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

• Correct answer
  A

  27 sq units
• B

  $\frac{13}{2}$ sq units
• C

  $\frac{15}{4} \mathrm{sq}$ units
• D

  45 sq units

Solution

We have, $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$ $\Rightarrow \quad a=3$ and $b=\sqrt{5}$ $\begin{aligned} \therefore \quad e &=\sqrt{1-\frac{b^{2}}{a^{2}}} \\ &=\sqrt{1-\frac{5}{9}}=\frac{2}{3} \end{aligned}$ $\therefore$ Foci $=(\pm a e, 0)$ $=\left(\pm 3 \times \frac{2}{3}, 0\right)=(\pm 2,0)$ $\therefore$ Ends of latusrectum $=\left(\pm 2, \pm \frac{5}{3}\right)$ Equation of tangent at the end of latusrectum $P$ is $\frac{x \times 2}{9}+\frac{y \times 5 / 3}{5}=1$ $\frac{2}{9} x+\frac{1}{3} y=1$ $2 x+3 y=9$ $O A=\frac{9}{2}$ and $O B=3$ $\therefore$ Area of quadrilateral $A B C D$ $$ =4 \times \text { Area of } \Delta O A B $$ $=4 \times \frac{1}{2} \times O A \times O B$ $=4 \times \frac{1}{2} \times \frac{9}{2} \times 3$ $=27 \mathrm{sq}$ units.

Hence, the correct answer is option: 'A'.