Question

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

• A. $27/4$ sq. units
• B. $9$ sq. units
• C. $27/12$ sq. units
• D. $27$ sq. units

Answer 1

Answer: (D)

Given equation of ellipse is $\frac{x^2}{9}+\frac{y^2}{5}=1$.

To find tangents at the end points of latus rectum we find
ae, i.e., ae $=\sqrt{a^2-b^2}=\sqrt{4}=2$
By symmetry the quadrilateral is rhombus.
So, area of rhombus is four times the area of the right angled triangle formed by the tangent and axes in the first quadrant.
$\Rightarrow$ Equation of tangent at $\left[ae,\sqrt{b^2\left(1-e^2\right)}\right]=\left(2,\frac{5}{3}\right)$ is
$\frac{2}{9}x+\frac{5}{3}\cdot \frac{y}{5}=1\Rightarrow 2x+3y=9$
$\therefore$ Area of quadrilateral $ABCD=4$(area of $\Delta AOB)$
$=4\underset{0}{\overset{9/2}{\int }}\frac{9-2x}{3}dx=\left|\frac{4}{3}{\left[9x-x^2\right]}_0^{9/2}\right|$
$=\frac{4}{3}\left[\frac{81}{2}-\frac{81}{4}\right]=27$ sq. units