The area (in s units) of the quadrilateral formed by the tangents at the end points of the latera
recta to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$, is:
$\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$ $a=3 $ $b=\sqrt{5}$ $e^{2}=1-\frac{b^{2}}{a^{2}}$ $=1-\frac{5}{9}=\frac{4}{9}$ $e=\frac{2}{3}$ now the quadrilateral formed will be a rhombu with area $=\frac{2 a^{2}}{e}$ $=\frac{2.9}{2} \times 3$ $=27$