JEE Main 2025 — 3 April, Morning Shift — Question 25

Given ellipse is $\frac{x^{2}}{36}+\frac{y^{2}}{25}=1$

Any point on line AB can be assumed as $\mathrm{Q}(\sqrt{5}+\mathrm{r} \cos \theta, \sqrt{5}+\mathrm{r} \sin \theta)$

Putting this in equation of ellipse, we get

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

Simplifying, we get $r^{2}\left(25 \cos ^{2} \theta+36 \sin ^{2} \theta\right)+2 \sqrt{5} r(25 \cos \theta+36 \sin \theta)-595=0$

$|\mathrm{r}|=\mathrm{PA}, \mathrm{PB}$

Thus, $\mathrm{PA} \cdot \mathrm{PB}=\frac{595}{25 \cos ^{2} \theta+36 \sin ^{2} \theta}=\frac{595}{25+11 \sin ^{2} \theta}$

$$=\text { maximum, if } \sin ^{2} \theta=0$$

This means line AB must be parallel to x -axis $\Rightarrow y_{A}=y_{B}=\sqrt{5}$

Putting $y=\sqrt{5}$ in equation of ellipse, we get $\frac{x^{2}}{36}+\frac{1}{5}=1 \Rightarrow x^{2}=36 \cdot \frac{4}{5}$

Hence, $\mathrm{PA}^{2}+\mathrm{PB}^{2}=\left(\sqrt{5}-\frac{12}{\sqrt{5}}\right)^{2}+\left(\sqrt{5}+\frac{12}{\sqrt{5}}\right)^{2}$

$=2\left(5+\frac{144}{5}\right)=\frac{338}{5}$

$5\left(\mathrm{PA}^{2}+\mathrm{PB}^{2}\right)=338$