Mathematics - Ellipse Question with Solution | TestHub

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
$25(\sqrt{5}+r\cos \theta {)}^2+36(\sqrt{5}+r\sin \theta {)}^2=900$
Simplifying, we get
$\begin{array}{rl}& {\mathrm{r}}^2\left(25{\cos }^2\theta +36{\sin }^2\theta \right)+2\sqrt{5}\mathrm{r}(25\cos \theta +36\sin \theta )-595=0\\ & |\mathrm{r}|=\mathrm{PA},\mathrm{PB}\end{array}$
$\text{ Thus, }\begin{array}{rl}\text{ PA }\cdot \mathrm{PB}& =\frac{595}{25{\cos }^2\theta +36{\sin }^2\theta }=\frac{595}{25+11{\sin }^2\theta }\\ & =\text{ maximum, if }{\sin }^2\theta =0\end{array}$
This means line $AB$ must be parallel to $x$-axis
$\Rightarrow y_A=y_B=\sqrt{5}$
Putting $\mathrm{y}=\sqrt{5}$ in equation of ellipse, we get
$\frac{x^2}{36}+\frac{1}{5}=1\Rightarrow x^2=36\cdot \frac{4}{5}$
Hence,
${\mathrm{PA}}^2+{\mathrm{PB}}^2={\left(\sqrt{5}-\frac{12}{\sqrt{5}}\right)}^2+{\left(\sqrt{5}+\frac{12}{\sqrt{5}}\right)}^2$
$\begin{array}{rl}& =2\left(5+\frac{144}{5}\right)=\frac{338}{5}\\ & 5\left({\mathrm{PA}}^2+{\mathrm{PB}}^2\right)=338\end{array}$