Question

The eccentric angle of a point on the ellipse
$\frac{x^2}{6}+\frac{y^2}{2}=1$ whose distance from the center of the ellipse is $\sqrt{5}$ is:

• A. $\frac{\pi }{6}$
• B. $\frac{5\pi }{6}$
• C. $\frac{7\pi }{6}$
• D. $\frac{11\pi }{6}$

Answer 1

Answer: (A), (B), (C), (D)

$\frac{11\pi }{6}$

$\mathrm{Any}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{given}\mathrm{ellipse}\mathrm{will}\mathrm{be}\mathrm{of}\mathrm{the}\mathrm{form}$
$P(\sqrt{6}\cos \theta ,\sqrt{2}\sin \theta ).\mathrm{The}\mathrm{center}\mathrm{of}\mathrm{ellipse}\mathrm{be}C(0,0).\mathrm{Then},\mathrm{CP}=\sqrt{5}\Rightarrow {\mathrm{CP}}^2=5\Rightarrow 6{\cos }^2\theta +2{\sin }^2\theta =5\Rightarrow 4{\sin }^2\theta =1\mathrm{or}\sin \theta =\pm \frac{1}{2}\Rightarrow \theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$