Question

The area of the quadrilateral formed by tangents at the end points of latus recta of the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ is

• A. $\frac{27}{4}uni{t}^2$
• B. $9uni{t}^2$
• C. $\frac{27}{2}uni{t}^2$
• D. $27uni{t}^2$

Answer 1

The correct option is D $27uni{t}^2$

By symmetry the quadrilateral is a rhombus

So area is four times the area of the right angled triangle formed by the tangent and axes in the 1st quadrant
Now $ae=\sqrt{a^2-b^2}\Rightarrow ae=2$

$\Rightarrow$ tangent( in first quadrant ) at the ned of latus rectum $(2,\frac{5}{3})$ is

$\frac{2}{9}x+\frac{5}{3}.\frac{y}{5}=1\Rightarrow \frac{x}{\frac{9}{2}}+\frac{y}{3}=1$

Area $=4\cdot \frac{1}{2}\cdot \frac{9}{2}\cdot 3=27$