Question

The area (in sq units) of the quadrilateral formed by the tangents at the endpoints of the latus recta to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ is___ .

• A. $\frac{27}{4}$
• B. 18
• C. $\frac{27}{2}$
• D. 27

Answer 1

The correct option is D

27

Given equation of ellipse is
$\frac{x^2}{9}+\frac{y^2}{5}=1\therefore a^2=9,b^2=5\Rightarrow a=3,b=\sqrt{5}Now,e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{5}{9}}=\frac{2}{3}Foci=(\pm ae,0)=(\pm 2,0)and\frac{b^2}{a}=\frac{5}{3}$

$\therefore$ Extermities of one of latusrectum are
$(2,\frac{5}{3})and(2,\frac{-5}{3})\therefore Equationoftangentat(2,\frac{5}{3})is\frac{x\left(2\right)}{9}+\frac{y\left(\frac{5}{3}\right)}{5}=1or2x+3y=9$
Since the tangent intersects X and Y axes at $(\frac{9}{2},0)$ and (0,3) respectively,
$\therefore$ Area of quadrilateral $=4\times Areaof\Delta POQ=4\times (\frac{1}{2}\times \frac{9}{2}\times 3)=27$