JEE Main 2026 — 22 January, Evening Shift — Question 11

$\quad \mathrm{P}(10,2 \sqrt{15})$ lies on $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$

$$\begin{gathered} \therefore \frac{100}{\mathrm{a}^{2}}-\frac{60}{\mathrm{~b}^{2}}=1 \end{gathered}$$

∵ length of latus rectum $=8$ $\begin{gathered} \frac{2 \cdot b^{2}}{a}=8 \Rightarrow \frac{b^{2}}{a}=4 \end{gathered}$

From & (2) $\frac{100}{\mathrm{a}^{2}}-\frac{60}{4 \mathrm{a}}=1$

$400-60 a=4 a^{2}$

$4 a^{2}+60 a-400=0$

$\mathrm{a}^{2}+15 \mathrm{a}-100=0$

$\mathrm{a}=5 \&-20$ (rejected)

$\Rightarrow \mathrm{b}=\sqrt{20}$

∴ Hyperbola is $\frac{\mathrm{x}^{2}}{25}-\frac{\mathrm{y}^{2}}{20}=1$

∴ Focal length $\mathrm{S}_{1} \mathrm{~S}_{2}=2 \mathrm{ae}=2.5 \cdot\left(\sqrt{1+\frac{4}{5}}\right)=6 \sqrt{5}$

∴ Area of $\Delta \mathrm{PS}_{1} \mathrm{~S}_{2}=\frac{1}{2} \cdot 6 \sqrt{5} \cdot 2 \sqrt{15}=30 \sqrt{3}=\mathrm{A}$

$\therefore \mathrm{A}^{2}=2700$