The area of the quadrilateral formed by the tangents at the end point of latusrectum to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ is

The quadrilateral formed by the tangents at the end points of the latusrectum is a rhombus. It is symmetrical about the axes.
So, total area is four times the area of the right triangle formed by the tangents and the axes in the first quadrant. Now, $\begin{aligned} & a e=\sqrt{a^2-b^2} \\ & a e=2 \end{aligned}$ $\therefore$ The co-ordinates of one end point latusrectum are $\left(2, \frac{5}{3}\right)$. The equation of tangent at that point is $\frac{x}{\frac{9}{2}}+\frac{y}{3}=1$, this meets the co-ordinate axes at $A\left(0, \frac{9}{2}\right)$ and $B(3,0)$. Area of $\triangle A O B=\frac{1}{2} \times \frac{9}{2} \times 3=\frac{27}{4}$ Hence, the area of rhombus $A B C D$ $=4 \times \frac{27}{4}=27 \text { sq units }$