Ellipse And Hyperbola(12th Grade > Mathematics ) Questions and answers for exam Preparation

12th Grade > Mathematics

ELLIPSE AND HYPERBOLA MCQs

Total Questions : 30 | Page 2 of 3 pages

Question 11. If the ellipse

\frac{x^{2}}{a^{2} - 3} + \frac{y^{2}}{a + 4} = 1

is inscribed in a square of side length

a \sqrt{2}

then a is

4
2
1
None of these

Discuss Question

Answer: Option D. -> None of these
:
D
Sides of the square will be perpendicular tangents to the ellipse so, vertices of the square will lie on director circle. So diameter of director circle is

2 \sqrt{(a^{2} - 3) + (a + 4)} = \sqrt{2 a^{2} + 2 a^{2}} 2 \sqrt{a^{2} + a + 1} = 2 a \Rightarrow a = - 1 B u t f o r e l l i p s e a^{2} > 3 & a > - 4

So there is no value of a which satisfy both.

Question 12. Tangents are drawn from any point on the circle

x^{2} + y^{2} = 41

to the Ellipse

\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1

then the angle between the two tangents is

Discuss Question

Answer: Option D. -> π2
:
D
The given circle is the director circle of the ellipse,so the angle between the tangents is

\frac{π}{2}

Question 13. The perimeter of a triangle is 20 and the points (-2, -3) and (-2, 3) are two of the vertices of it. Then the locus of third vertex is :

(x−2)249+y240=1
(x+2)249+y240=1
(x+2)240+y249=1
(x−2)240+y249=1

Discuss Question

Answer: Option C. -> (x+2)240+y249=1
:
C

A (- 2, - 3), B (- 2, 3)

Let the third vertex be C(x,y)
AB + BC+CA=20
BC+CA=14
So locus of C is ellipse with A,B as focii.
2a=14
2ae=6

e = \frac{3}{7}

∴

we get b as

\sqrt{40}

∴

Equation of ellipse is

\frac{(x + 2)^{2}}{40} + \frac{y^{2}}{49} = 1

Question 14. The equation of a hyperbola, conjugate to the hyperbola

x^{2} + 3 x y + 2 y^{2} + 2 x + 3 y + 1 = 0

, is

x2+3xy+2y2+2x+3y+1=0
x2+3xy+2y2+2x+3y+2=0
x2+3xy+2y2+2x+3y+3=0
x2+3xy+2y2+2x+3y+4=0

Discuss Question

Answer: Option B. -> x2+3xy+2y2+2x+3y+2=0
:
B
We know that,
Equation of hyperbola + Equation of conjugate
hyperbola = 2(Equation of asymptotes)

∴

Equation of conjugate hyperbola
= 2 Equation of asymptotes - Equation of hyperbola
C(x, y) = 2A(x, y) – H(x, y) …… (i)

∵

H (x, y) \equiv x^{2} + 3 x y + 2 y^{2} + 2 x + 3 y = 0

∴

Equation of asymptotes is

x^{2} + 3 x y + 2 y^{2} + 2 x + 3 y + λ = 0

∴

Δ = 0, a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0, t h e n λ = 1

∴

A (x, y) \equiv x^{2} + 3 x y + 2 y^{2} + 2 x + 3 y + 1 = 0

∴

C (x, y) \equiv x^{2} + 3 x y + 2 y^{2} + 2 x + 3 y + 2 = 0

[from equation (i)]

Question 15. If two tangents can be drawn to the different branches of hyperbola

\frac{x^{2}}{1} - \frac{y^{2}}{4} = 1

from the paint

(α, α^{2})

, then

αϵ(−2,0)
αϵ(−3,0)
αϵ(−∞,−2)
αϵ(−∞,−3)

Discuss Question

Answer: Option C. -> αϵ(−∞,−2)
:
C
Given that,

\frac{x^{2}}{1} - \frac{y^{2}}{4} = 1

If Two Tangents Can Be Drawn To The Different Branches Of Hy...

Since,

(α, α^{2})

lie on the parabola

y = x^{2}

, then

(α, α^{2})

must lie between the asymptotes of hyperbola

\frac{x^{2}}{1} - \frac{y^{2}}{4} = 1

in 1st and 2nd quadrants.
So, the asymptotes are

y = \pm 2 x

∴

2 α < α^{2}

\Rightarrow

α < 0 o r α > 2

and

- 2 α < α^{2}

α < - 2 o r - 2 α < 0

∴

α ϵ (- \infty, - 2) o r (2, \infty)

Question 16. The eccentricity of the rectangular hyperbola is

2
√2
0
None of these

Discuss Question

Answer: Option B. -> √2
:
B
In case of rectangular hyperbola,

a = b \Rightarrow b^{2} = a^{2} \Rightarrow a^{2} (e^{2} - 1) = a^{2} \Rightarrow e = \sqrt{2}

Question 17. A man running round a race course notes that the sum of the distances of two flag posts from him is 8 meters.The area of the path he encloses in square meters if the distance between flag posts is 4 is

15√3π
12√3π
18√3π
8√3π

Discuss Question

Answer: Option D. -> 8√3π
:
D
The path he runs is an ellipse whose major axis is 4m,e=

\frac{1}{2}

Area=

π

Question 18. If a variable tangent of the circle

x^{2} + y^{2} = 1

intersect the ellipse

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

at P and Q then the locus of the points of intersection of the tangents at P and Q is

A circle of radius 2 units
A parabola with fouc as (2, 3)
An ellipse with eccentricity √34
An ellipse with length of latus rectrum is 2 units

Discuss Question

Answer: Option D. -> An ellipse with length of latus rectrum is 2 units
:
D

x^{2} + y^{2} = 1

;

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

Let

R (x_{1}, y_{1})

is point of intersection of tangents drawn at P,Q to ellipse

\Rightarrow

PQ is chord of contact of

R (x_{1}, y_{1})

\Rightarrow x x_{1} + 2 y y_{1} - 4 = 0

This touches circle

\Rightarrow r^{2} (l^{2} + m^{2}) = n^{2} \Rightarrow 1 (x_{1}^{2} + 4 y_{1}^{2}) = 16 \Rightarrow x^{2} + 4 y^{2} = 16 i s e l l i p s e w i t h e c c e n t r i c i t y e = \frac{\sqrt{3}}{2} a n d l e n g t h o f l a t u s r e c t u m L L^{1} = 2

Question 19. The product of the perpendicular from any point on the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

to its asymptotes, is equal to

aba+b
a2b2a2+b2
2a2b2a2+b2
None of these

Discuss Question

Answer: Option B. -> a2b2a2+b2
:
B
Let

(a s e c θ, b t a n θ)

be the any point on the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

The equations of the asymptotes of the given hyperbola are

\frac{x}{a} - \frac{y}{b} = 0

and

\frac{x}{a} + \frac{y}{b} = 0

Now,

p_{1}

= length of the perpendicular from

(a s e c θ, b t a n θ)

\frac{x}{a} + \frac{y}{b} = 0

= \frac{s e c θ - t a n θ}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}}

and

p_{2}

= length of the perpendicular from

(a s e c θ, b t a n θ)

\frac{x}{a} + \frac{y}{b} = 0

= \frac{s e c θ - t a n θ}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}}

∴

p_{1} p_{2} = \frac{s e c^{2} θ - t a n^{2} θ}{\frac{1}{a^{2}} + \frac{1}{b^{2}}} = \frac{a^{2} b^{2}}{a^{2} + b^{2}}

Question 20. The locus of the middle points of chords of hyperbola

3 x^{2} - 2 y^{2} + 4 x - 6 y = 0

parallel to y = 2x, is

3x – 4y = 4
3y – 4x + 4 = 0
4x – 4y = 3
3x – 4y = 2

Discuss Question

Answer: Option A. -> 3x – 4y = 4
:
A
By T = S1, the equation of chord whose mid-point is

(α, β)

3 x α - 2 y β + (x + α) - 2 (y + β) = 0

\Rightarrow

x (3 α + 2) - y (2 β + 3) = 2

as it is parallel to y = 2x

∴

3 α - 4 β = 4

∴

Required locus is 3x – 4y = 4

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