Explanation

(d)
Given, $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1\$
Latusrectum of an ellipse be
$ae = \sqrt{a^{2} - b^{2}} = \sqrt{4} = 2$
[image:11_image88.emf]
By symmetry the quadrilateral is rhombus
$\Rightarrow \ Equation\ of\ tangent\ at\ \left( ae,\frac{b^{2}}{a} \right) = \left( 2,\ \frac{5}{3} \right)$
$ie,\ \frac{2}{9}\ x + \frac{5}{3}.\frac{y}{5} = 1$
$\Rightarrow \ \frac{x}{9/2} + \frac{y}{3} = 1$
$\therefore Area\ of\ quadrilateral\ ABCD = 4(area\ of\ \mathrm{\Delta}\ AOB\ )$
$= 4.\left\{ \frac{1}{2}.\frac{9}{2}.3 \right\}$
$= 27\$sq units