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$$\begin{aligned} & \frac{x-\sqrt{5}}{\cos \theta}=\frac{y-\sqrt{5}}{\sin \theta}=r \\ & x=r \cos \theta+\sqrt{5} \\ & y=r \sin \theta+\sqrt{5} \\ & 25 x^2+36 y^2=900,(x, y) \text { lie on } E \\ & r^2\left[25 \cos ^2 \theta+36 \sin ^2 \theta\right]+r[50 \sqrt{5} \cos \theta+72 \sqrt{5} \sin \theta] \\ & \quad+25[5]+36[5]=900 \\ & r_1 r_2=\frac{900-305}{25+11 \sin ^2 \theta},\left(r_1 r_2\right)_{\max }=\frac{595}{25}=\frac{119}{5} \\ & \left|r_1 \cdot r_2\right|_{\max }=\left(\frac{595}{25}\right)=\left(\frac{119}{5}\right) \text { at } \theta=0 \end{aligned}$$

JEE Main 2025 (Online) 3rd April Morning Shift Mathematics - Ellipse Question 20 English Explanation 2

$$\begin{aligned} & \frac{x^2}{36}+\frac{y^2}{25}=1 \\ & \text { At } y=\sqrt{5} \\ & \frac{x^2}{36}+\frac{1}{5}=1 \\ & \Rightarrow \frac{x^2}{36}=\frac{4}{5} \Rightarrow x= \pm \frac{12}{\sqrt{5}} \\ & P A^2+P B^2=(P A+P B)^2-2 P A \cdot P B=\left(\frac{24}{\sqrt{5}}\right)^2-2 \cdot \frac{119}{5} \\ & 5\left(P A^2+P B^2\right)=5\left(\frac{24^2}{5}-\frac{2.119}{5}\right)=24^2-238 \\ & =338 . \end{aligned}$$

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