Medium MCQ +4 / -1 PYQ · JEE Mains 2025

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 :

  1. A 12
  2. B 14
  3. C 15
  4. D 16 Correct answer

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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