If $\varphi \in \left[0,2\pi \right]$ is the eccentric angle of a point on the ellipse ${\mathrm{x}}^2+3{\mathrm{y}}^2=6$  at a distance 2 unit from origin , then sum of the values of $\varphi$ is  $\frac{2020\mathrm{\pi }}{\mathrm{k}}\mathrm{then} \mathrm{k}=$

Given ellipse is $\frac{{\mathrm{x}}^2}{6}+\frac{{\mathrm{y}}^2}{2}=1---\left(\mathrm{i}\right)$

a point on the ellipse (i) is $(\sqrt{6}\cos \mathrm{\varphi },\sqrt{2}\sin \mathrm{\varphi })$

Since, its distance from the origin is 2

$\begin{array}{l}\therefore 6{\cos }^2\mathrm{\varphi }+2{\sin }^2\mathrm{\varphi }=(2{)}^2\\ \Rightarrow 4{\cos }^2\mathrm{\varphi }+2=4\\ \Rightarrow 4{\cos }^2\mathrm{\varphi }=2\\ \Rightarrow {\cos }^2\mathrm{\varphi }=1/2\\ \Rightarrow 2{\cos }^2\mathrm{\varphi }=1\\ \cos 2\mathrm{\varphi }=0\\ \Rightarrow 2\mathrm{\varphi }=(2\mathrm{n}+1)\frac{\mathrm{\pi }}{2};\mathrm{n}\in \mathrm{I}\\ \text{ or } \mathrm{\varphi }=(2\mathrm{n}+1)\mathrm{\pi }/4;\mathrm{n}\in \mathrm{I}\Rightarrow \mathrm{\varphi }=\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\end{array}$