JEE Main 2025 — Mathematics Coordinate Geometry

Given ellipse is $\frac{x^2}{36}+\frac{y^2}{25}=1$36x2​+25y2​=1

Any point on line $A B$AB can be assumed as

$$\mathrm{Q}(\sqrt{5}+\mathrm{r} \cos \theta, \sqrt{5}+\mathrm{r} \sin \theta)$$Q(5​+rcosθ,5​+rsinθ)

Putting this in equation of ellipse, we get

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

Simplifying, we get

$\begin{aligned}
& \mathrm{r}^2\left(25 \cos ^2 \theta+36 \sin ^2 \theta\right)+2 \sqrt{5} \mathrm{r}(25 \cos \theta+36 \sin \theta)-595=0 \
& \quad|\mathrm{r}|=\mathrm{PA}, \mathrm{~PB}
\end{aligned}$

$\text { Thus, } \begin{aligned}
\text { PA } \cdot \mathrm{PB} & =\frac{595}{25 \cos ^2 \theta+36 \sin ^2 \theta}=\frac{595}{25+11 \sin ^2 \theta} \
& =\text { maximum, if } \sin ^2 \theta=0
\end{aligned}$

This means line $A B$AB must be parallel to $x$x-axis

$$\Rightarrow y_A=y_B=\sqrt{5}$$⇒yA​=yB​=5​

Putting $\mathrm{y}=\sqrt{5}$y=5​ in equation of ellipse, we get

$$\frac{x^2}{36}+\frac{1}{5}=1 \Rightarrow x^2=36 \cdot \frac{4}{5}$$36x2​+51​=1⇒x2=36⋅54​

Hence,

$$\mathrm{PA}^2+\mathrm{PB}^2=\left(\sqrt{5}-\frac{12}{\sqrt{5}}\right)^2+\left(\sqrt{5}+\frac{12}{\sqrt{5}}\right)^2$$PA2+PB2=(5​−5​12​)2+(5​+5​12​)2

$$\begin{aligned} & =2\left(5+\frac{144}{5}\right)=\frac{338}{5} \\ & 5\left(\mathrm{PA}^2+\mathrm{PB}^2\right)=338\end{aligned}$$​=2(5+5144​)=5338​5(PA2+PB2)=338​