Question

The area (in sq. units) of the quadrilateral formed by the tangent at the end points of the latera 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)

$27$

For the given ellipse

$e=\sqrt{1-\frac{5}{9}}=\frac{2}{3}$

The endpoint of one of the latus rectum will be $(2,\frac{5}{3})$

The tangents drawn at the end of the latera recta to the ellipse will intersect at the $x$ and the $y$ axis because of symmetry. The area of the quadrilateral thus formed will be $4$ times the area of the triangle formed by the tangent and the coordinate axis.
The equation of the tangent at $(2,\frac{5}{3})$ can be written as :
$\frac{2x}{9}+\frac{5y}{15}=1$

X intercept $=\frac{9}{2}$

Y intercept $=3$

Area of the triangle $=\frac{1}{2}\times \frac{9}{2}\times 3=\frac{27}{4}$

Hence, area of the quadrilateral $=4\times \frac{27}{4}=27$