Question

Fill in the blanks.
(i) The length of the latus rectum of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$ is P.
(ii) The equations of the hyperbola with vertices $\left(\pm 2, 0\right)$, foci $ \left(\pm3, 0\right)$ is Q.
(iii) If the distance between the foci of a hyperbola is $16$ and its eccentricity is $2$, then equation of the hyperbola is R.

	P	Q	R
(a)	$\frac{9}{2}$	$\frac{y^{2}}{4}-\frac{x^{2}}{5}=1$	$x^{2}-3y^{2}=48$
(b)	$\frac{3}{2}$	$\frac{y^{2}}{4}-\frac{x^{2}}{5}=1$	$3x^{2}-y^{2}=48$
(c)	$\frac{9}{2}$	$\frac{x^{2}}{4}-\frac{y^{2}}{5}=1$	$3x^{2}-y^{2}=48$
(d)	$\frac{9}{2}$	$\frac{x^{2}}{4}-\frac{y^{2}}{5}=1$	$x^{2}-y^{2}=48$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (C)

$\left(i\right)$ Given equation of hyperbola is
$\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$
Now, $a^{2} = 16$
$\Rightarrow a=4$ and $b^{2}=9$
$ \Rightarrow b=3$
Length of latus rectum $=\frac{2b^{2}}{a}=\frac{2\times9}{4}=\frac{9}{2}\cdot$
$\left(ii\right)$ Vertices are $\left(\pm2,0\right)$ which lie on $x$-axis.
So the equation of hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
Now, vertices are $\left(\pm2,0\right)$
$\Rightarrow a=2$
Foci are $\left(\pm3,0\right)$
$\Rightarrow ae=3$
$\Rightarrow e=\frac{3}{2}$
We know that
$b=a\sqrt{e^{2}-1}$
$ \Rightarrow b=2\sqrt{\frac{9}{4}}-1=2\frac{\sqrt{5}}{2}=\sqrt{5}$
Thus required equation of hyperbola is
$\frac{x^{2}}{\left(2\right)^{2}}-\frac{y^{2}}{\left(\sqrt{5}\right)^{2}}=1$
$\Rightarrow \frac{x^{2}}{4}-\frac{y^{2}}{5}=1$
$\left(iii\right)$ Since distance between the foci is $16$
$\Rightarrow 2ae - 16$
$\Rightarrow ae=8$
$\Rightarrow a=4 \, \left[\because\,e=2\right]$
Now, $b^{2} =a^{2} \left(e^{2}-1\right)=\left(4\right)^{2} \left[\left(2\right)^{2}-1\right]=16\times3=48$
Hence, the equation of hyperbola is $\frac{x^{2}}{16}-\frac{y^{2}}{48}=1$
$\Rightarrow 3x^{2}-y^{2}=48$