Let one focus of the hyperbola H:x^2a^2-y^2b^2=1 be at (√10,0) and the corresponding direc…

Question

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

Accepted Answer

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$

Directrix : $x=\frac{9}{\sqrt{10}}=\frac{a}{e}\quad\text{..... (i)}$

Focus : $(\sqrt{10}, 0) \equiv(a e, 0)$

$$a e=\sqrt{10}\quad\text{..... (ii)}$$

(i) $\times$ (ii)

$$\Rightarrow a^2=9 \Rightarrow a=3$$

Substitute in (ii)

$$e=\frac{\sqrt{10}}{3}$$

$$\begin{aligned}
& \text { Now } e^2=1+\frac{b^2}{a^2} \
& \frac{10}{9}=1+\frac{b^2}{a} \
& \Rightarrow b=1 \
& I=\frac{2 b^2}{a}=\frac{2 \times 1}{3}=\frac{2}{3} \
& a\left[e^2+l\right]=9\left[\frac{10}{9}+\frac{2}{3}\right]=10+6 \
& =16
\end{aligned}$$

Solutions · 1