JEE-Mains 2015 (21/30)

The quadrilateral in question is a rhombus formed by $4$ right triangle formed by one of the tangents with the two axes. Let us consider that triangle in the first quadrant.

The latus rectum has the same $x$ -coordinate as that of the focus $F(c,0)$ given by:

$c^2 = 9 - 5 \quad \Rightarrow c = 2$

The coordinates of the end of the semilatus rectum $P (c,y_P)$ are:

$\begin{aligned} \frac{c^2}{9} + \frac{y_P^2}{5} & = 1 \quad \Rightarrow \frac{4}{9} + \frac{y_P^2}{5} = 1 \quad \Rightarrow \frac{y_P^2}{5} = 1 - \frac{4}{9} \quad \Rightarrow y_P = \frac{5}{3} \end{aligned}$

Let us find the gradient of the ellipse at point $P \left(2,\frac{5}{3}\right)$ :

$\dfrac{x^2}{9} + \dfrac{y^2}{5} = 1 \\ \Rightarrow \dfrac{2x}{9} + \dfrac{2y}{5}\dfrac{dy}{dx} = 0 \quad \Rightarrow \dfrac{4}{9} + \dfrac{2}{3}\dfrac{dy}{dx} = 0 \quad \Rightarrow \dfrac{dy}{dx} = - \dfrac{2}{3} \\ \Rightarrow \dfrac{y-\frac{5}{2}}{x-2} = - \dfrac{2}{3} \quad \Rightarrow y = - \dfrac{2}{3}x + \dfrac{4}{3} + \dfrac{5}{3} \quad \Rightarrow y = - \dfrac{2}{3}x + 3$

Now find the $x$ - and $y$ -axis intercepts $x_0$ and $y_0$ of the tangent $y = - \dfrac{2}{3}x + 3$ :

$0 = - \dfrac{2}{3}x_0 + 3 \quad \Rightarrow x_0 = \dfrac{9}{2} \\ y_0 = - \dfrac{2}{3}(0) + 3 \quad \Rightarrow y_0 = 3$

The area of the quadrilateral:

$A = 4 \left( \dfrac{x_0y_0}{2} \right) = 4 \left( \dfrac{\frac{9}{2}(3)}{2} \right) = \boxed{27}$