The given ellipse is
$\frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{5}=1$
Then, a² = 9, b² = 5. Therefore,
$\mathrm{e}=\sqrt{1-\frac{5}{9}}=\frac{2}{3}$
Hence, the endpoint of latus rectum in first quadrant is L(2,5/3).
The equation of tangent at L is
$\frac{2\mathrm{x}}{9}+\frac{\mathrm{y}}{3}=1$
the tangent meets the x-axis at A(9/2, 0) and the y-axis at B(0,3). Therefore,
$\text{ Area of }\mathrm{\Delta OAB}=\frac{1}{2}\times \frac{9}{2}\times 3=\frac{27}{4}$

By symmetry
Area of quadrilateral = 4 x (Area of $∆$OAB)
$=4\times \frac{27}{4}=27\mathrm{sq}. \mathrm{units}$