Explanation

(d)
Given equation of ellipse is $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$
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To find tangents at the end points of latusrectum we find $ae$,
$ie,ae = \sqrt{a^{2} - b^{2}} = \sqrt{4} = 2$
By symmetry the quadrilateral is rhombus
So, area of rhombus is four times the area of the right angled formed by the tangent and axes in the first quadrant
$\Rightarrow$ Equation of tangent at $\left\lbrack ae,\ \sqrt{b\left( 1 - e^{2} \right)} \right\rbrack = \left( 2,\frac{5}{3} \right)$ is
$\frac{2}{9}x + \frac{5}{3} \bullet \frac{y}{5} = 1$
$\Rightarrow \ \frac{x}{9/2} + \frac{y}{3} = 1$
$\therefore$ Area of quadrilateral $ABCD = 4$(area of $\mathrm{\Delta}\ AOB)$
$= 4\left( \frac{1}{2} \bullet \frac{9}{2} \bullet 3 \right) = 27$ sq unit