Minimum distance from $2x^2 + 2xy + 4yz + z^2 = 1 $ to the origin Let Σ be the surface in $R^3$ given by
$2x^2 + 2xy + 4yz + z^2 = 1 $
By writing this equation as
$x^TAx$ = 1,
with A a real symmetric matrix, show that there is an orthonormal basis such that, if we use coordinates
$(u, v, w)$ with respect to this new basis, Σ takes the form
$λu^2 + µv^2 + νw^2 = 1.$
Find $λ$, $µ$ and $ν$ and hence find the minimum distance between the origin and Σ. 
Hint: it is not
necessary to find the basis explicitly.
I found matrix $A$ = $\begin{pmatrix} 
 2& 1 & 0 \\ 
1 & 0 & 2 \\ 
0 & 2 & 1  
\end{pmatrix} $ , and figured that  $λ$, $µ$ and $ν$ should be its eigenvalues, after changing the basis to the eigenvectors. The eigenvalues are $ 3, \sqrt{3}, -\sqrt{3}  $, so I have $3u^2 + \sqrt{3}^2 -\sqrt{3}w^2 = 1.$
I feel like I have made a mistake somewhere, I don't know how to proceed.

Apply Lagrange Multipliers Method to minimize $f(x,y,z)= x^2+y^2+z^2$ subject to $g(x,y,z)= 2x^2+2xy+4yz+z^2-1$, results in $x=y=z=\frac13$ or $x=y=z=-\frac13$ with the minimum value of $f(x,y,z)=\frac13$. Therefor the minimum distance is $\sqrt{\frac13}$.   
