Find eccentricity of ellipse $9x^2 +4xy+6y^2-22x-16y+9=0$ I tried to find the eccentricity by factorising $$9x^2 +4xy+6y^2-22x-16y+9=0$$ in the form of perpendicular distance from two lines but it is getting too lenghty as the $xy$ term is involved. 
Is there a way to do it using partial derivatives?

The first thing I would do is rotate the axes to eliminate the "4xy" term.  Write $9x^2+ 4xy+ 6y^2= \begin{bmatrix}x & y \end{bmatrix}\begin{bmatrix}9 & 2 \\ 2 & 6 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}$.  The characteristic equation for the matrix, $A= \begin{bmatrix}9 & 2 \\ 2 & 6 \end{bmatrix}$ is $\left|\begin{array}{cc}9- \lambda & 2 \\ 2 & 6- \lambda\end{array}\right|= (9- \lambda)(6- \lambda)- 4= \lambda^2- 15\lambda+ 50=$ $(\lambda- 5)(\lambda- 10)= 0$.  The eigenvalues are 5 and 10.
To find the eigenvectors corresponding to eigenvalue 5, solve $\begin{bmatrix}9 & 2 \\ 2 & 6 \end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}$$= \begin{bmatrix} 5x \\ 5y \end{bmatrix}$ so 9x+ 2y= 5x and 2x+ 6y= 5y.  Those both reduce to y= 2x. All eigenvectors corresponding to eigenvalue 5 are multiples of $\begin{bmatrix}1 \\ 2 \end{bmatrix}$. Similarly all eigenvectors corresponding to eigenvalue 10 are multiples of $\begin{bmatrix}-2 \\ 1 \end{bmatrix}$.
Form the matrix $P= \begin{bmatrix}1 & 2\\ -2 & 1\end{bmatrix}$ having those eigenvectors as columns.  Then $P^{-1}= \begin{bmatrix}\frac{1}{5} & -\frac{2}{5} \\ \frac{2}{5} & \frac{1}{5}\end{bmatrix}$ and we have $P^{-1}AP= \begin{bmatrix}5 & 0 \\ 0 & 10\end{bmatrix}$, the diagonal matrix having the eigenvalues on the diagonal.  
