Question

The area (in sq. units) of the quadrilateral formed by the tangents drawn at the end points of the latus recta to the ellipse $S\equiv \frac{x^2}{16}+\frac{y^2}{12}=1$ is

• A. 96
• B. 16
• C. 128
• D. 64

Answer 1

Answer: (D)

Given ellipse, $\frac{x^2}{16}+\frac{y^2}{12}=1$

End point of latus rectum $=\left(\pm ae,\frac{2b^2}{a}\right)=(2,3)$
Equation of tangent at $(2,3)$ of ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$ is, $\frac{2x}{16}+\frac{3y}{12}=1$
$\Rightarrow \frac{x}{8}+\frac{y}{4}=1$
Area of quadrilateral $ABCD=4\times \frac{1}{2}\times OA\times OB$
$=4\times \frac{1}{2}\times 8\times 4=64$