JEE Main 2024 — Hyperbola Question with Solution

Given equation of ellipse is, $\frac{x^2}{9}+\frac{y^2}{25}=1$

$\Rightarrow a=3, b=5$

We know that, $e=\sqrt{1-\frac{a^2}{b^2}}$

$\Rightarrow e=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$

Now, $\text{foci}=\left(0,\pm \mathrm{be}\right)$

$=\left(0,\pm 4\right)$

Since, eccentricity of hyperbola is given as $\frac{15}{8}$ same to that of ellipse, $\therefore e_H=\frac{4}{5}\times \frac{15}{8}=\frac{3}{2}$

Let equation of the hyperbola be $\frac{x^2}{A^2}-\frac{y^2}{B^2}=-1$.

$\Rightarrow B.e_H=4$

$\Rightarrow B=\frac{8}{3}$

$\Rightarrow A^2=B^2\left({e_H}^2-1\right)=\frac{64}{9}\left(\frac{9}{4}-1\right)$

$\Rightarrow A^2=\frac{80}{9}$

$\Rightarrow \frac{x^2}{{\displaystyle \frac{80}{9}}}-\frac{y^2}{{\displaystyle \frac{64}{9}}}=-1$

Directrix: $y=\pm \frac{B}{e_H}=\pm \frac{16}{9}$

$\Rightarrow PS=e·PM$

$\Rightarrow PS=\frac{3}{2}\left|\frac{14}{3}·\sqrt{\frac{2}{5}}-\frac{16}{9}\right|$

$\Rightarrow PS=7\sqrt{\frac{2}{5}}-\frac{8}{3}$