Free Objective Test 02 Practice Test - 11th and 12th

Question 1

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

A. ab

B. 4ab

C. 3ab

D. 2ab

SOLUTION

Solution : 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 $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$ 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)

B. (4, 6)

(6, - \infty)

D. [4, 6]

SOLUTION

Solution : 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 3

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

A. 4

B. 2

C. 1

D. None of these

SOLUTION

Solution : 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 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

SOLUTION

Solution : D

The path he runs is an ellipse whose major axis is 4m,e= $\frac{1}{2}$
Area= $π$ ab

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. a + b

B. a - b

a^{2} + b^{2}

a^{2} - b^{2}

SOLUTION

Solution : A

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

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}

SOLUTION

Solution : 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

SOLUTION

Solution : 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}

D. None of these

SOLUTION

Solution : 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

A. 16x – 75y = 418

B. 75x - 16y = 418

C. 25x - 4y = 400

D. None of these

SOLUTION

Solution : 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

SOLUTION

Solution : 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 11

The eccentricity of the rectangular hyperbola is

A. 2

\sqrt{2}

C. 0

D. None of these

SOLUTION

Solution : 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 12

Two concentric hyperbolas,whose axes meet at angle of $45^{\circ}, c u t$

A. at

45^{0}

B. at

90^{0}

C. Nothing can be said

D. None of these

SOLUTION

Solution : B

Let the equation to the rectangular hyperbola be
$x^{2} - y^{2} = a^{2}$   …… (i)
As the asymptotes of this are the axes of the other and vice-versa, hence the equation of the other hyperbola may be written as
$x y = c^{2}$   …… (ii)
Let equations (i) and (ii) meet at some point whose coordinates are
$(a s e c α, a t a n α)$
Then, the tangent at the point a $(s e c α, a t a n α)$ to equation (ii) is
$x - y s i n α = a c o s α$   …… (iii)
and the tangent at the point $(a s e c α, a t a n α)$ to equation (ii) is
$y + x s i n α = (\frac{2 c^{2}}{a}) c o s α$   …… (iv)
clearly, the slopes of the tangents given by equations (iv) and (iii) are respectively - $s e c α a n d \frac{1}{s i n α}$ , so their product is $- s e c α . (\frac{1}{s i n α}) = - 1$ Hence, the tangents are at right angle.

Question 13

If foci of hyperbola lie on y = x and one of the asymptotes is y = 2x, then equation of the hyperbola, given that it passes through (3, 4), is

x^{2} - y^{2} - \frac{5}{2} x y + 5 = 0

2 x^{2} - 2 y^{2} + 5 x y + 5 = 0

2 x^{2} + 2 y^{2} - 5 x y + 10 = 0

D. None of these

SOLUTION

Solution : C

Foci of hyperbola lie on y = x. So, the major axis is y = x. Major axis of hyperbola bisects the asymptote.
$\Rightarrow$ Equation of hyperbola is x = 2y
$\Rightarrow$ Equation of hyperbola is (y – 2x)(x – 2y) + k = 0
Given that, it passes through (3, 4)
$\Rightarrow$ Hence, required equation is
$2 x^{2} + 2 y^{2} - 5 x y + 10 = 0$

Question 14

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

α ϵ (- 2, 0)

α ϵ (- 3, 0)

α ϵ (- \infty, - 2)

α ϵ (- \infty, - 3)

SOLUTION

Solution : C

Given that,
$\frac{x^{2}}{1} - \frac{y^{2}}{4} = 1$

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 15

If a hyperbola passes through (2, 3) and has asymptotes 3x - 4y + 5 = 0 and 12x + 5y - 40 = 0, then the equation of its transverse axis is

A. 77x - 21y - 265 = 0

B. 21x - 77y + 265 = 0

C. 21x - 77y - 265 = 0

D. 21x + 77y - 265 = 0

SOLUTION

Solution : D

ransverse axis is the equation of the angle bisector passing containing point (2, 3), which is given by
$\frac{3 x - 4 y + 5}{5} = \frac{12 x + 5 y - 40}{13}$
$\Rightarrow$ $21 x + 77 y = 265$

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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