The distance of a point P on the ellipse ${\mathrm{x}}^2+3{\mathrm{y}}^2=12$ from the centre is $\sqrt{6}$ then the eccentric angle of P is

Given ellipse $\frac{{\mathrm{x}}^2}{12}+\frac{{\mathrm{y}}^2}{4}=1$

where ${\mathrm{a}}^2=12, {\mathrm{b}}^2=4$

Centre $\left(\mathrm{C}\right)=(0, 0)$

Let $\mathrm{P}=\left(2\sqrt{3} \mathrm{cos\theta }, 2\mathrm{sin\theta }\right)$ be any point on ellipse.

Given that $\mathrm{CP}=\sqrt{6}$

$\Rightarrow \sqrt{{\left(2\sqrt{3} \cos \theta \right)}^2+{\left(2\mathrm{sin\theta }\right)}^2}=\sqrt{6}$

squaring on both sides 

$\begin{array}{rcl}12{\cos }^2\mathrm{\theta }+4{\sin }^2\mathrm{\theta }& =& 6\\ & \Rightarrow & 12{\cos }^2\mathrm{\theta }+4-4{\cos }^2\mathrm{\theta }=6\\ & \Rightarrow & 8{\cos }^2\mathrm{\theta }=2\\ & \Rightarrow & {\cos }^2\mathrm{\theta }=\frac{1}{4}\\ & \Rightarrow & \mathrm{cos\theta }=\frac{1}{2}\\ & \Rightarrow & \mathrm{\theta }=\frac{\mathrm{\pi }}{3}\end{array}$