Question

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

• A. 27/4 sq units
• B. 9 sq units
• C. 27/2 sq uni
• D. 27 sq units

Answer 1

Answer: (D)

27 sq units

Answer 2

Given, $\frac{x^2}{9}+\frac{y^2}{5}=1$
To find tangents at the end points of latusrectum , we
find ae
i.e. ae = $\sqrt {a^2-b^2}$ = $\sqrt4 = 2$
and $\sqrt{b^2(1-e^2)}$ = $\sqrt{5\Big(1-\frac{4}{9}\Big)}$ = $\frac{5}{3}$
By symmetry, the quadrilateral is a rhombus
So, area is four times the area of the right angled
triangle formed by the tangent and axes in the 1st
quadrant.
$\therefore$ Equation of tangent at $\Big(2,\frac{5}{3}\Big)$ is
$\frac{2}{9} x + \frac{5}{3}.\frac{y}{5}$ = 1 $\Rightarrow \frac{3}{\frac{9}{2}}+\frac{y}{3}$ = 1
$therefore$ Area of quadrilateral ABCD
= 4 [area of $\Delta$ AOB]
= 27 sq units