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

The correct option is A $27$ sq. units

Let $L{L}^{\prime }$ and $M{M}^{\prime }$ are the latus rectum.


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

The equation of tangent at $(x_1,y_1)$ is $\frac{x{x}_1}{9}+\frac{y{y}_1}{5}=1$
The tangent passes through point $L(ae,\frac{b^2}{a})$
So, $\frac{x\left(ae\right)}{9}+\frac{y\left(\frac{b^2}{a}\right)}{5}=1$

$\Rightarrow ex+y=a$
$\Rightarrow \frac{x}{a/e}+\frac{y}{a}=1$
So, the tangent cuts the $x$-axis at point $A(\frac{a}{e},0)$ and $y$-axis at point $B(0,a)$

Area of quadrilateral $ABCD=4\times \text{area of}△OAB$
$=4\times \frac{1}{2}\times \frac{a}{e}\times a$
$=2\times \frac{a^2}{e}$
$=2\times \frac{9}{2/3}$
$=27\text{sq. units}$