Question

If $P$ and $Q$ are points with eccentric angles $\theta $ and $\left(\theta + \frac{\pi }{6}\right)$ on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1,$ then the area (in sq. units) of the triangle $OPQ$ (where $O$ is the origin) is equal to

Answer 1

Let, $P=(4 \cos \theta, 2 \sin \theta)$ and $Q=\left(4 \cos \left(\theta+\frac{\pi}{6}\right), 2 \sin \left(\theta+\frac{\pi}{6}\right)\right)$
$\Rightarrow Areaof\Delta OPQ=\frac{1}{2}\begin{vmatrix} 4cos\theta & 2sin\theta & 1 \\ 4cos\left(\theta + \frac{\pi }{6}\right) & 2sin\left(\theta + \frac{\pi }{6}\right) & 1 \\ 0 & 0 & 1 \end{vmatrix}$
$=4\left(cos \theta s i n \left(\theta + \frac{\pi }{6}\right) - c o s \left(\theta + \frac{\pi }{6}\right) sin \theta \right)$
$=4sin\left(\theta + \frac{\pi }{6} - \theta \right)=4sin\frac{\pi }{6}=2$ sq. units