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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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 9.
A bag contains 5 red, 4 blue and an unknown number m of green balls. If the probability of getting both the balls green, when two balls are selected at random is \(\frac{1}{7}\), find m.
Solution:
Total number of balls in the bag = 5 + 4 + m = 9 + m
Two balls are selected from (9 + m) balls in \({ }^{9+m} \mathrm{C}_{2}\) ways.
∴ n(S) = \({ }^{9+m} \mathrm{C}_{2}\)
Let event A: The two balls selected are green.
∴ 2 balls can be selected from m balls in \({ }^{\mathrm{m}} \mathrm{C}_{2}\) ways.
∴ n(A) = \({ }^{\mathrm{m}} \mathrm{C}_{2}\)

(9 + m)(8 + m) = 7m(m – 1)
72 + 9m + 8m + m2 = 7m2 – 7m
6m2 – 24m – 72 = 0
m2 – 4m – 12 = 0
(m – 6)(m + 2) = 0
m = 6 or m = -2
Since number of balls cannot be negative, m ≠ -2
∴ m = 6

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

(ii) Let event B: Exactly 2 men are selected.
∴ 2 men are selected from 4 men in \({ }^{4} \mathrm{C}_{2}\) ways, and remaining 2 persons are selected from 7 persons (i.e., 4 women and 3 children) in \({ }^{7} \mathrm{C}_{2}\) ways.

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

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 9 Probability Ex 9.5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Probability Ex 9.5

Question 1.
If odds in favour of X solving a problem are 4 : 3 and odds against Y solving the same problem are 2 : 3. Find the probability of:
(i) X solving the problem
(ii) Y solving the problem
Solution:
(i) Odds in favour of X solving a problem are 4 : 3.
∴ The probability of X solving the problem is
P(X) = \(\frac{4}{4+3}=\frac{4}{7}\)

(ii) Odds against Y solving the problem are 2 : 3.
∴ The probability of Y solving the problem is
P(Y) = 1 – P(Y’)
= 1 – \(\frac{2}{2+3}\)
= 1 – \(\frac{2}{5}\)
= \(\frac{3}{5}\)

Question 2.
The odds against John solving a problem are 4 to 3 and the odds in favour of Rafi solving the same problem are 7 to 5. What is the chance that the problem is solved when both of them try it?
Solution:
The odds against John solving a problem are 4 to 3.
Let event P(A’) = P (John does not solve the problem)
= \(\frac{4}{4+3}\)
= \(\frac{4}{7}\)
So, the probability that John solves the problem
P(A) = 1 – P(A’) = 1 – \(\frac{4}{7}\) = \(\frac{3}{7}\)
Similarly, Let P(B) = P(Rafi solves the problem)
Since the odds in favour of Rafi solving the problem are 7 to 5,
P(B) = \(\frac{7}{7+5}\) = \(\frac{7}{12}\)
Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Since A, B are independent events,
P(A ∩ B) = P(A) . P(B)
∴ Required probability = P(A) + P(B) – P(A) . P(B)

Question 3.
The odds against student X solving a statistics problem are 8 : 6 and odds in favour of student Y solving the same problem are 14 : 16. Find the chance that
(i) the problem will be solved if they try it independently.
(ii) neither of them solves the problem.
Solution:
The odds against X solving a problem are 8 : 6.
Let P(X’) = P(X does not solve the problem) = \(\frac{8}{8+6}\) = \(\frac{8}{14}\)
So, the probability that X solves the problem
P(X) = 1 – P(X’) = 1 – \(\frac{8}{14}\) = \(\frac{6}{14}\)
Similarly, let P(Y) = P(Y solves the problem)
Since odds in favour of Y solving the problem are 14 : 16,
P(Y) = \(\frac{14}{14+16}=\frac{14}{30}\)
So, the probability that Y does not solve the problem
P(Y’) = 1 – P(Y)
= 1 – \(\frac{14}{30}\)
= \(\frac{16}{30}\)
(i) Required probability
P(X ∪ Y) = P(X) + P(Y) – P(X ∩ Y)
Since X and Y are independent events,
P(X ∩ Y) = P(X) . P(Y)
∴ Required probability = P(X) + P(Y) – P(X) . P(Y)
= \(\frac{6}{14}+\frac{14}{30}-\frac{6}{14} \times \frac{14}{30}\)
= \(\frac{73}{105}\)

(ii) Required probability = P(X’ ∩ Y’)
Since X and Y are independent events, X’ and Y’ are also independent events.
∴ Required probability = P(X’) . P(Y’)
= \(\frac{8}{14} \times \frac{16}{30}\)
= \(\frac{32}{105}\)

Question 4.
The odds against a husband who is 60 years old, living till he is 85 are 7 : 5. The odds against his wife who is now 56, living till she is 81 are 5 : 3. Find the probability that
(i) at least one of them will be alive 25 years hence.
(ii) exactly one of them will be alive 25 years hence.
Solution:
The odds against her husband living till he is 85 are 7 : 5.
Let P(H’) = P(husband dies before he is 85) = \(\frac{7}{7+5}=\frac{7}{12}\)
So, the probability that the husband would be alive till age 85
P(H) = 1 – P(H’) = 1 – \(\frac{7}{12}\) = \(\frac{5}{12}\)
Similarly, P(W’) = P(Wife dies before she is 81)
Since the odds against wife will be alive till she is 81 are 5 : 3.
∴ P(W’) = \(\frac{5}{5+3}=\frac{5}{8}\)
So, the probability that the wife would be alive till age 81
P(W) = 1 – P(W’) = 1 – \(\frac{5}{8}\) = \(\frac{3}{8}\)
(i) Required probability
P(H ∪ W) = P(H) + P(W) – P(H ∩ W)
Since H and W are independent events,
P(H ∩ W) = P(H) . P(W)
∴ Required probability = P(H) + P(W) – P(H) . P(W)
= \(\frac{5}{12}+\frac{3}{8}-\frac{5}{12} \times \frac{3}{8}\)
= \(\frac{40+36-15}{96}\)
= \(\frac{61}{96}\)

(ii) Required probability = P(H ∩ W’) + P(H’ ∩ W)
Since H and W are independent events, H’ and W’ are also independent events.
∴ Required probability = P(H) . P(W’) + P(H’) . P(W)
= \(\frac{5}{12} \times \frac{5}{8}+\frac{7}{12} \times \frac{3}{8}\)
= \(\frac{25+21}{96}\)
= \(\frac{46}{96}\)
= \(\frac{23}{48}\)

Question 5.
There are three events A, B, and C, one of which must, and only one can happen. The odds against event A are 7 : 4 and odds against event B are 5 : 3. Find the odds against event C.
Solution:
Since odds against A are 7 : 4,
P(A) = \(\frac{4}{7+4}=\frac{4}{11}\)
Since odds against B are 5 : 3,
P(B) = \(\frac{3}{5+3}=\frac{3}{8}\)
Since only one of the events A, B and C can happen,
P(A) + P(B) + P(C) = 1
\(\frac{4}{11}\) + \(\frac{3}{8}\) + P(C) = 1
∴ P(C) = 1 – (\(\frac{4}{11}\) + \(\frac{3}{8}\))
= 1 – \(\left(\frac{32+33}{88}\right)\)
= \(\frac{23}{88}\)
∴ P(C’) = 1 – P(C)
= 1 – \(\frac{23}{88}\)
= \(\frac{65}{88}\)
∴ Odds against the event C are P(C’) : P(C)
= \(\frac{65}{88}\) : \(\frac{23}{88}\)
= 65 : 23

Question 6.
In a single toss of a fair die, what are the odds against the event that number 3 or 4 turns up?
Solution:
When a fair die is tossed, the sample space is
S = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
Let event A: 3 or 4 turns up.
∴ A = {3, 4}
∴ n(A) = 2
∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{2}{6}=\frac{1}{3}\)
P(A’) = 1 – P(A) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
∴ Odds against the event A are P(A’) : P(A)
= \(\frac{2}{3}: \frac{1}{3}\)
= 2 : 1

Question 7.
The odds in favour of A winning a game of chess against B are 3 : 2. If three games are to be played, what are the odds in favour of A’s winning at least two games out of the three?
Solution:
Let event A: A wins the game and event B: B wins the game.
Since the odds in favour of A winning a game against B are 3 : 2,
the probability of occurrence of event A and B is given by
P(A) = \(\frac{3}{3+2}=\frac{3}{5}\) and P(B) = \(\frac{2}{3+2}=\frac{2}{5}\)
Let event E: A wins at least two games out of three games.
∴ P(E) = P(A) . P(A) . P(B) + P(A) . P(B) . P(A) + P(B) . P(A) . P(A) + P(A) . P(A) . P(A)

∴ Odds in favour of A’s winning at least two games out of three are P(E) : P(E’)
= \(\frac{81}{125}: \frac{44}{125}\)
= 81 : 44

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

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

Question 1.
The means of two samples of sizes 60 and 120 respectively are 35.4 and 30.9 and the standard deviations are 4 and 5. Obtain the standard deviation of the sample of size 180 obtained by combining the two samples.
Solution:

Question 2.
For certain data, the following information is available.

Obtain the combined standard deviation.
Solution:

Question 3.
Calculate the coefficient of variation of marks secured by a student in the exam, where the marks are:
85, 91, 96, 88, 98, 82
Solution:

Question 4.
Find the coefficient of variation of a sample that has a mean equal to 25 and a standard deviation of 5.
Solution:

Question 5.
A group of 65 students of class XI has their average height as 150.4 cm with a coefficient of variance of 2.5%. What is the standard deviation of their heights?
Solution:
Given, n = 65, \(\bar{x}\) = 150.4, C.V. = 2.5%

∴ The standard deviation of students’ height is 3.76 cm.

Question 6.
Two workers on the same job show the following results:

(i) Regarding the time required to complete the job, which worker is more consistent?
(ii) Which worker seems to be faster in completing the job?
Solution:

(i) Since C.V. (P) < C.V.(Q), Worker P is more consistent regarding the time required to complete the job. (ii) Since \(\overline{\mathrm{p}}>\overline{\mathrm{q}}\),
i.e., the expected time for completing the job is less for worker Q.
∴ Worker Q seems to be faster in completing the job.

Question 7.
A company has two departments with 42 and 60 employees respectively. Their average weekly wages are ₹ 750 and ₹ 400. The standard deviations are 8 and 10 respectively.
(i) Which department has a larger bill?
(ii) Which department has larger variability in wages?
Solution:

(i) Since \(\bar{x}_{1}>\bar{x}_{2}\),
i.e., average weekly wages are more for the first department.
∴ The first department has a larger bill.

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

Question 8.
The following table gives the weights of the students of two classes. Calculate the coefficient of variation of the two distributions. Which series is more variable?

Solution:
Let x denote the data of class A and y denote the data of class B.
Calculation of S.D. for class A:



Since C.V. (Y) > C.V.(X),
C.V. (B) > C.V. (A)
∴ Series B is more variable.

Question 9.
Compute the coefficient of variation for team A and team B.

Which team is more consistent?
Solution:
Let f₁ denote no. of goals of team A and f₂ denote no. of goals of team B.




Since C.V. of team A > C.V. of team B,
Team B is more consistent.

Question 10.
Given below is the information about marks obtained in Mathematics and Statistics by 100 students in a class. Which subject shows the highest variability in marks?

Solution:

Since C.V. (S) > C.V. (M),
The subject statistics show higher variability in marks.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 9 Probability Ex 9.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Probability Ex 9.4

Question 1.
There are three bags, each containing 100 marbles. Bag 1 has 75 red and 25 blue marbles. Bag 2 has 60 red and 40 blue marbles and Bag 3 has 45 red and 55 blue marbles. One of the bags is chosen at random and marble is picked from the chosen bag. What is the probability that the chosen marble is red?
Solution:
Let event R: Chosen marble is red.
Let event B_i: ith bag is chosen.
∴ P(B_i) = \(\frac{1}{3}\)
If Bag 1 is chosen, it has 75 red and 25 blue marbles.
∴ Probability that the chosen marble is red under the condition that it is from Bag 1 = P(R/B₁)
= \(\frac{{ }^{75} \mathrm{C}_{1}}{{ }^{100} \mathrm{C}_{1}}\)
= \(\frac{75}{100}\)
= 0.75
Similarly we get,
P(R/B₂) = \(\frac{60}{100}\) = 0.60
P(R/B₃) = \(\frac{45}{100}\) = 0.45
∴ Required probability
P(R) = P(B₁) P(R/B₁) + P(B₂) P(R/B₂) + P(B₃) P(R/B₃)
= \(\frac{1}{3}\)(0.75) + \(\frac{1}{3}\)(0.60) + \(\frac{1}{3}\)(0.45)
= \(\frac{1}{3}\)(1.8)
= 0.60

Question 2.
A box contains 2 blue and 3 pink balls and another box contains 4 blue and 5 pink balls. One ball is drawn at random from one of the two boxes and it is found to be pink. Find the probability that it was drawn from
(i) first box
(ii) second box
Solution:
Let event A₁: The ball is drawn from 1st box and
event A₂: The ball is drawn from the 2nd box.
∴ P(A₁) = \(\frac{1}{2}\), P(A₂) = \(\frac{1}{2}\)
Let event B: The ball drawn is pink.
There are 5 balls in the 1st box, of which 3 are pink.
∴ P(B/A₁) = \(\frac{3}{5}\)
There are 9 balls in the 2nd box, of which 5 are pink.
∴ P(B/A₂) = \(\frac{5}{9}\)
(i) By Bayes’ theorem,
the probability that a pink ball is drawn from the first box, is given by

(ii) By Bayes’ theorem,
the probability that a pink ball is drawn from the second box, is given by

Question 3.
There is a working women’s hostel in a town, where 75% are from neighbouring town. The rest all are from the same town. 48% of women who hail from the same town are graduates and 83% of the women who have come from the neighbouring town are also graduates. Find the probability that a woman selected at random is a graduate from the same town.
Solution:
Let the total number of women be 100.
∴ n(S) = 100
Let event N: Women are from neighbouring town,
event W: Women are from same town and
event G: Women are graduates.
Number of women from neighbouring town,
n(N) = 75
Number of women from same town,
n(W) = 25
∴ P(N) = \(\frac{n(N)}{n(S)}=\frac{75}{100}\) and
P(W) = \(\frac{n(W)}{n(S)}=\frac{25}{100}\)
P(G/N), P(G/W) represent probabilities that woman is graduate given that she is from neighbouring town or same town respectively.
∴ P(G/N) = \(\frac{\mathrm{n}(\mathrm{G} / \mathrm{N})}{\mathrm{n}(\mathrm{S})}=\frac{83}{100}\) and
P(G/W) = \(\frac{\mathrm{n}(\mathrm{G} / \mathrm{W})}{\mathrm{n}(\mathrm{S})}=\frac{48}{100}\)
By Bayes’ theorem, the probability that a women selected at random is a graduate from the same town, is given by

Question 4.
If E₁ and E₂ are equally likely, mutually exclusive and exhaustive events and P(A/E₁) = 0.2, P(A/E₂) = 0.3. Find P(E₁/A).
Solution:
E₁ and E₂ are equally likely, mutually exclusive and exhaustive events.
∴ P(E₁) = P(E₂) = \(\frac{1}{2}\)

Question 5.
Jar I contains 5 white and 7 black balls. Jar II contains 3 white and 12 black balls. A fair coin is flipped; if it is Head, a ball is drawn from Jar I, and if it is Tail, a ball is drawn from Jar II. Suppose that this experiment is done and a white ball was drawn. What is the probability that this ball was in fact taken from Jar II?
Solution:
Let event J₁: Ball drawn from jar I,
event J₂: Ball drawn from jar II.
P(J₁) = P(head) = \(\frac{1}{2}\)
P(J₂) = P(tail) = \(\frac{1}{2}\)
Let event W: Ball drawn is white.
In Jar I, there are total 12 balls, out of which 5 balls are white.
∴ Probability that the ball drawn is white under the condtion that it is drawn from Jar I.
P(W/J₁) = \(\frac{{ }^{5} C_{1}}{{ }^{12} C_{1}}=\frac{5}{12}\)
Similarly, P(W/J₂) = \(\frac{{ }^{3} C_{1}}{{ }^{15} C_{1}}=\frac{3}{15}=\frac{1}{5}\)
Required probability = P(J₂/W)
By Bayes’ theorem,

Question 6.
A diagnostic test has a probability 0.95 of giving a positive result when applied to a person suffering from a certain disease, and a probability 0.10 of giving a (false) positive result when applied to a non-sufferer. It is estimated that 0.5% of the population are sufferers. Suppose that the test is now administered to a person about whom we have no relevant information relating to the disease (apart from the fact that he/she comes from this population). Calculate the probability that:
(i) given a positive result, the person is a sufferer.
(ii) given a negative result, the person is a non-sufferer.
Solution:
Let event T: Test positive
event S: Sufferer
P(S) = \(\frac{0.5}{100}\) = 0.005
∴ P(S’) = 1 – P(S) = 1 – 0.005 = 0.995
Since a probability of getting a positive result when applied to a person suffering from a disease is 0.95 and probability of getting positive result when applied to a non sufferer is 0.10.
∴ P(T/S) = 0.95 and P(T/S’) = 0.10
∴ P(T) = P(S) P(T/S) + P(S’) P(T/S’)
= 0.005 × 0.95 + 0.995 × 0.10
= 0.10425
∴ P(T’) = 1 – P(T) = 1 – 0.10425 = 0.8958
(i) Required probability = P(S/T)
By Bayes’ theorem,

(ii) P(T’/S’) = 1 – 0.1 = 0.9
Required probability = P(S’/T’)
By Bayes’ theorem

Question 7.
A doctor is called to see a sick child. The doctor has prior information that 80% of the sick children in that area have the flu, while the other 20% are sick with measles. Assume that there is no other disease in that area. A well-known symptom of measles is rash. From the past records, it is known that, chances of having rashes given that sick child is suffering from measles is 0.95. However occasionally children with flu also develop rash, whose chance are 0.08. Upon examining the child, the doctor finds a rash. What is the probability that child is suffering from measles?
Solution:
Let the total number of sick children be 100.
∴ n(S) = 100.
Let event A: The child is sick with flu,
event B: The child is sick with measles,
event C: The child is sick with rash.
∴ n(A) = 80 and n(B) = 20
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{80}{100}=\frac{4}{5}\)
P(B) = \(\frac{n(B)}{n(S)}=\frac{20}{100}=\frac{1}{5}\)
Since the chances of having rashes, if the child is suffering from measles is 0.95 and the chances of having rashes if the child has flu is 0.08,
P(C/B) = 0.95 = \(\frac{95}{100}\) and
P(C/A) = 0.08 = \(\frac{8}{100}\)
Required probability = P(B/C)
By Bayes’ theorem,

Question 8.
2% of the population have a certain blood disease of a serious form: 10% have it in a mild form; and 88% don’t have it at all. A new blood test is developed; the probability of testing positive is \(\frac{9}{10}\) if the subject has the
serious form, \(\frac{6}{10}\) if the subject has the mild form, and \(\frac{1}{10}\) if the subject doesn’t have the disease. A subject is tested positive. What is the probability that the subject has serious form of the disease?
Solution:
Let event A₁: Disease in serious form,
event A₂: Disease in mild form,
event A₃: Subject does not have disease,
event B: Subject tests positive.
P(A₁) = 0.02, P(A₂) = 0.1, P(A₃) = 0.88
The probability of testing positive is \(\frac{9}{10}\) if the subject has the serious form, \(\frac{6}{10}\) if the subject has the mild form, and \(\frac{1}{10}\) if the subject doesn’t have the disease.
∴ P(B/A₁) = 0.9, P(B/A₂) = 0.6, P(B/A₃) = 0.1
P(B) = P(A₁) P(B/A₁) + P(A₂) P(B/A₂) + P(A₃) P(B/A₃)
= 0.02 × 0.9 + 0.1 × 0.6 + 0.88 × 0.1
= 0.166
Required probability = P(A₁/B)
By Baye’s theorem

Question 9.
A box contains three coins: two fair coins and one fake two-headed coin. A coin is picked randomly from the box and tossed.
(i) What is the probability that it lands head up?
(ii) If happens to be head, what is the probability that it is the two-headed coin?
Solution:
Let event A: Fair coin is tossed,
event B: Fake coin is tossed
and event H: Head occur.
Clearly, a fair coin has one head.
∴ Probability that head occur under the condition that the fair coin is tossed = P(H/A) = \(\frac{1}{2}\)
Fake coin has two heads.
∴ Probability that head occur under the condition that the fake coin is tossed = P(H/B) = 1
n(A) = 2, n(B) = 1, n(S) = 3
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{2}{3}\)
P(B) = \(\frac{n(B)}{n(S)}=\frac{1}{3}\)
(i) Required probability
P(H) = P(A) P(H/A) + P(B) P(H/B)
= \(\frac{2}{3} \times \frac{1}{2}+\frac{1}{3} \times 1\)
= \(\frac{1}{3}+\frac{1}{3}\)
= \(\frac{2}{3}\)

(ii) Required probability = P(B/H)
By Baye’s theorem

Question 10.
There are three social media groups on a mobile: Group I, Group II and Group III. The probabilities that Group I, Group II and Group III sending the messages on sports are \(\frac{2}{5}\), \(\frac{1}{2}\) and \(\frac{2}{3}\) respectively. The probability of opening the messages by Group I, Group II and Group III are \(\frac{1}{2}\), \(\frac{1}{4}\) and \(\frac{1}{4}\) respectively. Randomly one of the messages is opened and found a message on sports. What is the probability that the message was from Group III.
Solution:
Let event A: Message sent on sports by group I,
event B: Message sent on sports by group II,
event C: Message sent on sports by group III,
event E: Message is opened.
Given that the probabilities that Group I, Group II and Group III sending the messages on sports are \(\frac{2}{5}\), \(\frac{1}{2}\) and \(\frac{2}{3}\) respectively and the probability of opening the messages by Group I, Group II and Group III are \(\frac{1}{2}\), \(\frac{1}{4}\) and \(\frac{1}{4}\) respectively.
∴ P(A) = \(\frac{2}{5}\)
P(B) = \(\frac{1}{2}\)
P(C) = \(\frac{2}{3}\)
P(E/A) = \(\frac{1}{2}\)
P(E/B) = \(\frac{1}{4}\)
P(E/C) = \(\frac{1}{4}\)
Required probability = P(C/E)
By Baye’s theorem

Question 11.
Mr. X goes to office by Auto, Car and train. The probabilities of him travelling by these modes are \(\frac{2}{7}\), \(\frac{3}{7}\), \(\frac{2}{7}\) respectively. The chances of him being late to the office are \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{4}\) respectively by Auto, Car and train. On one particular day he was late to the office. Find the probability that he travelled by car.
Solution:
Let A, C and T be the events that Mr. X goes to office by Auto, Car and Train respectively.
Let L be event that he is late.

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

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

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

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

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

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

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

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

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

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

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

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

Question 6.
For the parabola 3y² = 16x, find the parameter of the point:
(i) (3, -4)
(ii) (27, -12)
Solution:
Given the equation of the parabola is 3y² = 16x.
⇒ y² = \(\frac{16}{3}\)x
Comparing this equation with y² = 4ax, we get
⇒ 4a = \(\frac{16}{3}\)
⇒ a = \(\frac{4}{3}\)
If t is the parameter of the point P on the parabola, then
P(t) = (at², 2at)
i.e., x = at² and y = 2at ………(i)
(i) Given point is (3, -4)
Substituting x = 3, y = -4 and a = \(\frac{4}{3}\) in (i), we get
3 = \(\frac{4}{3}\) t² and -4 = 2(\(\frac{4}{3}\)) t

∴ The parameter of the given point is \(\frac{-3}{2}\)

(ii) Given point is (27, -12)
Substituting x = 27, y = -12 and a = \(\frac{4}{3}\) in (i), we get

∴ The parameter of the given point is \(\frac{-9}{2}\)

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

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

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

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

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

Question 11.
Find the area of the triangle formed by the line joining the vertex of the parabola x² = 12y to the endpoints of the latus rectum.
Solution:

Given the equation of the parabola is x² = 12y.
Comparing this equation with x² = 4by, we get
⇒ 4b = 12
⇒ b = 3
The co-ordinates of focus are S(0, b), i.e., S(0, 3)
End points of the latus-rectum are L(2b, b) and L'(-2b, b),
i.e., L(6, 3) and L'(-6, 3)
Also l(LL’) = length of latus-rectum = 4b = 12
l(OS) = b = 3
Area of ∆OLL’ = \(\frac{1}{2}\) × l(LL’) × l(OS)
= \(\frac{1}{2}\) × 12 × 3
Area of ∆OLL’ = 18 sq. units

Question 12.
If a parabolic reflector is 20 cm in diameter and 5 cm deep, find its focus.
Solution:

Let LOM be the parabolic reflector such that LM is the diameter and ON is its depth.
It is given that ON = 5 cm and LM = 20 cm.
LN = 10 cm
Taking O as the origin, ON along X-axis and a line through O ⊥ ON as Y-axis.
Let the equation of the reflector be y² = 4ax ……(i)
The point L has the co-ordinates (5, 10) and lies on parabola given by (i).
Substituting x = 5 and y = 10 in (i), we get
⇒ 10² = 4a(5)
⇒ 100 = 20a
⇒ a = 5
Focus is at (a, 0), i.e., (5, 0)

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

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

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

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

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

Question 16.
Two tangents to the parabola y² = 8x meet the tangents at the vertex in the points P and Q. If PQ = 4, prove that the equation of the locus of the point of intersection of two tangents is y² = 8(x + 2).
Solution:
Given equation of the parabola is y² = 8x
Comparing this equation with y² = 4ax, we get
⇒ 4a = 8
⇒ a = 2
Equation of tangent to given parabola at A(t₁) is y
t₁ = x + 2\(\mathrm{t}_{1}^{2}\) …….(i)
Equation of tangent to given parabola at B(t₂) is y
t₂ = x + 2\(\mathrm{t}_{2}^{2}\) …..(ii)

A tangent at the vertex is Y-axis whose equation is x = 0.
x-coordinate of points P and Q is 0.
Let P be(0, k₁) and Q be (0, k₂).
Then, from (i) and (ii), we get

∴ Equation of locus of R is y² = 8(x + 2).

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

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

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

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

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

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

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

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

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

(iv) Given equation of the hyperbola is 21x² – 4y² = 84.
\(\frac{x^{2}}{4}-\frac{y^{2}}{21}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = 4 and b² = 21
⇒ a = 2 and b = √21
Length of transverse axis = 2a = 2(2) = 4
Length of conjugate axis = 2b = 2√21
We know that

(v) Given equation of the hyperbola is 3x² – y² = 4.
\(\frac{x^{2}}{\left(\frac{4}{3}\right)}-\frac{y^{2}}{4}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = \(\frac{4}{3}\) and b² = 4

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

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

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

(ix) Given equation of the hyperbola is \(\frac{x^{2}}{100}-\frac{y^{2}}{25}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = 100 and b² = 25
⇒ a = 10 and b = 5
Length of transverse axis = 2a = 2(10) = 20
Length of conjugate axis = 2b = 2(5) = 10
We know that

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

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

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

Question 4.
If e and e’ are the eccentricities of a hyperbola and its conjugate hyperbola respectively, prove that \(\frac{1}{e^{2}}+\frac{1}{\left(e^{\prime}\right)^{2}}=1\).
Solution:
Let e be the eccentricity of a hyperbola

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

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

(iii) Let the required equation of hyperbola be \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\)
Given, eccentricity (e) = \(\frac{3}{2}\)
Distance between directrices = \(\frac{2a}{e}\)
Given, distance between directrices = \(\frac{8}{3}\)

(iv) Let the required equation of hyperbola be
\(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\) ……(i)
Length of conjugate axis = 2b
Given, length of conjugate axis = 12
⇒ 2b = 12
⇒ b = 6 …..(ii)
⇒ b² = 36
The hyperbola passes through (1, -2)
Substituting x = 1 and y = -2 in (i), we get

(v) Let the required equation of hyperbola be
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) ……(i)
The hyperbola passes through the points (6, 9) and (3, 0).

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

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

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

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

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

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

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

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

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

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

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

Maharashtra State Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9

(I) Select the correct answer from the given four alternatives.

Question 1.
There are 5 girls and 2 boys, then the probability that no two boys are sitting together for a photograph is
(A) \(\frac{1}{21}\)
(B) \(\frac{4}{7}\)
(C) \(\frac{2}{7}\)
(D) \(\frac{5}{7}\)
Answer:
(D) \(\frac{5}{7}\)
Hint:
There are 5 girls and 2 boys.
They can be arranged among themselves in \({ }^{7} \mathrm{P}_{7}\) = 7! ways.
∴ Girls can be arranged among themselves in \({ }^{5} \mathrm{P}_{5}\) = 5! ways.
No two boys should sit together.
Let girls be denoted by the letter G.
– G – G – G – G – G –
There are 6 places, marked by ‘-’ where boys can sit.
∴ Boys can be arranged in
\({ }^{6} \mathrm{P}_{2}=\frac{6 !}{(6-2) !}\)
= \(\frac{6 \times 5 \times 4 !}{4 !}\)
= 30 ways.
∴ Required probability = \(\frac{5 ! \times 30}{7 !}=\frac{5 ! \times 30}{7 \times 6 \times 5 !}=\frac{5}{7}\)

Question 2.
In a jar, there are 5 black marbles and 3 green marbles. Two marbles are picked randomly one after the other without replacement. What is the possibility that both the marbles are black?
(A) \(\frac{5}{14}\)
(B) \(\frac{5}{8}\)
(C) \(\frac{5}{7}\)
(D) \(\frac{5}{16}\)
Answer:
(A) \(\frac{5}{14}\)

Question 3.
Two dice are thrown simultaneously. Then the probability of getting two numbers whose product is even is
(A) \(\frac{3}{4}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{5}{7}\)
(D) \(\frac{1}{2}\)
Answer:
(A) \(\frac{3}{4}\)
Hint:
Two dice are thrown.
∴ n(S) = 36
Getting two numbers whose product is even, i.e., one of the two numbers must be even.
Let event A: Getting even number on first dice,
event B: Getting even number on second dice.
n(A) = 18, n(B) = 18, n(A ∩ B) = 9
Required probability = P(A ∩ B)
= \(\frac{n(A)+n(B)-n(A \cap B)}{n(S)}\)
= \(\frac{18+18-9}{36}\)
= \(\frac{3}{4}\)

Question 4.
In a set of 30 shirts, 17 are white and the rest are black. 4 white and 5 black shirts are tagged as ‘PARTY WEAR’. If a shirt is chosen at random from this set, the possibility of choosing a black shirt or a ‘PARTY WEAR’ shirt is
(A) \(\frac{11}{15}\)
(B) \(\frac{13}{30}\)
(C) \(\frac{9}{13}\)
(D) \(\frac{17}{30}\)
Answer:
(D) \(\frac{17}{30}\)
Hint:
17 white + 13 black = 30 shirts
4 white and 5 black are ‘PARTY WEAR’
A: Choosing a black shirt
∴ P(A) = \(\frac{{ }^{13} C_{1}}{{ }^{30} C_{1}}=\frac{13}{30}\)
B: Choosing a ‘PARTY WEAR’ shirt.
∴ P(B) = \(\frac{{ }^{9} \mathrm{C}_{1}}{{ }^{30} \mathrm{C}_{1}}=\frac{9}{30}\)
There are 5 black ‘PARTY WEAR’ shirts.
∴ P(A ∩ B) = \(\frac{{ }^{5} \mathrm{C}_{1}}{{ }^{30} \mathrm{C}_{1}}=\frac{5}{30}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{13}{30}\) + \(\frac{9}{30}\) – \(\frac{5}{30}\)
= \(\frac{17}{30}\)

Question 5.
There are 2 shelves. One shelf has 5 Physics and 3 Biology books and the other has 4 Physics and 2 Biology books. The probability of drawing a Physics book is
(A) \(\frac{9}{14}\)
(B) \(\frac{31}{48}\)
(C) \(\frac{9}{38}\)
(D) \(\frac{1}{2}\)
Answer:
(B) \(\frac{31}{48}\)
Hint:
Let event S₁: First shelve is selected,
event S₂: Second shelve is selected,
event P: Drawing a physics book.
∴ P(S₁) = \(\frac{1}{2}\) and P(S₂) = \(\frac{1}{2}\)
First shelve has 5 physics and 3 biology books, i.e., total 8 books.
∴ P(P/S₁) = \(\frac{{ }^{5} C_{1}}{{ }^{8} C_{1}}=\frac{5}{8}\)
Similarly, P(P/S₂) = \(\frac{{ }^{4} C_{1}}{{ }^{6} C_{1}}=\frac{4}{6}=\frac{2}{3}\)
∴ P(P) = P(S₁) . P(P/S₁) + P(S₂) . P(P/S₂)
= \(\frac{1}{2} \times \frac{5}{8}+\frac{1}{2} \times \frac{2}{3}\)
= \(\frac{31}{48}\)

Question 6.
Two friends A and B apply for a job in the same company. The chances of A getting selected is 2/5 and that of B is 4/7. The probability that both of them get selected is
(A) \(\frac{34}{35}\)
(B) \(\frac{1}{35}\)
(C) \(\frac{8}{35}\)
(D) \(\frac{27}{35}\)
Answer:
(C) \(\frac{8}{35}\)

Question 7.
The probability that a student knows the correct answer to a multiple-choice question is \(\frac{2}{3}\). If the student does not know the answer, then the student guesses the answer. The probability of the guessed answer being correct is \(\frac{1}{4}\). Given that the student has answered the question correctly, the probability that the student knows the correct answer is
(A) \(\frac{5}{6}\)
(B) \(\frac{6}{7}\)
(C) \(\frac{7}{8}\)
(D) \(\frac{8}{9}\)
Answer:
(D) \(\frac{8}{9}\)
Hint:
Let event A: Student knows the correct answer,
event A’: Student guesses the answer,
event B: Answer is correct.
∴ P(A) = \(\frac{2}{3}\), P(A’) = \(\frac{1}{3}\), P(B/A’) = \(\frac{1}{4}\)
Clearly, P(B/A) = 1
Required probability = P(A/B)
= \(\frac{\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B} / \mathrm{A})}{\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B} / \mathrm{A})+\mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}\left(\mathrm{B} / \mathrm{A}^{\prime}\right)}\)
= \(\frac{\frac{2}{3} \times 1}{\frac{2}{3} \times 1+\frac{1}{3} \times \frac{1}{4}}\)
= \(\frac{8}{9}\)

Question 8.
The bag I contain 3 red and 4 black balls while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. The probability that it was drawn from Bag II is
(A) \(\frac{33}{68}\)
(B) \(\frac{35}{69}\)
(C) \(\frac{34}{67}\)
(D) \(\frac{35}{68}\)
Answer:
(D) \(\frac{35}{68}\)

Question 9.
A fair die is tossed twice. What are the odds in favour of getting 4, 5, or 6 on the first toss and 1, 2, 3, or 4 on the second toss?
(A) 1 : 3
(B) 3 : 1
(C) 1 : 2
(D) 2 : 1
Answer:
(C) 1 : 2
Hint:
A fair dice is tossed twice.
∴ n(S) = 36
A: Getting 4, 5, or 6 on the first toss and Getting 1, 2, 3, or 4 on the second toss.
∴ A = {(4, 1), (4, 2), (4, 3), (4, 4), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4)}
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{12}{36}=\frac{1}{3}\)
∴ Required answer = P(A) : P(A’) = 1 : 2

Question 10.
The odds against an event are 5 : 3 and the odds in favour of another independent event are 7 : 5. The probability that at least one of the two events will occur is
(A) \(\frac{52}{96}\)
(B) \(\frac{71}{96}\)
(C) \(\frac{69}{96}\)
(D) \(\frac{13}{96}\)
Answer:
(B) \(\frac{71}{96}\)

(II) Solve the following.

Question 1.
The letters of the word ‘EQUATION’ are arranged in a row. Find the probability that
(i) all the vowels are together
(ii) arrangement starts with a vowel and ends with a consonant.
Solution:
The letters of the word EQUATION can be arranged in 8! ways.
∴ n(S) = 8!
There are 5 vowels and 3 consonants.
(i) A: all vowels are together we need to arrange (E, U, A, I, O), Q, T, N
Let us consider all vowels as one unit.
So, there are 4 units, which can be arranged in 4! ways.
Also, 5 vowels can be arranged among themselves in 5! ways.
∴ n(A) = 4! × 5!
Required probability = P(A)
= \(\frac{n(A)}{n(S)}\)
= \(\frac{4 ! \times 5 !}{8 !}\)
= \(\frac{1}{14}\)

(ii) B: arrangement start with a vowel and ends with a consonant.
First and last places can be filled in 5 and 3 ways respectively.
Remaining 6 letters are arranged in 6! Ways.
∴ n(B) = 5 × 3 × 6!
Required probability = P(B)
= \(\frac{n(B)}{n(S)}\)
= \(\frac{5 \times 3 \times 6 !}{8 !}\)
= \(\frac{15}{56}\)

Question 2.
There are 6 positive and 8 negative numbers. Four numbers are chosen at random, without replacement, and multiplied. Find the probability that the product is a positive number.
Solution:
Let event A: Four positive numbers are chosen,
event B: Four negative numbers are chosen,
event C: Two positive and two negative numbers are chosen.
Since four numbers are chosen without replacement,
n(A) = 6 × 5 × 4 × 3 = 360
n(B) = 8 × 7 × 6 × 5 = 1680
In event C, four numbers are to be chosen without replacement such that two numbers are positive and two numbers ate negative. This can be done in following ways:
+ + – – OR + – + – OR + – – + OR – + – + OR – – + + OR – + + –
∴ n(C) = 6 × 5 × 8 × 7 + 6 × 8 × 5 × 7 + 6 × 8 × 7 × 5 + 8 × 6 × 7 × 5 + 6 × 5 × 8 × 7 + 8 × 6 × 5 × 7
= 6 × (8 × 7 × 6 × 5)
=10080
Here, total number of numbers = 14
∴ n(S) = 14 × 13 × 12 × 11 = 24024
Since A, B, C are mutually exclusive events,
Required probability = P(A) + P(B) + P(C)

Question 3.
Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly, and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?
Solution:
S = {1, 2,…., 10}
∴ n(S) = 10
A: Number is more than 3.
A = {4, 5, 6, 7, 8, 9, 10}
∴ n(A) = 7
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{7}{10}\)
B: Number is even.
B = {2, 4, 6, 8, 10}
∴ A ∩ B = {4, 6, 8, 10}
∴ n(A ∩ B) = 4
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{4}{10}\)
Required probability = P(B/A)
= \(\frac{P(A \cap B)}{P(A)}\)
= \(\frac{\left(\frac{4}{10}\right)}{\left(\frac{7}{10}\right)}\)
= \(\frac{4}{7}\)

Question 4.
If A, B and C are independent events, P(A ∩ B) = \(\frac{1}{2}\), P(B ∩ C) = \(\frac{1}{3}\), P(C ∩ A) = \(\frac{1}{6}\), then find P(A), P(B) and P(C).
Solution:
Since A and B are independent events,
P(A ∩ B) = P(A) . P(B)
∴ P(A) . P(B) = \(\frac{1}{2}\) ……(i)
B and C are independent events.
∴ P(B ∩ C) = P(B) . P(C)
∴ P(B) . P(C) = \(\frac{1}{3}\) ……(ii)
A and C are independent events.
∴ P(A ∩ C) = P(A) . P(C)
∴ P(A) . P(C) = \(\frac{1}{6}\) ……(iii)
Dividing (i) by (ii), we get
\(\frac{\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})}{\mathrm{P}(\mathrm{B}) \mathrm{P}(\mathrm{C})}=\frac{\frac{1}{2}}{\frac{1}{3}}\)
P(A) = \(\frac{3}{2}\) P(C) ……(iv)
Substituting equation (iv) in (iii), we get
\(\frac{3}{2}\) P(C) . P(C) = \(\frac{1}{6}\)
[P(C)]² = \(\frac{1}{9}\)
∴ P(C) = \(\frac{1}{3}\)
Substituting P(C) = \(\frac{1}{3}\) in equation (ii), we get P(B) = 1
Substituting P(B) = 1 in equation (i), we get P(A) = \(\frac{1}{2}\)

Question 5.
If the letters of the word ‘REGULATIONS’ be arranged at random, what is the probability that there will be exactly 4 letters between R and E?
Solution:
There are 11 letters in the word ‘REGULATIONS’ which can be arranged among themselves in 11! ways.
∴ n(S) = 11!
Let event A: There will be exactly 4 letters between R and E.
R, E can occur at (1, 6), (2, 7), ….,(6, 11) positions. So, there are 6 possibilities.
Also, R and E can interchange their positions.
So, R, E can be arranged in 2 × 6 = 12 ways.
Remaining 9 letters can be arranged in 9! ways.
∴ n(A) = 12 × 9!
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{12 \times 9 !}{11 !}=\frac{12 \times 9 !}{11 \times 10 \times 9 !}=\frac{6}{55}\)

Question 6.
In how many ways can the letters of the word ARRANGEMENTS be arranged?
(i) Find the chance that an arrangement chosen at random begins with the letters EE.
(ii) Find the probability that the consonants are together.
Solution:
The word ‘ARRANGEMENTS’ has 12 letters in which 2A, 2E, 2N, 2R, G, M, T, S are there.
n(S) = Total number of arrangements = \(\frac{12 !}{2 ! 2 ! 2 ! 2 !}=\frac{12 !}{(2 !)^{4}}\)
(i) A: Arrangement chosen at random begins with the letters EE.
If the first and second places are filled with EE, there are 10 letters left in which 2A, 2N, 2R, G, M, T, S are there.
∴ n(A) = \(\frac{10 !}{2 ! 2 ! 2 !}=\frac{10 !}{(2 !)^{3}}\)

(ii) B: Consonants (G, M, T, S, 2N, 2R) are together.
2A, 2E, and the group containing consonants form total 5 units. Which can be arranged in \(\frac{5 !}{2 ! 2 !}\) ways.
Also, 8 consonants can be arranged among themselves in \(\frac{8 !}{2 ! 2 !}\) ways.

Question 7.
A letter is taken at random from the letters of the word ‘ASSISTANT’ and another letter is taken at random from the letters of the word ‘STATISTICS’. Find the probability that the selected letters are the same.
Solution:
Word ASSISTANT has 2A, I, N, 3S, 2T, and word STATISTICS has A, C, 2I, 3S, 3T.
C and N are uncommon letters.
In the words ASSISTANT, there are 9 letters out of which 2 letters are ‘A’, and in the word STATISTICS, there are 10 letters, out of which 1 letter is A.
∴ Probability of choosing A from both the letters = \(\frac{{ }^{2} C_{1}}{{ }^{9} C_{1}} \times \frac{{ }^{1} C_{1}}{{ }^{10} C_{1}}=\frac{2}{9} \times \frac{1}{10}=\frac{1}{45}\)
Similarly,
Probability of choosing I from both the letters = \(\frac{{ }^{1} \mathrm{C}_{1}}{{ }^{9} \mathrm{C}_{1}} \times \frac{{ }^{2} \mathrm{C}_{1}}{{ }^{10} \mathrm{C}_{1}}=\frac{1}{9} \times \frac{2}{10}=\frac{1}{45}\)
Probability of choosing S from both the letters = \(\frac{{ }^{3} \mathrm{C}_{1}}{{ }^{9} \mathrm{C}_{1}} \times \frac{{ }^{3} \mathrm{C}_{1}}{{ }^{10} \mathrm{C}_{1}}=\frac{3}{9} \times \frac{3}{10}=\frac{1}{10}\)
Probability of choosing T from both the letters = \(\frac{{ }^{2} C_{1}}{{ }^{9} C_{1}} \times \frac{{ }^{3} C_{1}}{{ }^{10} C_{1}}=\frac{2}{9} \times \frac{3}{10}=\frac{1}{15}\)
Required probability = \(\frac{1}{45}+\frac{1}{45}+\frac{1}{10}+\frac{1}{15}\) = \(\frac{19}{90}\)

Question 8.
A die is loaded in such a way that the probability of the face with j dots turning up is proportional to j for j = 1, 2,….., 6. What is the probability, in one roil of the die, that an odd number of dots will turn up?
Solution:
According to the given condition, the probability of the face with 1, 2, 3, 4, 5, 6 dots turning up is proportional to 1, 2, 3, 4, 5, 6.
Let k be the common ration of proportionality.
∴ The probabilities of the faces with 1, 2, 3, 4, 5, 6 dots turning up are 1k , 2k, 3k, 4k, 5k, 6k respectively.
Since sum of the probabilities = 1,
k(1 + 2+ ….. + 6) = 1
k(\(\frac{6 \times 7}{2}\)) = 1
k = \(\frac{1}{21}\)
Required probability = P(1) + P(3) + P(5)
= \(\frac{1}{21}+\frac{3}{21}+\frac{5}{21}\)
= \(\frac{9}{21}\)
= \(\frac{3}{7}\)

Question 9.
An urn contains 5 red balls and 2 green balls. A ball is drawn. If it’s green, a red ball is added to the urn, and if it’s red, a green ball is added to the urn. (The original ball is not returned to the urn). Then a second ball is drawn. What is the probability that the second ball is red?
Solution:
A: Event of drawing a red ball and placing a green ball in the urn
B: Event of drawing a green ball and placing a red ball
C: Event of drawing a red ball in the second draw
P(A) = \(\frac{5}{7}\)
P(B) = \(\frac{2}{7}\)
P(C/A) = \(\frac{4}{7}\)
P(C/B) = \(\frac{6}{7}\)
Required probability
P(C) = P(A) P(C/A) + P(B) P(C/B)
= \(\frac{5}{7} \times \frac{4}{7}+\frac{2}{7} \times \frac{6}{7}\)
= \(\frac{32}{49}\)

Question 10.
The odds against A solving a certain problem are 4 to 3 and the odds in favour of B solving the same problem are 7 to 5, find the probability that the problem will be solved.
Solution:
The odds against A solving the problems are 4 : 3.
Let P(A’) = P(A does not solve the problem) = \(\frac{4}{4+3}=\frac{4}{7}\)
So, the probability that A solves the problem = P(A) = 1 – P(A’)
= 1 – \(\frac{4}{7}\)
= \(\frac{3}{7}\)
Similarly, let P(B) = P(B solves the problem)
Since odds in favour of B solving the problem are 7 : 5.
∴ P(B) = \(\frac{7}{7+5}=\frac{7}{12}\)
Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Since A and B are independent events.
∴ P(A ∩ B) = P(A) . P(B)
∴ Required probability = P(A) + P(B) – P(A) . P(B)
= \(\frac{3}{7}+\frac{7}{12}-\frac{3}{7} \times \frac{7}{12}\)
= \(\frac{16}{21}\)

Question 11.
If P(A) = P(A/B) = \(\frac{1}{5}\), P(B/A) = \(\frac{1}{3}\), then find
(i) P(A’/B)
(ii) P(B’/A’)
Solution:
Since P(A) = P(A/B) = \(\frac{1}{5}\)
P(A) = \(\frac{1}{5}\)
and \(\frac{P(A \cap B)}{P(B)}=\frac{1}{5}\)
∴ P(A) = \(\frac{1}{5}\) ……(i)
P(B) = 5 P(A ∩ B) ……..(ii)
Since P(B/A) = \(\frac{1}{3}\)
\(\frac{P(A \cap B)}{P(A)}=\frac{1}{3}\)
∴ P(A) = 3 P(A ∩ B) ………(iii)

Question 12.
Let A and B be independent events with P(A) = \(\frac{1}{4}\) and P(A ∪ B) = 2P(B) – P(A). Find
(i) P(B)
(ii) P(A/B)
(iii) P(B’/A)
Solution:
A and B are independent events. .
∴ P(A ∩ B) = P(A) × P(B)
(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
∴ P(A ∪ B) = P(A) + P(B) – P(A) × P(B)
∴ 2P(B) – P(A) = P(A) + P(B) – P(A) × P(B) ……[∵ P(A ∪ B) = 2P(B) – P(A)]
∴ 2P(B) – \(\frac{1}{4}\) = \(\frac{1}{4}\) + P(B) – \(\frac{1}{4}\) × P(B)
∴ 2P(B) – P(B) + \(\frac{1}{4}\) P(B) = \(\frac{1}{4}\) + \(\frac{1}{4}\)

Question 13.
Find the probability that a year selected will have 53 Wednesdays.
Solution:
A leap year comes after 3 years.
∴ The probability of a year being a leap year = \(\frac{1}{4}\)
∴ Probability of a year being a non-leap year = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
In a non-leap year, there are 52 weeks and one extra day, whereas a leap year has 52 weeks and 2 extra days.
∴ 53rd Wednesday’s chance in a non-leap year = \(\frac{1}{7}\)
Two extra days of a leap year can be
(Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), (Sun, Mon)
∴ There are 2 possibilities of 53rd Wednesday in a leap year.
∴ 53rd Wednesday’s chance in a leap year = \(\frac{2}{7}\)
Required probability = P(a non-leap year and Wednesday) + P(a leap year and Wednesday)
= \(\frac{3}{4} \times \frac{1}{7}+\frac{1}{4} \times \frac{2}{7}\)
= \(\frac{5}{28}\)

Question 14.
The chances of P, Q and R, getting selected as principal of a college are \(\frac{2}{5}\), \(\frac{2}{5}\), \(\frac{1}{5}\) respectively. Their chances of introducing IT in the college are \(\frac{1}{2}\), \(\frac{1}{3}\), \(\frac{1}{4}\) respectively. Find the probability that
(a) IT is introduced in the college after one of them is selected as a principal.
(b) IT is introduced by Q.
Solution:
Let event P: P become principal,
event Q: Q become principal,
event R: R become principal,
event E: Subject IT is introduced.
Given, P(P) = \(\frac{2}{5}\)
P(Q) = \(\frac{2}{5}\)
P(R) = \(\frac{1}{5}\)
P(E/P) = \(\frac{1}{2}\)
P(E/Q) = \(\frac{1}{3}\)
P(E/R) = \(\frac{1}{4}\)
(a) Required probability
P(E) = P(P) P(E/P) + P(Q) P(E/Q) + P(R) P(E/R)
= \(\frac{2}{5} \times \frac{1}{2}+\frac{2}{5} \times \frac{1}{3}+\frac{1}{5} \times \frac{1}{4}\)
= \(\frac{1}{5}+\frac{2}{15}+\frac{1}{20}\)
= \(\frac{12+8+3}{60}\)
= \(\frac{23}{60}\)

(b) Required probability = P(Q/E)
By Bayes’ theorem,
P(Q/E) = \(\frac{P(Q) P(E / Q)}{P(E)}\)
= \(\frac{\frac{2}{5} \times \frac{1}{3}}{\frac{23}{60}}\)
= \(\frac{8}{23}\)

Question 15.
Suppose that five good fuses and two defective ones have been mixed up. To find the defective fuses, we test them one-by-one, at random and without replacement What is the probability that we are lucky and find both of the defective fuses in the first two tests?
Solution:
Number of fuses = 5 + 2 = 7
Testing two fuses one-by-one at random, without replacement from 7 can be done in \({ }^{7} \mathrm{C}_{1} \times{ }^{6} \mathrm{C}_{1}\) ways.
∴ n(S) = \({ }^{7} \mathrm{C}_{1} \times{ }^{6} \mathrm{C}_{1}\) = 7 × 6 = 42
Let event A: Getting defective fuses in the first two tests without replacement.
There are two defective fuses.
∴ n(A) = \({ }^{2} \mathrm{C}_{1} \times{ }^{1} \mathrm{C}_{1}\) = 2 × 1 = 2
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{2}{42}=\frac{1}{21}\)

Question 16.
For three events A, B and C, we know that A and C are independent, B and C are independent, A and B are disjoint, P(A ∪ C) = \(\frac{2}{3}\), P(B ∪ C) = \(\frac{3}{4}\), P(A ∪ B ∪ C) = \(\frac{11}{12}\). Find P(A), P(B) and P(C).
Solution:
Let P(A) = x, P(B) = y, P(C) = z
Since A, B are disjoint,
A ∩ B = Φ and A ∩ B ∩ C = Φ
∴ P(A ∩ B) = 0, P(A ∩ B ∩ C) = 0 ……(i)
Since A and C are independent,
P(A ∩ C) = P(A) P(C) = xz
Since B and C are independent,
P(B ∩ C) = P(B) P(C) = yz
P(A ∪ C) = P(A) + P(C) – P(A ∩ C)
∴ \(\frac{2}{3}\) = x + z – xz ……..(ii)
P(B ∪ C) = P(B) + P(C) – P(B ∩ C)
∴ \(\frac{3}{4}\) = y + z – yz ………(iii)
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C)
\(\frac{11}{12}\) = x + y + z – 0 – yz – zx + 0 …… [From(i)]
= (x + z – xz) + (y + z – yz) – z
= \(\frac{2}{3}+\frac{3}{4}\) – z ……. [From (ii) and (iii)]

Question 17.
The ratio of boys to girls in a college is 3 : 2 and 3 girls out of 500 and 2 boys out of 50 of that college are good singers. A good singer is chosen. What is the probability that the chosen singer is a girl?
Solution:
Let event S: The student is a good singer,
event B: The student is a boy,
event G: The student is a girl.
Since the ratio of boys to girls is 3 : 2 and 3 girls out of 500 and 2 boys out of 50 are good singers.

Question 18.
A and B throw a die alternatively till one of them gets a 3 and wins the game. Find the respective probabilities of winning. (Assuming A begins the game).
Solution:
Since P(getting 3) = \(\frac{1}{6}\),
P(not getting 3) = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
In 1st throw if A gets 3, A wins
∴ P(A win) = \(\frac{1}{6}\)
In 2nd throw by B (i.e., A does not get 3),
∴ P(B wins) = \(\frac{5}{6} \times \frac{1}{6}\)
In 3rd throw by A, P(A wins) = \(\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\)
(3rd throw by A shows that B has lost in 2nd throw) and so on.

Question 19.
Consider independent trials consisting of rolling a pair of fair dice, over and over. What is the probability that a sum of 5 appears before a sum of 7?
Solution:
When two dice are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let event A: The sum is 5 in a trial.
A = {(2, 3), (3, 2), (1, 4), (4, 1)}
∴ P(A) = \(\frac{4}{36}=\frac{1}{9}\)
Let event B: The sum is 7 in a trial.
B = {(2, 5), (5, 2), (3, 4), (4, 3), (1, 6), (6, 1)}
∴ P(B) = \(\frac{6}{36}=\frac{1}{6}\)
Let event C: Neither sum is 5 nor 7.
P(C) = 1 – P(A) – P(B)
= 1 – \(\frac{1}{9}\) – \(\frac{1}{6}\)
= \(\frac{26}{36}\)
Let the sum of 5 appear in the nth trial for the first time and the sum of 7 has not occurred in the first (n – 1) trials.

Question 20.
A machine produces parts that are either good (90%), slightly defective (2%), or obviously defective (8%). Produced parts get passed through an automatic inspection machine, which is able to detect any part that is obviously defective and discard it. What is the probability that the quality of the parts that make it through the inspection machine and get shipped?
Solution:
Let event G: The event that machine produces a good part,
event S: The event that machine produces a slightly defective part,
event D: The event that machine produces an obviously defective part.

Question 21.
Given three identical boxes, I, II, and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?
Solution:
Let event B₁: Select box I having two gold coins.
event B₂: Selecting box II having two silver coins,
event B₃: Selecting box III having one silver and one gold coin,
event G: Coin is gold.

To find the probability that the other can in the box is also gold. Which is possible only when it is drawn from the box I.
∴ Required probability = P(B₁/G)
By Bayes’ theorem,

Question 22.
In a factory which manufactures bulbs, machines A, B, and C manufacture respectively 25%, 35% and 40% of the bulbs. Of their outputs, 5, 4, and 2 percent are respectively defective bulbs. A bulb is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by machine B?
Solution:
Let event A: Bulb manufactured by machine A
event B: Bulb manufactured by machine B
event C: Bulb manufactured by machine C
event D: Bulb defective
∴ P(A) = \(\frac{25}{100}\)
P(B) = \(\frac{35}{100}\)
P(C) = \(\frac{40}{100}\)
Machines A, B and C manufacture respectively 25%, 35% and 40% of the bulbs.
Of their outputs, 5, 4, and 2 percent are respectively defective bulbs.

Required probability = P(B/D)
By Bayes’ theorem,

Question 23.
A family has two children. One of them is chosen at random and found that the child is a girl. Find the probability that
(i) both the children are girls.
(ii) both the children are girls given that at least one of them is a girl.
Solution:
A family has two children.
∴ Sample space S = {BB, BG, GB, GG}
(i) A: First child is a girl.
∴ A = {GB, GG}
∴ P(A) = \(\frac{2}{4}=\frac{1}{2}\)
B: Second child is a girl.
∴ B = {BG, GG}
∴ A ∩ B = {GG}
∴ P(A ∩ B) = \(\frac{1}{4}\)
Required probability
P(B/A) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}\)

(ii) A: At least one of the children is a girl.
∴ A = {GG, GB, BG}
∴ P(A) = \(\frac{3}{4}\)
B: both children are girls.
B = {GG}
∴ P(B) = \(\frac{1}{4}\)
Also, A ∩ B = B