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 $=\left(\pm a e, \frac{2 b^{2}}{a}\right)=(2,3)$
Equation of tangent at $(2,3)$ of ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$ is, $\frac{2 x}{16}+\frac{3 y}{12}=1$
$\Rightarrow \frac{x}{8}+\frac{y}{4}=1$
Area of quadrilateral $A B C D=4 \times \frac{1}{2} \times O A \times O B$
$=4 \times \frac{1}{2} \times 8 \times 4=64$