Question

The area of the quadrilateral formed by the tangent at the endpoints of latus recta to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ is -

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

Answer 1

Answer: (C)

$27$ sq. units

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

$\Rightarrow a=3,b=\sqrt{5}$

$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{5}{9}}=\sqrt{\frac{4}{5}}=\frac{2}{3}$

Foci $=(\pm ae,0)$

$=(\pm 3\times \frac{2}{3},0)=(\pm 2,0)$

Ends of latus rectum $=(\pm ae,\frac{b^2}{a})$

$=(\pm 2,\pm \frac{5}{3})$

Let the tangent at point $(2,\frac{5}{3})$ be $\frac{x{x}_1}{9}+\frac{y{y}_1}{5}=1$

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

It cuts the coordinates axes at $P(\frac{9}{2},0)$ and $Q(0,3)$

By symmetryarea of quadrilateral $=4(\frac{1}{2}\times \frac{9}{2}\times 3)$

$=4\times \frac{27}{4}=27$ sq. units