Question

The eccentric angle of a point on the ellipse $x^2 + 3y^2 = 6$ lying at a distance of 2 units from its centre is

• A. $\dfrac{\pi}{6}$
• B. $\dfrac{\pi}{4}$
• C. $\dfrac{\pi}{3}$
• D. $\dfrac{\pi}{2}$

Hint:

Use parametric form of ellipse and Pythagoras to relate eccentric angle to distance.

Answer 1

The Correct Option is B

Given ellipse: $\dfrac{x^2}{6} + \dfrac{y^2}{2} = 1$
Parametric form: $x = \sqrt{6} \cos\theta$, $y = \sqrt{2} \sin\theta$
Distance from centre = $\sqrt{x^2 + y^2} = \sqrt{6 \cos^2\theta + 2 \sin^2\theta} = 2$
$\Rightarrow 6 \cos^2\theta + 2 \sin^2\theta = 4$
Using $\sin^2\theta = 1 - \cos^2\theta$, simplify:
$6\cos^2\theta + 2(1 - \cos^2\theta) = 4 \Rightarrow 4\cos^2\theta = 2 \Rightarrow \cos^2\theta = \dfrac{1}{2}$
$\Rightarrow \theta = \dfrac{\pi}{4}$