \(\frac{3 x}{4}\)
Substituting y = \(\frac{3 x}{4}\) in x² + y² – 2x – 4y + 4 = 0, we get

∴ Point of contact is \(\left(\frac{8}{5}, \frac{6}{5}\right)\) and equation of common tangent is 3x – 4y = 0.

Question 14.
Find the length of the tangent segment drawn from the point (5, 3) to the circle x² + y² + 10x – 6y – 17 = 0.
Solution:
Given equation of circle is x² + y² + 10x – 6y – 17 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = 10, 2f = -6, c = -17
⇒ g = 5, f = -3, c = -17
Centre of circle = (-g, -f) = (-5, 3)

In right angled ∆ABC,
BC² = AB² + AC² …..[Pythagoras theorem]
⇒ (10)² = AB²+ (√51)²
⇒ AB² = 100 – 51 = √49
⇒ AB = 7
∴ Length of the tangent segment from (5, 3) is 7 units.

Alternate method:
Given equation of circle is x² + y² + 10x – 6y – 17 = 0
Here, g = 5, f = -3, c = -17
Length of the tangent segment to the circle x² + y² + 2gx + 2fy + c = 0 from the point (x₁, y₁) is \(\sqrt{x_{1}^{2}+y_{1}^{2}+2 g x_{1}+2 f y_{1}+c}\)
Length of the tangent segment from (5, 3)
= \(\sqrt{(5)^{2}+(3)^{2}+10(5)-6(3)-17}\)
= \(\sqrt{25+9+50-18-17}\)
= √49
= 7 units

Question 15.
Find the value of k, if the length of the tangent segment from the point (8, -3) to the circle x² + y² – 2x + ky – 23 = 0 is √10.
Solution:
Given equation of the circle is x² + y² – 2x + ky – 23 = 0
Here, g = -1, f = \(\frac{\mathrm{k}}{2}\), c = -23
Length of the tangent segment to the circle x² + y² + 2gx + 2fy + c = 0 from the point (x₁, y₁) is \(\sqrt{x_{1}^{2}+y_{1}^{2}+2 g x_{1}+2 f y_{1}+c}\)
Length of the tangent segment from (8, -3) = √10
⇒ \(\sqrt{8^{2}+(-3)^{2}-2(8)+k(-3)-23}=\sqrt{10}\)
⇒ 64 + 9 – 16 – 3k – 23 = 10 …..[Squaring both the sides]
⇒ 34 – 3k = 10
⇒ 3k = 24
⇒ k = 8

Question 16.
Find the equation of tangent to circle x² + y² – 6x – 4y = 0, at the point (6, 4) on it.
Solution:
Given equation of the circle is x² + y² – 6x – 4y = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -6, 2f = -4, c = 0
⇒ g = -3, f = -2, c = 0
The equation of a tangent to the circle x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
the equation of the tangent at (6, 4) is
x(6) + y(4) – 3(x + 6) – 2(y + 4) + 0 = 0
⇒ 6x + 4y – 3x – 18 – 2y – 8 = 0
⇒ 3x + 2y – 26 = 0

Alternate method:
Given equation of the circle is x² + y² – 6x – 4y = 0
x(x – 6) + y(y – 4) = 0, which is in diameter form where (0, 0) and (6, 4) are endpoints of diameter.

Slope of OP = \(\frac{4-0}{6-0}=\frac{2}{3}\)
Since, OP is perpendicular to the required tangent.
Slope of the required tangent = \(\frac{-3}{2}\)
the equation of the tangent at (6, 4) is
y – 4 = \(\frac{-3}{2}\) (x – 6)
⇒ 2(y – 4) = 3(x – 6)
⇒ 2y – 8 = -3x + 18
⇒ 3x + 2y – 26 = 0

Question 17.
Fihd the equation of tangent to circle x² + y² = 5, at the point (1, -2) on it.
Solution:
Given equation of the circle is x² + y² = 5
Comparing this equation with x² + y² = r², we get
r² = 5
The equation of a tangent to the circle x² + y² = r² at (x₁, y₁) is xx₁ + yy₁ = r²
the equation of the tangent at (1, -2) is
x(1) + y(-2) = 5
⇒ x – 2y = 5

Question 18.
Find the equation of tangent to circle x = 5 cos θ, y = 5 sin θ, at the point θ = \(\frac{\pi}{3}\) on it.
Solution:
The equation of a tangent to the circle x² + y² = r² at P(θ) is x cos θ + y sin θ = r
Here, r = 5, θ = \(\frac{\pi}{3}\)
the equation of the tangent at P(\(\frac{\pi}{3}\)) is
x cos \(\frac{\pi}{3}\) + y sin \(\frac{\pi}{3}\) = 5
⇒ \(x\left(\frac{1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=5\)
⇒ x + √3y = 10

Question 19.
Show that 2x + y + 6 = 0 is a tangent to x² + y² + 2x – 2y – 3 = 0. Find its point of contact.
Solution:
Given equation of circle is
x² + y² + 2x – 2y – 3 = 0 ….(i)
Given equation of line is 2x + y + 6 = 0
y = -6 – 2x ……(ii)
Substituting y = -6 – 2x in (i), we get
x + (-6 – 2x)² + 2x – 2(-6 – 2x) – 3 = 0
⇒ x² + 36 + 24x + 4x² + 2x + 12 + 4x – 3 = 0
⇒ 5x² + 30x + 45 = 0
⇒ x² + 6x + 9 = 0
⇒ (x + 3)² = 0
⇒ x = -3
Since, the roots are equal.
∴ 2x + y + 6 = 0 is a tangent to x² + y² + 2x – 2y – 3 = 0
Substituting x = -3 in (ii), we get
y = -6 – 2(-3) = -6 + 6 = 0
Point of contact = (-3, 0)

Question 20.
If the tangent at (3, -4) to the circle x² + y² = 25 touches the circle x² + y² + 8x – 4y + c = 0, find c.
Solution:
The equation of a tangent to the circle
x² + y² = r² at (x₁, y₁) is xx₁ + yy₁ = r²
Equation of the tangent at (3, -4) is
x(3) + y(-4) = 25
⇒ 3x – 4y – 25 = 0 ……(i)
Given equation of circle is x² + y² + 8x – 4y + c = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = 8, 2f = -4
⇒ g = 4, f = -2
∴ C = (-4, 2) and r = \(\sqrt{4^{2}+(-2)^{2}-c}=\sqrt{20-c}\)
Since line (i) is a tangent to this circle also, the perpendicular distance from C(-4, 2) to line (i) is equal to radius r.

Question 21.
Find the equations of the tangents to the circle x² + y² = 16 with slope -2.
Solution:
Given equation of the circle is x² + y² = 16
Comparing this equation with x² + y² = a², we get
a² = 16
Equations of the tangents to the circle x² + y² = a² with slope m are
\(y=m x \pm \sqrt{a^{2}\left(1+m^{2}\right)}\)
Here, m = -2, a² = 16
the required equations of the tangents are
y = \(-2 x \pm \sqrt{16\left[1+(-2)^{2}\right]}\)
⇒ y = \(-2 x \pm \sqrt{16(5)}\)
⇒ y = -2x ± 4√5
⇒ 2x + y ± 4√5 = 0

Question 22.
Find the equations of the tangents to the circle x² + y² = 4 which are parallel to 3x + 2y + 1 = 0.
Solution:
Given equation of the circle is x² + y² = 4
Comparing this equation with x² + y² = a², we get
a² = 4
Given equation of the line is 3x + 2y + 1 = 0
Slope of this line = \(\frac{-3}{2}\)
Since, the required tangents are parallel to the given line.
Slope of required tangents (m) = \(\frac{-3}{2}\)
Equations of the tangents to the circle x² + y² = a² with slope m are
y = mx ± \(\sqrt{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}\)
the required equations of the tangents are

Question 23.
Find the equations of the tangents to the circle x² + y² = 36 which are perpendicular to the line 5x + y = 2.
Solution:
Given equation of the circle is x² + y² = 36
Comparing this equaiton with x² + y² = a², we get
a² = 36
Given equation of line is 5x + y = 2
Slope of this line = -5
Since, the required tangents are perpendicular to the given line.
Slope of required tangents (m) = \(\frac{1}{5}\)
Equations of the tangents to the circle x² + y² = a² with slope m are
y = mx ± \(\sqrt{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}\)
the required equations of the tangents are

Question 24.
Find the equations of the tangents to the circle x² + y² – 2x + 8y – 23 = 0 having slope 3.
Solution:
Let the equation of the tangent with slope 3 be y = 3x + c.
3x – y + c = 0 ……(i)
Given equation of circle is x² + y² – 2x + 8y – 23 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -2, 2f = 8, c = -23
g = -1, f = 4, c = -23
The centre of the circle is C(1, -4)
and its radius = \(\sqrt{1+16+23}\)
= √40
= 2√10
Since line (i) is a tangent to this circle the perpendicular distance from C(1, -4) to line (i) is equal to radius r.
\(\left|\frac{3(1)+4+c}{\sqrt{9+1}}\right|\) = 2√10
⇒ \(\left|\frac{7+c}{\sqrt{10}}\right|\) = 2√10
⇒ (7 + c) = ± 20
⇒ 7 + c = 20 or 7 + c = -20
⇒ c = 13 or c = – 27
∴ Equations of the tangents are 3x – y + 13 = 0 and 3x – y – 21 = 0

Question 25.
Find the equation of the locus of a point, the tangents from which to the circle x² + y² = 9 are at right angles.
Solution:
Given equation of the circle is x² + y² = 9
Comparing this equation with x² + y² = a², we get
a² = 9
The locus of the point of intersection of perpendicular tangents is the director circle of the given circle.
The equation of the director circle of the circle x² + y² = a² is x² + y² = 2a².
the required equation is
x² + y² = 2(9)
x² + y² = 18

Alternate method:
Given equation of the circle is x² + y² = 9
Comparing this equation with x² + y² = a², we get a² = 9
Let P(x₁, y₁) be a point on the required locus.
Equations of the tangents to the circle x² + y² = a² with slope m are
y = mx ± \(\sqrt{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}\)
∴ Equations of the tangents are
y = mx ± \(\sqrt{9\left(\mathrm{~m}^{2}+1\right)}\)
⇒ y = mx ± 3\(\sqrt{1+m^{2}}\)
Since, these tangents pass through (x₁, y₁).
y₁ = mx₁ ± 3\(\sqrt{1+m^{2}}\)
⇒ y₁ – mx₁ = ± 3\(\sqrt{1+m^{2}}\)
⇒ (y₁ – mx₁)² = 9(1 + m²) ……[Squaring both the sides]
⇒ \(y_{1}^{2}-2 m x_{1} y_{1}+m^{2} x_{1}^{2}=9+9 m^{2}\)
⇒ \(\left(x_{1}^{2}-9\right) \mathrm{m}^{2}-2 \mathrm{~m} x_{1} y_{1}+\left(y_{1}^{2}-9\right)=0\)
This is a quadratic equation which has two roots m₁ and m₂.
m₁m₂ = \(\frac{y_{1}^{2}-9}{x_{1}^{2}-9}\)
Since, the tangents are at right angles.
m₁m₂ = -1
⇒ \(\frac{y_{1}^{2}-9}{x_{1}^{2}-9}=-1\)
⇒ \(y_{1}^{2}-9=9-x_{1}^{2}\)
⇒ \(x_{1}^{2}+y_{1}^{2}=18\)
Equation of the locus of point P is x² + y² = 18.

Question 26.
Tangents to the circle x² + y² = a² with inclinations, θ₁ and θ₂ intersect in P. Find the locus of P such that
(i) tan θ₁ + tan θ₂ = 0
(ii) cot θ₁ + cot θ₂ = 5
(iii) cot θ₁ . cot θ₂ = c
Solution:
Let P(x₁, y₁) be a point on the required locus.
Equations of the tangents to the circle x² + y² = a² with slope m are
y = mx ± \(\sqrt{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}\)
Since, these tangents pass through (x₁, y₁).

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

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:

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)

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

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)

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

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

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)

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 \(\left(\frac{-12}{5}, \frac{6}{5}\right)\)

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

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.

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)

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 = \(\frac{a}{x_{1}}\)

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

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)

∴ 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 = \(\frac{2 a}{e}\) = \(\frac{2(5)}{\frac{4}{5}}\) = \(\frac{25}{2}\)

(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)

(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

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 \(\frac{x^{2}}{64}+\frac{y^{2}}{55}=1\)

(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]

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.

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:

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.

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)

p₂ = length of perpendicular segment from S'(-3, 0) to the tangent (i)

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 \(\frac{x^{2}}{16}-\frac{y^{2}}{20}=1\)

(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

(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)

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.

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.
The 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

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 expressed 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.

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%

(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 is the frequency distribution of weekly wages of 400 workers. Find the range.

Solution:
Here, largest value (L) = 40, smallest value (S) = 10
∴ Range = L – S
= 40 – 10
= 30

Question 4.
Find the range of the following data.

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:

Question 6.
Find variance and S.D. for the following set of numbers.
125, 130, 150, 165, 190, 195, 210, 230, 245, 260
Solution:

Question 7.
Following data gives no. of goals scored by a team in 90 matches. Compute the standard deviation.

Solution:

Question 8.
Compute the variance and S.D. for the following data:

Solution:

Question 9.
Calculate S.D. from the following data.

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:

Question 10.
Given below is the frequency distribution of marks obtained by 100 students. Compute arithmetic mean and S.D.

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:


∴ Var (X) = h² var (u)
= (10)² × 1.4875
= 100 × 1.4875
= 148.75
∴ S.D. = σ_x = √Var(X)
= √148.75
= 12.2

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, σ_x = 5 …..(given)
\(\bar{x}=\frac{1}{\mathrm{n}} \sum x_{\mathrm{i}}\)
∴ \(\sum x_{\mathrm{i}}=\mathrm{n} \bar{x}\) = 20 × 20 = 400
Corrected Σx_i = Σx_i – (incorrect observation) + (correct observation)
= 400 – 30 + 13
= 383

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

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:

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

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.

Solution:

Question 15.
Calculate the coefficient of variation of the following data.
23, 27, 25, 28, 21, 14, 16, 12, 18, 16
Solution:

Question 16.
The following data relates to the distribution of weights of 100 boys and 80 girls in a school.

Which of the two is more variable?
Solution:

Here, C.V. of boys > C.V. of girls
∴ The Series of boys is more variable.

Question 17.
The mean and standard deviations of the two brands of watches are given below:

Calculate the coefficient of variation of the two brands and interpret the results.
Solution:

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:

Solution:

Question 19.
Calculate the coefficient of variation for the data given below:

Solution:
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-6500}{1000}\)
Calculation of variance of u:

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.

Solution:


∴ 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.

Which company shows greater consistency?
Solution:
Let f₁ denote no. of cars of company U and f₂ denote no. of cars of company V.



Here, C.V. (U) < C.V. (V)
∴ Company U shows greater consistency in performance.

Question 22.
The means and S.D. of weights and heights of 100 students of a school are as follows.

Which shows more variability, weights, or heights?
Solution:

Here, C.V. for weight < C.V. for height
∴ Height shows more variability.