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 $\left(\pm 3,0\right)$ 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 $=\frac{2b^2}{a}=\frac{2\times 9}{4}=\frac{9}{2}\cdot$
$\left(ii\right)$ Vertices are $\left(\pm 2,0\right)$ 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 $\left(\pm 2,0\right)$
$\Rightarrow a=2$
Foci are $\left(\pm 3,0\right)$
$\Rightarrow ae=3$
$\Rightarrow e=\frac{3}{2}$
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, $b^2=a^2\left(e^2-1\right)={\left(4\right)}^2\left[{\left(2\right)}^2-1\right]=16\times 3=48$
Hence, the equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{48}=1$
$\Rightarrow 3x^2-y^2=48$