Question

The area of the quadrilateral formed by the tangents at the endpoints of the latus rectum to the ellipse $\frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{5}=1$, is

• A. $\frac{27}{4}$sq. units
• B. $9$sq. units
• C. $\frac{27}{2}$sq. units
• D. $27$sq. units

Answer 1

The correct option is D

$27$sq. units

Explanation for the correct option:

Given equation of the ellipse, $\frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{5}=1$

The equation of the tangent at $\mathrm{L}\left(\mathrm{ae},\frac{{\mathrm{b}}^2}{\mathrm{a}}\right)$ becomes,

$$\begin{gathered} \Rightarrow \frac{{\mathrm{xx}}_1}{{\mathrm{a}}^2}+\frac{{\mathrm{yy}}_1}{{\mathrm{b}}^2}=1 \\ \Rightarrow \frac{\mathrm{xae}}{{\mathrm{a}}^2}+\frac{\mathrm{y}\left({\displaystyle \frac{{\mathrm{b}}^2}{\mathrm{a}}}\right)}{{\mathrm{b}}^2}=1 \\ \Rightarrow \mathrm{ex}+\mathrm{y}=\mathrm{a} \end{gathered}$$

Area of quadrilateral,

$=4\times$area of triangle $\mathrm{OAB}$

$$\begin{gathered} =4\times \left(\frac{1}{2}\right)\left(\frac{\mathrm{a}}{\mathrm{e}}\right)\times \mathrm{a} \\ =\frac{\left[2\times 9\right]}{\left[\sqrt{1-{\displaystyle \frac{5}{9}}}\right]} \\ =27\mathrm{sq}.\mathrm{units} \end{gathered}$$

Hence, option $\left(\mathrm{D}\right)$ is the correct.