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

$$\begin{gathered} \mathrm{Given} \mathrm{ellipse} \frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{5}=1 \\ \mathrm{e}=\frac{2}{3} \\ \mathrm{one} \mathrm{end} \mathrm{of} \mathrm{latusrectum}\left(\mathrm{L}\right)=\left(2, \frac{5}{3}\right) \\ \mathrm{eq}uation \mathrm{of} \mathrm{Tangent} \mathrm{at} \mathrm{L} \mathrm{is} \\ \frac{\mathrm{x}\left(2\right)}{9}+\frac{\mathrm{y}(5/3)}{5}=1 \\ \frac{2\mathrm{x}}{9}+\frac{\mathrm{y}}{3}=1 \end{gathered}$$

The tangents drawn at the end of the laterarecta to the ellipse will intersect at the x and the y axis because of symmetry. The area of the quadrilateral thus formed will be 4 times the area of the triangle formed by the tangent and the coordinate axes

$$\begin{gathered} Area of quadrilateral=4\left(\frac{1}{2\left|{\displaystyle \frac{2}{9}}\times {\displaystyle \frac{1}{3}}\right|}\right) \\ =27 \end{gathered}$$