JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Ellipse

Take a point on the line through $P(\sqrt{5},\sqrt{5})$ as $(\sqrt{5}+r\cos \theta ,\sqrt{5}+r\sin \theta )$ and substitute into $\frac{x^2}{36}+\frac{y^2}{25}=1$. This yields $r^2(25{\cos }^2\theta +36{\sin }^2\theta )+2\sqrt{5}r(25\cos \theta +36\sin \theta )-595=0$, so $PA\cdot PB=\frac{595}{25+11{\sin }^2\theta }$, which is greatest when ${\sin }^2\theta =0$, i.e. the chord is horizontal. With $y=\sqrt{5}$, $\frac{x^2}{36}+\frac{1}{5}=1$ gives $x^2=\frac{144}{5}$, so $x=\pm \frac{12}{\sqrt{5}}$ and the distances from $P$ are $\frac{7}{\sqrt{5}}$ and $\frac{17}{\sqrt{5}}$. Then $P{A}^2+P{B}^2=\frac{49+289}{5}=\frac{338}{5}$, so $5(P{A}^2+P{B}^2)=338$.