Question

The distance of a point P on the ellipse $\frac{x^2}{12}+\frac{y^2}{4}=1$ from centre is $\sqrt{6}$ then the ecentric angle of P is

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

Answer 1

Answer: (A), (D)

$\frac{\pi }{3}$
$\frac{2\pi }{3}$

Ellipse$:\frac{x^2}{12}+\frac{y^2}{4}=1$

Centre$:C(0,0)$

Given that $PC=\sqrt{6}$

Parametric coordinates $:y=b\sin \varphi ,x=a\cos \varphi$

$\varphi$ is eccentric angle.

$\Rightarrow 6=(2\sqrt{3}\cos \varphi -0{)}^2+(2\sin \varphi -0{)}^2$

$\Rightarrow 6=12\cos {}^2\varphi +4\sin {}^2\varphi$

$\Rightarrow 2=8{\cos }^2\varphi$

$\Rightarrow {\cos }^2\varphi =\frac{1}{4}$

$\Rightarrow \cos \varphi =\pm \frac{1}{2}$

$\Rightarrow \varphi =\frac{\pi }{3}$

$\Rightarrow \varphi =\frac{2\pi }{3}$