Question

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

• A. $27$
• B. $\frac{15}{4}$
• C. $\frac{13}{2}$
• D. 45

Answer 1

Answer: (A)

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
$L\left(2,\frac{5}{3}\right),M\left(-2,\frac{5}{3}\right),M^{\prime }\left(-2,\frac{-5}{3}\right)$ and $L^{\prime }\left(2,-\frac{5}{3}\right)$
The equation of tangents at these points are
$2x+3y-9=0\dots$(i)
$-2x+3y-9=0\dots$(ii)
$2x+3y+9=0\dots$(iii)
$-2x+3y+9=0\dots$(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, $AC=\sqrt{(3+3{)}^2}=\sqrt{6^2}=6$
and $BD=\sqrt{{\left(\frac{9}{2}+\frac{9}{2}\right)}^2}=\sqrt{9^2}=9$
Now, area of quadrilateral $=\frac{1}{2}\times 6\times 9$
$=27$ sq. units