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}=1 is $

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