Find locus of points by finding eigenvalues 
Let $\boldsymbol{x}=\left(\begin{matrix}x\\ y\end{matrix}\right)$ be a vector in two-dimensional real space. By finding the eigenvalues and eigenvectors of $\boldsymbol{M}$, sketch the locus of points $\boldsymbol{x}$ that satisfy $$ \boldsymbol{x^TMx}=4$$
  given that 
  $$\boldsymbol{M}=\left(\begin{matrix}&5 &\sqrt{3}\\ &\sqrt{3} &3\end{matrix}\right). $$ 

I found two eigenvalues to be $\lambda_1 = 6$ and $\lambda_2=2$, and the corresponding eigenvectors are 
$$ \boldsymbol{v}_1=\left(\begin{matrix}\sqrt{3}\\ 1\end{matrix}\right)\quad\text{ and }\quad \boldsymbol{v}_2=\left(\begin{matrix}1\\ -\sqrt{3}\end{matrix}\right)$$
(if I'm not mistaken :) ), but... what now? Frankly, I can't figure out how to make this helpful to find $\boldsymbol{x^TMx}=4$. 
Any hints?  

The eigenvectors are orthogonal and span $\Bbb R^2$. This means $\mathbf v=\begin{bmatrix}x\\y\end{bmatrix}=c_1\mathbf x_1+c_2\mathbf x_2$. 
$\mathbf v^TM\mathbf v=(c_1\mathbf x_1^T+c_2\mathbf x_2^T)M(c_1\mathbf x_1+c_2\mathbf x_2)\\=(c_1\mathbf x_1^T+c_2\mathbf x_2^T)(c_1\lambda_1\mathbf x_1+c_2\lambda_2\mathbf x_2)\\=c_1^2\lambda_1||\mathbf x_1||^2+c_2^2\lambda_2||\mathbf x_2||^2\\=24c_1^2+8c_2^2=4$
$\implies6c_1^2+2c_2^2=1$
