We have,
$ \frac{x^{2}}{9}+\frac{y^{2}}{5}=1 $
$\Rightarrow a=3 $ and $ b=\sqrt{5} $
$\therefore e =\sqrt{1-\frac{b^{2}}{a^{2}}} $
$=\sqrt{1-\frac{5}{9}}=\frac{2}{3} $
$\therefore $ Foci $=(\pm a e, 0) $
$=\left(\pm 3 \times \frac{2}{3}, 0\right)=(\pm 2,0) $
$\therefore $ Ends of latusrectum $=\left(\pm 2, \pm \frac{5}{3}\right)$
Equation of tangent at the end of latusrectum $P$ is
$\frac{x \times 2}{9}+\frac{y \times 5 / 3}{5}=1$
$\Rightarrow \frac{2}{9} x+\frac{1}{3} y=1 $
$\Rightarrow 2 x+3 y=9 $
$\therefore O A= \frac{9}{2} $ and $O B=3$
$\therefore $ Area of quadrilateral $A B C D$
$=4 \times$ Area of $\triangle O A B$
$=4 \times \frac{1}{2} \times O A \times O B$
$=4 \times \frac{1}{2} \times \frac{9}{2} \times 3$
$=27$ sq units.
