MCQ

MCQ 11 Mark

If the focus of the parabola is $(0, -3)$, its directrix is y $= 3$, then its equation is ________

✓
$x^2=-12 y$
B
$x^2=12 y$
C
$y^2=12 x$
D
$y^2=-12 x$

Answer

Correct option: A.

$x^2=-12 y$

$S P^2= PM ^2 $
$ \Rightarrow(x-0)^2+(y+3)^2=\left|\frac{y-3}{\sqrt{1}}\right|^2 $
$\Rightarrow x^2+y^2+6 y+9=y^2-6 y+9 $
$\Rightarrow x^2=-12 y$

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MCQ 21 Mark

The length of latus rectum of the parabola $x^2-4 x-8 y+12=0$ is

A
$4$
B
$6$
✓
$8$
D
$10$

Answer

Correct option: C.

$8$

$8$
Hint:
Given equation of parabola is
$x^2-4 x-8 y+12=0$
$ \Rightarrow x^2-4 x=8 y-12$
$\Rightarrow x^2-4 x+4=8 y-12+4 $
$\Rightarrow(x-2)^2=8(y-1)$
Comparing this equation with $(x-h)^2=4 b(y-k)$, we get
4b = 8
∴ Length of latus rectum $= 4b = 8$

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MCQ 31 Mark

The line $y=m x+1$ is a tangent to the parabola $y^2=4 x$, if $m$ is

✓
$1$
B
$2$
C
$3$
D
$4$

Answer

Correct option: A.

$1$

$1$
hint:
$y^2=4 x$
Compare with $y^2=4 a x$
$∴ a = 1$ Equation of tangent is $y = mx + 1$
Compare with $y=m x+\frac{a}{m}$
$\frac{a}{m}=1 $
$ \therefore a=m=1$

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MCQ 41 Mark

The foci of hyperbola $4 x^2-9 y^2-36=0$ are

✓
(±√13, 0)
B
(±√11, 0)
C
(±√12, 0)
D
(0, ±√12)

Answer

Correct option: A.

(±√13, 0)

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MCQ 51 Mark

If the line $2 x-y=4$ touches the hyperbola $4 x^2-3 y^2=24$, the point of contact is

A
(1, 2)
B
(2, 3)
✓
(3, 2)
D
(-2, -3)

Answer

Correct option: C.

(3, 2)

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MCQ 61 Mark

Centre of the ellipse $9 x^2+5 y^2-36 x-50 y-164=0$ is at

✓
$(2, 5)$
B
$(1, -2)$
C
$(-2, 1)$
D
$(0, 0)$

Answer

Correct option: A.

$(2, 5)$

(2, 5)
$9 x^2+5 y^2-36 x-50 y-164=0 $
$ \Rightarrow 9(x-2)^2+5(y-5)^2=325$
$ \Rightarrow \frac{(x-2)^2}{\frac{325}{9}}+\frac{(y-5)^2}{65}=1$
⇒ centre of the ellipse = (2, 5)

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MCQ 71 Mark

Eccentricity of the hyperbola $16 x^2-3 y^2-32 x-12 y-44=0$ is

A
$\sqrt{\frac{17}{3}}$
B
$\sqrt{\frac{19}{3}}$
✓
$\frac{\sqrt{19}}{3}$
D
$\frac{\sqrt{17}}{3}$

Answer

Correct option: C.

$\frac{\sqrt{19}}{3}$

$\sqrt{\frac{19}{3}}$

$16 x^2-3 y^2-32 x-12 y-44=0$

$\Rightarrow 16(x-1)^2-3(y+2)^2=48$

$\Rightarrow \frac{(x-1)^2}{3}-\frac{(y+2)^2}{16}=1$

Here, $a^2=3$ and $b^2=16$

$e=\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{3+16}}{\sqrt{3}}=\sqrt{\frac{19}{3}}$

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MCQ 81 Mark

The equation of the tangent to the ellipse $4 x^2+9 y^2=36$ which is perpendicular to $3 x+4 y$

= 17 is

A
y = 4x + 6
B
3y + 4x = 6
✓
3y = 4x + 6√5
D
3y = x + 25

Answer

Correct option: C.

3y = 4x + 6√5

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MCQ 91 Mark

The equation of the ellipse is $16 x^2+25 y^2=400$. The equations of the tangents making an

angle of 180° with the major axis are

A
x = 4
✓
y = ±4
C
x = -4
D
x = ±5

Answer

Correct option: B.

y = ±4

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MCQ 101 Mark

If the line $4 x-3 y+k=0$ touches the ellipse $5 x^2+9 y^2=45$, then the value of $k$ is

A
21
✓
±3√21
C
3
D
3(21)

Answer

Correct option: B.

±3√21

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MCQ 111 Mark

The equation of the ellipse having eccentricity $\frac{\sqrt{3}}{2}$ and passing through $(-8,3)$ is

A
$4 x^2+y^2=4$
✓
$x^2+4 y^2=100$
C
$4 x^2+y^2=100$
D
$x^2+4 y^2=4$

Answer

Correct option: B.

$x^2+4 y^2=100$

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MCQ 121 Mark

The equation of the ellipse having one of the foci at $(4,0)$ and eccentricity $\frac{1}{3}$ is

A
$9 x^2+16 y^2=144$
B
$144 x^2+9 y^2=1296$
✓
$128 x^2+144 y^2=18432$
D
$144 x^2+128 y^2=18432$

Answer

Correct option: C.

$128 x^2+144 y^2=18432$

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MCQ 131 Mark

The eccentricity of rectangular hyperbola is

A
$\frac{1}{2}$
B
$\frac{1}{2^{\frac{1}{2}}}$
✓
$2^{\frac{1}{2}}$
D
$\frac{1}{3^{\frac{1}{2}}}$

Answer

Correct option: C.

$2^{\frac{1}{2}}$

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MCQ 141 Mark

If the parabola $y ^2=4$ ax passes through $(3,2)$, then the length of its latus rectum is

A
$\frac{2}{3}$
✓
$\frac{4}{3}$
C
$\frac{1}{3}$
D
$4$

Answer

Correct option: B.

$\frac{4}{3}$

$\frac{4}{3}$Length of latus rectum = 4a The given parabola passes through $(3, 2)$
$\therefore(2)^2=4 a(3)$
$ \therefore 4 a=\frac{4}{3}$

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MCQ 151 Mark

The equation of the parabola having (2, 4) and (2, -4) as end points of its latus rectum is ________

A
$y^2=4 x$
✓
$y^2=8 x$
C
$y^2=-16 x$
D
$x^2=8 y$

Answer

Correct option: B.

$y^2=8 x$

The given points lie in the 1st and 4th quadrants.

$\therefore$ Equation of the parabola is $y ^2=4 ax$

End points of latus rectum are (a, 2a) and (a, -2a)

∴ a = 2

∴ required equation of parabola is y = 8x

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MCQ 161 Mark

If $P \left(\frac{\pi}{4}\right)$ is any point on the ellipse $9 x^2+25 y^2=225, S$ and $S ^{\prime}$ are its foci, then $SP . S ^{\prime} P =$________

A
$13$
B
$14$
✓
$17$
D
$19$

Answer

Correct option: C.

$17$

$17$
$9 x^2+25 y^2=225$
$ \frac{x^2}{25}+\frac{y^2}{9}=1$
Here, $a = 5, b = 3$
$\text { Eccentricity }(e)=\frac{4}{5} $
$\therefore \frac{ a }{ e }=\frac{5}{\left(\frac{4}{5}\right)}=\frac{25}{4}$
Coordinates of foci are S(4, 0) and S'(-4, 0)
P(θ) = (a cos θ, b sin θ)
$\therefore \quad P \left(\frac{\pi}{4}\right)=\left(5 \cos \frac{\pi}{4}, 3 \sin \frac{\pi}{4}\right)=\left(\frac{5}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right)$.
$ PM =\frac{25}{4}-\frac{5}{\sqrt{2}}=\frac{25-10 \sqrt{2}}{4} $
$ SP = ePM =\frac{4}{5}\left(\frac{25-10 \sqrt{2}}{4}\right)=5-2 \sqrt{2}$
$\text { Similarly, } S^{\prime} P=5+2 \sqrt{2}$
$\therefore \quad \text { SP.S'P }=(5-2 \sqrt{2})(5+2 \sqrt{2})=25-8=17$

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MCQ 171 Mark

The area of the triangle formed by the lines joining the vertex of the parabola $x^2=12 y$ to the endpoints of its latus rectum is ________

A
$22$ sq. units
B
$20$ sq. units
✓
$18$ sq. units
D
$14$ sq. units

Answer

Correct option: C.

$18$ sq. units

$18$ sq. units
$x^2=12 y $
$ 4 b=12 $
$ b=3$

Area of triangle $=\frac{1}{2} \times A B \times O S$
$=\frac{1}{2} \times 4 a \times a $
$ =\frac{1}{2} \times 12 \times 3$
$ =18 sq . \text { units }$

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MCQ 181 Mark

Equation of the parabola with vertex at the origin and directrix with equation x + 8 = 0 is ________

A
$y^2=8 x$
✓
$y^2=32 x$
C
$y^2=16 x$
D
$x^2=32 y$

Answer

Correct option: B.

$y^2=32 x$

Since directrix is parallel to Y-axis,

The X-axis is the axis of the parabola.

Let the equation of parabola be $y^2=4 a x$.

Equation of directrix is x + 8 = 0

∴ a = 8

$\therefore$ required equation of parabola is $y ^2=32 x$

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MCQ 191 Mark

The end points of latus rectum of the parabola $y^2=24 x$ are

✓
(6, ±12)
B
(12, ±6)
C
(6, ±6)
D
none of these

Answer

Correct option: A.

(6, ±12)

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MCQ 201 Mark

The co-ordinates of a point on the parabola $y ^2=8 x$ whose focal distance is 4 are

A
$\left(\frac{1}{2}, \pm 2\right)$
B
$(1, \pm 2 \sqrt{2})$
✓
$(2, \pm 4)$
D
none of these

Answer

Correct option: C.

$(2, \pm 4)$

(2, ±4)

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MCQ 212 Marks

The equation of the director circle of the hyperbola $\frac{x^2}{16}-\frac{y^2}{4}=1$ is given by

A
$x^2+y^2=16$
B
$x^2+y^2=4$
C
$x^2+y^2=20$
✓
$x^2+y^2=12$

Answer

Correct option: D.

$x^2+y^2=12$

(D)
Equation of 'director-circle' of hyperbola is
$x^2+y^2= a ^2- b ^2$.
Here $a^2=16, b^2=4$
$\therefore x^2+y^2=12$ is the required 'director circle'

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MCQ 222 Marks

If the straight line $x \cos \alpha+y \sin \alpha= p$ be a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then

A
$a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha=p^2$
✓
$a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2$
C
$a^2 \sin ^2 \alpha+b^2 \cos ^2 \alpha=p^2$
D
$a^2 \sin ^2 \alpha-b^2 \cos ^2 \alpha=p^2$

Answer

Correct option: B.

$a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2$

(B)
$\begin{array}{l}x \cos \alpha+y \sin \alpha=p \\
\Rightarrow y=-\cot \alpha \cdot x+p \operatorname{cosec} \alpha\end{array}$
It is tangent to the hyperbola $\frac{x^2}{ a ^2}-\frac{y^2}{b^2}=1$
Therefore, $p ^2 \operatorname{cosec}^2 \alpha= a ^2 \cot ^2 \alpha- b ^2$
$\Rightarrow a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2$

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MCQ 232 Marks

The equation of the locus of the point intersection of the tangents to the hyperbola $\frac{x^2}{36}-\frac{y^2}{25}=1$, such that, the product of their slopes is 2 is

A
$x^2-2 y^2=90$
B
$2 x^2-y^2=104$
✓
$2 x^2-y^2=97$
D
$x^2-2 y^2=70$

Answer

Correct option: C.

$2 x^2-y^2=97$

(C)
Equation of a tangent with slope $m$ to the hyperbola $\frac{x^2}{ a ^2}-\frac{y^2}{b^2}-1$ is
$y=m x \pm \sqrt{a^2 m^2-b^2}$
$\Rightarrow y=m x \pm \sqrt{36 m^2-25}$
$\Rightarrow(y-m x)^2=36 m^2-25$
$\Rightarrow m^2\left(x^2-36\right)-2 m x y+\left(y^2+25\right)=0 \quad\ldots(i)$
Given, product $m _1 m _2=2$
From (i),
$\frac{y^2+25}{x^2-36}=2 $
$\therefore y^2+25=2 x^2-72$
$\therefore 2 x^2-y^2=97$ which is the equation of locus of point of intersection.

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MCQ 242 Marks

Equation of tangents drawn from point $(-2, -1)$, to the hyperbola $2 x^2-3 y^2=6$ are

A
3x - y + 7 = 0 and x - y + 1 = 0
B
$3 x \pm y+7=0$
✓
3x - y + 5 = 0 and x - y + 1 = 0
D
3y - x - 7 = 0 and x - y + 7 = 0

Answer

Correct option: C.

3x - y + 5 = 0 and x - y + 1 = 0

(C)
$\begin{array}{l}\frac{x^2}{3}-\frac{y^2}{2}=1 \text { is hyperbola }
\\ \therefore a^2=3, b^2=2\end{array}$
Any tangent is $y= mx \pm \sqrt{ a ^2 m^2- b ^2}$
It passes through $(-2,-1)$
$\therefore -1=-2 m \pm \sqrt{a^2 m^2-b^2}$
$\begin{array}{l}\therefore(2 m-1)^2=3 m^2-2 \\ \therefore m^2-4 m+3=0 \\ \therefore m=3 \text { or } 1 \\ \therefore \text { tangents are }\end{array}$
$\begin{array}{l}3 x-y \pm 5=0 \\ \text {and } x-y \pm 1=0\end{array}$

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MCQ 252 Marks

Find the equation of axis of the given hyperbola $\frac{x^2}{3}-\frac{y^2}{2}=1$ which is equally inclined to the axes?

✓
y = x + 1
B
y = x - 1
C
y = x + 2
D
y = x - 2

Answer

Correct option: A.

y = x + 1

(A)
We have, $\frac{x^2}{3}-\frac{y^2}{2}=1$
Since, equation of tangent are equally inclined to the axis i.e., $\tan \theta=1=m$.
Equation of tangent in slope form is
$y= m x+\sqrt{ a ^2 m^2- b ^2}$
Here, $a ^2=3, b^2=2$
$\therefore y=1 \cdot x+\sqrt{3 \times(1)^2-2}$
$\Rightarrow y=x+1$

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MCQ 262 Marks

The equation of the tangents to the hyperbola $3 x^2-4 y^2=12$ which cuts equal intercepts from the axes are

A
$y+x= \pm 2$
✓
$y-x= \pm 1$
C
$x-y= \pm 3$
D
$x+y= \pm 4$

Answer

Correct option: B.

$y-x= \pm 1$

(B)
The tangent at $( h , k )$ is $\frac{x}{\frac{4}{h}}-\frac{y}{\frac{3}{k}}=1$
Given condition
$\Leftrightarrow \frac{4}{h}=\frac{3}{k} \Rightarrow \frac{h}{k}=\frac{4}{3} \quad\ldots(i)$
and $3 h^2-4 k ^2=12 \quad\ldots(ii)$
As point ( $h , k$ ) lies on it, using (i) and (ii), we get the tangents as $y-x= \pm 1$.

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MCQ 272 Marks

A line parallel to the straight line $2 x-y=0$ is tangent to the hyperbola $\frac{x^2}{4}-\frac{y^2}{2}=1$ at the point $\left(x_1, y_1\right)$. Then $x_1^2+5 y_1^2$ is equal to

A
10
B
5
✓
6
D
8

Answer

Correct option: C.

(C)
The equation of the tangent at $\left(x_1, y_1\right)$ is
$\frac{x x_1}{4}-\frac{y y_1}{2}=1$
Slope of tangent $=\frac{x_1}{2 y_1}$
Slope of the line $2 x-y=0$ is 2
$\Rightarrow \frac{x_1}{2 y_1}=2 \Rightarrow x_1=4 y_1 \quad\ldots(i)$
$\left(x_1, y_1\right)$ lies on the hyperbola $\frac{x^2}{4}-\frac{y^2}{2}=1$
$\Rightarrow \frac{x_1^2}{4}-\frac{y_1^2}{2}=1$
$\begin{array}{l}\Rightarrow \frac{\left(4 y_1\right)^2}{4}-\frac{y_1^2}{2}=1 \quad \ldots[\text { From (i) }] \\ \Rightarrow y_1^2=\frac{2}{7} \\ x_1^2+5 y_1^2=\left(4 y_1\right)^2+5 y_1^2=21 y_1^2=21\left(\frac{2}{7}\right)=6\end{array}$

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MCQ 282 Marks

The equation of the tangent to the hyperbola $2 x^2-3 y^2=6$ which is parallel to the line $y=3 x+4$, is

A
y = 3x + 5 only
B
y = 3x - 5 only
✓
y = 3x + 5 and y = 3x - 5
D
None of these

Answer

Correct option: C.

y = 3x + 5 and y = 3x - 5

(C)
Let tangent be $y=3 x+ c$
$\begin{array}{l}c= \pm \sqrt{a^2 m^2-b^2}= \pm \sqrt{3(9)-2}= \pm 5 \\\Rightarrow y=3 x \pm 5\end{array}$

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MCQ 292 Marks

The equation of the tangents to the conic $3 x^2-y^2=3$ perpendicular to the line $x+3 y=2$ is

✓
$y=3 x \pm \sqrt{6}$
B
$y=6 x \pm \sqrt{3}$
C
$y=x \pm \sqrt{6}$
D
$y=3 x \pm 6$

Answer

Correct option: A.

$y=3 x \pm \sqrt{6}$

(A)
Slope of $x+3 y-2=0$ is $-\frac{1}{3}$.
$\therefore$ Slope of required tangent $=3$
Tangent to $\frac{x^2}{1}-\frac{y^2}{3}=1$ and perpendicular to
$x+3 y-2=0$ is given by
$y=3 x \pm \sqrt{9-3}=3 x \pm \sqrt{6}$

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MCQ 302 Marks

Let the eccentricity of the hyperbola $\frac{x^2}{ a ^2}-\frac{y^2}{b^2}=1$ be the reciprocal to that of the ellipse $x^2+4 y^2=4$. If the hyperbola passes through a focus of the ellipse, then a focus of the hyperbola is at

✓
(2,0)
B
(0,2)
C
(3,0)
D
(0,3)

Answer

Correct option: A.

(2,0)

(A)
The equation of the ellipse is $\frac{x^2}{4}+\frac{y^2}{1}=1$
Let e be its eccentricity.
Then, $e =\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$
The foci of the ellipse are $S(\sqrt{3}, 0)$ and $S ^{\prime}(-\sqrt{3}, 0)$.
Eccentricity of the hyperbola $=\frac{1}{ e }=\frac{2}{\sqrt{3}}$
$\therefore b^2=a^2\left(\frac{4}{3}-1\right)=\frac{a^2}{3}$
The hyperbola passes through $S (\sqrt{3}, 0)$.
$\therefore \frac{3}{a^2}-0=1 \Rightarrow a^2=3 \Rightarrow a=\sqrt{3}$
$\therefore$ the co-ordinates of the foci of hyperbola are $( \pm 2,0)$.

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MCQ 312 Marks

Let the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be reciprocal to that of the ellipse $x^2+9 y^2=9$, then the ratio $a^2: b^2$ equals

✓
(8 : 1)
B
(1 : 8)
C
(9 : 1)
D
(1 : 9)

Answer

Correct option: A.

(8 : 1)

(A)
Eccentricity of ellipse $x^2+9 y^2=9$
i.e. $\frac{x^2}{9}+\frac{y^2}{1}=1$
$\left(e_1\right)^2=1-\frac{1}{9}=\frac{8}{9}$
Now, eccentricity of hyperbola $\frac{x^2}{ a ^2}-\frac{y^2}{b^2}=1$ is,
$\begin{aligned}& \left(e_2\right)^2=1+\frac{b^2}{a^2}=\frac{9}{8} \\\therefore & \frac{b^2}{a^2}=\frac{9}{8}-1 \\\therefore & \frac{a^2}{b^2}=\frac{8}{1}\end{aligned}$

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MCQ 322 Marks

If the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide, then the value of $b^2$ is

A
1
B
5
✓
7
D
9

Answer

Correct option: C.

(C)
Hyperbola is $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$
$\begin{array}{l}a=\sqrt{\frac{144}{25}}, b=\sqrt{\frac{81}{25}} \\\therefore e_1=\sqrt{1+\frac{81}{144}}=\sqrt{\frac{225}{144}}=\frac{15}{12}=\frac{5}{4} \\
\therefore f o c i=\left(a e_1, 0\right)=\left(\frac{12}{5} \times \frac{5}{4}, 0\right)=(3,0)\end{array}$
$\therefore$ focus of ellipse $=(4 e , 0)$
Since focus of ellipse and hyperbola is same
$\begin{array}{l}\therefore(4 e, 0)=(3,0) \\\Rightarrow e=\frac{3}{4}\end{array}$
Hence, $b^2=16\left(1-\frac{9}{16}\right)=7$

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MCQ 332 Marks

$x^2-4 y^2-2 x+16 y-40=0$ represents

A
a pair of straight lines
B
an ellipse
✓
a hyperbola
D
a parabola

Answer

Correct option: C.

a hyperbola

(C)
$\begin{array}{l}x^2-2 x-4 y^2+16 y-40=0 \\ \Rightarrow\left(x^2-2 x\right)-4\left(y^2-4 y\right)-40=0 \\ \Rightarrow(x-1)^2-1-4\left[(y-2)^2-4\right]-40=0 \\ \Rightarrow(x-1)^2-4(y-2)^2=25 \\ \Rightarrow \frac{(x-1)^2}{25}-\frac{(y-2)^2}{\frac{25}{4}}=1, \text { which is a hyperbola. }\end{array}$

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MCQ 342 Marks

If t is a parameter, then $x= a \left( t +\frac{1}{ t }\right), y= b \left( t -\frac{1}{ t }\right)$ represents

A
an ellipse
B
a circle
C
a pair of straight lines
✓
a hyperbola

Answer

Correct option: D.

a hyperbola

(D)
$\begin{array}{l}\text { Given, } x=a\left(t+\frac{1}{t}\right) \\\Rightarrow \frac{x}{a}=t+\frac{1}{t} \quad\ldots(i) \\\text { and } y=b\left(t-\frac{1}{t}\right) \\\Rightarrow \frac{y}{b}=t-\frac{1}{t} \quad\ldots(ii)\end{array}$
Squaring and subtracting equation (ii) from (i), we get
$\begin{array}{l}\frac{x^2}{a^2}-\frac{y^2}{b^2}=t^2+2+\frac{1}{t^2}-t^2+2-\frac{1}{t^2} \\\Rightarrow \frac{x^2}{a^2}-\frac{y^2}{b^2}=4\end{array}$
which represents a hyperbola.

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MCQ 352 Marks

The equation $\frac{x^2}{12-k}+\frac{y^2}{8-k}=1$ represents

A
a hyperbola if k < 8
B
an ellipse if k > 8
✓
a hyperbola if 8 < k < 12
D
none of these

Answer

Correct option: C.

a hyperbola if 8 < k < 12

(C)
$\begin{array}{l}\frac{x^2}{12-k}+\frac{y^2}{8-k}=1 \\\Rightarrow \frac{x^2}{12-k}-\frac{y^2}{k-8}=1 \\\therefore12>k \text { and } k>8 \\\Rightarrow 8 < k < 12\end{array}$
$\therefore$ the given equation represents a hyperbola, if $8 < k< 12$.

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MCQ 362 Marks

If $\frac{x^2}{36}-\frac{y^2}{k^2}=1$ is a hyperbola, then which of the following statements can be true?

A
(-3,1) lies on the hyperbola
B
(3, 1) lies on the hyperbola
✓
(10, 4) lies on the hyperbola
D
(5, 2) lies on the hyperbola

Answer

Correct option: C.

(10, 4) lies on the hyperbola

(C)
$\frac{x^2}{36}-\frac{y^2}{ k ^2}=1$ is a hyperbola $\Rightarrow k ^2>0$
Now, $\frac{y^2}{ k ^2}=\frac{x^2}{36}-1=\frac{x^2-36}{36}$
$\begin{array}{l}\Rightarrow k^2=\frac{36 y^2}{x^2-36}>0 \Rightarrow x^2-36>0 \\\Rightarrow x^2>36\end{array}$
This is true only for point $(10,4)$.
$\therefore(10,4)$ lies on given hyperbola

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MCQ 372 Marks

The locus of the point of intersection of the lines $ax$ $\sec \theta+b y \tan \theta=a$ and $a x \tan \theta+b y \sec \theta=b$, where $\theta$ is the parameter, is

A
a straight line
B
a circle
C
an ellipse
✓
a hyperbola

Answer

Correct option: D.

a hyperbola

(D)
Squaring and subtracting, we get $a^2 x^2-b^2 y^2=a^2-b^2$, which is the equation of hyperbola.

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MCQ 382 Marks

If $e$ and $e^{\prime}$ are eccentricities of two conics of same type and $e ^2+ e ^{\prime 2}=3$, then they must be

A
Parabola
B
Ellipse
✓
Hyperbola
D
None of these

Answer

Correct option: C.

Hyperbola

(C)
For ellipse, $e <1$ and also $e ^{\prime}<1$
$\therefore e^2+e^{\prime 2}<2$
For parabola, $e =1$ and $e ^{\prime}=1$
$\therefore e^2+e^{\prime 2}=2$
For hyperbola, $e >1$ and $\therefore e^{\prime}>1$
$\therefore e^2+e^{\prime 2}>2$
Hence, it can be 3 in case of hyperbola.

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MCQ 392 Marks

The distance between the directrices of a rectangular hyperbola is 10 units, then distance hetween its foci is

A
$10 \sqrt{2}$
B
5
C
$5 \sqrt{2}$
✓
20

Answer

Correct option: D.

(D)
Since distance between directrices $=\frac{2 a}{e}$ and
eccentricity of rectangular hyperbola $=\sqrt{2}$.
$\therefore$ Distance between directrices $=\frac{2 a }{\sqrt{2}}$
Given, $\frac{2 a }{\sqrt{2}}=10$
$\Rightarrow 2 a=10 \sqrt{2}$
Now, distance between foci $=2 ae$
$\begin{array}{l}=(10 \sqrt{2})(\sqrt{2}) \\=20\end{array}$

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MCQ 402 Marks

If $e$ and $e^{\prime}$ are the eccentricities of the ellipse $5 x^2+9 y^2=45$ and the hyperbola $5 x^2-4 y^2=45$, then $ee ^{\prime}=$

A
9
B
5
C
-4
✓
1

Answer

Correct option: D.

(D)
$e ^2=1-\frac{ b ^2}{ a ^2}=\frac{4}{9}$
$\Rightarrow e =\frac{2}{3}$
$\text {and } e ^{\prime 2}=1+\frac{ b ^2}{ a ^2}=1+\frac{\frac{45}{4}}{9}=\frac{9}{4}$
$ \Rightarrow e ^{\prime}=\frac{3}{2}$
$\therefore ee^{\prime}=1$

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MCQ 412 Marks

If the eccentricities of the hyperbolas $\frac{x^2}{ a ^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{b^2}-\frac{x^2}{ a ^2}=1$ be e and $e_1$, then $\frac{1}{e^2}+\frac{1}{e^2}=$

✓
1
B
2
C
3
D
4

Answer

Correct option: A.

(A)
Eccentricity of hyperbola $\frac{x^2}{ a ^2}-\frac{y^2}{b^2}=1$ is
$\begin{array}{l}e=\sqrt{1+\frac{b^2}{a^2}} \\\Rightarrow e^2=\frac{a^2+b^2}{a^2} \quad\ldots(i)\end{array}$
Eccentricity of hyperbola $\frac{y^2}{b^2}-\frac{x^2}{ a ^2}=1$ is
$\begin{array}{l} e_1=\sqrt{1+\frac{a^2}{b^2}} \\\Rightarrow e_1^2=\frac{b^2+a^2}{b^2}\quad\ldots(ii)\end{array}$
From (i) and (ii), we get
$\frac{1}{e_1^2}+\frac{1}{e^2}=1$

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MCQ 422 Marks

The eccentricity of the hyperooia whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is

A
$\frac{4}{3}$
B
$\frac{4}{\sqrt{3}}$
✓
$\frac{2}{\sqrt{3}}$
D
$\sqrt{3}$

Answer

Correct option: C.

$\frac{2}{\sqrt{3}}$

(C)
Given length of $L R=8$
$\Rightarrow \frac{2 b^2}{a}=8$
Also, $2 b=\frac{1}{2}(2 ae )$
$\begin{array}{l}\Rightarrow 4 b^2=a^2 e^2 \Rightarrow 4 a^2\left(e^2-1\right)=a^2 e^2 \\\Rightarrow 4 e^2-e^2=4 \\\Rightarrow e=\frac{2}{\sqrt{3}}\end{array}$

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MCQ 432 Marks

A hyperbola, centred at the origin, has transverse axis 2 a . If it passes through a given point $\left(x_1, y_1\right)$, then its eccentricity is

A
$\sqrt{\frac{x_1^2-y_1^2- a ^2}{x_1^2-y_1^2}}$
✓
$\sqrt{\frac{ a ^2-x_1^2-y_1^2}{ a ^2-x_1^2}}$
C
$\sqrt{\frac{ a ^2+x_1^2+y_1^2}{ a ^2-x_1^2}}$
D
None of these

Answer

Correct option: B.

$\sqrt{\frac{ a ^2-x_1^2-y_1^2}{ a ^2-x_1^2}}$

(B)
Equation of hyperbola passes through
$\begin{array}{l}\left(x_1, y_1\right) \\\Rightarrow \frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1 \Rightarrow \frac{x_1^2}{a^2}-1=\frac{y_1^2}{b^2} \\
\Rightarrow \frac{x_1^2-a^2}{a^2}=\frac{y_1^2}{b^2} \Rightarrow \frac{b^2}{a^2}=\frac{y_1^2}{x_1^2-a^2}\end{array}$
Now, $\frac{b^2}{a^2}=e^2-1$
$\begin{array}{l}\Rightarrow \frac{y_1^2}{x_1^2-a^2}=e^2-1 \\
\Rightarrow e^2=\frac{y_1^2+\left(x_1^2-a^2\right)}{x_1^2-a^2} \\
\Rightarrow e=\sqrt{\frac{x_1^2-a^2+y_1^2}{x_1^2-a^2}}=\sqrt{\frac{a^2-x_1^2-y_1^2}{a^2-x_1^2}}\end{array}$

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MCQ 442 Marks

The latus rectum of the hyperbola $9 x^2-16 y^2+72 x-32 y-16=0$ is

✓
$\frac{9}{2}$
B
$-\frac{9}{2}$
C
$\frac{32}{3}$
D
$-\frac{32}{3}$

Answer

Correct option: A.

$\frac{9}{2}$

(A)
Given, equation of hyperbola is
$\begin{array}{l}9 x^2-16 y^2+72 x-32 y-16=0 \\\Rightarrow 9\left(x^2+8 x\right)-16\left(y^2+2 y\right)-16=0 \\\Rightarrow 9(x+4)^2-16(y+1)^2=144 \\\Rightarrow \frac{(x+4)^2}{16}-\frac{(y+1)^2}{9}=1\end{array}$
$\therefore \text { Latus rectum }=\frac{2 b^2}{a}=2 \times \frac{9}{4}=\frac{9}{2}$

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MCQ 452 Marks

The equation of the directrices of the conic $x^2+2 x-y^2+5=0$ are

A
$x= \pm 1$
B
$y= \pm 2$
✓
$y= \pm \sqrt{2}$
D
$x= \pm \sqrt{3}$

Answer

Correct option: C.

$y= \pm \sqrt{2}$

(C)
$\begin{array}{l}(x+1)^2-y^2-1+5=0 \\\Rightarrow-\frac{(x+1)^2}{4}+\frac{y^2}{4}=1\end{array}$
Equation of directrices of $\frac{y^2}{b^2}-\frac{x^2}{ a ^2}=1$ are
$y= \pm \frac{b}{e}$
Here, $b =2, e =\sqrt{1+1}=\sqrt{2}$
Hence, $y= \pm \frac{2}{\sqrt{2}}\$\Rightarrow y= \pm \sqrt{2}$

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MCQ 462 Marks

The vertices of the hyperbola $9 x^2-16 y^2-36 x+96 y-252=0$ are

A
(6, 3) and (-6, 3)
✓
(6, 3) and (-2, 3)
C
(-6, 3) and (-6, -3)
D
(3, 6) and (-3, 2)

Answer

Correct option: B.

(6, 3) and (-2, 3)

(B)
$\begin{array}{l}9 x^2-16 y^2-36 x+96 y-252=0 \\\Rightarrow \frac{(x-2)^2}{16}-\frac{(y-3)^2}{9}=1 \\\Rightarrow \frac{X^2}{16}-\frac{Y^2}{9}=1\end{array}$
$\therefore$ Vertices are ( $X = \pm a , Y =0$ )
i.e., $(x-2= \pm 4, y-3=0)$
$\therefore$ The vertices of the hyperbola are $(6,3)$ and $(-2,3)$

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MCQ 472 Marks

The equation of a hyperbola with foci at (6,5) and (-4,5) and eccentricity $=\frac{5}{4}$ is

✓
$\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=1$
B
$\frac{(x-5)^2}{16}-\frac{(y-5)^2}{9}=1$
C
$\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=-1$
D
$\frac{(x-1)^2}{9}-\frac{(y-5)^2}{16}=1$

Answer

Correct option: A.

$\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=1$

(A)
Centre of the hyperbola is midpoint of foci.
Hence, its centre is $(1,5)$.
Also, distance between foci is 2ae = 10
$\Rightarrow a=4 \quad\ldots[\because e =\frac{5}{4}]$
$\Rightarrow a^2=16$
$\text {Now,}b^2 =a^2\left(e^2-1\right)$
$=a^2 e^2-a^2=25-16 \Rightarrow b^2=9$
Hence, equation of hyperbola is
$\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=1$

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MCQ 482 Marks

The equation of the hyperbola with vertices ( $0, \pm 15$ ) and foci $(0, \pm 20)$ is

A
$\frac{x^2}{175}-\frac{y^2}{225}=1$
B
$\frac{x^2}{625}-\frac{y^2}{125}=1$
C
$\frac{y^2}{225}-\frac{x^2}{125}=1$
✓
$\frac{y^2}{225}-\frac{x^2}{175}=1$

Answer

Correct option: D.

$\frac{y^2}{225}-\frac{x^2}{175}=1$

(D)
Vertices $=(0, \pm 15)$, foci $=(0, \pm 20)$
$\therefore b =15 \text { and } b e=20 \Rightarrow e=\frac{4}{3}$
$a^2 =b^2\left(e^2-1\right)$
$=15^2\left(\frac{16}{9}-1\right)$
$=175$
$\therefore$ The equation of hyperbola is
$\frac{-x^2}{175}+\frac{y^2}{225}=1$
$\Rightarrow \frac{y^2}{225}-\frac{x^2}{175}=1$

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MCQ 492 Marks

The equation of hyperbola whose coordinates of the foci are $( \pm 8,0)$ and the length of latus rectum is 24 units, is

✓
$3 x^2-y^2=48$
B
$4 x^2-y^2=48$
C
$x^2-3 y^2=48$
D
$x^2-4 y^2=48$

Answer

Correct option: A.

$3 x^2-y^2=48$

(A)
Given, $ae =8$ and $\frac{2 b^2}{ a }=24$
$\Rightarrow b ^2=12 a$
Now, $b^2=a^2\left(e^2-1\right)$
$\Rightarrow 12 a=a^2 e^2-a^2$
$\Rightarrow 12 a=64-a^2$
$\Rightarrow a^2+12 a-64=0$
$\Rightarrow a=4 \quad\ldots[\because a>0]$
$\therefore b^2=12(4)=48$
$\therefore $ The equation of hyperbola is
$\frac{x^2}{16}-\frac{y^2}{48}=1 \Rightarrow 3 x^2-y^2=48$

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MCQ 502 Marks

The distance between the foci of a hyperbola is 16 and its eccentricity is $\sqrt{2}$. Its equation is

✓
$x^2-y^2=32$
B
$\frac{x^2}{4}-\frac{y^2}{9}=1$
C
$2 x^2-3 y^2=7$
D
$y^2-x^2=32$

Answer

Correct option: A.

$x^2-y^2=32$

(A)
$\begin{array}{l}\text {Given, } 2 a e=16, e=\sqrt{2} \\ \Rightarrow a e=8, e=\sqrt{2} \\ \therefore a=4 \sqrt{2}\end{array}$
$\begin{array}{l}\text {Now, } b^2=a^2\left(e^2-1\right) \\\Rightarrow b^2=32(2-1) \\\Rightarrow b^2=32\end{array}$
$\therefore$ The equation of hyperbola is $\frac{x^2}{32}-\frac{y^2}{32}=1$
$\therefore x^2-y^2=32$

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MCQ 512 Marks

The equation of the hyperbola whose foci are (-2,0) and (2,0) and eccentricity is 2 is given by

A
$x^2-3 y^2=3$
✓
$3 x^2-y^2=3$
C
$-x^2+3 y^2=3$
D
$-3 x^2+y^2=3$

Answer

Correct option: B.

$3 x^2-y^2=3$

(B)
Given, $ae =2, e =2$
$\therefore$ (a=1)
Now, $b ^2= a ^2\left( e ^2-1\right)$
$\begin{array}{l}\Rightarrow b^2=1(4-1) \\\Rightarrow b^2=3\end{array}$
$\therefore$ The equation of hyperbola is
$\begin{array}{l}\frac{x^2}{1}-\frac{y^2}{3}=1 \\\Rightarrow 3 x^2-y^2=3\end{array}$

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MCQ 522 Marks

The equation of the hyperbola whose conjugent axis is 5 and the distance between the foci is 13 , is

✓
$25 x^2-144 y^2=900$
B
$144 x^2-25 y^2=900$
C
$144 x^2+25 y^2=900$
D
$25 x^2+144 y^2=900$

Answer

Correct option: A.

$25 x^2-144 y^2=900$

(A)
Conjugate axis is 5 and distance between $foci =13$
$\begin{array}{l}\Rightarrow 2 b=5 \text { and } 2 ae=13 \\\text { Also, } b^2=a^2\left(e^2-1\right) \\\Rightarrow \frac{25}{4}=\frac{(13)^2}{4 e^2}\left(e^2-1\right) \\\Rightarrow \frac{25}{4}=\frac{169}{4}-\frac{169}{4 e^2} \\\Rightarrow e=\frac{13}{12} \\\Rightarrow a=6, b=\frac{5}{2}\end{array}$
Hence, the required equation of hyperbola is
$\frac{x^2}{36}-\frac{y^2}{\frac{25}{4}}=1$
i.e., $25 x^2-144 y^2=900$

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MCQ 532 Marks

If the latus rectum of an hyperbola be 8 and eccentricity be $\frac{3}{\sqrt{5}}$, then the equation of the hyperbola is

✓
$4 x^2-5 y^2=100$
B
$5 x^2-4 y^2=100$
C
$4 x^2+5 y^2=100$
D
$5 x^2+4 y^2=100$

Answer

Correct option: A.

$4 x^2-5 y^2=100$

(A)
$\begin{array}{l}\text {Given, } \frac{2 b^2}{a}=8 \text { and } \frac{3}{\sqrt{5}}=\sqrt{1+\frac{b^2}{a^2}} \\\Rightarrow \frac{4}{5}=\frac{b^2}{a^2} \\\Rightarrow a=5, b=2 \sqrt{5}\end{array}$
Hence, the required equation of hyperbola is
$\begin{array}{l}\frac{x^2}{25}-\frac{y^2}{20}=1 \\\text { i.e., } 4 x^2-5 y^2=100\end{array}$

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MCQ 542 Marks

Equation of the hyperbola with eccentricity $\frac{3}{2}$ and foci $( \pm 2,0)$ is

✓
$\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}$
B
$\frac{x^2}{9}-\frac{y^2}{9}=\frac{4}{9}$
C
$\frac{x^2}{4}-\frac{y^2}{9}=1$
D
$\frac{x^2}{5}-\frac{y^2}{9}=1$

Answer

Correct option: A.

$\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}$

(A)
$\begin{array}{l}\text {Given that, } e =\frac{3}{2} \\ foci =( \pm 2,0)=( \pm ae , 0) \\ \Rightarrow ae =2 \\ \Rightarrow a =\frac{4}{3} \\ \Rightarrow a ^2=\frac{16}{9}\end{array}$
Now, condition for eccentricity is
$\begin{aligned}b^2 & =a^2\left(e^2-1\right) \\ \therefore b^2 & =\frac{16}{9}\left(\frac{9}{4}-1\right)=\frac{16}{9}\left(\frac{5}{4}\right)=\frac{20}{9}\end{aligned}$
Now, equation of hyperbola is $\frac{x^2}{ a ^2}-\frac{y^2}{b^2}=1$
$\begin{array}{l}\Rightarrow \frac{9 x^2}{16}-\frac{9 y^2}{20}=1 \\\Rightarrow \frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}\end{array}$

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MCQ 552 Marks

If the centre, vertex and focus of a hyperbola be (0, 0), (4, 0) and (6, 0) respectively, then the equation of the hyperbola is

A
$4 x^2-5 y^2=8$
B
$4 x^2-5 y^2=80$
✓
$5 x^2-4 y^2=80$
D
$5 x^2-4 y^2=8$

Answer

Correct option: C.

$5 x^2-4 y^2=80$

(C)
Centre $(0,0)$, vertex $(4,0) \Rightarrow a =4$ and focus $(6,0)$
$\Rightarrow ae =6$
$\Rightarrow e =\frac{3}{2}$
Also, $b^2 =a^2\left(e^2-1\right)$
$=20$
Hence, required equation is $\frac{x^2}{16}-\frac{y^2}{20}=1$
i.e., $5 x^2-4 y^2=80$

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MCQ 562 Marks

The eccentricity of a hyperbola passing through the points $(3,0),(3 \sqrt{2}, 2)$ will be

A
$\sqrt{13}$
✓
$\frac{\sqrt{13}}{3}$
C
$\frac{\sqrt{13}}{4}$
D
$\frac{\sqrt{13}}{2}$

Answer

Correct option: B.

$\frac{\sqrt{13}}{3}$

(B)
Let the equation of hyperbola be $\frac{x^2}{ a ^2}-\frac{y^2}{b^2}=1$
Since hyperbola passes through the points $(3,0),(3 \sqrt{2}, 2)$
$\therefore \quad \frac{9}{a^2}-0=1$ and $\frac{18}{a^2}-\frac{4}{b^2}=1$
$\Rightarrow a^2=9$ and $\frac{4}{b^2}=\frac{18}{a^2}-1$
$\Rightarrow a^2=9$ and $\frac{4}{b^2}=\frac{18}{9}-1$
$\Rightarrow a^2=9$ and $b^2=4$
Now, $e =\sqrt{1+\frac{ b ^2}{ a ^2}}=\sqrt{1+\frac{4}{9}}=\frac{\sqrt{13}}{3}$

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MCQ 572 Marks

The length of the transverse axis of a hyperbola is 7 and it passes through the point $(5,-2)$. The equation of the hyperbola is

A
$\frac{5}{49} x^2-\frac{196}{51} y^2=1$
B
$\frac{49}{4} x^2-\frac{51}{196} y^2=1$
✓
$\frac{4}{49} x^2-\frac{51}{196} y^2=1$
D
$\frac{51}{4} x^2-\frac{49}{196} y^2=1$

Answer

Correct option: C.

$\frac{4}{49} x^2-\frac{51}{196} y^2=1$

(C)
$2 a=7 \Rightarrow a=\frac{7}{2}$
Also $(5,-2)$ satisfies $\frac{4}{49}(25)-\frac{51}{196}(4)=1$
and $a^2=\frac{49}{4}$
$\Rightarrow a =\frac{7}{2}$
$\therefore$ option $( C )$ is the correct answer.

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MCQ 582 Marks

The equations of the tangent to the hyperbola $x^2-\frac{y^2}{4}=1$, having slope $=-3$ are

A
$x+3 y= \pm \sqrt{5}$
B
$x-3 y= \pm \sqrt{3}$
✓
$3 x+y= \pm \sqrt{5}$
D
$3 x-y= \pm 3$

Answer

Correct option: C.

$3 x+y= \pm \sqrt{5}$

(C)
Equation of the hyperbola is $\frac{x^2}{1}-\frac{y^2}{4}=1$
$a ^2=1$ and $b ^2=4$
slope $m=-3$
Equation of the tangent to the hyperbola is
$y=m x \pm \sqrt{a^2 m^2-b^2}$
$y=-3 x \pm \sqrt{1(9)-4}$
$\therefore 3 x+y= \pm \sqrt{5}$

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MCQ 592 Marks

If the line $y=2 x+\lambda$ be a tangent to the hyperbola $36 x^2-25 y^2=3600$, then $\lambda=$

A
16
B
-16
✓
$\pm 16$
D
$+6$

Answer

Correct option: C.

$\pm 16$

(C)
If $y=2 x+\lambda$ is tangent to given hyperbola, then
$\lambda= \pm \sqrt{a^2 m^2-b^2}$
$= \pm \sqrt{(100)(4)-144}$
$= \pm \sqrt{256}= \pm 16$

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MCQ 602 Marks

The line $l x+ my + n =0$ will be a tangent to the hyperbola $\frac{x^2}{ a ^2}-\frac{y^2}{b^2}=1$, if

A
$a^2 l^2+b^2 m^2=n^2$
B
$a^2 l^2-b^2 m^2=n^2$
✓
$a m^2-b^2 n^2=a^2 l^2$
D
None of these

Answer

Correct option: C.

$a m^2-b^2 n^2=a^2 l^2$

(C)
Given equation of line is
$\begin{array}{l}l x+m y+n=0 \\\Rightarrow y=\left(\frac{-l}{m}\right) x-\frac{n}{m} \quad\ldots(i)\end{array}$
Since the line $y= m x+ c$ touches the hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \text { if } c^2=a^2 m^2-b^2$
$\therefore$ Line (i) will be a tangent to the hyperbola, if
$\begin{array}{l}\left(\frac{- n }{ m }\right)^2= a ^2\left(-\frac{l}{m}\right)^2- b ^2 \\ \Rightarrow n ^2= a ^2 l^2- b ^2 m^2\end{array}$

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MCQ 612 Marks

The point of contact of the tangent $y=x+2$ to the hyperbola $5 x^2-9 y^2=45$ is

A
$\left(\frac{9}{2}, \frac{5}{2}\right)$
✓
$\left(\frac{5}{2}, \frac{9}{2}\right)$
C
$\left(-\frac{9}{2},-\frac{5}{2}\right)$
D
$\left(-\frac{5}{2}, \frac{-9}{2}\right)$

Answer

Correct option: B.

$\left(\frac{5}{2}, \frac{9}{2}\right)$

(B)
Hyperbola is $\frac{x^2}{9}-\frac{y^2}{5}=1$ and the line is $y=x+2$.
Here, $a^2=9, b^2=5, m=1, c=2$
$\begin{aligned}\text {Point of contact } & \equiv\left(\frac{-9(1)}{2}, \frac{-5}{2}\right)
\\& \equiv\left(-\frac{9}{2}, \frac{-5}{2}\right)\end{aligned}$
Alternate method:
Substitute $y=x+2$ in $5 x^2-9 y^2=45$ to get
$\begin{array}{l}(2 x+9)^2=0
\\\Rightarrow x=-\frac{9}{2} \Rightarrow y=-\frac{5}{2} .\end{array}$
The point of contact is $\left(-\frac{9}{2},-\frac{5}{2}\right)$

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MCQ 622 Marks

The equation of the tangent to the hyperthy $9 x^2-16 y^2=144$ at $(4 \sec \theta, 3 \tan \theta)$ is

A
$3 x \sec \theta+y \tan \theta=1$
B
$x \sec \theta-2 y \tan \theta=1$
✓
$3 x \sec \theta-4 y \tan \theta=12$
D
$3 x \sec \theta-y \tan \theta=4$

Answer

Correct option: C.

$3 x \sec \theta-4 y \tan \theta=12$

(C)
Equation of hyperbola is $9 x^2-16 y^2=144$
$\Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1$
Equation of tangent to $\frac{x^2}{16}-\frac{y^2}{9}=1$ at
$(4 \sec \theta, 3 \tan \theta)$ is
$\frac{4(\sec \theta) x}{16}-\frac{3(\tan \theta) y}{9}=1$
$\therefore\frac{(\sec \theta) x}{4}-\frac{(\tan \theta) y}{3}=1$
$\therefore 3 x \sec \theta-4 y \tan \theta=12$

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MCQ 632 Marks

The equation of the tangent to the hyperbola $4 y^2=x^2-1$ at the point $(1,0)$ is

✓
x = 1
B
y = 1
C
y = 4
D
x = 4

Answer

Correct option: A.

x = 1

(A)
The equation of the tangent to $4 y^2=x^2-1$ at
$\begin{array}{l}(1,0) \text { is } 4(y \times 0)=x \times 1-1
\\\Rightarrow x-1=0 \\\Rightarrow x=1\end{array}$

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MCQ 642 Marks

The auxiliary equation of circle of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, is

✓
$x^2+y^2=a^2$
B
$x^2+y^2= b ^2$
C
$x^2+y^2=a^2+b^2$
D
$x^2+y^2=a^2-b^2$

Answer

Correct option: A.

$x^2+y^2=a^2$

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MCQ 652 Marks

The equation $x=\frac{ e^t+ e^{-t}}{2}, y=\frac{ e^t- e^{-t}}{2}, t \in R$,represents represents

A
an ellipse
B
a parabola
✓
a hyperbola
D
a circle

Answer

Correct option: C.

a hyperbola

(C)
$2 x=e^{t}+e^{-t} \text { and } 2 y=e^{t}-e^{-t}$
$\Rightarrow 4 x^2=e^{2 t}+2+e^{-2 t} \quad\ldots(i)$
$\text {and } 4 y^2=e^{2 t}-2+e^{-2 t} \quad\ldots(ii)$
Subtracting (ii) from (i), we get
$4 x^2-4 y^2=4$
$\Rightarrow x^2-y^2=1$
The equation represents hyperbola.

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MCQ 662 Marks

The eccentricity of the hyperbola $5 x^2-4 y^2+20 x+8 y=4$ is

A
$\sqrt{2}$
✓
$\frac{3}{2}$
C
2
D
3

Answer

Correct option: B.

$\frac{3}{2}$

(B) Given equation of hyperbola is
$\begin{array}{l}5 x^2-4 y^2+20 x+8 y=4 \\\Rightarrow 5\left(x^2+4 x+4\right)-4\left(y^2-2 y+1\right)=4+20-4 \\\Rightarrow 5(x+2)^2-4(y-1)^2=20 \\\Rightarrow \frac{(x+2)^2}{4}-\frac{(y-1)^2}{5}=1 \\ \Rightarrow a^2=4, b^2=5 \\e=\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{4+5}}{2}=\frac{3}{2}\end{array}$

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MCQ 672 Marks

The distance between the directrices of the hyperbola $x=8 \sec \theta, y=8 \tan \theta$ is

A
$16 \sqrt{2}$
B
$\sqrt{2}$
✓
$8 \sqrt{2}$
D
$4 \sqrt{2}$

Answer

Correct option: C.

$8 \sqrt{2}$

(C) Equation of the hyperbola
$x=8 \sec \theta, y=8 \tan \theta$
$\Rightarrow \frac{x}{8}=\sec \theta, \frac{y}{8}=\tan \theta$
$\because \sec ^2 \theta-\tan ^2 \theta=1$
$\Rightarrow \frac{x^2}{8^2}-\frac{y^2}{8^2}=1$
Here, $a=8, b=8$
Here, $a = b$
$\therefore $ it is a rectangular hyperbola
Eccentricity of the rectangular hyperbola $=\sqrt{2}$.
$\Rightarrow e=\sqrt{2}$
$\begin{aligned} \therefore \text {Distance between directrices } & =\frac{2 a }{ e } \\ & =\frac{2 \times 8}{\sqrt{2}} \\ & =8 \sqrt{2}\end{aligned}$

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MCQ 682 Marks

A point on the curve $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$ is

A
$( A \cos \theta, B \sin \theta)$
✓
$( A \sec \theta, B \tan \theta)$
C
$\left( A \cos ^2 \theta, B \sin ^2 \theta\right)$
D
None of these

Answer

Correct option: B.

$( A \sec \theta, B \tan \theta)$

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MCQ 692 Marks

The directrix of the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ is

✓
$x=\frac{9}{\sqrt{13}}$
B
$y=\frac{9}{\sqrt{13}}$
C
$x=\frac{6}{\sqrt{13}}$
D
$y=\frac{6}{\sqrt{13}}$

Answer

Correct option: A.

$x=\frac{9}{\sqrt{13}}$

(A)
$\frac{x^2}{9}-\frac{y^2}{4}=1$
$\Rightarrow a^2=9, b^2=4$
$e=\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{9+4}}{3}=\frac{\sqrt{13}}{3}$
Directrix of hyperbola is $x= \pm \frac{a}{e}$
$\Rightarrow x= \pm \frac{3}{\frac{\sqrt{13}}{3}} \Rightarrow x= \pm \frac{9}{\sqrt{13}}$

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MCQ 702 Marks

The length of transverse axis of the hyperbola $3 x^2-4 y^2=32$ is

✓
$\frac{8 \sqrt{2}}{\sqrt{3}}$
B
$\frac{16 \sqrt{2}}{\sqrt{3}}$
C
$\frac{3}{32}$
D
$\frac{64}{3}$

Answer

Correct option: A.

$\frac{8 \sqrt{2}}{\sqrt{3}}$

(A)
The given equation can be written as
$\begin{array}{l}\frac{x^2}{\frac{32}{3}}-\frac{y^2}{8}=1 \\\Rightarrow \frac{x^2}{\left(\frac{4 \sqrt{2}}{\sqrt{3}}\right)^2}-\frac{y^2}{(2 \sqrt{2})^2}=1 \\\Rightarrow a=\frac{4 \sqrt{2}}{\sqrt{3}}\end{array}$
$\therefore$ Length of transverse axis
$=2 a=2 \times \frac{4 \sqrt{2}}{\sqrt{3}}=\frac{8 \sqrt{2}}{\sqrt{3}}$

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MCQ 712 Marks

The length of the latus rectum of the hyperbola $3 x^2-y^2=4$ is

A
$8 \sqrt{3}$
✓
$4 \sqrt{3}$
C
16
D
32

Answer

Correct option: B.

$4 \sqrt{3}$

(B)
$\text {Length of latus rectum } =\frac{2 b^2}{a}$
$=\frac{2 \times 4}{\frac{2}{\sqrt{3}}}=4 \sqrt{3}$

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MCQ 722 Marks

The latus-rectum of the hyperbola $16 x^2-9 y^2=144$ is

A
$\frac{16}{3}$
✓
$\frac{32}{3}$
C
$\frac{8}{3}$
D
$\frac{4}{3}$

Answer

Correct option: B.

$\frac{32}{3}$

(B)
The given equation of hyperbola is
$\begin{array}{l}16 x^2-9 y^2=144 \Rightarrow \frac{x^2}{9}-\frac{y^2}{16}=1 \\\text { L.R. }=\frac{2 b^2}{a}=\frac{2 \times 16}{3}=\frac{32}{3}\end{array}$

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MCQ 732 Marks

The difference of the focal distance of any point on the hyperbola $9 x^2-16 y^2=144$, is

✓
8
B
7
C
6
D
4

Answer

Correct option: A.

(A)
The hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}=1$
Difference of the focal distances on the hyperbola $=2 a =$ length of transverse axis
Difference of focal distance $=8$

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MCQ 742 Marks

If the length of the transverse and conjugate axes of a hyperbola be 8 and 6 respectively, then the difference of focal distances of any point of the hyperbola will be

✓
8
B
6
C
14
D
2

Answer

Correct option: A.

(A)
$2 a=8,2 b=6$
Difference of the focal distances on the hyperbola $=2 a =$ length of transverse axis
Difference of focal distances of any point of the hyperbola $=8$

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MCQ 752 Marks

If $(4,0)$ and $(-4,0)$ be the vertices and $(6,0)$ and $(-6,0)$ be the foci of a hyperbola, then its eccentricity is

A
$\frac{5}{2}$
B
2
✓
$\frac{3}{2}$
D
$\sqrt{2}$

Answer

Correct option: C.

$\frac{3}{2}$

(C)
$\text {Vertices }( \pm 4,0) \equiv( \pm a, 0)$
$\Rightarrow a=4$
$\text {Foci }( \pm 6,0) \equiv( \pm a e, 0)$
$\Rightarrow e=\frac{6}{4}=\frac{3}{2}$

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MCQ 762 Marks

Eccentricity of hyperbola $\frac{x^2}{ k }+\frac{y^2}{ k ^2}=1( k <0)$ is

A
$\sqrt{1+k}$
B
$\sqrt{1-k}$
C
$\sqrt{1+\frac{1}{k}}$
✓
$\sqrt{1-\frac{1}{k}}$

Answer

Correct option: D.

$\sqrt{1-\frac{1}{k}}$

(D)
$\frac{y^2}{ k ^2}-\frac{x^2}{- k }=1$
$\text { Also }, a ^2= b ^2\left( e ^2-1\right)$
$\Rightarrow- k = k ^2\left( e ^2-1\right)$
$\Rightarrow-\frac{1}{ k }= e ^2-1$
$\Rightarrow e ^2=1-\frac{1}{ k }$
$\Rightarrow e =\sqrt{1-\frac{1}{ k }}$

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MCQ 772 Marks

The eccentricity of the hyperbola $x^2-y^2=25$ is

✓
$\sqrt{2}$
B
$\frac{1}{\sqrt{2}}$
C
2
D
$1+\sqrt{2}$

Answer

Correct option: A.

$\sqrt{2}$

(A)
$\begin{array}{l}\frac{x^2}{25}-\frac{y^2}{25}=1
\\ \text { Eccentricity }=\sqrt{2} \text { as } a = b .\end{array}$

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MCQ 782 Marks

The eccentricity of the hyperbola can never be equal to

A
$\sqrt{\frac{9}{5}}$
✓
$2 \sqrt{\frac{1}{9}}$
C
$2 \sqrt{\frac{1}{3}}$
D
2

Answer

Correct option: B.

$2 \sqrt{\frac{1}{9}}$

(B)
Since e >1 always for hyperbola and $\frac{2}{3} < 1$.

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MCQ 792 Marks

The foci of the hyperbola $9 x^2-16 y^2=144$ are

A
$( \pm 4,0)$
B
$(0, \pm 4)$
✓
$( \pm 5,0)$
D
$(0, \pm 5)$

Answer

Correct option: C.

$( \pm 5,0)$

(C)
The equation of hyperbola is
$\begin{array}{l}9 x^2-16 y^2=144 \\\Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1 \\e=\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{16+9}}{4}=\frac{5}{4}\end{array}$
Hence, foci are $( \pm ae , 0) \equiv\left( \pm 4 \times \frac{5}{4}, 0\right)$
$\equiv( \pm 5,0)$

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MCQ 802 Marks

A tangent to the ellipse $3 x^2+4 y^2=12$ with slope $=\frac{-1}{2}$ intersects X -axis and Y -axis at points A and B . If O is the origin, then the area of triangle OAB is

A
8 sq. units
✓
4 sq. units
C
9 sq. units
D
3 sq. units

Answer

Correct option: B.

4 sq. units

(B)
$3 x^2+4 y^2=12$
$\Rightarrow \frac{x^2}{4}+\frac{y^2}{3}=1$
$\therefore a^2=4, b^2=3$
Equation of tangent in terms of its slope is
$y=m x+\sqrt{a^2 m^2+b^2}$
Since, $m =\frac{-1}{2}$
$\therefore y=-\frac{1}{2} x+\sqrt{4\left(\frac{-1}{2}\right)^2+3}=-\frac{1}{2} x+2$
A is on X -axis, hence put $y=0$
$0=-\frac{1}{2} x+2 \Rightarrow x=4$
$\therefore A(4,0)$ and $O A=4$
B is on Y -axis, thence put $x=0$
$y=\left(-\frac{1}{2}\right) 0+2=2 $
$B(0,2) \text { and } OB=2$
$A(\Delta OAB)=\frac{1}{2} \times 4 \times 2=4 \text { sq. units }$

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MCQ 812 Marks

The locus of the point of intersection of the perpendicular tangents to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is

A
$x^2+y^2=9$
B
$x^2+y^2=4$
✓
$x^2+y^2=13$
D
$x^2+y^2=5$

Answer

Correct option: C.

$x^2+y^2=13$

(C)
The locus of point of intersection of two perpendicular tangents drawn on the ellipse is $x^2+y^2= a ^2+ b ^2$ which is called 'director- circle'.
Given ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1$,
$\therefore$ Locus is $x^2+y^2=13$

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MCQ 822 Marks

The equations of the tangents of the ellipse $9 x^2+16 y^2=144$ which passes through the point (2,3) is

✓
y = 3, x + y = 5
B
y = - 3, x - y = 5
C
y = 4, x + y = 3
D
y = - 4, x - y = 3

Answer

Correct option: A.

y = 3, x + y = 5

(A)
Equation of the ellipse is
$\frac{x^2}{16}+\frac{y^2}{9}=1$
The equation of the tangent to the ellipse is
$y=m x \pm \sqrt{16 m^2+9} \quad\ldots(i)$
The equation of a line passing through $(2,3)$, having slope $m$ is
$\begin{array}{l}(y-3)=m(x-2) \\y-m x=3-2 m \quad\ldots(ii) \end{array}$
Comparing (i) and (ii),
$\frac{1}{1}=\frac{1}{1}=\frac{\sqrt{16 m^2+9}}{3-2 m}$
$\therefore 16 m^2+9=(3-2 m)^2$
$=9+4 m^2-12 m$
$\therefore 12 m^2= -12 m$
$\Rightarrow m= 0 \text { or } m=-1$
$\therefore$ The tangents are $y=3$ and $x+y=5$
Alternate method :
We know, $\left(x_1^2- a ^2\right) m ^2-2 x_1 y_1 m+\left(y_1^2- b ^2\right)=0$
$\therefore (4-16) m^2-2(2)(3) m+(9-9)=0$
$\therefore m=0 \text { or } m=-1$
$\therefore$ Equations are
$y-3=0 .(x-2), \quad y-3=-1 \cdot(x-2)$
$\therefore y=3, \quad \therefore y-3=-x+2 $
$\quad\quad\quad\quad\therefore x+y=5 .$

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MCQ 832 Marks

The equation of the tangents drawn at the ends of the major axis of the ellipse $9 x^2+5 y^2-30 y=0$, are

A
$y= \pm 3$
B
$x= \pm \sqrt{5}$
✓
$y=0, y=6$
D
$x=0, x=6$

Answer

Correct option: C.

$y=0, y=6$

(C)
$\begin{array}{l}9 x^2+5 y^2-30 y=0
\\\Rightarrow 9 x^2+5\left(y^2-6 y\right)=0
\\\Rightarrow 9 x^2+5\left(y^2-6 y+9\right)=45
\\\Rightarrow \frac{x^2}{5}+\frac{(y-3)^2}{9}=1\end{array}$
$\because \quad a^2 < b^2$, so axis of ellipse on $Y$-axis.
At Y - axis, put $x=0$, so we can obtain vertex.
Then $0+5 y^2-30 y=0 \Rightarrow y=0, y=6$
Therefore, tangents of vertex $y=0, y=6$

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MCQ 842 Marks

A man running round a race-course notes that the sum of the distance of two flag-posts from him is always 10 metres and the distance between the flag-posts is 8 metres. The area of the path he encloses in square metres is

✓
$15 \pi$
B
$12 \pi$
C
$18 \pi$
D
$8 \pi$

Answer

Correct option: A.

$15 \pi$

(A)
Here, $2 a =10 m$ and $2 ae =8 m$
$\therefore e=\frac{4}{5}, a=5 m$
Now, $b ^2= a ^2\left(1- e ^2\right)=9$
$\Rightarrow b=3$
Thus, required area $=\pi a b=15 \pi$ sq. metre.

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MCQ 852 Marks

Let $P$ be a variable point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with foci $F_1$ and $F_2$. If $A$ is the area of the triangle $PF _1 F_2$, then maximum value of A is

A
ab
B
abe
C
$\frac{ e }{ ab }$
D
$\frac{ ab }{ e }$

Answer

(B)
We have, $\frac{x^2}{ a ^2}+\frac{y^2}{b^2}=1$
$\therefore y=\frac{b}{a} \sqrt{a^2-x^2} \quad\ldots(i)$

Now, area of $\triangle PF _1 F_2=\frac{1}{2}\left(F_1 F_2\right) \times PL$
$=\frac{1}{2}(2 ae ) \times y$
$= ae \cdot \frac{ b }{ a } \sqrt{ a ^2-x^2} \quad \ldots .[$ From $( i )]$
$\therefore A = eb \sqrt{ a ^2-x^2}$, which is maximum when
$x=0$.
Thus, the maximum value of A is abe.

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MCQ 862 Marks

B is an extremity of the minor axis of an ellipse whose foci are S and $S ^{\prime}$. If $\angle SBS ^{\prime}$ is a right angle, then the eccentricity of the ellipse is

A
$\frac{1}{2}$
B
$\frac{1}{\sqrt{2}}$
C
$\frac{2}{3}$
D
$\frac{1}{3}$

Answer

(B)

$S =( ae , 0), S ^{\prime}=(- ae , 0)$ and $b = ae$
Now, $a ^2 e ^2= a ^2\left(1- e ^2\right)$
$\Rightarrow e =\frac{1}{\sqrt{2}}$

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MCQ 872 Marks

The curve represented by $x=3(\cos t+\sin t)$, $y=4(\cos t-\sin t)$ is

✓
an ellipse
B
a parabola
C
a hyperbola
D
a circle

Answer

Correct option: A.

an ellipse

(A)
$x =3(\cos t +\sin t ), y=4(\cos t -\sin t )$
$ \Rightarrow \frac{x}{3}=\cos t +\sin t , \frac{y}{4}=\cos t -\sin t$
$\Rightarrow \frac{x^2}{9}=1+\sin 2 t , \frac{y^2}{16}=1-\sin 2 t $
$\Rightarrow \frac{x^2}{9}+\frac{y^2}{16}=2 $
$\Rightarrow \frac{x^2}{18}+\frac{y^2}{32}=1 \text { which is an ellipse. }$

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MCQ 882 Marks

If the eccentricity of the two ellipse $\frac{x^2}{169}+\frac{y^2}{25}=1$ and $\frac{x^2}{ a ^2}+\frac{y^2}{b^2}=1$ are equal, then the value of $\frac{a}{b}$ is

A
$\frac{5}{13}$
B
$\frac{6}{13}$
✓
$\frac{13}{5}$
D
$\frac{13}{6}$

Answer

Correct option: C.

$\frac{13}{5}$

(C)
According to the given condition,
$\sqrt{1-\frac{25}{169}}=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow \frac{144}{169}=1-\frac{b^2}{a^2}$
$\Rightarrow \frac{b^2}{a^2}=\frac{25}{169}$
$\Rightarrow \frac{b}{a}=\frac{5}{13} \quad\ldots[\because a>0, b>0]$
$\Rightarrow \frac{a}{b}=\frac{13}{5}$

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MCQ 892 Marks

The equation $\frac{x^2}{2-r}+\frac{y^2}{r-5}+1=0$ represents an ellipse, if

A
r > 2
✓
2 < r < 5
C
r > 5
D
None of these

Answer

Correct option: B.

2 < r < 5

(B)
$\frac{x^2}{2- r }+\frac{y^2}{ r -5}+1=0$
$\Rightarrow \frac{x^2}{ r -2}+\frac{y^2}{5- r }=1$
$\text {Hence, } r >2 \text { and } r <5$
$\Rightarrow 2< r <5$

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MCQ 902 Marks

The locus of a variable point whose distance from (-2,0) is $\frac{2}{3}$ times its distance from the $\operatorname{linc} x=-\frac{9}{2}$, is

✓
Ellipse
B
Parabola
C
Hyperbola
D
None of these

Answer

Correct option: A.

Ellipse

(A)
Let point $P \left(x_1, y_1\right)$
$\begin{array}{l}\therefore \sqrt{\left(x_1+2\right)^2+y_1^2}=\frac{2}{3}\left(x_1+\frac{9}{2}\right)
\\ \Rightarrow\left(x_1+2\right)^2+y_1^2=\frac{4}{9}\left(x_1+\frac{9}{2}\right)^2
\\ \Rightarrow 9\left(x_1^2+y_1^2+4 x_1+4\right)=4\left(x_1^2+\frac{81}{4}+9 x_1\right) \\ \Rightarrow 5 x_1^2+9 y_1^2=45
\\ \Rightarrow \frac{x_1^2}{9}+\frac{y_1^2}{5}=1\end{array}$
∴ Locus of $\left(x_1, y_1\right)$ is $\frac{x^2}{9}+\frac{y^2}{5}=1$, which is equation of an ellipse.

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MCQ 912 Marks

An ellipse is described by using an endless string which is passed over two pins. If the axes are 6 cm and 4 cm , the necessary length of the string and the distance between the pins respectively in cm , are

A
$6,2 \sqrt{5}$
B
$6, \sqrt{5}$
C
$4,2 \sqrt{5}$
✓
$6+2 \sqrt{5}, 2 \sqrt{5}$

Answer

Correct option: D.

$6+2 \sqrt{5}, 2 \sqrt{5}$

(D)
Given, $2 a =6,2 b=4$
i.e., $a=3, b=2$
Also, $e ^2=1-\frac{ b ^2}{ a ^2}=\frac{5}{9}$
$\Rightarrow e =\frac{\sqrt{5}}{3}$
∴ Distance between the pins $=2 ae =2 \sqrt{5} cm$
and length of string $=2 a +2 ae =6+2 \sqrt{5} cm$

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MCQ 922 Marks

Eccentricity of the ellipse whose latus rectum is equal to the distance between two focus points, is

A
$\frac{\sqrt{5}+1}{2}$
✓
$\frac{\sqrt{5}-1}{2}$
C
$\frac{\sqrt{5}}{2}$
D
$\frac{\sqrt{3}}{2}$

Answer

Correct option: B.

$\frac{\sqrt{5}-1}{2}$

(B)
We have, $\frac{2 b^2}{ a }=2 ae$
$\Rightarrow b ^2= a ^2 e$
$\Rightarrow e=\frac{b^2}{a^2} \quad\ldots(i)$
Also, $e =\sqrt{1-\frac{ b ^2}{ a ^2}}$
$\Rightarrow e ^2=1- e \quad\ldots[From (i)]$
$\Rightarrow e ^2+ e -1=0$
$\therefore e =\frac{-1 \pm \sqrt{5}}{2}$
$\therefore e=\frac{\sqrt{5-1}}{2}$

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MCQ 932 Marks

If the distance between $a$ focus and corresponding directrix of an ellipse be 8 and the eccentricity be $\frac{1}{2}$, then length of the minor axis is

A
3
B
$4 \sqrt{2}$
C
6
✓
$\frac{16 \sqrt{3}}{3}$

Answer

Correct option: D.

$\frac{16 \sqrt{3}}{3}$

(D)
Given that, $\frac{ a }{ e }- ae =8$ and $e =\frac{1}{2}$
$\begin{array}{l}\Rightarrow a=\frac{8 e}{\left(1-e^2\right)} \\=\frac{8.4}{2(3)}=\frac{16}{3} \\ \therefore b=\frac{16}{3} \sqrt{\left(1-\frac{1}{4}\right)}=\frac{16}{3} \frac{\sqrt{3}}{2}=\frac{8 \sqrt{3}}{3}\end{array}$
Hence, the length of minor axis is $\frac{16 \sqrt{3}}{3}$

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MCQ 942 Marks

The equation of ellipse whose distance between the foci is equal to 8 and distance between the directrix is 18 , is

A
$5 x^2-9 y^2=180$
B
$9 x^2+5 y^2=180$
C
$x^2+9 y^2=180$
✓
$5 x^2+9 y^2=180$

Answer

Correct option: D.

$5 x^2+9 y^2=180$

(D)
We have, $2 ae =8, \frac{2 a }{ e }=18$
$\Rightarrow ae \times \frac{a}{e}=4 \times 9$
$\begin{array}{l}\Rightarrow a=\sqrt{4 \times 9}=6 \text { and } e=\frac{2}{3} \\\text { Also, } b=a \sqrt{1-e^2} \\\Rightarrow b=6 \sqrt{1-\frac{4}{9}}=\frac{6}{3} \sqrt{5}=2 \sqrt{5}\end{array}$
Hence, the required equation is $\frac{x^2}{36}+\frac{y^2}{20}=1$
i.e., $5 x^2+9 y^2=180$

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MCQ 952 Marks

The eccentricity of an ellipse, with its centre at the origin, is $\frac{1}{2}$. If one of the directrices is $x=4$, then the equation of the ellipse is

A
$4 x^2+3 y^2=1$
✓
$3 x^2+4 y^2=12$
C
$4 x^2+3 y^2=12$
D
$3 x^2+4 y^2=1$

Answer

Correct option: B.

$3 x^2+4 y^2=12$

(B)
Since, directrix is parallel to Y -axis, hence axes of the ellipse are parallel to X -axis. Let the equation of the ellipse be
$\frac{x^2}{ a ^2}+\frac{y^2}{b^2}=1,( a > b )$
Now, $e ^2=1-\frac{ b ^2}{ a ^2}$
$\Rightarrow \frac{ b ^2}{ a ^2}=1- e ^2=1-\frac{1}{4}$
$\Rightarrow \frac{b^2}{a^2}=\frac{3}{4}$
Also, one of the directrices is $x=4$
$\Rightarrow \frac{a}{a}=4 \Rightarrow a=4 e=4 \times \frac{1}{2}=2$
$b^2=\frac{3}{4} a^2=\frac{3}{4} \times 4=3$
$\therefore$ Required equation of ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$
$\Rightarrow 3 x^2+4 y^2=12$

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MCQ 962 Marks

The equation of the ellipse whose one of the vertices is (0,7) and the corresponding directrix is y = 12, is

A
$95 x^2+144 y^2=4655$
✓
$144 x^2+95 y^2=4655$
C
$95 x^2+144 y^2=13680$
D
$144 x^2+95 y^2=13680$

Answer

Correct option: B.

$144 x^2+95 y^2=4655$

(B)
Vertex $(0,7)$ and directrix $y=12$
$\therefore \quad b =7$ and $\frac{ b }{ e }=12$
$\Rightarrow e =\frac{7}{12}$
Also, $a = b \sqrt{1- e ^2}$
$\Rightarrow a =7 \sqrt{\frac{95}{144}}$
$\Rightarrow a ^2=\frac{4655}{144}$
Hence, equation of ellipse is
$\frac{x^2}{4655 / 144}+\frac{y^2}{49}=1 \text { i.e, } 144 x^2+95 y^2=4655$

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MCQ 972 Marks

The equation of the ellipse whose foci are ( $\pm 5,0$ ) and one of its directrix is 5x = 36 is

A
$\frac{x^2}{36}+\frac{y^2}{11}=1$
B
$\frac{x^2}{6}+\frac{y^2}{\sqrt{11}}=1$
C
$\frac{x^2}{6}+\frac{y^2}{11}=1$
D
$\frac{x^2}{11}+\frac{y^2}{6}=1$

Answer

(A)
Foci $( \pm 5,0) \equiv( \pm ae , 0),$
$\Rightarrow a e=5 \quad\ldots(i)$
Equation of directrix is $(x=\frac{ a }{ e }$)
Given, $x=\frac{36}{5}$
$

\Rightarrow \frac{a}{e}=\frac{36}{5} \quad\ldots(ii)$
$\Rightarrow a =6$ and $e =\frac{5}{6} \quad \ldots [$ From (i) and (ii) $]$
$\therefore b=a \sqrt{1-e^2}=6 \sqrt{1-\frac{25}{36}}=\sqrt{11}$
Hence, equation is $\frac{x^2}{36}+\frac{y^2}{11}=1$

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MCQ 982 Marks

If the centre, one of the foci and semi-major axis of an ellipse be (0 ,0),(0 ,3) and 5 , then its equation

✓
$\frac{x^2}{16}+\frac{y^2}{25}=1$
B
$\frac{x^2}{25}+\frac{y^2}{16}=1$
C
$\frac{x^2}{9}+\frac{y^2}{25}=1$
D
$\frac{x^2}{25}+\frac{y^2}{9}=1$

Answer

Correct option: A.

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

(A)
Given, centre $(0,0)$, focus $(0,3), b =5$
Focus $(0,3)$
$\Rightarrow be =3$
$\Rightarrow e =\frac{3}{5}$
Also, $a = b \sqrt{1- e ^2}=5 \sqrt{1-\frac{9}{25}}=4$
Hence, the required equation is $\frac{x^2}{16}+\frac{y^2}{25}=1$

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MCQ 992 Marks

The latus rectum of an ellipse is 10 and the minor axis is equal to the distance between the foci. The equation of the ellipse is

✓
$x^2+2 y^2=100$
B
$x^2+\sqrt{2} y^2=10$
C
$x^2-2 y^2=100$
D
$\sqrt{2} x^2+y^2-10$

Answer

Correct option: A.

$x^2+2 y^2=100$

(A)
Given, $\frac{2 b^2}{a}=10 \Rightarrow b^2=5 a$ and
$2 b=2 ae \Rightarrow \frac{ b }{ a }= e$
Also, $b^2=a^2\left(1-e^2\right)$
$\Rightarrow e^2=\left(1-e^2\right) \quad \ldots .\left[\because e=\frac{b}{a}\right]$
$\Rightarrow e =\frac{1}{\sqrt{2}}$
$\Rightarrow b =\frac{ a }{\sqrt{2}}$
$\Rightarrow b =5 \sqrt{2}, a =10$
Hence, equation of ellipse is
$\frac{x^2}{(10)^2}+\frac{y^2}{(5 \sqrt{2})^2}=1$
i.e., $x^2+2 y^2=100$

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MCQ 1002 Marks

An ellipse passes through the point (-3,1) and its eccentricity is $\sqrt{\frac{2}{5}}$. The equation of the ellipse is

✓
$3 x^2+5 y^2=32$
B
$3 x^2+5 y^2=25$
C
$3 x^2+y^2=4$
D
$3 x^2+y^2=9$

Answer

Correct option: A.

$3 x^2+5 y^2=32$

(A)
Let the equation of ellipse be
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\because$ It passes through $(-3,1)$
$\therefore\frac{9}{a^2}+\frac{1}{b^2}=1 $
$\Rightarrow9+\frac{a^2}{b^2}=a^2 \quad\ldots(i)$
Given, eccentricity is $\sqrt{\frac{2}{5}}$
$\therefore\frac{2}{5}=1-\frac{b^2}{a^2}$
$\Rightarrow \frac{b^2}{a^2}=\frac{3}{5} \quad\ldots(ii)$
From equation (i) and (ii), we get
$a^2=\frac{32}{3}, b^2=\frac{32}{5}$
Hence, required equation of ellipse is
$3 x^2+5 y^2=32$

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MCQ 1012 Marks

The equation of the ellipse whose centre is (2, -3), one of the foci is (3, -3) and the corresponding vertex is (4, -3), is

A
$\frac{(x-2)^2}{3}+\frac{(y+3)^2}{4}=1$
✓
$\frac{(x-2)^2}{4}+\frac{(y+3)^2}{3}=1$
C
$\frac{x^2}{3}+\frac{y^2}{4}=1$
D
$\frac{x^2}{4}+\frac{y^2}{3}=1$

Answer

Correct option: B.

$\frac{(x-2)^2}{4}+\frac{(y+3)^2}{3}=1$

(B)
$\begin{array}{l}\text { Foci }=(3,-3) \\\Rightarrow ae=3-2=1 \\\text { Vertex }=(4,-3) \\\Rightarrow a=4-2=2 \\\Rightarrow e=\frac{1}{2} \\\Rightarrow b=2 \sqrt{\left(1-\frac{1}{4}\right)}=\frac{2}{2} \sqrt{3}=\sqrt{3}\end{array}$
Therefore, equation of ellipse with centre
$(2,-3) \text { is } \frac{(x-2)^2}{4}+\frac{(y+3)^2}{3}=1$

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MCQ 1022 Marks

If the foci of an ellipse are $( \pm \sqrt{5}, 0)$ and its eccentricity is $\frac{\sqrt{5}}{3}$, then the equation of the ellipse is

A
$9 x^2+4 y^2=36$
✓
$4 x^2+9y^2=36$
C
$36 x^2+9y^2=4$
D
$9 x^2+36y^2=4$

Answer

Correct option: B.

$4 x^2+9y^2=36$

(B)
We have, $ae = \pm \sqrt{5}$
$\Rightarrow a = \pm \sqrt{5}\left(\frac{3}{\sqrt{5}}\right) \quad \ldots\left[\because e =\frac{\sqrt{5}}{3}\right]$
$\Rightarrow a = \pm 3$
$\Rightarrow a^2=9$
Now, $b^2=a^2\left(1-e^2\right)=9\left(1-\frac{5}{9}\right)=4$
Hence, equation of ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1$
$\Rightarrow 4 x^2+9 y^2=36$

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MCQ 1032 Marks

The equation of the ellipse whose vertices art $( \pm 5,0)$ and foci are $( \pm 4,0)$ is

✓
$9 x^2+25 y^2=225$
B
$25 x^2+9y^2=225$
C
$3 x^2+4y^2=192$
D
$4 x^2+3y^2=192$

Answer

Correct option: A.

$9 x^2+25 y^2=225$

(A)
Vertices $( \pm 5,0) \equiv( \pm a, 0)$
$\Rightarrow a=5$
Foci $( \pm 4,0) \equiv( \pm a e, 0)$
$\Rightarrow e =\frac{4}{5}$
$\therefore \quad b=\sqrt{a^2\left(1-e^2\right)}=\sqrt{25 \times \frac{9}{25}}=3$
Hence, equation is $\frac{x^2}{25}+\frac{y^2}{9}=1$
i.e., $9 x^2+25 y^2=225$

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MCQ 1042 Marks

The equation of the ellipse whose centre is at origin and which passes through the points (-3, 1) and (2, -2) is

A
$5 x^2+3 y^2=32$
✓
$3 x^2+5 y^2=32$
C
$5 x^2-3 y^2=32$
D
$3 x^2+5 y^2+32= 0$

Answer

Correct option: B.

$3 x^2+5 y^2=32$

(B)
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Since, it passes through $(-3,1)$ and $(2,-2)$
$\therefore \quad \frac{9}{ a ^2}+\frac{1}{b^2}=1$ and $\frac{1}{ a ^2}+\frac{1}{b^2}=\frac{1}{4}$
$\Rightarrow a^2=\frac{32}{3}, b^2=\frac{32}{5}$
Hence, required equation of ellipse is $3 x^2+5 y^2=32$

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MCQ 1052 Marks

The equations of the directrices of the ellipse $16 x^2+25 y^2=400$ are

A
$2 x= \pm 25$
B
$5 x= \pm 9$
C
$3 x= \pm 10$
✓
$3 x= \pm 25$

Answer

Correct option: D.

$3 x= \pm 25$

(D)
$16 x^2+25 y^2=400$
$\Rightarrow \frac{x^2}{25}+\frac{y^2}{16}=1$
$\Rightarrow e =\sqrt{1-\frac{16}{25}}=\frac{3}{5}$
Therefore, directrices are $x= \pm \frac{5}{3}$ or $3 x= \pm 25$

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MCQ 1062 Marks

The length of the axes of the continue $9 x^2+4 y^2-6 x+4 y+1=0$ are

A
$\frac{1}{2}, 9$
B
$3, \frac{2}{5}$
✓
$1, \frac{2}{3}$
D
3, 2

Answer

Correct option: C.

$1, \frac{2}{3}$

(C)
$9 x^2+4 y^2-6 x+4 y+1=0$
$\Rightarrow 9\left(x^2-\frac{2}{3} x+\frac{1}{9}\right)+4\left(y^2+y+\frac{1}{4}\right)=-1+1+1$
$\Rightarrow 9\left(x-\frac{1}{3}\right)^2+4\left(y+\frac{1}{2}\right)^2=1$
$\Rightarrow \frac{\left(x-\frac{1}{3}\right)^2}{\frac{1}{9}}+\frac{\left(y+\frac{1}{2}\right)^2}{\frac{1}{4}}=1$
Here, $a =\frac{1}{3}, b=\frac{1}{2}$
$\Rightarrow 2 a =\frac{2}{3}, 2 b=1$
∴ Length of axes are $1, \frac{2}{3}$.

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MCQ 1072 Marks

The co-ordinates of the foci of the ellipse $3 x^2+4 y^2-12 x-8 y+4=0$, are

A
(1, 2), (3, 4)
B
(1, 4), (3, 1)
✓
(1, 1), (3, 1)
D
(2, 3), (5, 4)

Answer

Correct option: C.

(1, 1), (3, 1)

(C)
$3 x^2-12 x+4 y^2-8 y=-4$
$\Rightarrow 3\left(x^2-4 x+4\right)+4\left(y^2-2 y+1\right)=-4+12+4$
$\Rightarrow 3(x-2)^2+4(y-1)^2=12$
$\Rightarrow \frac{(x-2)^2}{4}+\frac{(y-1)^2}{3}=1$
$\Rightarrow \frac{ X ^2}{4}+\frac{ Y ^2}{3}=1$
$\therefore e =\sqrt{1-\frac{3}{4}}=\frac{1}{2}$
Foci are $X= \pm ae , Y=0$
$\Rightarrow x-2= \pm 1, y-1=0$
$\Rightarrow x=3$ or $1, y=1$
i.e., $(3,1)$ and $(1,1)$

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MCQ 1082 Marks

The eccentricity of the ellipse $\frac{(x-1)^2}{2}+\left(y+\frac{3}{4}\right)^2=\frac{1}{16}$ is

✓
$\frac{1}{\sqrt{2}}$
B
$\frac{1}{2 \sqrt{2}}$
C
$\frac{1}{2}$
D
$\frac{1}{4}$

Answer

Correct option: A.

$\frac{1}{\sqrt{2}}$

(A)
$\frac{(x-1)^2}{2}+\left(y+\frac{3}{4}\right)^2=\frac{1}{16}$
$\Rightarrow \frac{(x-1)^2}{2\left(\frac{1}{16}\right)}+\frac{\left(y+\frac{3}{4}\right)^2}{\left(\frac{1}{16}\right)}=1$
$\Rightarrow \frac{(x-1)^2}{\left(\frac{1}{8}\right)}+\frac{\left(y+\frac{3}{4}\right)^2}{\left(\frac{1}{16}\right)}=1$
$\therefore a^2=\frac{1}{8}, b^2=\frac{1}{16}$
$b^2=a^2\left(1-e^2\right)$
$\Rightarrow \frac{1}{16}=\frac{1}{8}\left(1- e ^2\right)$
$\Rightarrow e ^2=1-\frac{1}{2}$
$\Rightarrow e =\frac{1}{\sqrt{2}}$

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MCQ 1092 Marks

The eccentricity of the ellipse $4 x^2+9 y^2+8 x+36 y+4=0$ is

A
$\frac{5}{6}$
B
$\frac{3}{5}$
C
$\frac{\sqrt{2}}{3}$
✓
$\frac{\sqrt{5}}{3}$

Answer

Correct option: D.

$\frac{\sqrt{5}}{3}$

(D)
$4\left(x^2+2 x+1\right)+9\left(y^2+4 y+4\right)=36$
$\Rightarrow 4(x+1)^2+9(y+2)^2=36$
$\Rightarrow \frac{(x+1)^2}{9}+\frac{(y+2)^2}{4}=1$
$\therefore \quad a^2=9, b^2=4$
$b^2=a^2\left(1-e^2\right)$
$\Rightarrow 4=9\left(1- e ^2\right)$
$\Rightarrow e ^2=1-\frac{4}{9}=\frac{5}{9}$
$\Rightarrow e =\frac{\sqrt{5}}{3}$

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MCQ 1102 Marks

The equation of the tangent to the ellipse $9 x^2+16 y^2=144$ at the point $(4 \cos \theta, 3 \sin \theta)$ is

A
$3 x \cos \theta+y \sin \theta=12$
✓
$3 x \cos \theta+4 y \sin \theta=12$
C
$x \cos \theta+y \sin \theta=12$
D
$3 x+4 y=12$

Answer

Correct option: B.

$3 x \cos \theta+4 y \sin \theta=12$

(B)
Equation of ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$
$a^2=16, b^2=9$
Equation of the tangent to the ellipse at
$(4 \cos \theta, 3 \sin \theta)$ is
$\frac{x(4 \cos \theta)}{16}+\frac{y(3 \sin \theta)}{9}=1$
$\Rightarrow \frac{x \cos \theta}{4}+\frac{y \sin \theta}{3}=1$
$\Rightarrow 3 x \cos \theta+4 y \sin \theta=12$

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MCQ 1112 Marks

The equation of the tangent to the ellipse $2 x^2+3 y^2=30$ at the point $(-3,2)$ is

A
y = x + 3
B
y = x + 2
C
x = y + 5
✓
y = x + 5

Answer

Correct option: D.

y = x + 5

(D)
Equation of the ellipse is $\frac{x^2}{15}+\frac{y^2}{10}=1$
$a^2=15, b^2=10$
Equation of the tangent to the ellipse
$\frac{x^2}{ a ^2}+\frac{y^2}{b^2}=1$ at $P \left(x_1, y_1\right)$ is
$\frac{x x_1}{a^2}+\frac{y y_1}{b^2}=1$
$\therefore \text { Required tangent is } \frac{x(-3)}{15}+\frac{y(2)}{10}=1$
$\therefore \frac{-x}{5}+\frac{y}{5}=1$
$\therefore -x+y=5 \text { i.e. } y=x+5$

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MCQ 1122 Marks

If the line y = 2x + c be a tangent to the ellipse $\frac{x^2}{8}+\frac{y^2}{4}=1$, then $c=$

A
$\pm 4$
✓
$\pm 6$
C
$\pm 1$
D
$\pm 8$

Answer

Correct option: B.

$\pm 6$

(B)
$c= \pm \sqrt{b^2+a^2 m^2}$
$= \pm \sqrt{4+8(4)}$
$= \pm 6$

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MCQ 1132 Marks

If a bar of given length moves with its extremities on two fixed straight lines at right angles, then the locus of any point on bar marked on the bar describes a/an

A
Circle
B
Parabola
✓
Ellipse
D
Hyperbola

Answer

Correct option: C.

Ellipse

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MCQ 1142 Marks

The distance of the point ' $\theta$ ' on the ellipse $\frac{x^2}{ a ^2}+\frac{y^2}{b^2}=1$ from a focus is

A
$a ( e +\cos \theta)$
B
$a ( e -\cos \theta)$
✓
$a(1+e \cos \theta)$
D
$a(1+2 e \cos \theta)$

Answer

Correct option: C.

$a(1+e \cos \theta)$

(C)
Focal distance of any point $P (x, y)$ on the ellipse is equal to $S P=a+e x$.
Here, $x= a \cos \theta$
$\therefore \quad S P=a+a e \cos \theta$
$= a (1+ e \cos \theta)$

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MCQ 1152 Marks

The distance between the foci of the ellipse $x=3 \cos \theta, y=4 \sin \theta$ is

✓
$2 \sqrt{7}$
B
$7 \sqrt{2}$
C
$\sqrt{7}$
D
$3 \sqrt{7}$

Answer

Correct option: A.

$2 \sqrt{7}$

(A)
$x^2=9 \cos ^2 \theta$ and $y^2=16 \sin ^2 \theta$
$\Rightarrow \frac{x^2}{9}+\frac{y^2}{16}=1$
$e=\frac{\sqrt{7}}{4}$
Distance between foci $=2 be =2 \sqrt{7}$

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MCQ 1162 Marks

Equation x = a $\cos \theta$, y = b $\sin \theta$ (a > b) represent a conic section whose eccentricity e is given by

A
$e^2=\frac{a^2+b^2}{a^2}$
B
$e^2=\frac{a^2+b^2}{b^2}$
✓
$e ^2=\frac{ a ^2- b ^2}{ a ^2}$
D
$e^2=\frac{a^2-b^2}{b^2}$

Answer

Correct option: C.

$e ^2=\frac{ a ^2- b ^2}{ a ^2}$

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MCQ 1172 Marks

If the latus rectum of an ellipse be equal to half of its minor axis, then its eccentricity is

A
$\frac{3}{2}$
✓
$\frac{\sqrt{3}}{2}$
C
$\frac{2}{3}$
D
$\frac{\sqrt{2}}{3}$

Answer

Correct option: B.

$\frac{\sqrt{3}}{2}$

(B)
Given, $\frac{2 b^2}{a}=b$
$\Rightarrow \frac{ b }{ a }=\frac{1}{2}$
$\Rightarrow \frac{b^2}{a^2}=\frac{1}{4}$
Hence, $e=\sqrt{1-\frac{b^2}{a^2}}=\frac{\sqrt{3}}{2}$

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MCQ 1182 Marks

If the eccentricity of an ellipse be $\frac{5}{8}$ and the distance between its foci be 10, then its latus rectum is

✓
$\frac{39}{4}$
B
12
C
15
D
$\frac{37}{2}$

Answer

Correct option: A.

$\frac{39}{4}$

(A)
We have, $2 ae =10 \Rightarrow a =\frac{10}{\frac{2 \times 5}{8}}=8$
Also, $b^2=a^2\left(1-e^2\right)$
$\Rightarrow b =8 \sqrt{1-\frac{25}{64}}=\sqrt{39}$
Now, Latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 39}{8}=\frac{39}{4}$

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MCQ 1192 Marks

If the eccentricity of an ellipse be $\frac{1}{\sqrt{2}}$ then its latus rectum is equal to its

A
minor axis
B
semi-minor axis
C
major axis
✓
semi-major axis

Answer

Correct option: D.

semi-major axis

(D)
We have, $e =\frac{1}{\sqrt{2}}$
$\therefore \quad$ Latus rectum $=\frac{2 b^2}{ a }=\frac{2}{ a } \times a ^2\left(1- e ^2\right)$
$=2 a\left(1-\frac{1}{2}\right)=a$
i.e., semi-major axis

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MCQ 1202 Marks

The length of the latus rectum of an ellipse is $\frac{1}{3}$ of the major axis. Its eccentricity is

A
$\frac{2}{3}$
✓
$\sqrt{\frac{2}{3}}$
C
$\frac{5 \times 4 \times 3}{7^3}$
D
$\left(\frac{3}{4}\right)^4$

Answer

Correct option: B.

$\sqrt{\frac{2}{3}}$

(B)
Latus rectum $=\frac{1}{3}$ (major axis)
$\Rightarrow \frac{2 b^2}{ a }=\frac{2 a }{3}$
$\Rightarrow a^2=3 b^2$
$\Rightarrow a^2=3 a^2\left(1-e^2\right)$
$\Rightarrow e =\sqrt{\frac{2}{3}}$

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MCQ 1212 Marks

If the distance between the foci of an ellipse be equal to its minor axis, then its eccentricity is

A
$\frac{1}{2}$
✓
$\frac{1}{\sqrt{2}}$
C
$\frac{1}{3}$
D
$\frac{1}{\sqrt{3}}$

Answer

Correct option: B.

$\frac{1}{\sqrt{2}}$

(B)
Foci are ( ±ae, 0)
∴ According to the condition, $2 ae =2 b$
$\Rightarrow ae = b \quad\ldots(i)$
Also, $b^2=a^2\left(1-e^2\right)$
$\Rightarrow e ^2=\left(1- e ^2\right) \quad\ldots[From (i)]$
$\Rightarrow e =\frac{1}{\sqrt{2}}$

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MCQ 1222 Marks

The distance between the foci of an ellipse is 16 and eccentricity is $\frac{1}{2}$ . Length of the major axis of the ellipse is

A
8
B
64
C
16
✓
32

Answer

Correct option: D.

(D)
Distance between the foci $=2 ae =16$ and
$e=\frac{1}{2}$
∴ Length of the major axis of the ellipse
$=2 a =\frac{2 ae }{ e }=\frac{16}{\frac{1}{2}}=32$

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MCQ 1232 Marks

If distance between the directrices be thrice the distance between the foci, then eccentricity of ellipse is

A
$\frac{1}{2}$
B
$\frac{2}{3}$
✓
$\frac{1}{\sqrt{3}}$
D
$\frac{4}{5}$

Answer

Correct option: C.

$\frac{1}{\sqrt{3}}$

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MCQ 1242 Marks

If the foci and vertices of an ellipse be (±1,0) and (±2, 0), then the minor axis of the ellipse is

A
$2 \sqrt{5}$
B
2
C
4
✓
$2 \sqrt{3}$

Answer

Correct option: D.

$2 \sqrt{3}$

(D)
We have, $ae =1$ and $a =2$
$\Rightarrow e =\frac{1}{2}$
Also, $b^2=a^2\left(1-e^2\right)$
$\Rightarrow b =2 \sqrt{1-\frac{1}{4}}=\sqrt{3}$
$\therefore \quad$ Minor axis $=2 b=2 \sqrt{3}$

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MCQ 1252 Marks

The equation $5 x^2+y^2+y=8$ represents

✓
an ellipse
B
a parabola
C
a hyperbola
D
a circle

Answer

Correct option: A.

an ellipse

(A)
$5 x^2+y^2+y=8$
$\Rightarrow 5 x^2+y^2+y+\frac{1}{4}=8+\frac{1}{4}$
$\Rightarrow 5 x^2+\left(y+\frac{1}{2}\right)^2=\frac{33}{4}$
$\Rightarrow \frac{x^2}{\left(\frac{33}{20}\right)}+\frac{\left(y+\frac{1}{2}\right)^2}{\left(\frac{33}{4}\right)}=1$
The equation represent an ellipse.

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MCQ 1262 Marks

The eccentricity of an ellipse is $\frac{2}{3}$ latus rectum is 5 and centre is (0, 0). The equation of the ellipse is

A
$\frac{x^2}{81}+\frac{y^2}{45}=1$
✓
$\frac{4 x^2}{81}+\frac{4 y^2}{45}=1$
C
$\frac{x^2}{9}+\frac{y^2}{5}=1$
D
$\frac{x^2}{81}+\frac{y^2}{45}=5$

Answer

Correct option: B.

$\frac{4 x^2}{81}+\frac{4 y^2}{45}=1$

(B)
Since $e^2=1-\frac{b^2}{a^2}$
$\Rightarrow\left(\frac{2}{3}\right)^2=1-\frac{b^2}{a^2}$
$\Rightarrow b^2=\frac{5 a^2}{9} \quad\ldots(i)$
Given length of latus rectum $=5$
$\Rightarrow \frac{2 b^2}{a}=5$
$\Rightarrow b^2=\frac{5 a}{2} \quad\ldots(ii)$
From (i) and (ii), we get
$a^2=\frac{81}{4}, b^2=\frac{45}{4}$
∴ Equation of ellipse is $\frac{4 x^2}{81}+\frac{4 y^2}{45}=1$

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MCQ 1272 Marks

The equation of the ellipse whose one focus is at $(4,0)$ and whose eccentricity is $\frac{4}{5}$, is

A
$\frac{x^2}{3^2}+\frac{y^2}{5^2}=1$
✓
$\frac{x^2}{5^2}+\frac{y^2}{3^2}=1$
C
$\frac{x^2}{5^2}+\frac{y^2}{4^2}=1$
D
$\frac{x^2}{4^2}+\frac{y^2}{5^2}=1$

Answer

Correct option: B.

$\frac{x^2}{5^2}+\frac{y^2}{3^2}=1$

(B)
Here, $ae =4$ and $e =\frac{4}{5}$
$\Rightarrow a=5$
Now, $b^2=a^2\left(1-e^2\right)$
$\Rightarrow b^2=25\left(1-\frac{16}{25}\right)=9$
∴ Equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{9}=1$.

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MCQ 1282 Marks

The lengths of major and minor axis of an ellipse are 10 and 8 respectively and its major axis is along the Y -axis. The equation of the ellipse referred to its centre as origin is

A
$\frac{x^2}{25}+\frac{y^2}{16}=1$
✓
$\frac{x^2}{16}+\frac{y^2}{25}=1$
C
$\frac{x^2}{100}+\frac{y^2}{64}=1$
D
$\frac{x^2}{64}+\frac{y^2}{100}=1$

Answer

Correct option: B.

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

(B)
Here, given that $2 b=10,2 a=8$
$\Rightarrow b=5, a=4$
Hence, the required equation is $\frac{x^2}{16}+\frac{y^2}{25}=1$

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MCQ 1292 Marks

If $P \equiv(x, y), F _1 \equiv(3,0), F _2 \equiv(-3,0)$ and $16 x^2+25 y^2=400$, then $PF _1+ PF _2$ equals

A
8
B
6
✓
10
D
12

Answer

Correct option: C.

(C)
Equation of the curve is $\frac{x^2}{5^2}+\frac{y^2}{4^2}=1$
Here, $a=5, b=4$
$\therefore \quad PF _1+ PF _2=2 a =2 \times 5=10$

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MCQ 1302 Marks

The sum of focal distances of any point on the ellipse with major and minor axes as 2 a and 2 b respectively, is equal to

✓
2a
B
$\frac{2 a }{ b }$
C
$\frac{2 b}{a}$
D
$\frac{ b ^2}{ a }$

Answer

Correct option: A.

(A)
Sum of focal distances of any point on an ellipse is always equal to length of major axis of that ellipse.

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MCQ 1312 Marks

The centre of the ellipse
$4 x^2+9 y^2-16 x-54 y+61=0 \text { is }$

A
(1, 3)
✓
(2, 3)
C
(3, 2)
D
(3,1)

Answer

Correct option: B.

(2, 3)

(B)
$4 x^2+9 y^2-16 x-54 y+61=0$
$\Rightarrow 4 x^2-16 x+9 y^2-54 y=-61$
$\Rightarrow 4\left(x^2-4 x+4-4\right)+9\left(y^2-6 y+9-9\right)=-61$
$\Rightarrow 4(x-2)^2+9(y-3)^2=36$
$\Rightarrow \frac{(x-2)^2}{9}+\frac{(y-3)^2}{4}=1$
Hence, the centre is $(2,3)$

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MCQ 1322 Marks

The distance between the directrices of the ellipse $\frac{x^2}{36}+\frac{y^2}{20}=1$ is

A
8
B
12
✓
18
D
24

Answer

Correct option: C.

(C)
$a=6, b=2 \sqrt{5}$
Now, $b^2=a^2\left(1-e^2\right)$
$\Rightarrow \frac{20}{36}=\left(1- e ^2\right)$
$\Rightarrow e =\sqrt{\frac{16}{36}}=\frac{2}{3}$
Distance between directrix $=\frac{2 a }{ e }=\frac{2 \times 6}{\frac{2}{3}}=18$

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MCQ 1332 Marks

For the ellipse $3 x^2+4 y^2=12$, the length of latus rectum is

A
$\frac{3}{2}$
✓
3
C
$\frac{8}{3}$
D
$\sqrt{\frac{3}{2}}$

Answer

Correct option: B.

(B)
$3 x^2+4 y^2=12$
$\Rightarrow \frac{x^2}{4}+\frac{y^2}{3}=1$
$\therefore \quad$ Latus rectum $=\frac{2 b^2}{a}=3$

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MCQ 1342 Marks

The distance between the foci of the ellipse $3 x^2+4 y^2=48$ is

A
2
✓
4
C
6
D
8

Answer

Correct option: B.

(B)
$3 x^2+4 y^2=48$
$\Rightarrow \frac{x^2}{16}+\frac{y^2}{12}=1$
$\therefore \quad a^2=16, b^2=12$
$\Rightarrow e =\sqrt{1-\frac{ b ^2}{ a ^2}}=\sqrt{1-\frac{12}{16}}=\frac{1}{2}$
∴ Distance between the foci $=2 ae$
$\begin{array}{l}=2 \times 4 \times \frac{1}{2}
\\ =4\end{array}$

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MCQ 1352 Marks

The foci of $16 x^2+25 y^2=400$ are

✓
$( \pm 3,0)$
B
$(0, \pm 3)$
C
(3, -3)
D
(-3, 3)

Answer

Correct option: A.

$( \pm 3,0)$

(A)
The equation of the ellipse is
$16 x^2+25 y^2=400 \Rightarrow \frac{x^2}{25}+\frac{y^2}{16}=1$
Here, $a^2=25, b^2=16$
$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{16}{25}} \Rightarrow e=\frac{3}{5}$
Hence, the foci are $( \pm ae , 0)=( \pm 3,0)$

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MCQ 1362 Marks

Eccentricity of the conic $16 x^2+7 y^2=112$ is

A
$\frac{3}{\sqrt{7}}$
B
$\frac{7}{16}$
✓
$\frac{3}{4}$
D
$\frac{4}{3}$

Answer

Correct option: C.

$\frac{3}{4}$

(C)
$\frac{x^2}{\frac{112}{16}}+\frac{y^2}{\frac{112}{7}}=1$
$\therefore \quad e=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{112}{16} \cdot \frac{7}{112}}=\frac{3}{4}$

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MCQ 1372 Marks

Two tangents of the parabola $y^2=8 x$ meet its tangent at vertex at points P and Q . If $PQ =4$, then the locus of the point of intersection of these two tangents is

A
$y^2=8(x-2)$
✓
$y^2=8(x+2)$
C
$x^2=8(y-2)$
D
$x^2=8(y+2)$

Answer

Correct option: B.

$y^2=8(x+2)$

(B)
$y^2=8 x$
$\Rightarrow y^2=4 \times 2 \times x$
$\Rightarrow a =2$
Let equations of tangents to the parabola at P and $Q$ be respectively.
$y t _1=x+ a t _1^2 \quad\ldots(i) \ and$
$y t _2=x+ a t _2^2 \quad\ldots(ii)$
Tangent at the vertex is Y -axis
$\Rightarrow x=0$
$\therefore y=\frac{a t_1^2}{t_1}$
$\Rightarrow y= at _1$
and $y=\frac{ a t _2^2}{ t _2}$
$\Rightarrow y= at _2$
$\therefore $ The co-ordinates of P and Q are $P \left(0, a t_1\right)$ and $Q \left(0, a t _2\right)$
$PQ =4$
$\Rightarrow PQ ^2=16$
$\Rightarrow\left(a t_1-a t_2\right)^2=16$
$\Rightarrow a^2\left[\left(t_1+t_2\right)^2-4 t_1 t_2\right]=16$
Solving (i) and (ii), $t _1+ t _2=\frac{y}{ a }$ and $t _1 t _2=\frac{x}{ a }$
∴ locus of the required point is
$a^2\left(\frac{y^2}{a^2}-\frac{4 x}{a}\right]=16$
$\Rightarrow y^2-4 x a=16$
$\Rightarrow y^2=8 x+16 \quad\ldots [\because a =2]$
$\Rightarrow y^2=8(x+2)$

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MCQ 1382 Marks

The tangent drawn at any point P to the parabola $y^2=4 a x$ meets the directrix at the point K , then the angle which KP subtends at its focus is

A
$30^{\circ}$
B
$45^{\circ}$
C
$60^{\circ}$
D
$90^{\circ}$

Answer

(D)
Here, $P\left(a t^2, 2 a t\right)$ and $S(a, 0)$.
If the tangent at $P , t y=x+ at ^2$, meets the directrix
$x=- a$ at K , then $K =\left(- a , \frac{ at ^2- a }{ t }\right)$
Now, $m_1=$ slope of $S P=\frac{2 a t}{a\left(t^2-1\right)}$ and $m _2=$ slope of $SK =\frac{ a \left( t ^2-1\right)}{-2 at }$
$\Rightarrow m _1 m_2=-1$
$\therefore \quad \angle PSK =90^{\circ}$
$x=- a$

$x = – a$

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MCQ 1392 Marks

If $y=m x+4$ is a tangent to both the parabolas, $y^2=4 x$ and $x^2=2 b y$, then b is equal to

A
-64
B
128
✓
-128
D
-32

Answer

Correct option: C.

-128

(C)
The equation of tangent to $y^2=4 x$ is $y= m x+\frac{1}{m}$
It is given that the tangent to $y^2=4 x$ is $y=m x+4$
$\Rightarrow m =\frac{1}{4}$
$\Rightarrow y=\frac{1}{4} x+4$
Also, $y=\frac{1}{4} x+4$ is a tangent to the parabola
$x^2=2 b y$
Substituting $y=\frac{1}{4} x+4$ in $x^2=2 b y$, we get
$x^2=2 b\left(\frac{1}{4} x+4\right)$
$\Rightarrow 2 x^2- b x-16 b=0$ must have equal roots
$\Rightarrow(-b)^2-4 \times 2 \times(-16 b)=0$
$\Rightarrow b ^2+128 b=0$
But, $b=0$ is not possible
$\Rightarrow b =-128$

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MCQ 1402 Marks

The equation of the common tangent of the parabolas $x^2=108 y$ and $y^2=32 x$, is

A
$2 x+3 y=36$
✓
$2 x+3 y+36=0$
C
$3 x+2 y=36$
D
$3 x+2 y+36=0$

Answer

Correct option: B.

$2 x+3 y+36=0$

(B)
Here, $a=8$ and $b=27$
Equation of common tangent to the parabolas $y^2=4 a x$ and $x^2=4 b y$ is $y b^{\frac{1}{3}}+x a ^{\frac{1}{3}}+( ab )^{\frac{2}{3}}=0$.
$2 x+3 y+36=0$

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MCQ 1412 Marks

The locus of a foot of perpendicular drawn to the tangent of parabola $y^2=4 a x$ from focus, is

✓
x = 0
B
y = 0
C
$y^2=2 a (x+ a )$
D
$x^2+y^2(x+ a )=0$

Answer

Correct option: A.

x = 0

(A)
Tangent to parabola is,
$y=m x+\frac{a}{m} \quad\ldots(i)$
A line perpendicular to tangent and passing from focus $( a , 0)$ is,
$y=-\frac{x}{m}+\frac{a}{m} \quad\ldots(ii)$
Solving both lines (i) and (ii), we get
$x=0$

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MCQ 1422 Marks

The product of the slopes of the tangents drawn from the point $(4,10)$ to the parabola $y^2=16 x$ is

✓
1
B
-1
C
2
D
$-\frac{1}{2}$

Answer

Correct option: A.

(A)
$y=m x+\frac{a}{m} ; y=m x+\frac{4}{m}$
It passes through $(4,10)$
$\therefore \quad 10=4 m+\frac{4}{m}$
$\Rightarrow 4 m^2-10 m+4=0$
$\Rightarrow 2 m^2-5 m+2=0$
$\Rightarrow m =2$ or $\frac{1}{2}$
$\therefore \quad m_1 m_2=\frac{2}{2}=1$

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MCQ 1432 Marks

The tangent to the parabola $y^2=4 ax$ at the point (a, 2a) makes with X -axis an angle equal to

A
$\frac{\pi}{3}$
✓
$\frac{\pi}{4}$
C
$\frac{\pi}{2}$
D
$\frac{\pi}{6}$

Answer

Correct option: B.

$\frac{\pi}{4}$

(B)
The equation of the tangent at point $( a , 2 a )$ of the parabola $y^2=4 a x$ is $y y_1=2 a \left(x+x_1\right)$
$\Rightarrow 2 a y=2 a (x+ a ) \Rightarrow y=x+ a$
This line makes an angle of $\frac{\pi}{4}$ with the
X -axis, as $m =\tan \theta=1$.

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MCQ 1442 Marks

The point on the curve $y^2=x$, the tangent at which makes an angle $45^{\circ}$ with $X$-axis will be

A
$\left(\frac{1}{2}, \frac{1}{4}\right)$
B
$\left(\frac{1}{2}, \frac{1}{2}\right)$
C
(2, 4)
✓
$\left(\frac{1}{4}, \frac{1}{2}\right)$

Answer

Correct option: D.

$\left(\frac{1}{4}, \frac{1}{2}\right)$

(D)
Given, slope $m =\tan 45^{\circ}=1$
Equation of parabola is $y^2=x$
$\Rightarrow y^2=4 \cdot \frac{1}{4} \cdot x$
$\Rightarrow a =\frac{1}{4}$
Point of contact $\equiv\left(\frac{ a }{ m ^2}, \frac{2 a }{ m }\right) \equiv\left(\frac{\frac{1}{4}}{1^2}, \frac{2 \times \frac{1}{4}}{1}\right)$
$\equiv\left(\frac{1}{4}, \frac{1}{2}\right)$

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MCQ 1452 Marks

The equation of the tangent to the parabola $y^2=16 x$, which is perpendicular to the line $y=3 x+7$ is

A
y - 3x + 4 = 0
B
3y - x + 36 = 0
C
3y + x - 36 = 0
✓
3y + x + 36 = 0

Answer

Correct option: D.

3y + x + 36 = 0

(D)
Line perpendicular to given line, $3 y+x=\lambda$
$y=\frac{-1}{3} x+\frac{\lambda}{3}$
Here, $m =\frac{-1}{3}, c =\frac{\lambda}{3}$
Comparing $y^2=16 x$ with $y^2=4 a x$, we get
$a =4$
Condition for tangency is,
$c =\frac{ a }{ m }$
$\Rightarrow \frac{\lambda}{3}=\frac{4}{(-1 / 3)}$
$\Rightarrow \lambda=-36$
Required equation is $x+3 y+36=0$

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MCQ 1462 Marks

The equation of the tangent to the parabola $y^2=4 x+5$ parallel to the line $y=2 x+7$ is

A
2x - y - 3 = 0
✓
2x - y + 3 = 0
C
2x + y + 3 = 0
D
2x - y + 5 = 0

Answer

Correct option: B.

2x - y + 3 = 0

(B)
Equation of parabola is $Y^2=4 X$, where
$X =x+\frac{5}{4}$
Tangent parallel to $Y =2 X +7$ is
$Y = mX +\frac{ a }{ m }$
$\Rightarrow y=2\left(x+\frac{5}{4}\right)+\frac{1}{2}$
$\Rightarrow y=2 x+3$ i.e., $2 x-y+3=0$.

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MCQ 1472 Marks

The two parabolas $x^2=4 y$ and $y^2=4 x$ meet in two distinct points. One of these is the origin and the other is

A
(2, 2)
B
(4, -4)
✓
(4, 4)
D
(-2,2)

Answer

Correct option: C.

(4, 4)

(C)
Given, $x^2=4 y \quad\ldots(i)$
$y^2=4 x \quad\ldots(ii)$
$\Rightarrow \frac{x^4}{16}=4 x \quad\ldots[From \ (i) \ and \ (ii)]$
$\Rightarrow x^4=64 x$
$y=0,4$
∴ Other point is $(4,4)$.

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MCQ 1482 Marks

For the parabola $y^2+6 y-2 x=-5$
(I) the vertex is (-2,-3)
(II) the directrix is $y+3=0$
Which of the following is correct?

A
Both I and II are correct
✓
I is true, II is false
C
Both I and II are false
D
I is false, II is true

Answer

Correct option: B.

I is true, II is false

(B)
$y^2+6 y-2 x=-5$
$\Rightarrow y^2+6 y+9=2 x-5+9$
$\Rightarrow(y+3)^2=2(x+2)$
$\therefore \quad$ vertex $=(-2,-3)$
Here, $a =\frac{1}{2}$
∴ Equation of directrix is $x+2=-\frac{1}{2}$
$\Rightarrow 2 x+5=0$

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MCQ 1492 Marks

The equation of the directrix of the parabola $x^2+4 x+2 y=0$ is

A
$2 y+5=0$
B
$3 y=2$
✓
$2 y=5$
D
$3 y+2=0$

Answer

Correct option: C.

$2 y=5$

(C)
$x^2+4 x+2 y=0$
$\Rightarrow x^2+4 x+4=-2 y+4$
$\Rightarrow(x+2)^2=-2(y-2)$
∴ Equation of directrix is $y-2=\frac{1}{2}$
$\Rightarrow y=\frac{5}{2}$
$\Rightarrow 2 y=5$

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MCQ 1502 Marks

The distance between the vertex and the focus of the parabola $x^2-2 x+3 y-2=0$ is

A
$\frac{4}{5}$
✓
$\frac{3}{4}$
C
$\frac{1}{2}$
D
$\frac{5}{6}$

Answer

Correct option: B.

$\frac{3}{4}$

(B)
Given equation of parabola
$x^2-2 x+3 y-2=0$
$\Rightarrow x^2-2 x+1=-3 y+2+1$
$\Rightarrow(x-1)^2=-3(y-1)$
$\therefore \quad$ vertex $=(h, k)=(1,1)$,
focus $=(h, k+b)=\left(1, \frac{1}{4}\right)$
∴ distance between focus and vertex
$=\sqrt{0+\left(1-\frac{1}{4}\right)^2}$
$=\sqrt{\left(\frac{3}{4}\right)^2}$
$=\frac{3}{4}$

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MCQ 1512 Marks

The focus of the parabola $y=2 x^2+x$ is

A
(0, 0)
B
$\left(\frac{1}{2}, \frac{1}{4}\right)$
✓
$\left(-\frac{1}{4}, 0\right)$
D
$\left(-\frac{1}{4}, \frac{1}{8}\right)$

Answer

Correct option: C.

$\left(-\frac{1}{4}, 0\right)$

(C)
The given equation of parabola is
$y=2 x^2+x$
$\Rightarrow x^2+\frac{x}{2}=\frac{y}{2}$
$\Rightarrow\left(x+\frac{1}{4}\right)^2=\frac{y}{2}+\frac{1}{16} \Rightarrow\left(x+\frac{1}{4}\right)^2=\frac{1}{2}\left(y+\frac{1}{8}\right)$
Let $X ^2=\frac{1}{2} Y \quad\ldots(i)$
Here $A=\frac{1}{8}$, focus of (i) is $\left(0, \frac{1}{8}\right)$
i.e., $X=0, Y=\frac{1}{8}$
$\Rightarrow x+\frac{1}{4}=0, y+\frac{1}{8}=\frac{1}{8}$
$\Rightarrow x=-\frac{1}{4}, y=0$
∴ focus of given parabola is $\left(-\frac{1}{4}, 0\right)$.

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MCQ 1522 Marks

The focus of the parabola $y^2-4 y-x+3=0$ is

A
$\left(\frac{3}{4}, 2\right)$
B
$\left(\frac{3}{4},-2\right)$
C
$\left(2, \frac{-3}{4}\right)$
✓
$\left(\frac{-3}{4}, 2\right)$

Answer

Correct option: D.

$\left(\frac{-3}{4}, 2\right)$

(D)
$y^2-4 y-x+3=0$
$\Rightarrow y^2-4 y+4-x+3-4=0$
$\Rightarrow(y-2)^2-(x+1)=0$
$\Rightarrow(y-2)^2=(x+1)$
Comparing with $Y ^2=4 aX$, we get
$a =\frac{1}{4}, Y =y-2, X =x+1$
Focus of the parabola is $X = a , Y =0$
$\Rightarrow x+1=\frac{1}{4}, y-2=0$
$\Rightarrow x=\frac{-3}{4}, y=2$
∴ focus $=\left(\frac{-3}{4}, 2\right)$

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MCQ 1532 Marks

A point on the parabola whose focus is $S(1,-1)$ and whose vertex is $A(1,1)$ is

✓
$\left(3, \frac{1}{2}\right)$
B
(1, 2)
C
$\left(2, \frac{1}{2}\right)$
D
(2, 2)

Answer

Correct option: A.

$\left(3, \frac{1}{2}\right)$

(A)
Equation of parabola having vertex ( $p, q$ ) and
focus $(p, b+q)$ is given by
$(x-p)^2=4 b(y-q)$
Given, vertex $A=(1,1)$ and focus $S=(1,-1)$
$\therefore p=1, q=1, b=-2$
∴ Equation of parabola is
$(x-1)^2=4(-2)(y-1)$
i.e. $x^2-2 x+8 y-7=0$
Only $\left(3, \frac{1}{2}\right)$ satisfies the above equation of parabola.

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MCQ 1542 Marks

The equation of the parabola whose vertex and focus lies on the X -axis at distance a and $a ^{\prime}$ from the origin, is

✓
$y^2=4\left(a^{\prime}-a\right)(x-a)$
B
$y^2=4\left(a^{\prime}-a\right)(x+a)$
C
$y^2=4\left(a^{\prime}+a\right)(x-a)$
D
$y^2=4\left(a^{\prime}+a\right)(x+a)$

Answer

Correct option: A.

$y^2=4\left(a^{\prime}-a\right)(x-a)$

(A)
Equation will be of the form $y^2=4 A(x-a)$,
where $A =\left( a ^{\prime}- a \right)$ or $y^2=4\left( a ^{\prime}- a \right)(x- a )$.

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MCQ 1552 Marks

The equation of parabola whose vertex and focus are (0, 4) and (0, 2) respectively, is

A
$y^2-8 x=32$
B
$y^2+8 x=32$
✓
$x^2+8 y=32$
D
$x^2-8 y=32$

Answer

Correct option: C.

$x^2+8 y=32$

(C)
$\begin{array}{l}\text { Vertex }=(0,4), \text { focus }=(0,2) \\ \Rightarrow a =2\end{array}$
Hence, equation of parabola is
$(x-0)^2=-4 \times 2(y-4)$
i.e., $x^2+8 y=32$

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MCQ 1562 Marks

A parabola has the origin as its focus and the line $x=2$ as the directrix. Then the vertex of the parabola is at

✓
(1, 0)
B
(0, 1)
C
(2, 0)
D
(0, 2)

Answer

Correct option: A.

(1, 0)

(A)
Since vertex is the midpoint of focus and directrix.
$\therefore \quad$ vertex $=\left(\frac{0+2}{2}, \frac{0+0}{2}\right)=(1,0)$

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MCQ 1572 Marks

If $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ are the end points of focal chord of the parabola $y^2=5 x$, the $4 x_1 x_2+y_1 y_2=$

A
25
B
5
✓
$0$
D
$\frac{5}{4}$

Answer

Correct option: C.

$0$

(C)
Given $y^2=5 x$
Here $\left(x_1, y_1\right)=(a, 2 a)$ and $\left(x_2, y_2\right)=(a,-2 a)$
$\therefore \quad x_1 x_2=a^2=\left(\frac{25}{16}\right)$
$\therefore \quad a=\frac{5}{4}$
$y_1 y_2=-4 a^2=\frac{-25}{4}$
$\therefore \quad 4 x_1 x_2+y_1 y_2=0$

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MCQ 1582 Marks

The parametric representation $\left(2+t^2, 2 t+1\right)$ represents

A
a parabola with focus (1, 2)
✓
a parabola with vertex (2, 1)
C
a parabola with vertex (0, 0)
D
a parabola with vertex (1, 2)

Answer

Correct option: B.

a parabola with vertex (2, 1)

(B)
$x=2+t^2$ and $y=2 t+1$
$\Rightarrow t ^2=(x-2)$ and $t ^2=\left(\frac{y-1}{2}\right)^2=\frac{(y-1)^2}{4}$
$\Rightarrow \frac{(y-1)^2}{4}=(x-2) \Rightarrow(y-1)^2=4(x-2)$
$\therefore$ Vertex is $(2,1)$

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MCQ 1592 Marks

The length of the latus-rectum of the parabola whose focus is $\left(\frac{ u ^2}{2 g} \sin 2 \alpha,-\frac{ u ^2}{2 g} \cos 2 \alpha\right)$ and directrix is $y=\frac{ u ^2}{2 g}$, is

A
$\frac{ u ^2}{g} \cos ^2 \alpha$
B
$\frac{ u ^2}{g} \cos 2 \alpha$
C
$\frac{2 u ^2}{g} \cos ^2 2 \alpha$
D
$\frac{2 u ^2}{g} \cos ^2 \alpha$

Answer

(D)

According to the figure, the length of latus rectum is
$2( SM )=2 \times \frac{ u ^2}{2 g}(1+\cos 2 \alpha)=\frac{2 u ^2 \cos ^2 \alpha}{g}$

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MCQ 1602 Marks

The latus rectum of a parabola whose directrix is $x+y-2=0$ and focus is $(3,-4)$. is

A
$2 \sqrt{2}$
✓
$3 \sqrt{2}$
C
$6 \sqrt{2}$
D
$\frac{3}{\sqrt{2}}$

Answer

Correct option: B.

$3 \sqrt{2}$

(B)
Distance between focus and directrix is
$=\left|\frac{3-42}{\sqrt{2}}\right|=\frac{3}{\sqrt{2}}$
Hence, latus rectum is $3 \sqrt{2}$
....[Since latus rectum is two times the distance between focus and directrix]

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MCQ 1612 Marks

The equation of latus rectum of a parabola is $x+y=8$ and the equation of the tangent at the vertex is $x+y=12$, then length of the latus rectum is

A
$4 \sqrt{2}$
B
$2 \sqrt{2}$
C
8
✓
$8 \sqrt{2}$

Answer

Correct option: D.

$8 \sqrt{2}$

(D)
Since $a =$ distance between tangent at vertex and latus rectum
$\therefore \quad a=\left|\frac{-8-(-12)}{\sqrt{1+1}}\right|=\frac{4}{\sqrt{2}}$
Length of latus rectum $=4 a=4 \times \frac{4}{\sqrt{2}}=8 \sqrt{2}$

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MCQ 1622 Marks

The axis of the parabola $9 y^2-16 x-12 y-57=0$, is

✓
3y = 2
B
x + 3y = 3
C
2x = 3
D
y = 3

Answer

Correct option: A.

3y = 2

(A)
Since $9 y^2-16 x-12 y-57=0$
$\Rightarrow y^2-\frac{16}{9} x-\frac{4}{3} y-\frac{57}{9}=0$
$\Rightarrow y^2-\frac{4}{3} y+\frac{4}{9}=\frac{16}{9} x+\frac{57}{9}+\frac{4}{9}$
$\Rightarrow\left(y-\frac{2}{3}\right)^2=\frac{16}{9}\left(x+\frac{61}{16}\right)$
This equation can be written as
$Y ^2=4\left(\frac{4}{9}\right) X$
Axis of the parabola is $Y =0$
$\Rightarrow y-\frac{2}{3}=0$
$\Rightarrow 3 y=2$

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MCQ 1632 Marks

If the line $x-1=0$ is the directrix of the parabola $y^2-k x+8=0$, then one of the values of k is

A
$\frac{1}{8}$
B
$0$
✓
4
D
$\frac{1}{4}$

Answer

Correct option: C.

(C)
Given, equation can be written as
$y^2=\frac{4 k }{4}\left(x-\frac{8}{ k }\right)$
The standard equation of parabola is
$y^2=4 a x$
$\Rightarrow a =\frac{ k }{4}$
∴ Equation of directrix is $X +\frac{ k }{4}=0$
$\Rightarrow x-\frac{8}{k}+\frac{k}{4}=0$
But the given equation of directrix is $x-1=0$. Since both the equations are same.
$\therefore \quad \frac{8}{k}-\frac{k}{4}=1$
$\Rightarrow 32-k^2=4 k \Rightarrow k=-8,4$

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MCQ 1642 Marks

If the vertex of the parabola $y=x^2-8 x+ c$ lies on $X$-axis, then the value of c is

A
-16
B
-4
C
4
✓
16

Answer

Correct option: D.

(D)
The given equation can be written as
$(x-4)^2=1[y-(c-16)]$
Therefore, the vertex of the parabola is $(4, c -16)$.
The point lies on X -axis.
$\Rightarrow c-16=0$
$\Rightarrow c=16$

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MCQ 1652 Marks

The equation of parabola is $y^2+8 x-12 y+20=0$, then which of the following is correct?

✓
Vertex (2,6)
B
Focus (0,6)
C
Latus rectum = 14
D
Axis y = 4

Answer

Correct option: A.

Vertex (2,6)

(A)
$y^2+8 x-12 y+20=0$
$\Rightarrow y^2-12 y=-8 x-20$
$\Rightarrow y^2-12 y+36=-8 x-20+36$
$\Rightarrow(y-6)^2=-8 x+16$
$\Rightarrow(y-6)^2=-8(x-2)$
⇒ Vertex is $(2,6)$

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MCQ 1662 Marks

Vertex of the parabola $9 x^2-6 x+36 y+9=0$ is

✓
$\left(\frac{1}{3},-\frac{2}{9}\right)$
B
$\left(\frac{-1}{3}, \frac{-1}{2}\right)$
C
$\left(\frac{-1}{3}, \frac{1}{2}\right)$
D
$\left(\frac{1}{3}, \frac{1}{2}\right)$

Answer

Correct option: A.

$\left(\frac{1}{3},-\frac{2}{9}\right)$

(A)
$9 x^2-6 x+36 y+9=0$
$\Rightarrow x^2-\frac{2}{3} x+4 y+1=0$
$\Rightarrow x^2-\frac{2}{3} x+\frac{1}{9}+4 y+1-\frac{1}{9}=0$
$\Rightarrow\left(x-\frac{1}{3}\right)^2=-4\left(y+\frac{2}{9}\right)$
Hence, the vertex is $\left(\frac{1}{3},-\frac{2}{9}\right)$

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MCQ 1672 Marks

The equation of the parabola whose vertex is at (2, -1) and focus at (2, -3) is

A
$x^2+4 x-8 y-12=0$
B
$x^2-4 x+8 y+12=0$
C
$x^2+8 y=12$
D
$x^2-4 x+12=0$

Answer

(B)
$a = VS =\sqrt{(2-2)^2+(-3+1)^2}$
$=2$
Since
$(x- h )^2=-4 a (y- k )$
$\therefore \quad(x-2)^2=-4 \times 2(y+1)$
$\Rightarrow(x-2)^2=-8(y+1)$
$\Rightarrow x^2+4-4 x=-8 y-8$
$\Rightarrow x^2-4 x+8 y+12=0$

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MCQ 1682 Marks

The equation of the parabola whose vertex is (-1, -2), axis is vertical and which passes through the point (3, 6), is

✓
$x^2+2 x-2 y-3=0$
B
$2 x^2=3 y$
C
$x^2-2 x-y+3=0$
D
$3 x^2=2 y$

Answer

Correct option: A.

$x^2+2 x-2 y-3=0$

(A)
Since the axis of parabola is vertical and it's vertex is $(-1,-2)$.
∴ equation of parabola is
$(x+1)^2=4 a(y+2)$
Also, it passes through $(3,6)$
$\therefore \quad 16=4 a \times 8$
$\Rightarrow a =\frac{1}{2} \Rightarrow(x+1)^2=2(y+2)$
$\Rightarrow x^2+2 x-2 y-3=0$

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MCQ 1692 Marks

The equation of the parabola with its vertex at the origin, axis on the Y -axis and passing through the point (6, -3), is

A
$y^2=12 x+6$
B
$x^2=12 y$
✓
$x^2=-12 y$
D
$y^2=-12 x+6$

Answer

Correct option: C.

$x^2=-12 y$

(C)
Since the axis of parabola is Y -axis.
∴ Equation of parabola is $x^2=4 a y$
Since it passes through $(6,-3)$
$\therefore 36=-12 a$
$\Rightarrow a=-3$
∴ Equation of parabola is $x^2=-12 y$

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MCQ 1702 Marks

The locus of the point of intersection of the perpendicular tangents to the parabola $x^2=4 a y$ is

A
Axis of the parabola
✓
Directrix of the parabola
C
Focal chord of the parabola
D
Tangent at vertex to the parabola

Answer

Correct option: B.

Directrix of the parabola

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MCQ 1712 Marks

The line $y=2 x+ c$ is a tangent to the parabola $y^2=16 x$, if c equals

A
-2
B
-1
C
$0$
✓
2

Answer

Correct option: D.

(D)
According to the given condition,
$c =\frac{4}{2}=2 \quad \ldots\left[\because c =\frac{ a }{ m }\right]$

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MCQ 1722 Marks

The line $x-y+2=0$ touches the parabola $y^2=8 x$ at the point

A
(2, -4)
B
$(1,2 \sqrt{2})$
C
$(4,-4 \sqrt{2})$
✓
(2, 4)

Answer

Correct option: D.

(2, 4)

(D)
The line $y= mx + c$ touches the parabola
$y^2=4 a x$ at the point $\left(\frac{ a }{ m ^2}, \frac{2 a }{ m }\right)$.
Here, $a =2$ and $m =1$
⇒ The point is $(2,4)$

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MCQ 1732 Marks

The point at which the line $y= m x+ c$ touches the parabola $y^2=4 ax$ is

✓
$\left(\frac{ a }{ m ^2}, \frac{2 a }{ m }\right)$
B
$\left(\frac{ a }{ m ^2}, \frac{-2 a }{ m }\right)$
C
$\left(-\frac{ a }{ m ^2}, \frac{2 a }{ m }\right)$
D
$\left(-\frac{ a }{ m ^2},-\frac{2 a }{ m }\right)$

Answer

Correct option: A.

$\left(\frac{ a }{ m ^2}, \frac{2 a }{ m }\right)$

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MCQ 1742 Marks

The equation of the tangent to the parabola $y^2=4 a x$ at point $\left(\frac{ a }{ t ^2}, \frac{2 a }{ t }\right)$ is

✓
$t y=x t^2+a$
B
$t y=x+a t^2$
C
$y= tx + at ^2$
D
$y=t x+\left(\frac{a}{t^2}\right)$

Answer

Correct option: A.

$t y=x t^2+a$

(A)
Equation of the tangent to the parabola,
$y^2=4 ax$ at $\left(x_1, y_1\right)$ is $y y_1=2 a \left(x+x_1\right)$
$\Rightarrow y \cdot \frac{2 a }{ t }=2 a \left(x+\frac{ a }{ t ^2}\right)$
$\Rightarrow \frac{y}{ t }=\left(x+\frac{ a }{ t ^2}\right)$
$\Rightarrow \frac{y}{t}=\frac{t^2 x+a}{t^2}$
$\Rightarrow t y= t ^2 x+ a$

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MCQ 1752 Marks

If the parabola $y^2=4 ax$ passes through the point $(1,-2)$, then the tangent at this point is

A
x + y - 1 = 0
B
x - y - 1 = 0
✓
x + y + 1 = 0
D
x - y + 1 = 0

Answer

Correct option: C.

x + y + 1 = 0

(C)
Since parabola passes through the point $(1,-2)$
$\Rightarrow 4=4 a$
$\Rightarrow a =1$
Required tangent is $y y_1=2 a \left(x+x_1\right)$
$\Leftrightarrow-2 y=2(x+1)$
$\Leftrightarrow x+y+1=0$

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MCQ 1762 Marks

The equation of the tangent to the parabola, $y^2=16 x$ at $(1,4)$ is

A
x = 4y
B
2x + y + 2 = 0
C
y = 4x
✓
2x - y + 2 = 0

Answer

Correct option: D.

2x - y + 2 = 0

(D)
Equation of tangent to the parabola
$y^2=16 x$ at $(1,4)$ is
$y y_1=2 \times 4\left(x+x_1\right)$
$\therefore 4 y=8(x+1)$
$\therefore 4 y=8 x+8$
$\therefore y=2 x+2$
$\therefore 2 x-y+2=0$

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MCQ 1772 Marks

$x-2= t ^2, y=2 t$ are the parametric equations of the parabola

A
$y^2=4 x$
B
$y^2=-4 x$
C
$x^2=-4 y$
✓
$y^2=4(x-2)$

Answer

Correct option: D.

$y^2=4(x-2)$

(D)
Here, $\frac{y}{2}= t$ and $x-2= t ^2$
$\Rightarrow(x-2)=\left(\frac{y}{2}\right)^2$
$\Rightarrow y^2=4(x-2)$

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MCQ 1782 Marks

The co-ordinates of a point on $2 y^2=7 x$ whose parameter is -2 are

✓
$\left(\frac{7}{2},-\frac{7}{2}\right)$
B
$\left(\frac{1}{2}, \frac{1}{3}\right)$
C
(1,-2)
D
(1,-3)

Answer

Correct option: A.

$\left(\frac{7}{2},-\frac{7}{2}\right)$

(A)
Given $t=-2$
$2 y^2=7 x$
$\Rightarrow y^2=\frac{7}{2} x$
Comparing with $y^2=4 a x$, we get
$4 a=\frac{7}{2} \Rightarrow a=\frac{7}{8}$
∴ the point is
$P=\left(a t^2, 2 a t\right)=\left(\frac{7}{8} \times 4,2 \times \frac{7}{8} \times(-2)\right)$
$\therefore \quad P =\left(\frac{7}{2}, \frac{-7}{2}\right)$

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MCQ 1792 Marks

Eccentricity of the parabola $x^2-4 x-4 y+4=0$ is

A
e = 0
✓
e = 1
C
e > 4
D
e = 4

Answer

Correct option: B.

e = 1

(B)
Eccentricity of parabola is always 1 i.e., $e =1$.

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MCQ 1802 Marks

The length of the latus rectum of the parabola $x^2-4 x-8 y+12=0$ is

A
4
B
6
✓
8
D
10

Answer

Correct option: C.

(C)
The given equation of parabola is $x^2-4 x-8 y+12=0$
$x^2-4 x-8 y+12=0$
$\Rightarrow x^2-4 x=8 y-12$
$\Rightarrow x^2-4 x+4=8 y-12+4$
$\Rightarrow(x-2)^2=8(y-1)$
Hence, the length of latus rectum $=4 a=8$.

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MCQ 1812 Marks

Axis of the parabola $x^2-4 x-3 y+10=0$ is

A
y + 2 = 0
B
x + 2 = 0
C
y - 2 = 0
✓
x - 2 = 0

Answer

Correct option: D.

x - 2 = 0

(D)
The parabola is $(x-2)^2=(3 y-6)$
Hence, axis is $x-2=0$.

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MCQ 1822 Marks

Vertex of the parabola $y^2+2 y+x=0$ lies in the

A
First quadrant
B
Second quadrant
C
Third quadrant
✓
Fourth quadrant

Answer

Correct option: D.

Fourth quadrant

(D)
$y^2+2 y+x=0$
$\Rightarrow y^2+2 y+1=-x+1$
$\Rightarrow(y+1)^2=-(x-1)$
Hence, vertex is $(1,-1)$, which lies in IV ${ }^{\text {th }}$ quadrant.

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MCQ 1832 Marks

If $(2,-8)$ is at one end of a focal chord of the parabola $y^2=32 x$, then the other end of the chord is

A
(2, -8)
✓
(2, 8)
C
(32, 32)
D
(32, -32)

Answer

Correct option: B.

(2, 8)

(B)
Since $(a,-2 a) \equiv(2,-8)$
$\therefore \quad$another end $(a, 2 a) \equiv(2,8)$

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MCQ 1842 Marks

Consider the equation of a parabola $y^2 + 4ax=0$, where a > 0. Which of the following is false?

A
Vertex of the parabola is at the origin.
✓
Focus of the parabola is at $(a, 0)$.
C
Directrix of the parabola is $x=a$.
D
Tangent at the vertex is $x=0$.

Answer

Correct option: B.

Focus of the parabola is at $(a, 0)$.

(B)
Parabola is $y^2=-4 a x$ (left handed parabola).
$\therefore \quad$ its focus is $(-a, 0)$.
$\therefore \quad$option (B) is false.

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MCQ 1852 Marks

Focus and directrix of the parabola $x^2=-8 ay$ are

✓
$(0,-2 a)$ and $y=2 a$
B
$(0,2 a )$ and $y=-2 a$
C
$(2 a, 0)$ and $x=-2 a$
D
$(-2 a , 0)$ and $x=2 a$

Answer

Correct option: A.

$(0,-2 a)$ and $y=2 a$

(A)
Given, equation is $x^2=-8 a y$
Here, $A =2 a$
Focus of parabola $(0,- A )$ i.e., $(0,-2 a )$
Directrix $y= A$ i.e., $y=2 a$

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MCQ 1862 Marks

If $(2,0)$ is the vertex and Y -axis the directrix of a parabola, then its focus is

A
(2, 0)
B
(-2, 0)
✓
(4, 0)
D
(-4, 0)

Answer

Correct option: C.

(4, 0)

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MCQ 1872 Marks

The equation of the parabola with focus $(3,0)$ and the directrix $x+3=0$ is

A
$y^2=3 x$
B
$y^2=2 x$
C
$y^2=12 x$
D
$y^2=6 x$

Answer

(C)

$\because SP ^2= PM ^2$
$\Rightarrow(x-3)^2+y^2=\left|\frac{x+3}{\sqrt{1}}\right|^2$
$\Rightarrow x^2+9-6 x+y^2=x^2+9+6 x$
$\Rightarrow y^2=12 x$

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MCQ 1882 Marks

The point on the parabola $y^2=36 x$ whose ordinate is three times the abscissa is

✓
(4, 12)
B
(6, 2)
C
(2, 6)
D
(1, 3)

Answer

Correct option: A.

(4, 12)

(A)
According to the given condition,
$y=3 x$
$y^2=36 x$
$∴(3 x)^2=36 x$
$\Rightarrow x=4 \text { and } y=12$
∴ Required point is $(4,12)$.

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MCQ 1892 Marks

The points on the parabola $y^2=12 x$ whose focal distance is 4 , are

A
$(2, \sqrt{3}),(2,-\sqrt{3})$
✓
$(1,2 \sqrt{3}),(1,-2 \sqrt{3})$
C
(1, 2)
D
None of these

Answer

Correct option: B.

$(1,2 \sqrt{3}),(1,-2 \sqrt{3})$

(B)
$y^2=12 x$
$\therefore \quad a=3$
⇒ abscissa is $4-3=1$ and $y^2=12$,
$y= \pm 2 \sqrt{3}$
Hence, points are $(1,2 \sqrt{3}),(1,-2 \sqrt{3})$

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MCQ 1902 Marks

The end points of the latus rectum of the parabola $x^2+5 y=0$ is

✓
$\left( \pm \frac{5}{2},-\frac{5}{4}\right)$
B
$\left( \pm \frac{2}{5}, \pm \frac{4}{5}\right)$
C
$\left( \pm \frac{4}{5}, \pm \frac{4}{5}\right)$
D
$\left( \pm \frac{5}{4},-\frac{5}{2}\right)$

Answer

Correct option: A.

$\left( \pm \frac{5}{2},-\frac{5}{4}\right)$

(A)
$x^2+5 y=0$
$\Rightarrow x^2=-5 y$
On comparing with $x^2=-4 a y$, we get
$a=\frac{5}{4}$
End points of latus rectum of the parabola are
$( \pm 2 a,-a)=\left( \pm \frac{5}{2}, \frac{-5}{4}\right)$

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MCQ 1912 Marks

The co-ordinates of end points of the latus rectum of the parabola $5 y^2=4 x$ are

A
$\left(\frac{1}{5}, \frac{2}{5}\right),\left(-\frac{1}{5}, \frac{2}{5}\right)$
✓
$\left(\frac{1}{5}, \frac{2}{5}\right),\left(\frac{1}{5},-\frac{2}{5}\right)$
C
$\left(\frac{1}{5}, \frac{4}{5}\right),\left(\frac{1}{5},-\frac{4}{5}\right)$
D
$\left(\frac{1}{5}, \frac{4}{5}\right),\left(-\frac{1}{5}, \frac{4}{5}\right)$

Answer

Correct option: B.

$\left(\frac{1}{5}, \frac{2}{5}\right),\left(\frac{1}{5},-\frac{2}{5}\right)$

(B)
$y^2=4 \cdot \frac{1}{5} x$
$\Rightarrow a =\frac{1}{5}$
Co-ordinates of latus rectum are (a, 2a) and (a, -2a)
i.e., $\left(\frac{1}{5}, \frac{2}{5}\right)$ and $\left(\frac{1}{5}, \frac{-2}{5}\right)$

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MCQ 1922 Marks

A parabola passing through the point (-4, -2) has its vertex at the origin and Y-axis as its axis. The latus rectum of the parabola is

A
6
✓
8
C
10
D
12

Answer

Correct option: B.

(B)
Let the equation of parabola be $x^2=-4 ay$
Since parabola passes through the point
$(-4,-2)$.
$\therefore(-4)^2=-4 a(-2)$
$\Rightarrow a=2$
$\therefore $ Equation becomes $x^2=-8 y$ and
latus rectum $=4 a=8$

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MCQ 1932 Marks

If the vertex of a parabola be at origin and directrix be x + 5 = 0 then its latus rectum is