JEE MAIN - Mathematics (2025 - 2nd April Morning Shift - No. 10)
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 :
12
14
15
16
Explanation
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$
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)
$$e=\frac{\sqrt{10}}{3}$$
$$\begin{aligned} & \text { Now } e^2=1+\frac{b^2}{a^2} \\ & \frac{10}{9}=1+\frac{b^2}{a} \\ & \Rightarrow b=1 \\ & I=\frac{2 b^2}{a}=\frac{2 \times 1}{3}=\frac{2}{3} \\ & a\left[e^2+l\right]=9\left[\frac{10}{9}+\frac{2}{3}\right]=10+6 \\ & =16 \end{aligned}$$
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