The area of the quadrilateral formed by the tangents at the end points of the latus recta of the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ is (in sq. units)

Here $a=3, b=\sqrt{5},$ so that 

$\begin{array}{l}{b}^2=a^2\left(1-e^2\right)\Rightarrow 5=9\left(1-e^2\right)\\ \Rightarrow 9e^2=4\Rightarrow e^2=4/9\Rightarrow e=2/3\end{array}$

Thus coordinates of foci are $(\pm ae, 0)=(\pm 2, 0)$

When $x=2, y^2=5\left(1-\frac{4}{9}\right)=\frac{25}{9}$

$\Rightarrow y=\pm 5/3$

$\therefore$ Coordinates of A are (2, 5/3).

Similarly, coordinates of $B, C, D$ are respectively $(-2, 5/3), (-2, -5/3), (2, -5/3).$ Equation of tangents at $A, B, C, D$ are given

 $\pm \frac{x}{9}\left(2\right)\pm \frac{y}{5}\left(\frac{5}{3}\right)=1$ or $\pm \frac{2}{9}x\pm \frac{1}{3}y=1$

 These lines meet the axes in $(\pm 9/2, \pm 3).$ Area of quadrilateral $PQRS$

Area of quadrilateral $PQRS$

  $=4$[Area of $∆OPQ$]

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