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)

27

Answer 2

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, E (ii) intersects X and Y-axes at $\bigg(\frac{9}{2} , 0 \bigg)$
and (0,3), respectively.
$\therefore$ Area of quadrilateral = 4 x Area of APOQ
$= 4 \times \bigg( \frac{1}{2} \times \frac{9}{2} \times 3 \bigg ) = 27$ sq units