Question

Find the eccentric angle of a point on the ellipse $\frac{x^2}{6}+\frac{y^2}{2}=1$ whose distance from centre is 2.

• A. ${45}^{\circ }$
• B. ${30}^{\circ }$
• C. ${135}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

Answer: (A), (C)

${135}^{\circ }$

From the equation $\frac{x^2}{6}+\frac{y^2}{2}=1$ we have $a=\sqrt{6}andb=\sqrt{2}$. The parametric representation of point will be
$(\sqrt{6}cos\theta ,\sqrt{2}sin\theta )Given,{\left(\sqrt{6}cos\theta \right)}^2+{\left(\sqrt{2}sin\theta \right)}^2=2^{2}4co{s}^2\theta +2(co{s}^2\theta +si{n}^2\theta )=4co{s}^2\theta =\frac{1}{2}\Rightarrow \theta =\pm {45}^{\circ },\pm {135}^{\circ }\therefore Eccentricanglecanbe+{45}^{\circ },+{135}^{\circ }$