Explanation:
D Given the ellipse \(\frac{x^2}{9}+\frac{y^2}{5}=1\)
Here, \(a^2=9, b^2=5\)
\(\therefore 1-\mathrm{e}^2=\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{5}{9} \Rightarrow \mathrm{e}^2=1-\frac{5}{9}=\frac{4}{9}\)
\(\therefore \mathrm{e}=\frac{2}{3}\)
ae \(=3 \times \frac{2}{3}=2\)
Let the co-ordinate of end point of latus rectum is (ae, \(\mathrm{x})=(2, \mathrm{k})\)
This point lies on the ellipse.
\(\frac{4}{9}+\frac{\mathrm{k}^2}{5}=1\)
\(\text { or } \frac{\mathrm{k}^2}{5}=1-\frac{4}{9}=\frac{5}{9}\)
\(\therefore \mathrm{k}^2=\frac{25}{9}\)
\(\therefore \mathrm{k}= \pm \frac{5}{3}\)
\(\therefore\) One point on the curve is \(\left(2, \frac{5}{3}\right)\)
Differentiating the curve (ellipse) w.r.t \(\mathrm{x}\), we get -
\(\quad \frac{2 x}{9}+\frac{2 y}{5} \cdot \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-2 x / 2 y \times \frac{5}{9}=\frac{-5 x}{9 y}\)
\(\left.\therefore \quad \frac{d y}{d x}\right|_{2, \frac{5}{3}}=\frac{-5 \times 2}{9 \times \frac{5}{3}}=\frac{-2}{3}\)
\(\therefore\) The equation of one side of the quadrilateral is
\(\mathrm{y}=\frac{-2}{3} \mathrm{x}+\mathrm{c}\)
Since, it passes through \(\left(2, \frac{5}{3}\right)\)
\(\therefore \quad \frac{5}{3}=\frac{-2}{3} \times 2+\mathrm{c}\)
\(\therefore \quad \mathrm{c}=\frac{5}{3}+\frac{4}{3}=\frac{9}{3}=3\)
\(\therefore\) The equation of the line is \(y=\frac{-2}{3} x+3\)
Now, \(x\) intercept is obtain by putting \(y=0\)
\(\therefore \quad 0=\frac{-2}{3} \mathrm{x}+3\)
\(\therefore \quad \mathrm{x}=\frac{9}{2}\)
\(\mathrm{y}\)-intercept is obtained by putting \(\mathrm{x}=0\)
\(\therefore \quad \mathrm{y}=0+3=3\)
\(\therefore\) Area of the quadrilateral
\(=4 \times\left[\frac{1}{2} \times 3 \times \frac{9}{2}\right]=27\)