Question

The area (in sq. units) of the of the quadrilateral formed by the tangents at the end points of the later recta to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1,$ is :

• A. $\frac{27}{2}$
• B. $\frac{27}{4}$
• C. $18$
• D. $27$

Answer 1

The correct option is D $27$

In the question it is asked to find the area of the quadrilateral formed by the tangents at the end points of the latus recta of the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$.

or, $\frac{x^2}{3^2}+\frac{y^2}{(\sqrt{5}{)}^2}=1$

So, $a=3,b=\sqrt{5}$

$e=\sqrt{1-\frac{5}{9}}=\sqrt{\frac{4}{9}}=\frac{2}{3}$

Focii$=(\pm ae,0)=(\pm 2,0)$

Ends of Latus rectum $=(\pm ae,\pm \frac{b^2}{a})=(\pm 2,\pm \frac{5}{3})$

Equation of tangent at $(2,\frac{5}{3})$ is given by $\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$ $(x_1=2,y_1=\frac{5}{3})$

$\therefore \frac{2x}{9}+\frac{5}{3}\times \frac{y}{3}=1$

$\Rightarrow \frac{2x}{9}+\frac{y}{3}=1$

$\therefore$ Co-ordinates of point P will be $(\frac{9}{2},0)$ and Q will be $(0,3)$

By symmetry of area of quadrilateral PQRS,

$=4\left(Areaof△POQ\right)=4\times (\frac{1}{2}\times \frac{9}{2}\times 3)=27uni{t}^2$