Consider a hyperbola $\mathrm{H}$ having centre at the origin and foci on the $\mathrm{x}$-axis. Let…
Consider a hyperbola $\mathrm{H}$ having centre at the origin and foci on the $\mathrm{x}$-axis. Let $\mathrm{C}_1$ be the circle touching the hyperbola $\mathrm{H}$ and having the centre at the origin. Let $\mathrm{C}_2$ be the circle touching the hyperbola $\mathrm{H}$ 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 $\mathrm{H}$ is
$\frac{14}{3}$
$\frac{28}{3}$
$\frac{11}{3}$
$\frac{10}{3}$
Solution
Let $\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \quad\left(\mathrm{~b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\right)$
$\begin{aligned}
& \therefore \mathrm{eq}^{\mathrm{n}} \text { of } \mathrm{C}_1 \\
& \text { Ar. }=36 \pi \\
& \pi \mathrm{a}^2=36 \pi \\
& \mathrm{a}=6
\end{aligned}$
$\therefore \mathrm{eq}^{\mathrm{n}} \text { of } \mathrm{C}_1=\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2$ Now radius of $\mathrm{C}_2$ can be $\mathrm{a}(\mathrm{e}-1)$ or $\mathrm{a}(\mathrm{e}+1)$ for $\mathrm{r}=\mathrm{a}(\mathrm{e}-1)$ for $\mathrm{r}=\mathrm{a}(\mathrm{e}+1)$ Ar. $=4 \pi$ $\pi r^2=4 \pi$
$\pi \mathrm{a}^2(\mathrm{e}-1)^2=4 \pi$
$\mathrm{a}^2(\mathrm{e}+1)^2=4$
$36 \pi(e-1)^2=4 \pi$
$36(e+1)^2=4$
$\mathrm{e}-1=\frac{1}{3}$
$\mathrm{e}+1=\frac{1}{3}$
$\mathrm{e}=\frac{4}{3}$
$-\frac{2}{3}$
Not possible
$\begin{aligned}
& \therefore \mathrm{b}^2=36\left(\frac{16}{9}-1\right)=28 \\
& \therefore L R=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{2 \times 28}{6}=\frac{28}{3}
\end{aligned}$