The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latus ractum to…

$\frac{x^2}{9}+\frac{y^2}{5}=1$

$e=\sqrt{1-\frac{5}{9}}=\frac{2}{3}$

$a^2=9, {\mathit{b}}^2=5$

Foci $=\left(\pm \mathit{ae}\mathit{,}0\right)=\left(\pm 2,0\right)$

Ends of latus rectum$=\left(\pm \mathit{ae},\pm \frac{b^2}{a}\right)$

$=\left(\pm 2,\pm \frac{5}{3}\right)$

Tangent at '$L$' is $\mathit{T}=0$

$\frac{2\times x}{9}+\frac{5}{3}\times \frac{y}{5}=1$

It cuts coordinate axes at $\mathit{P}\left(\frac{9}{2}\mathit{,}0\right) \& \mathit{Q}\left(0,3\right)$

Area of quadrilateral $PQRS=4$ (Area of triangle $OPQ$)

$=\mathrm{4 }\left(\frac{1}{2}\times \frac{9}{2}\times 3\right)=27$ sq. units