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)

$27$ sq.units

At the end point $(\sqrt{a^2-b^2},\frac{b^2}{a})$ equation of tangent is $\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$

So, the intercepts are $(\frac{a^2}{x_1},0)$ and $(0,\frac{b^2}{y_1})$

Now, the area of quadrilateral $=4\times$ area of triangle formed by the intercepts and origin.

$\Rightarrow 4\times \frac{1}{2}\times \frac{a^2{b}^2}{\sqrt{a^2-b^2}\times \frac{b^2}{a}}=2\times \frac{27}{2}=27$

So, option A is correct.