How to deal with translations, with change of coordinates for Quadratic Forms: $x^TAx=k$ ? When working with a change of coordinates using
$x^TAx=k$ 
how and when do we deal with translations?  I'm comfortable with setting up the formula $x^TAx$ where A is the matrix whose diagonal entries are the coefficients of (say in $R^2$) x and y and whose other entries correspond to half of the $xy$ term in said equation.  Then finding eigenvectors that translate our coordinate system and sketching our graph from there.  What I'm struggling with is say we have:
$x^2 - 2xy + y^2 + 6x - 4 = 0$
I know that if I were to say, complete the square I would have something that looks like:
$(x+3)^2 - 2xy + y^2 = 13$
Do I then just move forward and still build my matrix A as \begin{align}
 A = \begin{pmatrix} 1 & -1  \\ -1 & 1 \\ \end{pmatrix}
\end{align}
Or do I need to account for the shift when building this matrix?  Again, I'm comfortable with the math up until dealing with the translation. Do I complete the process of sketching the graph and then account for my translation or does the translation play a role throughout the process?  
 A: You should complete the square the "matrix" way:
$$
x^2 - 2xy + y^2 + 6x - 4
= [x,y]\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix} + [6,0]\begin{bmatrix}x\\ y\end{bmatrix} + 4.
$$
In general, for a quadratic form $\mathbf{x}^TA\mathbf{x}+2\mathbf{b}^T\mathbf{x}+c$ where $A$ is symmetric, if $\mathbf{b}=A\mathbf{u}$ for some vector $\mathbf{u}$, then we can kill the linear term by a translation because
$$
\mathbf{x}^TA\mathbf{x}+2\mathbf{b}^T\mathbf{x}+c
\ =\ \mathbf{x}^TA\mathbf{x}+2\mathbf{u}^TA\mathbf{x}+c
\ =\ (\mathbf{x}+\mathbf{u})^TA(\mathbf{x}+\mathbf{u})+\left(c-u^TAu\right).
$$
Now, since $(6,0)^T$ does not reside in the column space of $A=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$, there is no way to kill all the linear terms, no matter what translation you apply.
A: If you center your $(p,q)$ coordinates at $\left(\frac{7}{24}, \frac{43}{24} \right)$ and rotate $135^\circ,$ so
$$  x =  \frac{7}{24} - \frac{p}{\sqrt 2} - \frac{q}{\sqrt 2},   $$
$$  y =  \frac{43}{24} + \frac{p}{\sqrt 2} - \frac{q}{\sqrt 2},   $$
you see the evident parabola in
$$ q = \left(\frac{\sqrt 2}{3} \right) \, p^2.  $$ 
It is not that difficult to check this, as your original equation is
$$ (y-x)^2 + 6 x - 4 = 0  $$
and $y-x$ is not so bad in terms of $p,q.$
