JEE Main 2025 — Ellipse Question with Solution

$$\begin{array}{rl}& \frac{x-\sqrt{5}}{\cos \theta }=\frac{y-\sqrt{5}}{\sin \theta }=r\\ & x=r\cos \theta +\sqrt{5}\\ & y=r\sin \theta +\sqrt{5}\\ & 25x^2+36y^2=900,(x,y)\text{ lie on }E\\ & r^2\left[25{\cos }^2\theta +36{\sin }^2\theta \right]+r[50\sqrt{5}\cos \theta +72\sqrt{5}\sin \theta ]\\ & +25[5]+36[5]=900\\ & r_1{r}_2=\frac{900-305}{25+11{\sin }^2\theta },{\left(r_1{r}_2\right)}_{\max }=\frac{595}{25}=\frac{119}{5}\\ & {\left|r_1\cdot r_2\right|}_{\max }=\left(\frac{595}{25}\right)=\left(\frac{119}{5}\right)\text{ at }\theta =0\end{array}$$

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