Question

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

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

Answer 1

The correct option is D $27$

The quadrilateral formed by the tangents at the end points of the latusrectum is a rhombus.

It is symmetrical about the axes.

So, total area is four times the area of the right triangle formed by the tangents and the axes in the first quadrant.
Now, $ae=\sqrt{a^2-b^2}$
$\Rightarrow ae=2$
Therefore, the co-ordinates of one end point of latusrectum are $(2,\frac{5}{3})$.
The equation of tangent at that point is $\frac{x}{\frac{9}{2}}+\frac{y}{3}=1$, this meets the co-ordinate axes at $A(0,\frac{9}{2})$ and $B(3,0)$.
Area of $△AOB=\frac{1}{2}\times \frac{9}{2}\times 3=\frac{27}{4}$
Hence, the area of rhombus $ABCD$
$=4\times \frac{27}{4}=27$ sq. units