MCQ

MCQ 11 Mark

Equation of the hyperbola whose vertices are $( \pm 3,0)$ and foci at $( \pm 5,0)$, is

✓
$16 x^2-9 y^2=144$
B
$9 x^2-16 y^2=144$
C
$25 \mathrm{x}^2-9 \mathrm{y}^2=225$
D
$9 x^2-25 y^2=81$

Answer

Correct option: A.

$16 x^2-9 y^2=144$

$16 x^2-9 y^2=144$

Solution:
The vertices of the hyperbola are $( \pm 3,0)$ and foci are $( \pm 5,0)$.
Thus, the value of a and ae are 3 and 5 , respectively.
Now, using the relation $b^2=a^2\left(e^2-1\right)$, we get:
$b^2=25-9$
$\Rightarrow b^2=16$
Equation of hyperbola is given below:
$\frac{x^2}{9}-\frac{y^2}{16}=1$
$16 x^2-9 y^2=144$

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MCQ 21 Mark

If $e_1$ and $e_2$ are respectively the eccentricities of the ellipse $\frac{\text{x}^2}{18}+\frac{\text{y}^2}{4}=1$ and the hyperbola $\frac{\text{x}^2}{9}-\frac{\text{y}^2}{4}=1,$ then the relation between $e_1$ and $e_2$ is

A
$3\text{e}_1^2 + \text{e}_2^2 = 2$
B
$\text{e}_1^2 + 2\text{e}_2^2 = 3$
✓
$2\text{e}_1^2 +\text{e}_2^2 = 3$
D
$\text{e}_1^2 + 3\text{e}_2^2 = 2$

Answer

Correct option: C.

$2\text{e}_1^2 +\text{e}_2^2 = 3$

$2\text{e}_1^2 +\text{e}_2^2 = 3$

Solution:
The standard from of the ellipse is $\frac{\text{x}^2}{18}+\frac{\text{y}^2}{4}=1,$ where $a^2 = 18$ and $b^2 = 4$.
So, the eccentricity is calculated in the following way:
$\text{b}2 = \text{a}2 (1 - \text{e}_1^2)$
$\Rightarrow4 = 18 (1 - \text{e}_1^2)$
$\Rightarrow\frac{2}{9}=1-\text{e}_1^2$
$\Rightarrow\text{e}_1^2=\frac{7}{9}$
The standard from of the hyperbola is $\frac{\text{x}^2}{9}-\frac{\text{y}^2}{4}=1,$ where $a^2 = 9$ and $b^2 = 4$.
So, the eccentricity is calculated in the following way:
$\text{b}^2 = \text{a}^2(\text{e}_2^2 - 1)$
$\Rightarrow4 = 9(\text{e}_2^2 - 1)$
$\Rightarrow\frac{4}{9}=\text{e}_2^2-1$
$\Rightarrow\text{e}_2^2=\frac{13}{9}$
$\therefore2\text{e}_1^2+\text{e}_2^2=\frac{2\times7}{9}+\frac{13}{9}$
$=\frac{27}{9}$
$=3$

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MCQ 31 Mark

The eccentricity of the hyperbola $x^2 - 4y^2 = 1$

A
$\frac{\sqrt3}{2}$
✓
${\frac{\sqrt5}{2}}$
C
${\frac{2}{\sqrt3}}$
D
$\frac{2}{\sqrt5}$

Answer

Correct option: B.

${\frac{\sqrt5}{2}}$

${\frac{\sqrt5}{2}}$

Solution:
The equation of the hyperbola is $x^2 - 4y^2 = 1$.
This can be rewritten in the following way:
$\frac{\text{x}^2}{1}-\frac{\text{y}^2}{\frac{1}{4}}=1$
This is the standard form of a hyperbola, where a = 1 and $\text{b}^2=\frac{1}{4}.$
The value of eccentricity is calculated in the following way:
$\text{b}^2=\text{a}^2(\text{e}^2-1)$
$\Rightarrow\frac{1}{4}=(\text{e}^2-1)$
$\Rightarrow\text{e}^2=\frac{5}{4}$
$\Rightarrow\text{e}=\frac{\sqrt5}{4}$

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MCQ 41 Mark

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

✓
$8\sqrt2$
B
$16\sqrt2$
C
$4\sqrt2$
D
$6\sqrt2$

Answer

Correct option: A.

$8\sqrt2$

$8\sqrt2$

Solution:
We have:
$\text{x}=8\sec\theta,\text{y}=8\tan\theta$
On squaring and subtracting:
$\text{x}^2-\text{y}^2=8\sec^2\theta-8\tan^2\theta$
$\Rightarrow\text{x}^2-\text{y}^2=8$
$\Rightarrow\frac{\text{x}^2}{8}-\frac{\text{y}^2}{8}=1$
$\therefore\text{a}=\text{b}=\text{c}$
Distance between the directrices of the hyperbola $=\frac{2\text{a}^2}{\sqrt{\text{a}^2+\text{b}^2}}$
Distance between the directrices $=\frac{2\times64}{\sqrt{64+64}}$
$=\frac{128}{8\sqrt2}$
$=\frac{16}{\sqrt2}$
$=8\sqrt2$

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MCQ 51 Mark

The equation of the conic with focus at $(1, -1)$ directrix along $x - y + 1 = 0$ and eccentricity $\sqrt2$ is

A
$x y=1$
B
$2 x y+4 x-4 y-1=0$
C
$x^2-y^2=1$
✓
$2 x y-4 x+4 y+1=0$

Answer

Correct option: D.

$2 x y-4 x+4 y+1=0$

d. $2 x y-4 x+4 y+1=0$
Solution:
Let $P(x, y)$ be any point on the hyperbola.
Then, the distance of any point from the focus is eccentricity times the distance from the directrix.
$\therefore \sqrt{(x-1)^2+(y+1)^2}=\sqrt{2}\left|\frac{x-y+1}{\sqrt{2}}\right|$
Squaring both the sides, we get:
$(x-1)^2+(y+1)^2=(x-y+1)^2$
$x^2-2 x+1+y^2+1+2 y=x^2+y^2+1-2 x y-2 y+2 x$
$2 x y-4 x+4 y+1=0$

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MCQ 61 Mark

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

A
$x^2+y^2=32$
B
$x^2-y^2=16$
C
$x^2+y^2=16$
✓
$x^2-y^2=32$

Answer

Correct option: D.

$x^2-y^2=32$

$x^2-y^2=32$

Solution:
The distance between the foci is 2ae.
$\therefore$ 2ae = 16
⇒ ae = 8
$\text{e}=\sqrt2$
$\therefore\text{a}\sqrt2=8$
$\Rightarrow\text{a}=4\sqrt2$
Also, $b^2 = a^2(e^2 − 1)$
$\Rightarrow b^2 = 32(2 − 1)$
$\Rightarrow b^2 = 32$
Standard form of the hyperbola is given below:
$\frac{\text{x}^2}{32}-\frac{\text{y}^2}{32}=1$
$\text{x}^2-\text{y}^2=32$

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MCQ 71 Mark

The eccentricity of the hyperbola whose latus-rectum is half of its transverse axis, is

A
$\frac{1}{\sqrt2}$
B
$\sqrt{\frac{2}{3}}$
✓
$\sqrt{\frac{3}{2}}$
D
None of these.

Answer

Correct option: C.

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

The lengths of the latus rectum and the transverse axis are $\frac{2\text{b}^2}{\text{a}}\text{ and }2\text{a},$ respectively.
According to the given statement, length of the latus rectum is half of its transverse axis.
$\therefore\frac{2\text{b}^2}{\text{a}}=\frac{1}{2}\times2\text{a}$
$\Rightarrow\frac{2\text{b}^2}{\text{a}}=\text{a}$
$\Rightarrow2\text{b}^2=\text{a}$
Eccentricity, $\text{e}=\frac{\sqrt{\text{a}^2+\text{b}^2}}{\text{a}}$
Substituting the value $\text{b}^2=\frac{\text{a}^2}{2},$ we get:
$\text{e}=\frac{\sqrt{\text{a}^+\frac{\text{a}}{2}}}{\text{a}}$
$=\frac{\text{a}\sqrt{\frac{3}{2}}}{\text{a}}$
$=\sqrt{\frac{3}{2}}$
$\therefore$ Eccentricity is $\sqrt{\frac{3}{2}}$

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MCQ 81 Mark

If $e_1$ is the eccentricity of the conic $9 x^2+4 y^2=36$ and $e_2$ is the eccentricity of the conic $9 x^2-4 y^2=36$, then

A
$\text{e}_1^2-\text{e}_2^2=2$
✓
$2<\text{e}_2^2-\text{e}_1^2<3$
C
$\text{e}_2^2-\text{e}_1^2=2$
D
$\text{e}_2^2-\text{e}_1^2>3$

Answer

Correct option: B.

$2<\text{e}_2^2-\text{e}_1^2<3$

$2<\text{e}_2^2-\text{e}_1^2<3$

Solution:
The conic $9x^2 + 4y^2 = 36$ can rewritten in the following way:
$\frac{9\text{x}^2}{36}+\frac{4\text{y}^2}{36}=1$
$\Rightarrow\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}=1$
This is the standard equation of an ellipse.
$\therefore b^2 = a^2(1−e_1)^2$
$\Rightarrow9=4(1-\text{e}_1)^2$
$\Rightarrow(\text{e}_1)^2=\frac{-5}{4}$
The conic $9x^2 − 4y^2 = 36$ can rewritten in the following way:
$\frac{9\text{x}^2}{36}-\frac{4\text{y}^2}{36}=1$
$\Rightarrow\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}=1$
This is the standard equation of a hyperbola.
$\therefore b^2 = a^2(e_2^2 − 1)$
$\Rightarrow9=4(\text{e}_2^2-1)$
$\Rightarrow(\text{e}_2)^2=\frac{13}{4}$
$\therefore\text{e}_2^2-\text{e}_1^2=\frac{13}{4}+\frac{5}{4}=2.5$

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MCQ 91 Mark

If the eccentricity of the hyperbola $x^2-y^2 \sec ^2 \alpha=5$ is $\sqrt{3}$ times the eccentricity of the ellipse $x^2 \sec ^2 \alpha+y^2=25$, then $\alpha=$

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

Answer

Correct option: B.

$\frac{\pi}{4}$

$\frac{\pi}{4}$

Solution:
The hyperbola $\text{x}^2 − \text{y}^2 \sec^2\alpha = 5$ can be rewritten in the following way:
$\frac{\text{x}^2}{5}-\frac{\text{y}^2}{5\cos^2\text{a}}=1$
This is the standard form of a hyperbola, where $ \text{a}^2 = 5 \text{ and } \text{b}^2 = 5\cos^2\alpha.$
$\Rightarrow\text{b}^2 = \text{a}^2(\text{e}_1^2 − 1)$
$⇒ 5\cos^2\alpha=5(\text{e}_1^2−1)$
$\Rightarrow\text{e}_1^2=\cos^2\alpha+1...(1)$
The ellipse $\text{x}^2\sec^2\alpha+\text{y}^2=25$ can be rewritten in the following way:
$\frac{\text{x}^2}{25\cos^2\alpha}+\frac{\text{y}^2}{25}=1$
This is the standard form of an ellipse, where $\text{a}^2=25\text{ and }\text{b}^2=25\cos^2\alpha$
$\text{b}^2=\text{a}^2(1-\text{e}_2^2)$
$\Rightarrow\text{e}_2^2=1-\cos^2\alpha$
$\Rightarrow\text{e}_2^2=\sin^2\alpha...(2)$
According to the question,
$\cos^2\alpha+1=3(\sin^2\alpha)$
$\Rightarrow2=4\sin^2\alpha$
$\Rightarrow\sin\alpha=\frac{1}{\sqrt2}$
$\Rightarrow\alpha=\frac{\pi}{4}$

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MCQ 101 Mark

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

A
$(\pm4,0)$
B
$(0,\pm4)$
✓
$(\pm5,0)$
D
$(0,\pm5)$

Answer

Correct option: C.

$(\pm5,0)$

$(\pm5,0)$

Solution:
The equation of the hyperbola is given below:
$9x^2 − 16y^2 = 144$
This equation can be rewritten in the following way:
$\frac{9\text{x}^2}{144}-\frac{16\text{y}^2}{144}=1$
$\Rightarrow\frac{\text{x}^2}{16}-\frac{\text{y}^2}{9}=1$
This is the standard equation of a hyperbola, where $a^2 = 16$ and $b^2 = 9$.
The eccentricity is calculated in the following way:
$b^2 = a^2(e^2 − 1)$
$\Rightarrow 9 = 16(e^2 − 1)$
$\Rightarrow\frac{9}{16}=\text{e}^2-1$
$\Rightarrow\text{e}=\frac{5}{4}$
$\text{Foci}=(\pm\text{ae},0)=(\pm5,0)$

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MCQ 111 Mark

The equation of the conic $9x^2 - 16y^2 = 144$ is

✓
$\frac{5}{4}$
B
$\frac{4}{3} $
C
$\frac{4}{5}$
D
$\sqrt7$

Answer

Correct option: A.

$\frac{5}{4}$

$\frac{5}{4}$

Solution:
Standard form of a hyperbola $=\frac{\text{x}^2}{16}-\frac{\text{y}^2}{9}=1$
Here, $a^2 = 16$ and $y^2 = 9$
The eccentricity is calculated in the following way:
$b^2 = a^2(e^2 - 1)$
$\Rightarrow 9 = 16(e^2 - 1)$
$\Rightarrow\text{e}^2-1=\frac{9}{16} $
$\Rightarrow\text{e}^2=\frac{25}{16}$
$\Rightarrow\text{e}=\frac{5}{4}$

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MCQ 121 Mark

The difference of the focal distances of any point on the hyperbola is equal to

A
length of the conjugate axis.
B
eccentricity.
✓
length of the transverse axis.
D
Latus-rectum.

Answer

Correct option: C.

length of the transverse axis.

Let P(x,y) be any point on the hyperbola, and S, S' be the focus with coordinates $(\pm\text{ae},0).$
⇒ S'P − SP = 2a
Thus, the difference of the focal distances of any point on the hyperbola is equal to the length of the transverse axis.

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MCQ 131 Mark

A point moves in a plane so that its distances PA and PB from two fixed points A and B in the plane satisfy the relation PA − PB = k (k ≠ 0), then the locus of P is

✓
A hyperbola.
B
A branch of the hyperbola.
C
A parabola.
D
An ellipse.

Answer

Correct option: A.

A hyperbola.

A hyperbola.

Solution:
Let P(x, y) be any point on the hyperbola $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1.$
By definition, we have:
$\text{PA}=\text{e}\big(\text{x}-\frac{\text{a}}{\text{e}}\big)=\text{ex}-\text{a}$
$\text{and }\text{PB}=\text{e}\big(\text{x}+\frac{\text{a}}{\text{e}}\big)=\text{ex}+\text{a}$
$\therefore\text{PB}−\text{PA}=\text{(ex+a)}−\text{(ex}−\text{a})=\text{2a = k}$

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