Question

The area of the parallelogram formed by the tangents at the points whose eccentric angles are $\theta ,\theta +\frac{\pi }{2},\theta +\pi ,\theta +\frac{3\pi }{2}$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

• A. ab
• B. 4ab
• C. 3ab
• D. 2ab

Answer 1

The correct option is B 4ab


$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Area=$\pi ab$
Let P $=(acos\theta ,bsin\theta )$
S=(ae,0)
M(h,k) be the midpoint of PS
$(h,k)=(\frac{ae+acos\theta }{2},\frac{bsin\theta }{2})$
$\frac{{(h-\frac{ae}{2})}^2}{{\left(\frac{a}{2}\right)}^2}+\frac{k^2}{{\left(\frac{b}{2}\right)}^2}=1$
$Area=\frac{\pi ab}{4}$
$Ratio=\frac{1}{4}$