Hyperbola Question - None

Question

Consider a hyperbola $H$ having centre at the origin and foci and the $x$-axis. Let $C_{1}$ be the circle touching the hyperbola $H$ and having the centre at the origin. Let $C_{2}$ be the circle touching the hyperbola $H$ at its vertex and having the centre at one of its foci. If areas (in sq. units) of $C_{1}$ and $C_{2}$ are $36 \pi$ and $4 \pi$, respectively, then the length (in units) of latus rectum of $H$ is

Accepted Answer

Equation of hyperbola is
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \left(\because b^{2}=a^{2}\left(e^{2}-1 ight) ight)$
$ herefore$ eqn${ }^{n}$ of $C_{1}=x^{2}+y^{2}=a^{2}$
Area $=36 \pi \Rightarrow \pi a^{2}=36 \pi$
$\Rightarrow a=6$
Now radius of $C_{2}$ can be $a(e-1)$ or $a(e+1)$
For $r=a(e-1)$
Area $=4 \pi$
$\Rightarrow \pi a^{2}(e-1)^{2}=4 \pi$
$\Rightarrow 36 \pi(e-1)^{2}=4 \pi$
$\Rightarrow e-1=\frac{1}{3} \Rightarrow e=\frac{4}{3}$
For $r=a(e+1)$
$\Rightarrow \pi r^{2}=4 \pi$
$\Rightarrow a^{2}(e-1)^{2}=4$
$\Rightarrow 36(e+1)^{2}=4$
$\Rightarrow e+1=\frac{1}{3}=-\frac{2}{3}$
Not possible
$ herefore b^{2}=36\left(\frac{16}{9}-1 ight)=28$
$ herefore$ Latus Rectum $=\frac{2 b^{2}}{a}$
$=\frac{2 imes 28}{6}=\frac{28}{3}$

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