Question

Tangents are drawn to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ at the ends of both latus rectum. The area of the quadrilateral so formed is

• A. $27$ sq. units
• B. $\frac{13}{2}$ sq. units
• C. $\frac{15}{4}$ sq. units
• D. $45$ sq. units

Answer 1

Answer: (A)

We have,
$\frac{x^2}{9}+\frac{y^2}{5}=1$
$\Rightarrow a=3$ and $b=\sqrt{5}$
$\therefore e=\sqrt{1-\frac{b^2}{a^2}}$
$=\sqrt{1-\frac{5}{9}}=\frac{2}{3}$
$\therefore$ Foci $=(\pm ae,0)$
$=\left(\pm 3\times \frac{2}{3},0\right)=(\pm 2,0)$
$\therefore$ Ends of latusrectum $=\left(\pm 2,\pm \frac{5}{3}\right)$

Equation of tangent at the end of latusrectum $P$ is
$\frac{x\times 2}{9}+\frac{y\times 5/3}{5}=1$
$\Rightarrow \frac{2}{9}x+\frac{1}{3}y=1$
$\Rightarrow 2x+3y=9$
$\therefore OA=\frac{9}{2}$ and $OB=3$
$\therefore$ Area of quadrilateral $ABCD$
$=4\times$ Area of $△OAB$
$=4\times \frac{1}{2}\times OA\times OB$
$=4\times \frac{1}{2}\times \frac{9}{2}\times 3$
$=27$ sq units.