Question

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

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

Answer 1

The correct option is D $27sq.units$

Given $\frac{x^2}{9}+\frac{y^2}{5}=1$
To find tangent at end points of latusrectum, we find
$ae=\sqrt{a^2-b^2}=\sqrt{4}=2$
and $\sqrt{b^2(1-e^2)}=\sqrt{5(1-\frac{4}{9})}=\frac{5}{3}$
So, area is four times of the right angles triangle formed by the tangent and axes in the $I^{st}$ quadrant
Therefore equation of tangent at $(2,\frac{5}{3})$ is
$\frac{2}{9}x+\frac{5}{3},\frac{y}{5}=1\Rightarrow \frac{x}{\frac{9}{2}}+\frac{y}{3}=1$
Therefore area of quadrilateral $ABCD=4($area of triangle $AOB)$
$=4(\frac{1}{2}\times \frac{9}{2}\times 3)=27$