JEE Main 2024 — 8 April, Shift 1 — Question 20

$\quad \mathrm{H}: \frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}-\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}=1$

$e=\sqrt{1+\frac{a^{2}}{b^{2}}}=\sqrt{3}$

$\mathrm{e}=\sqrt{3}$

$\Rightarrow \frac{a^{2}}{b^{2}}=2$

$a^{2}=2 b^{2}$

length of L.R.

$=\frac{2 \mathrm{a}^{2}}{\mathrm{~b}}=4 \sqrt{3}$

$a=\sqrt{6}$

$P(\alpha, 6)$ lie on $\frac{y^{2}}{3}-\frac{x^{2}}{6}=1$

$12-\frac{\alpha^{2}}{6}=1 \Rightarrow \alpha^{2}=66$

$Foci=\left( 0,\pm be \right)=\left( 0,3 \right)\And \left( 0,-3 \right)$

Let $\mathrm{d}_{1} \& \mathrm{~d}_{2}$ be focal distances of $\mathrm{P}(\alpha, 6)$

$d_{1}=\sqrt{\alpha^{2}+(6+b e)^{2}}$

$d_{2}=\sqrt{\alpha^{2}+(6-b e)^{2}}$

$d_{1}=\sqrt{66+81}$

$d_{2}=\sqrt{66+9}$

$\beta=d_{1} d_{2}=\sqrt{147 \times 75}=105$

$\alpha^{2}+\beta=66+105=171$