JEE Main 2024 — 4 April, Shift 2 — Question 14

Let $\mathrm{H}: \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1 \quad\left(\mathrm{~b}^{2}=\mathrm{a}^{2}\left(\mathrm{e}^{2}-1\right)\right)$

$\therefore \mathrm{eq}^{\mathrm{n}}$ of $\mathrm{C}_{1}=\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2}$

Ar. $=36 \pi$

$\pi \mathrm{a}^{2}=36 \pi$

$\mathrm{a}=6$

Now radius of $\mathrm{C}_{2}$ can be $\mathrm{a}(\mathrm{e}-1)$ or $\mathrm{a}(\mathrm{e}+1)$

for $r=a(e-1) \quad$

for $r=a(e+1)$

Ar. $=4 \pi$

$\pi r^{2}=4 \pi$

$\pi \mathrm{a}^{2}(\mathrm{e}-1)^{2}=4 \pi$

$\mathrm{a}^{2}(\mathrm{e}+1)^{2}=4$

$36 \pi(e-1)^{2}=4 \pi$

$36(e+1)^{2}=4$

$\mathrm{e}-1=\frac{1}{3}$

$e+1=\frac{1}{3}$

$\mathrm{e}=\frac{4}{3}$

$-\frac{2}{3}$

Not possible

$\therefore \mathrm{b}^{2}=36\left(\frac{16}{9}-1\right)=28$

$\therefore \mathrm{LR}=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=\frac{2 \times 28}{6}=\frac{28}{3}$