JEE Main 2025 Apr 3 Shift 1, Mathematics Q6: Maxima & Minima

Parametrise points on a line through $P(\sqrt{5},\sqrt{5})$ as $(\sqrt{5}+r\cos \theta ,\sqrt{5}+r\sin \theta )$. Substitute into $25x^2+36y^2=900$:

$25(\sqrt{5}+r\cos \theta {)}^2+36(\sqrt{5}+r\sin \theta {)}^2=900.$

This yields a quadratic in $r$ whose roots are the signed distances $\pm PA,\pm PB$. The product of the roots, by Vieta, is

$PA\cdot PB=\left|\frac{\text{const}}{25{\cos }^2\theta +36{\sin }^2\theta }\right|=\frac{595}{25+11{\sin }^2\theta }.$

This is maximised when $\sin \theta =0$, i.e. the chord is horizontal through $P$, giving $y=\sqrt{5}$.

Setting $y=\sqrt{5}$ in the ellipse: $\frac{x^2}{36}+\frac{1}{5}=1\Rightarrow x^2=\frac{144}{5}$, so $x=\pm \frac{12}{\sqrt{5}}.$

$P{A}^2+P{B}^2={\left(\sqrt{5}-\frac{12}{\sqrt{5}}\right)}^2+{\left(\sqrt{5}+\frac{12}{\sqrt{5}}\right)}^2=2\left(5+\frac{144}{5}\right)=\frac{338}{5}.$

Therefore $5(P{A}^2+P{B}^2)=338.$