Free Objective Test 01 Practice Test - 11th and 12th

Question 1

If a tangent of slope 2 of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is normal to the circle $x^{2} + y^{2} + 4 x + 1 = 0$ then the maximum value of ab is

A. 2

B. 4

C. 6

D. Can’t be found

SOLUTION

Solution : B

A tangent of slope 2 is $y = 2 x \pm \sqrt{4 a^{2} + b^{2}}$ ,this is normal to $x^{2} + y^{2} + 4 x + 1 = 0$
then $0 = - 4 \pm \sqrt{4 a^{2} + b^{2}} \Rightarrow 4 a^{2} + b^{2} = 16$
$u s i n g A . M . \geq G . M ., W e g e t a b \leq 4$

Question 2

The normal at an end of a latus rectum of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ passes through an end of the minor axis if

e^{4} + e^{2} = 1

e^{3} + e^{2} = 1

e^{2} + e = 1

e^{3} + e = 1

SOLUTION

Solution : A

Given ellipse equation is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Let $P (a e, \frac{b^{2}}{a})$ be one end of latus rectum.
Slope of normal at $P (a e, \frac{b^{2}}{a}) = \frac{1}{e}$
Equation of normal is
$y - \frac{b^{2}}{a} = \frac{1}{e} (x - a e)$
It passes through $´ B (0, b)$ then
$b - \frac{b^{2}}{a} = - a$
$a^{2} - b^{2} = - a b$
$a^{4} e^{4} = a^{2} b^{2}$
$e^{4} + e^{2} = 1$

Question 3

Consider two points A and B on the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$
circles are drawn having segments of tangents at A and B in between tangents at the two ends of major axis of ellipse as diameter, then the length of common chord of the circles is

A. 8

B. 6

C. 10

4 \sqrt{2}

SOLUTION

Solution : A

All such circles pass through foci $∴$ The common chord is of the length 2ae = $10 \times \frac{4}{5} = 8$

Question 4

If PSQ is a focal chord of the ellipse $16 x^{2} + 25 y^{2} = 400$ such that SP = 8 then the length of SQ =

A. 2

\frac{11}{3}

C. 16

D. 25

SOLUTION

Solution : A

$\frac{1}{S P} + \frac{1}{S Q} = \frac{2 a}{b^{2}}$
a= 5
b=4
Solving we get,
SQ=2.

Question 5

The ratio of the area enclosed by the locus of mid-point of PS and area of the ellipse where P is any point on the ellipse and S is the focus of the ellipse, is

\frac{1}{2}

\frac{1}{3}

\frac{1}{5}

\frac{1}{4}

SOLUTION

Solution : D

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, A r e a = π a b L e t P = (a c o s θ, b s i n θ) S = (a e, 0) M (h, k) m i d p o i n t o f P S \Rightarrow h = \frac{a e + a c o s θ}{2}; k = \frac{b s i n θ}{2} \Rightarrow {(\frac{h - \frac{a e}{2}}{\frac{a}{2}})}^{2} + {(\frac{k}{\frac{b}{2}})}^{2} = 1$
Area = $\frac{1}{4} π a b$

Question 6

If $(\sqrt{3}) b x + a y = 2 a b$ is tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , then eccentric angle $θ$ is

\frac{π}{4}

\frac{π}{6}

\frac{π}{2}

\frac{π}{3}

SOLUTION

Solution : B

Equation of tangent at a point $(a c o s θ, b s i n θ) i s \frac{x}{a} c o s θ + \frac{y}{b} s i n θ = 1$
But, it is the same as $\frac{x}{a} \frac{\sqrt{3}}{2} + \frac{y}{b} . \frac{1}{2} = 1$
$∴ c o s θ = \frac{\sqrt{3}}{2}, s i n θ = \frac{1}{2} \Rightarrow θ = \frac{π}{6}$

Question 7

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. A circle of radius 2 units

B. A parabola with fouc as (2, 3)

C. An ellipse with eccentricity

\frac{\sqrt{3}}{4}

D. An ellipse with length of latus rectrum is 2 units

SOLUTION

Solution : 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 8

How many tangents to the circle $x^{2} + y^{2} = 3$ are there which are normal to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$

A. 3

B. 2

C. 1

D. 0

SOLUTION

Solution : D

Equation of normal at $p (3 c o s θ, 2 s i n θ)$ is $3 x s e c θ - 2 y c o s e c θ = 5$
$\frac{5}{\sqrt{9 s e c^{2} θ + 4 c o s e c^{2} θ}} = \sqrt{3} B u t m i n i m u m v a l u e o f 9 s e c^{2} θ + 4 c o s e c^{2} θ = 25 ∴ n o s u c h θ^{- 1} e x i s t s$
So the number of tangents to the circle $x^{2} + y^{2} = 3$ which are normal to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ is 0.

Question 9

The equation of a tangent parallel to y = x drawn to $\frac{x^{2}}{3} - \frac{y^{2}}{2} = 1$ , is

A. x -y + 1 = 0

B. x - y + 2 = 0

C. x + y - 1 = 0

D. x + y – 2 = 0

SOLUTION

Solution : A

Let y = x + c be a tangent to $\frac{x^{2}}{3} - \frac{y^{2}}{2} = 1$ , then $c^{2} = 3 - 2 = 1$ $\Rightarrow$ $c = \pm 1$ So, the required tangents are $y = x \pm 1$

Question 10

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

A. 3x – 4y = 4

B. 3y – 4x + 4 = 0

C. 4x – 4y = 3

D. 3x – 4y = 2

SOLUTION

Solution : A

By T = S1, the equation of chord whose mid-point is $(α, β)$ is $3 x α - 2 y β + (x + α) - 2 (y + β) = 0$
$\Rightarrow$ $x (3 α + 2) - y (2 β + 3) = 2$
as it is parallel to y = 2x
$∴$ $3 α - 4 β = 4$
$∴$ Required locus is 3x – 4y = 4

Question 11

If x = 9 is the chord of contact of the hyperbola $x^{2} - y^{2} = 9$ , then the equation of the corresponding pair of tangents is

9 x^{2} - 8 y^{2} + 18 x - 9 = 0

9 x^{2} - 8 y^{2} - 18 x + 9 = 0

9 x^{2} - 8 y^{2} - 18 x - 9 = 0

9 x^{2} - 8 y^{2} + 18 x + 9 = 0

SOLUTION

Solution : B

If x = 9 meets the hyperbola at $(9, 6 \sqrt{2})$ and $(9, - 6 \sqrt{2})$ .
Then, equations of tangent at these points are $3 x - 2 \sqrt{2} y - 3 = 0$ and $3 x + 2 \sqrt{2} y - 3 = 0$ .
The combined equation of these two is $9 x^{2} - 8 y^{2} - 18 x + 9 = 0$ .

Question 12

The locus of the point which is such that the chord of contact of tangents drawn from it to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ forms a triangle of constant area with the coordinate axes, is

A. A straight line

B. A hyperbola

C. An ellipse

D. A circle

SOLUTION

Solution : B

The chord of contact of tangents from $(x_{1} y_{1}) i s \frac{x x_{1}}{a^{2}} + \frac{y y_{1}}{b^{2}} = 1$
It meets the axes at the points $(\frac{a^{2}}{x}, 0)$ and $(0, \frac{b^{2}}{y_{1}})$
Area of the triangle is $\frac{1}{2} . \frac{a^{2}}{x_{1}} . \frac{b^{2}}{y_{1}}$ [constant]
$\Rightarrow$ $x_{1} y_{1} = \frac{a^{2} b^{2}}{2 k} = c^{2}$ where c is a constant.
$\Rightarrow$ $x y = c^{2}$ is the required locus.

Question 13

From any point on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , tangents are drawn to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 2$ . Then, area cut-off by the chord of contact on the asymptotes is equal to

A. a/2 sq unit

B. ab sq unit

C. 2ab sq unit

D. 4ab sq unit

SOLUTION

Solution : D

Let P $(x_{1}, y_{1})$ be a point on the hyperbola $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
The chord of contact of tangents from P to the hyperbola is given by $\frac{x x_{1}}{a^{2}} + \frac{y y_{1}}{b^{2}} = 1$ …… (i)
The equation of the asymptotes are $\frac{x}{a} - \frac{y}{b} = 0$
and $\frac{x}{a} + \frac{y}{b} = 0$
The points of intersection of Equation (i) with the two asymptotes are given by
$x_{1} = \frac{2 a}{\frac{x_{1}}{a} + \frac{y_{1}}{b}}, y_{1} = \frac{2 a}{\frac{x_{1}}{a} + \frac{y_{1}}{b}}$
$x_{2} = \frac{2 a}{\frac{x_{1}}{a} + \frac{y_{1}}{b}}, y_{2} = \frac{2 a}{\frac{x_{1}}{a} + \frac{y_{1}}{b}}$
Area of the triangle = $\frac{1}{2} (x_{1} x_{2} - x_{2} y_{1})$
$= \frac{1}{2} ∣ ∣ ∣ ∣ ∣ ⎛ ⎜ ⎝ - \frac{4 a b \times 2}{\frac{x_{1}^{2}}{a^{2}} - \frac{y_{1}^{2}}{b^{2}}} ⎞ ⎟ ⎠ ∣ ∣ ∣ ∣ ∣$

Question 14

A rectangular hyperbola whose centre is C, is cut by any circle of radius r in four points P, Q, R and S. Then, $C P^{2} + C Q^{2} + C R^{2} + C S^{2}$ is equal to

$r^{2}$

$2 r^{2}$

$3 r^{2}$

$4 r^{2}$

SOLUTION

Solution : D

Let equation of the rectangular hyperbola be $x y = c^{2}$ and equation of circle be $x^{2} + y^{2} = r^{2}$
Put $y = \frac{c^{2}}{x}$ in equation (ii), we get $x^{2} + \frac{c^{4}}{x^{2}} r^{2}$
$x^{4} - r^{2} x^{2} + c^{4} = 0 N o w, C P^{2} + C Q^{2} + C R^{2} + C S^{2}$
$= x_{1}^{2} + y_{1}^{2} + x_{2}^{2} + y_{2}^{2} + x_{3}^{2} + y_{3}^{2} + x_{4}^{2} + y_{4}^{2}$
$= {(\sum_{i = 1}^{4} x_{i})}_{i}^{2} - 2 \sum x_{1} x_{2} + = {(\sum_{i = 1}^{4} y_{i})}_{i}^{2} - 2 \sum y_{1} y_{2}$
$= 2 r^{2} + 2 r^{2} = 4 r^{2}$ [from equation (iii)]

Question 15

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

\frac{a b}{a + b}

\frac{a^{2} b^{2}}{a^{2} + b^{2}}

\frac{2 a^{2} b^{2}}{a^{2} + b^{2}}

D. None of these

SOLUTION

Solution : 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 $(a s e c θ, b t a n θ)$ on $\frac{x}{a} + \frac{y}{b} = 0$
$= \frac{s e c θ - t a n θ}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}}$
and $p_{2}$ = length of the perpendicular from $(a s e c θ, b t a n θ)$ $\frac{x}{a} + \frac{y}{b} = 0$
$= \frac{s e c θ - t a n θ}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}}$
$∴$ $p_{1} p_{2} = \frac{s e c^{2} θ - t a n^{2} θ}{\frac{1}{a^{2}} + \frac{1}{b^{2}}} = \frac{a^{2} b^{2}}{a^{2} + b^{2}}$

Free Objective Test 01 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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