Question

The area of the parallelogram formed by the tangents at the points whose eccentric angles $\theta ,\theta +\frac{\pi }{2},\theta +\pi ,\theta +\frac{3\pi }{2}$ 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 D $2ab$

Given eccentric angles are $\theta ,\theta +\frac{\pi }{2},\theta +\pi ,\theta +\frac{3\pi }{2}$

so, points are $(a\cos \theta ,bsin\theta ),(-asin\theta ,b\cos \theta ),(-a\cos \theta ,-bsin\theta ),(asin\theta ,-b\cos \theta )$

Area of parallelogram $=\frac{1}{2}\left|\begin{array}{ccccc}a\cos \theta & -asin\theta & -a\cos \theta & asin\theta & a\cos \theta \\ bsin\theta & b\cos \theta & -bsin\theta & -b\cos \theta & bsin\theta \end{array}\right|$

$=\frac{1}{2}|(ab{\cos }^2\theta +ab{\sin }^2\theta )+(ab{\sin }^2\theta +ab{\cos }^2\theta )+(ab{\cos }^2\theta +ab{\sin }^2\theta )+(ab{\sin }^2\theta +ab{\cos }^2\theta )|$

$=2ab$ sq.units