If tangents are drawn to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ at the ends of latus recta, then the area of the quadrilateral thus formed is

The given equation of the ellipse is,

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

The eccentricity of the ellipse is,

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

The coordinates of the endpoints of latus rectum are

$L\left(2,\frac{5}{3}\right),M\left(-2,\frac{5}{3}\right),M^{\text{'}}\left(-2,\frac{-5}{3}\right)$ and $L^{\text{'}}\left(2,\frac{-5}{3}\right)$

The equations of tangents at these points are,

$2 x+3 y-9=0$
$-2 x+3 y-9=0$
$2 x+3 y+9=0$
$-2 x+3 y+9=0$

By solving the above equations,

$A\left(0,3\right)$

$B\left(-\frac{9}{2},0\right)$

$C\left(0,-3\right)$
$D\left(\frac{9}{2},0\right)$

And
$AC=\sqrt{(3+3{)}^2}=6$

$BD=\sqrt{{\left(\frac{9}{2}+\frac{9}{2}\right)}^2}=9$

The area of the quadrilateral is,
$A=\frac{1}{2}\times 9\times 6$

$=27 \mathrm{sq}.\mathrm{units}$