Let one focus of the hyperbola $\mathrm{H}: \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$ 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 $9\left(e^2+l\right)$ is equal to :
Solution
<p>$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$</p>
<p>Directrix : $x=\frac{9}{\sqrt{10}}=\frac{a}{e}\quad\text{..... (i)}$</p>
<p>Focus : $(\sqrt{10}, 0) \equiv(a e, 0)$</p>
<p>$a e=\sqrt{10}\quad\text{..... (ii)}$</p>
<p>(i) $\times$ (ii)</p>
<p>$\Rightarrow a^2=9 \Rightarrow a=3$</p>
<p>Substitute in (ii)</p>
<p>$e=\frac{\sqrt{10}}{3}$</p>
<p>$$\begin{aligned}
& \text { Now } e^2=1+\frac{b^2}{a^2} \\
& \frac{10}{9}=1+\frac{b^2}{a} \\
& \Rightarrow b=1 \\
& I=\frac{2 b^2}{a}=\frac{2 \times 1}{3}=\frac{2}{3} \\
& a\left[e^2+l\right]=9\left[\frac{10}{9}+\frac{2}{3}\right]=10+6 \\
& =16
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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