Question

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

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

Answer 1

The correct option is D $27$

Given equation of ellipse is $\frac{x^2}{9}+\frac{y^2}{5}=1$ .....$\left(1\right)$

$\therefore a^2=9,b^2=5$

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

Now, $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{5}{9}}=\sqrt{\frac{9-5}{9}}=\frac{2}{3}$

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

and $\frac{b^2}{a}=\frac{5}{3}$

$\therefore$ Extremities of one of latus rectum are $(2,\frac{5}{3})$ and $(2,\frac{-5}{3})$

$\therefore$ Equation of tangent at $(2,\frac{5}{3})$ is

$\frac{x\left(2\right)}{9}+\frac{y\left(\frac{5}{3}\right)}{5}=1$

or $2x+3y=9$ ........$\left(2\right)$

Eqn$\left(2\right)$ intersects $X$ and $Y-$ axes at $(\frac{9}{2},0)$ and $(0,3)$ respectively.

$\therefore$ Area of quadrilateral$=4\times$Area of $△POQ$

$=4\times \frac{1}{2}\times \frac{9}{2}\times 3=27$sq.units.