Thank you for reporting, we will resolve it shortly
Q.
If tangents are drawn to the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$ at the ends ot latus recta, then the area of the quadrilateral thus formed is
We have, $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$
Let $e$ be the eccentricity of the ellipse. Then,
$e^{2}=1-\frac{5}{9} \Rightarrow e=\frac{2}{3}$
The coordinates of the end-points of latusrectum are
$L\left(2, \frac{5}{3}\right), M\left(-2, \frac{5}{3}\right), M'\left(-2, \frac{-5}{3}\right) $ and $ L'\left(2,-\frac{5}{3}\right)$
The equation of tangents at these points are
$2 x+3 y-9=0 \dots$(i)
$-2 x+3 y-9=0 \dots$(ii)
$2 x+3 y+9=0 \dots$(iii)
$-2 x+3 y+9=0 \dots$(iv)
By solving above equations, we get
$A(0,3), B\left(-\frac{9}{2}, 0\right), C(0,-3)$ and $D\left(\frac{9}{2}, 0\right)$
Now, $A C=\sqrt{(3+3)^{2}}=\sqrt{6^{2}}=6$
and $B D=\sqrt{\left(\frac{9}{2}+\frac{9}{2}\right)^{2}}=\sqrt{9^{2}}=9$
Now, area of quadrilateral $=\frac{1}{2} \times 6 \times 9$
$=27$ sq. units
