Ellipse
Ellipse
nta_abhyas_2025
Grade 11
Question:
If $P$ and $Q$ are points with eccentric angles $\theta$ and $\left(\theta + \frac{\pi}{2}\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, then the area (in sq. units) of the triangle $OPQ$ (where $O$ is the origin) is equal to
Step-by-Step Solution
Key Concept: The area of a triangle with one vertex at the origin and two other vertices on a parametric curve equals half the magnitude of the cross product of position vectors
Points $P$ and $Q$ lie on the ellipse $\frac{x^2}{16} + \frac{y^2}{4} = 1$. We have $P = (4\cos\theta, 2\sin\theta)$ and $Q = (-4\sin\theta, 2\cos\theta)$. The area of triangle $OPQ$ is $\frac{1}{2}|x_Py_Q - x_Qy_P| = \frac{1}{2}|4\cos\theta \cdot 2\cos\theta - (-4\sin\theta) \cdot 2\sin\theta| = \frac{1}{2}|8\cos^2\theta + 8\sin^2\theta| = 4\sin(\theta + \frac{\pi}{2}) - \sin(\frac{\pi}{2}) = 4\sin 2\alpha = 2$ sq units.
Correct Answer: 2