$\text{ The area of the rectangle formed by the perpendiculars from centre of }\frac{x^2}{9}+\frac{y^2}{4}=1\text{ to the tangent and normal at }$

$\text{ the point whose eccentric angle is }\frac{\pi }{4}\text{ is }A\text{ . The value of 13A is}$

$\text{ Equation of tangent at point }P\left(\frac{\pi }{4}\right)\text{ is }$

$\begin{array}{l}\frac{x}{3}\cos \frac{\pi }{4}+\frac{y}{2}\sin \frac{\pi }{4}=1\\ or 2x+3y-6\sqrt{2}=0\end{array}$

Equation of normal at P is

$3\sqrt{2}x-2\sqrt{2}y-5=0$

$\text{ Distance of centre from the tangent is }\frac{6\sqrt{2}}{\sqrt{13}}\text{ . }$

$\text{ Distance of centre from the normal is }\frac{5}{\sqrt{2}\sqrt{13}}\text{ . }$

$\therefore \text{ Area of required rectangle }=\frac{6\sqrt{2}}{\sqrt{13}}\times \frac{5}{\sqrt{2}\sqrt{13}}=\frac{30}{13}\text{ sq. units. }$