## Rajasthan Board RBSE Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise

Question 1.

Radius of a circle

9x^{2} +y^{2} + 8x = 4(x^{2} – y^{2}) is :

(A) 1

(B) 2

(C) 4 / 5

(D) 5 / 4

Solution:

Equation of circle,

9x^{2} + y^{2} + 8x = 4(x^{2} – y^{2})

⇒ 9x^{2} + y^{2} + 8x = 4x^{2} – 4y^{2}

⇒ 9x^{2} – 4x^{2} + 8x + y^{2} + 4y^{2} = 0

⇒ 5x^{2} + 5y^{2} + 8x = 0

Question 2.

Equation of a circle whose centre is in I quadrant as (α, β) and touches x-axis will be :

(A) x^{2} + y^{2} – 2αx – 2βy + α^{2} = 0

(B) x^{2} + y^{2} + 2αx – 2βy + α^{2} = 0

(C) x^{2} + y^{2} – 2αx + 2βy + α^{2} = 0

(D) x^{2} + y^{2} + 2αx + 2βy + α^{2} = 0

Solution:

Centre of circle,

(- g, -f) = α, β

g = – α

Question 3.

If liney = mx + c touches the circle x^{2} + y^{2} = 4y, then value of c is :

Solution:

Equation of line

y = mx + c …..(i)

and equation of circle

x^{2} + y^{2} = 4y

⇒ x^{2} + y^{2} – 4y = 0

⇒ x^{2} + y^{2} – 2.2.y + 4 = 4

⇒ x^{2} + (y – 2)^{2} = 4

⇒ x^{2} + y^{2} = 2^{2}

where X = x, Y – y – 2, a = 2

According to condition of tangency

Thus option (C) is correct.

Question 4.

Line 3x + 4y = 25 touches the circle x^{2 }+ y^{2} = 25 at the point :

(A) (4, 3)

(B) (3, 4)

(C) (-3, -4)

(D) (3, -4)

Solution:

Equation of line,

3x + 4y = 25

and equation of circle

Question 5.

A conic section will be parabola if:

(A) e = 0

(B) e < 0

(C) e > 0

(D) e = 1

Solution:

Option (C) is correct.

Question 6.

Equation of directrix of parabola x^{2} =- 8y is :

(A) y = -2

(B) y = 2

(C) x = 2

(D) x = – 2

Solution:

Equation of parabola

x^{2} = – 8y

⇒ x^{2} = -4 × 2 × y

Here a = 2

By equation of directrix y = a

y = 2

Thus, option (B) is correct.

Question 7.

Vertex of parabola x^{2} + 4x + 2y = 0 is:

(A) (0, 0)

(B) (2, – 2)

(C) (- 2, -2)

(D) (-2, 2)

Solution:

Equation of parabola is :

x^{2} + 4x + 2y = 0

⇒ x^{2} + 4x + 4 = – 2y + 4

⇒ (x + 2)^{2} = – 2(y – 2)

⇒ x^{2} = – 2y

Here X = x + 2

and Y = y – 2

Vertex of parabola is (0, 0).

then x = x + 2 = 0

⇒ y = -2

y = y -2 = 0

⇒ y = 2

So vertex is (- 2, 2).

Thus, option (D) is correct.

Question 8.

If focus of any parabola is (-3,0) and directrix is x + 5 = 0, then its equation will be :

(A) y^{2} = 4(x + 4)

(B) y^{2} + 4x + 16 = 0

(C) y^{2} + 4x = 16

(D) x^{2} = 4(y + 4)

Solution:

Let P(x, y) be any point on parabola. By definition of parabola,

Distance between (x, y) and focus (- 3, 0)

= Length of ⊥ drawn from P(x, y) to directrix x + 5 = 0.

Question 9.

If vertex and focus of any parabola are (2,0) and (5, 0) respectively, then its equation will be :

(A) y^{2} = 12x + 24

(B) y^{2} = 12x – 24

(C) y^{2} = – 12x – 24

(D) y^{2} = – 12x + 24

Solution:

Vertex = (2, 0),

Focus = (5, 0)

Distance between vertex and focus

∴ X = x – 2

and Y = y

Thus equation of parabola

Y^{2} = 4aX

⇒ y^{2} = 4 × 3 × (x – 2)

⇒ y^{2} = 12x – 24

Thus, optional (B) is correct.

Question 10.

Focus of parabola x^{2} = – 8y is :

(A) (2, 0)

(B) (0, 2)

(C) (-2, 0)

(D) (0, – 2)

Solution:

Equation of parabola,

x^{2} = – 8y

⇒ x^{2} = -4 × 2 × y

Coordinates of focus = (0, – a) – (0, – 2)

Thus, optional (D) is correct.

Question 11.

Equation of any tangent at parabola y^{2} = x is :

(A) y = mx + 1/m

(B) y = mx + 1 /4m

(C) y = mx + 4/m

(D) y = mx + 4m

Solution:

Equation of parabola,

y^{2} = x

Thus, optional (B) is correct.

Question 12.

If line 2y – x = 2 touches the parabola y^{2} = 2x then tangent point is :

(A) (4, 3)

(B) (-4, 1)

(C) (2, 2)

(D) (1, 4)

Solution:

Equation of line,

2y – x = 2

x = 2y – 2 ….(i)

Equation of parabola,

y^{2} = 2x

From equation (i) and (ii),

y^{2} = 2(2y – 2)

⇒ y^{2} = 4y – 4

⇒ y^{2} – 4y + 4 = 0

⇒ (y – 2)^{2} = 0

⇒ y – 2 = 0

⇒ y = 2

Put value ofy in equation (i),

x = 2 × 2 – 2 = 2

Thus tangent point, (x, y) = (2, 2)

Thus, optional (C) is correct.

Question 13.

Tangent equation of parabola x^{2} = 8y parallel to line x + 2y + 1 = 0 is :

(A) x + 2y + 1 = 0

(B) x – 2y + 1 = 0

(C) x + 2y – 1 = 0

(D) x – 2y – 1 = 0

Solution:

Equation of parabola,

x^{2} = 8y

⇒ x^{2} = 4 × 2 × y

here a = 2

Equation of line,

x + 2y + 1 = 0

Equation of its parallel line.

x + 2y + λ = 0

According to condition of tandency

Thus, equation of tangent line x + 2y – 4 = 0

Thus, optional (A) is correct.

Question 14.

A normal of parabola y^{2} = 4x is :

(A) y = x + 4

(B) y + x = 3

(C) y + x = 2

(D) y + x = 1

Solution:

Equation of parabola

y^{2} = 4x

y^{2} = 4 × 1 × x

Here a = 1

To find normal, a point should be given which is not given here. So Question is incomplete.

Question 15.

Length of semi latus rectum of ellipse x^{2} + 4y^{2} = 12 will be :

Solution:

Education of ellipse

3x^{2} + 4y^{2} =12

Thus option (A) is correct.

Question 16.

Eccentricity of ellipse 3x^{2} + 4y^{2} = 12 will be :

(A) -2

(B) \(\frac { 1 }{ 2 } \)

(C) 1

(D) 2

Solution:

Question 17.

If line y = mx + c touches the ellipse

then value of c will be:

Solution:

Option (C) is correct.

Question 18.

Coordinates of focus of ellipse

(A) (± ae, 0)

(B) (± be, 0)

(C) (0, ± ae)

(D) (0, ± be)

Solution:

Option (D) is correct.

Question 19.

Eccentricity of rectangular hyperbola will be:

(A) 0

(B) 1

(C) \(\sqrt { 2 }\)

(D) 2

Solution:

Option (C) is correct.

Question 20.

Eccentricity of hyperbola 9×2 – 16y2 = 144 will be :

(A) 1

(B) 0

(C) 5/16

(D) 5 / 4

Solution:

Equation of hyperbola,

9x^{2} – 16y^{2} = 144

Thus Option (D) is correct.

Question 21.

Write the equation of circle whose centre is (a cos α, a sin α) and radius is a.

Solution:

Centre of circle

(- g, -f) – (a cos α, a sin α)

or (g,f) = (-a cos α, – a sin α)

Then equation of circle,

x^{2} + y^{2} + 2gx + 2fy + c = 0

⇒ x^{2} + y^{2} – 2a cos α . x – 2a sin α. y = 0

This is required equation.

Question 22.

If tangents at points (x_{1}, y_{1}) and (x_{2}, y_{2}) of circle x^{2} + y^{2} + 2gx + 2fy + c = 0 are perpendicular to each other than prove that

x_{1}x_{2} + y_{1}y_{2} + g(x_{1} + x_{2}) + f(y_{1} + y_{2}) + g^{2} + f^{2} = 0

Solution:

Given equation of circle

x^{2} + y^{2} + 2gx + 2fy + c = 0 …..(i)

Equation of tangent at point (x_{1}, y_{1})

xx_{1} + yy_{1} + g(x + x_{1}) +f(y + y_{1}) + c = o

⇒ x_{6}x + y_{6}y + g(x_{1} + x) +f(y_{1} + y) + c = 0 …..(ii)

Question 23.

Find the equation of circle with radius r, whose centre lies in 1st quadrant and touches y-axis at a distance of h from the origin. Find the equation of other tangent which passes through origin.

Solution:

Centre of circle = (r, h)

Radius of circle = r

Thus, equation of circle

(x – r)^{2} + (y – h)^{2} = r^{2}

Let tangent OB touches the circle at B. Tangent passes through origin. Let tangent touches the circle at point (x_{1},y_{1}).

∴ Equation of tangent at point of circle,

xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

xx_{1} + yy_{1} – r(x + x_{1}) + h(y + y_{1}) + h^{2} = 0

Since the line passes through origin.

∴ x × 0 + y × 0 – r(x + 0) – h(y + 0) + h^{2} = 0

⇒ – rx – hy + h^{2} = 0

⇒ rx + hy – h^{2} = 0

This is required equation.

Question 24.

Tangents drawn at point (α, β) of circle x^{2} + y^{2} = a^{2} meets the axis at points A and B respectively. Prove that area of ∆OAB will be,

where O is origin.

Solution:

Tangent equation of circle

x^{2} + y^{2} = a^{2}

Equation of tangent at point (α, β)

xx_{1} + yy_{1} = a^{2}

At x-axis x × a + y × β = a^{2
}

Question 25.

Find the equation of tangent at circle x^{2} + y^{2} = a^{2} which makes a triangle of area a^{2} with axis.

Solution:

Equation of circle,

x^{2} + y^{2} = a^{2}

Area of ∆ABO = a^{2
}

Question 26.

Write the coordinates of the focus of parabola x^{2} – 4x – 8y = 4.

Solution:

Equation of parabola,

x^{2} – 4x – 8y = 4

⇒ x^{2} – 2.2x + (2)^{2} = 8y + 4 + 2^{2}

⇒ (x – 2)^{2} = 8y + 8

= 8(y + 1)

(x-2)^{2} = 4.2.(y + 1)

X^{2} = 4.2.Y

Where x = x – 2 and Y = y + 1

a = 2

Coordinates of focus = (0, a)

x = x- 2 = 0

⇒ x = 2

Y = y + 1 = 2

⇒ y = 1

Thus coordinates of focus = (2, 1)

Question 27.

Write the eccentricity of parabola

x^{2} – 4x – 4y + 4 = 0.

Solution:

Eccentricity of parabola is 1.

Question 28.

Write the condition for which line lx + my + n = 0 touches the parabola y^{2} = 4ax

Solution:

Required condition: In = am^{2}

Question 29.

Write the equation of parabola whose vertex is (0, 0) and focus is (0, – a).

Solution:

Vertex (0, 0) and focus is at (0, – a) which lies at y-axis.

Thus axis of parabola is -ve y-axis.

Equation of parabola is of the form x^{2} = – 4ay

Thus equation of parabola,

x^{2} = 4(-a)y

⇒ x^{2} = – 4ay

This is required equation.

Question 30.

Write the equation of axis of parabola

9y^{2} – 16x – 12y – 57 = 0.

Solution:

Equation of parabola,

Question 31.

Write the coordinates of centre of ellipse

Solution:

Equation of ellipse

Question 32.

Write the condition for which line x cos α + y sin α = p touches the ellipse

Solution:

Putting value ofy form line x cos a + y sin α =p in the ellipse.

or x^{2}(α^{2} cos^{2} α + b^{2} sin^{2} α) – 2a^{2} px cos α

+ (α^{2}p^{2} – a^{2}b^{2} sin^{2} α) = 0 …(i)

Given line will touch ellipse if eq^{n}. (i) has equal roots.

(- 2a^{2} p cos α)^{2} – 4(α^{2} cos^{2} α + b^{2} sin^{2} α)

(α^{2}p^{2} – α^{2}b^{2} sin^{2} α) = 0

or 4α^{2}b^{2} sin^{2} α[α^{2} cos^{2} α – p^{2} + b^{2} sinz α) = 0

Thus p^{2} = a^{2} cos^{2} α + b^{2} sin^{2} α

Question 33.

Write the equation of hyperbola whose trans¬verse axis and conjugate axis are 4 and 5 respectively.

Solution:

Equation of hyperbola,

Question 34.

Write the coordinates of centre of hyperbola

Solution:

Equation of hyperbola

Comparing eq^{n}. (1) by standard equation of hyperbola, coordinates of centre.

X = x – 1 = 0

⇒ x = 1

Y = y + 2 = 0

⇒ y = – 2

Thus, coordinates of centre = (1, – 2).

## Leave a Reply