Let one focus of the hyperbola $\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ be at $(\sqrt{10}, 0)$ and the corresponding directrix be $\mathrm{x}=\frac{9}{\sqrt{10}}$. If e and $l$ respectively are the eccentricity and the length of the latus rectum of H ,…

Question

Let one focus of the hyperbola $\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ be at $(\sqrt{10}, 0)$ and the corresponding directrix be $\mathrm{x}=\frac{9}{\sqrt{10}}$. If e and $l$ respectively are the eccentricity and the length of the latus rectum of H , then $9\left(\mathrm{e}^2+l\right)$ is equal to:

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{ae}=\sqrt{10} 	ext { and } \frac{\mathrm{a}}{\mathrm{e}}=\frac{9}{10} \ & \Rightarrow \mathrm{a}^2=9 	ext { and } \mathrm{e}=\frac{\sqrt{10}}{3} \ & 	ext { Now } \quad \begin{aligned} & (\mathrm{ae})^2=\mathrm{a}^2+\mathrm{b}^2 \ & 10=9+\mathrm{b}^2 \Rightarrow \mathrm{~b}^2=1 \ & \ & \ell=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{2(1)}{3} \ \Rightarrow & 9\left(\mathrm{e}^2+\ellight) \ & =9\left(\frac{10}{9}+\frac{2}{3}ight) \ & =10+6 \ & =16\end{aligned}\end{aligned}$

Let one focus of the hyperbola $\mathrm{H}:…

Solution