Ellipse And Hyperbola(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

ELLIPSE AND HYPERBOLA MCQs

Total Questions : 30 | Page 1 of 3 pages

The area of the parallelogram formed by the tangents at the points whose eccentric angles are $θ, θ + \frac{π}{2}, θ + π, θ + \frac{3 π}{2}$ on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is

Discuss Question

Answer: Option B. -> 4ab
:
B

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

Area=

π a b

Let P

= (a c o s θ, b s i n θ)

S=(ae,0)
M(h,k) be the midpoint of PS

(h, k) = (\frac{a e + a c o s θ}{2}, \frac{b s i n θ}{2})

\frac{{(h - \frac{a e}{2})}^{2}}{{(\frac{a}{2})}^{2}} + \frac{k^{2}}{{(\frac{b}{2})}^{2}} = 1

A r e a = \frac{π a b}{4}

R a t i o = \frac{1}{4}

Question 2.

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 3.

If the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$ meet the ellipse $\frac{x^{2}}{1} + \frac{y^{2}}{a^{2}} = 1$ in four distinct points and $a = b^{2} - 10 b + 25$ , then the value b does not satisfy

$(- \infty, 4)$
(4, 6)
$(6, - \infty)$
[4, 6]

Discuss Question

Answer: Option D. -> [4, 6]
:
D
For the two ellipse to intersect at four different points

1 < a^{2} \Rightarrow 1 < (b - 5)^{2} \Rightarrow b / ϵ [4, 6]

Question 4.

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

$15 \sqrt{3} π$
$12 \sqrt{3} π$
$18 \sqrt{3} π$
$8 \sqrt{3} π$

Discuss Question

Answer: Option D. ->

8 \sqrt{3} π

:
D
The path he runs is an ellipse whose major axis is 4m,e=

\frac{1}{2}

Area=

π

Question 5.

A tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ cuts the axes in M and N. Then the least length of MN is

a + b
a - b
$a^{2} + b^{2}$
$a^{2} - b^{2}$

Discuss Question

Answer: Option A. -> a + b
:
A
Equation of tangent at

(θ)

\frac{x}{a} c o s θ + \frac{y}{b} s i n θ = 1 I t m e e t s a x e s a t M = (\frac{a}{c o s θ}, 0) a n d N = (0, \frac{b}{s i n θ}) ∴ M N = \sqrt{a^{2} s e c^{2} θ + b^{2} c o s e c^{2} θ} = \sqrt{a^{2} + b^{2} + a^{2} c o t^{2} θ + b^{2} t a n^{2} θ} ∴ M i n i m u m v a l u e o f a^{2} c o t^{2} θ + b^{2} t a n^{2} θ i s 2 a b (∵ A . M \geq G . M) ∴ M i n i m u m v a l u e o f M N = a + b

Question 6.

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

$\frac{π}{4}$
$\frac{π}{3}$
$\frac{π}{6}$
$\frac{π}{2}$

Discuss Question

Answer: Option D. ->

\frac{π}{2}

:
D
The given circle is the director circle of the ellipse,so the angle between the tangents is

\frac{π}{2}

Question 7.

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 :

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

Discuss Question

Answer: Option C. ->

\frac{(x + 2)^{2}}{40} + \frac{y^{2}}{49} = 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 8.

The line lx + my + n = 0 will be a normal to the hyperbola $b^{2} x^{2} - a^{2} y^{2} = a^{2} b^{2}$ , if

$\frac{a^{2}}{l^{2}} + \frac{b^{2}}{m^{2}} = \frac{(a^{2} + b^{2})^{2}}{n^{2}}$
$\frac{a^{2}}{l^{2}} + \frac{b^{2}}{m^{2}} = \frac{(a^{2} - b^{2})^{2}}{n^{2}}$
$\frac{a^{2}}{l^{2}} + \frac{b^{2}}{m^{2}} = \frac{(a^{2} + b^{2})^{2}}{n}$
None of these

Discuss Question

Answer: Option D. -> None of these
:
D
The equation of the normal at

(a s e c ϕ, b t a n ϕ)

to the hyperbola

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

is
ax

s i n ϕ + b y = (a^{2} + b^{2}) t a n ϕ

......(i)
and the equation of the line is
lx + my + n =0 …… (ii)
now, equations (i) and (ii) represent the same line, therefore

\frac{2 s i n ϕ}{l} = \frac{b}{m} = \frac{(a^{2} + b^{2}) t a n ϕ}{- n} = \frac{(a^{2} + b^{2}) s i n ϕ}{- n c o s ϕ}

∴

s i n ϕ = \frac{b l}{a m} a n d c o s ϕ = \frac{(a^{2} + b^{2}) l}{- n a}

\Rightarrow

\frac{a^{2}}{l^{2}} - \frac{b^{2}}{m^{2}} = \frac{(a^{2} + b^{2})^{2}}{- n^{2}}

Question 9.

Equation of the chord of the hyperbola $25 x^{2} - 16 y^{2} = 400$ which is bisected at the point (6, 2), is

16x – 75y = 418
75x - 16y = 418
25x - 4y = 400
None of these

Discuss Question

Answer: Option B. -> 75x - 16y = 418
:
B
Equation of hyperbola is

25 x^{2} - 16 y^{2} = 400

\Rightarrow

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

Chord is bisected at (6, 2)

∴

\frac{6 x}{16} - \frac{2 y}{25} = \frac{6^{2}}{16} - \frac{2^{2}}{25}

\Rightarrow

75 x - 16 y = 418

Question 10.

The equation of a hyperbola, conjugate to the hyperbola $x^{2} + 3 x y + 2 y^{2} + 2 x + 3 y + 1 = 0$ , is

$x^{2} + 3 x y + 2 y^{2} + 2 x + 3 y + 1 = 0$
$x^{2} + 3 x y + 2 y^{2} + 2 x + 3 y + 2 = 0$
$x^{2} + 3 x y + 2 y^{2} + 2 x + 3 y + 3 = 0$
$x^{2} + 3 x y + 2 y^{2} + 2 x + 3 y + 4 = 0$

Discuss Question

Answer: Option B. ->

x^{2} + 3 x y + 2 y^{2} + 2 x + 3 y + 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)]

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