Explanation

(d)
The quadrilateral formed by the tangents at the end points of latusrectum is a rhombus. It is symmetrical about the axes.
[image:11_image27.emf]
So, total area is four times the area of the right angled triangle formed by the tangent and axes in the Ist quadrant
Now, $ae = \sqrt{a^{2} - b^{2}} \Rightarrow ae = 2$
$\therefore$ Coordinates of one end point of 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 equation meets the coordinate axes at a point $A\left( 0,\ \frac{9}{2} \right)$ and $B(3,\ 0)$
In $\mathrm{\Delta}AOB,$ Area$= \frac{1}{2} \times \frac{9}{2} \times 3 = \frac{27}{4}$
Total area of rhombus $ABCD = 4 \times area\ of\ \mathrm{\Delta}AOB$
$= 4 \times \frac{27}{4} = 27$ sq unit