Question

The area (in sq units) of the quadrilateral formed by the tangents at the end points of the latusrectum recta to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1is$

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

Answer 1

Answer: (D)

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 $\frac{b^2}{a}=\frac{5}{3}$

$\therefore$ Extremities of one of latusrectum are
$(2,\frac{5}{3})$ and $(2,\frac{-5}{3})$
$\therefore$ Equation of tangent at $(2,\frac{5}{3})$ is
$\frac{x(2)}{9}+\frac{y(5/3)}{5}=1$ or $2x+3y=9$
Since, Eq. (ii) intersects $X$ and $Y$-axes at $(\frac{9}{2},0)$ and $(0,3)$, respectively.
$\therefore$ Area of quadrilateral $=4\times$ Area of $APOQ$
$=4\times (\frac{1}{2}\times \frac{9}{2}\times 3)=27$ sq units