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$$\begin{aligned} & \frac{x^2}{9}-\frac{y^2}{4}=1 \\ & a^2=9, b^2=4 \\ & b^2=a^2\left(e^2-1\right) \Rightarrow e^2=1+\frac{b^2}{a^2} \\ & e^2=1+\frac{4}{9}=\frac{13}{9} \\ & e=\frac{\sqrt{13}}{3} \Rightarrow s_1 s_2=2 \mathrm{ae}=2 \times 3 \times \sqrt{\frac{13}{3}}=2 \sqrt{13} \end{aligned}$$

$$\begin{aligned} & \text { Area of } \triangle \mathrm{PS}_1 \mathrm{~S}_2=\frac{1}{2} \times \beta \times \mathrm{S}_1 \mathrm{~S}_2=2 \sqrt{13} \\ & \Rightarrow \frac{1}{2} \times \beta \times(2 \sqrt{13})=2 \sqrt{13} \Rightarrow \beta=2 \\ & \begin{aligned} & \frac{\alpha^2}{9}-\frac{\beta^2}{4}=1 \Rightarrow \frac{\alpha^2}{9}-1=1 \Rightarrow \alpha^2=18 \Rightarrow \alpha=3 \sqrt{2} \\ & \text { Distance of P from origin }=\sqrt{\alpha^2+\beta^2} \\ &=\sqrt{18+4}=\sqrt{22} \end{aligned} \end{aligned}$$

19:T577,

JEE Main 2024 (Online) 30th January Evening Shift Mathematics - Hyperbola Question 29 Hindi Explanation

$$\begin{aligned} & \text { का क्षेत्रफल } \triangle \mathrm{PS}_1 \mathrm{~S}_2=\frac{1}{

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