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?
 A: 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}$$= \begin{bmatrix} 5x \\ 5y \end{bmatrix}$ so 9x+ 2y= 5x and 2x+ 6y= 5y.  Those both reduce to y= 2x. All eigenvectors corresponding to eigenvalue 5 are multiples of $\begin{bmatrix}1 \\ 2 \end{bmatrix}$. Similarly all eigenvectors corresponding to eigenvalue 10 are multiples of $\begin{bmatrix}-2 \\ 1 \end{bmatrix}$.
Form the matrix $P= \begin{bmatrix}1 & 2\\ -2 & 1\end{bmatrix}$ having those eigenvectors as columns.  Then $P^{-1}= \begin{bmatrix}\frac{1}{5} & -\frac{2}{5} \\ \frac{2}{5} & \frac{1}{5}\end{bmatrix}$ and we have $P^{-1}AP= \begin{bmatrix}5 & 0 \\ 0 & 10\end{bmatrix}$, the diagonal matrix having the eigenvalues on the diagonal.  
A: I can’t think of any good way to compute it directly (finding a focus/directrix pair is at least as much work as what’s described below and in other answers), but it’s not terribly difficult to find the ellipse’s semiaxis lengths and compute it using those. Here’s a method that uses partial derivatives as requested.  
We have $f(x,y)=9x^2+4xy+6y^2-22x-16y+9$. First, find the center $(x_0,y_0)$ of the ellipse by solving $\nabla f=0$ for $x$ and $y$. At the points where the ellipse and its axes intersect, the normal to the ellipse is parallel to the vector from the center to that point. This condition can be expressed as $$\det\begin{bmatrix}x-x_0 & y-y_0 \\ f_x(x,y) & f_y(x,y)\end{bmatrix} = 0.$$ This is the equation of a pair of lines, so you can either try to factor it directly or use the method for splitting a degenerate conic described in step 4 here and then find the intersections of these lines with the ellipse, or bash out a solution to the system of equations. You will then have the endpoints of the ellipse’s semiaxes. I’m not sure that this is any simpler or less tedious than the methods described in other answers.  
Another approach the uses partial derivatives along the way is to use a theorem of Appolonius: if $c_1$ and $c_2$ are the half-lengths of a pair of conjugate diameters and $a$ and $b$ are the semiaxis lengths, then $c_1^2+c_2^1=a^2+b^2$ and the area of the triangle formed by the center and the endpoint of the half-diameters is equal to $\frac12ab$. So, once you have the center of the ellipse, find a point at which, say, the line $y=y_0$ intersects the ellipse. The distance to the center from this point is $c_1$, and the normal to the ellipse at that point is also normal to the conjugate diameter of $y=y_0$. Work out the distance from the center to the intersections of that line with the ellipse for $c_1$. You then have a pair of equations that you can solve for $a$ and $b$. This seems like even more work than any of the other methods so far, though.  
If you’re comfortable working with matrices, a way to solve this is to write the equation in matrix form: $$\mathbf p^TC\mathbf p = \begin{bmatrix}x&y&1\end{bmatrix} \begin{bmatrix}9&2&-11\\2&6&-8\\-11&-8&9\end{bmatrix} \begin{bmatrix}x\\y\\1\end{bmatrix} = 0.$$ Let $S=\det C$ and $\lambda_1$, $\lambda_2$ be the eigenvalues of the upper-right $2\times2$ submatrix (the quadratic part of the equation), that is, the roots of $\lambda^2+15\lambda-50$. The semiaxis lengths are then the square roots of $-S/(\lambda_1\lambda_2^2)$ and $-S/(\lambda_1^2\lambda_2)$.
A: First of all you must find the center $(a,b)$ of the ellipse: substitute into the equation $x\to x+a$ and $y\to y+b$, choosing then $a$ and $b$ such that linear terms vanish. The new equation you get is that of the ellipse, translated with its center at $(0,0)$.
To compute the semi-axes of the ellipse, consider a generic line through the origin $y=mx$ and find its intersections $(x(m), y(m))$ with the translated ellipse as a function of $m$: minimum and maximum of $x(m)^2+y(m)^2$ will give you the squares of the semi-axes.
Of course you could do the same by solving a constrained max/min problem in two variables: find minimum and maximum of $x^2+y^2$, with a constraint represented by the equation of the translated ellipse.
