JEE Main 2024 — Hyperbola Question with Solution

The given equation of hyperbola is,  $\mathrm{H}:\frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{9}=1$

$\Rightarrow {\mathrm{e}}_1=\sqrt{1+\frac{3^2}{4^2}}=\sqrt{\frac{16+9}{16}}$

$\Rightarrow {\mathrm{e}}_1=\frac{5}{4}$

Foci of this hyperbola is given by, $\mathrm{F}\left(\pm \mathrm{ae},0\right)$.

$\Rightarrow \mathrm{F}\left(\pm 5,0\right)$

It is given that, ${\mathrm{e}}_1{\mathrm{e}}_2=1$

$\Rightarrow {\mathrm{e}}_2=\frac{4}{5}$

Also, ellipse is passing through $(\pm 5,0)$

$\Rightarrow \frac{5^2}{{\mathrm{a}}^2}+\frac{0}{{\mathrm{b}}^2}=1$

$\Rightarrow \frac{25}{{\mathrm{a}}^2}=1$

$\Rightarrow {\mathrm{a}}^2=25$

$\Rightarrow \mathrm{a}=5$

Now, ${\mathrm{e}}_2=\sqrt{1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}$

$\Rightarrow \frac{4}{5}=\sqrt{1-\frac{{\mathrm{b}}^2}{25}}$

$\Rightarrow \frac{16}{25}=1-\frac{{\mathrm{b}}^2}{25}$

$\Rightarrow \frac{{\mathrm{b}}^2}{25}=\frac{9}{25}$

$\Rightarrow {\mathrm{b}}^2=9$

$\therefore \mathrm{a}=5$ and $\mathrm{b}=3$

$\mathrm{E}:\frac{{\mathrm{x}}^2}{25}+\frac{{\mathrm{y}}^2}{9}=1$

Putting, $\mathrm{y}=2$

$\Rightarrow \frac{{\mathrm{x}}^2}{25}+\frac{4}{9}=1$

$\Rightarrow \frac{{\mathrm{x}}^2}{25}=\frac{5}{9}$

$\Rightarrow {\mathrm{x}}^2=\frac{125}{9}$

$\Rightarrow \mathrm{x}=\pm \frac{5\sqrt{5}}{3}$

So, end points of chord are $\left(\pm \frac{5\sqrt{5}}{3},2\right)$

Thus, length of chord $\mathrm{PQ}$- is given by,

${\mathrm{L}}_{\mathrm{PQ}}=2\times \frac{5\sqrt{5}}{3}$

$\Rightarrow {\mathrm{L}}_{\mathrm{PQ}}=\frac{10\sqrt{5}}{3}$