Identifying the quadrant in which the parabola lies (or finding the vertex of parabola) 
In which quadrant does the  graph of the parabola $x^2 + y^2 - 2xy - 8x - 8y +32 = 0 $ lie? 

One easy way to solve this question would be to simply graph the parabola and check but that;s time consuming when you are provided with around 5 minutes,  a pen and a paper and no desmos. 
The parabola doesn't intersect the axes. This can be proven by substituting $x=0$ and $y=0$ and confirming that the equation has no real roots.
It can also be represented as $(x-4)^2+ (y-4)^2 = 2xy$. But I am unable to find the quadrant of the parabola through this equation. 
I guess if we find the vertex of the parabola, then it's quadrant will be the quadrant of the parabola but how do I find the vertex?
What makes this question difficult is that there are both $x^2$ and $y^2$ present together. 
Edit: I see symmetry arguments in the answers, what if it was not so easy to identify the line of symmetry of the parabola? Which method should I employ then?
 A: The question makes it clear that we haven't to check that the curve, let us name it (P), is a parabola.
1st solution: Using the form you have obtained
$$(x-4)^2+ (y-4)^2 = 2xy\tag{1}$$
The LHS being positive, we must have $xy>0$; therefore, quadrants 2 and 4 are excluded.
Besides, (P) doesn't cross the $x$ axis, because setting $y=0$
in (1) would give $16=0$. For the same reason, (P) doesn't cross the $y$ axis. Therefore (P) is entirely situated in either one of the quadrants, 1 or 3. As $(x,y)=(2,2)$ belongs to (P), only quadrant 1 is possible.
2nd solution:
3 successive facts:

*

*exchanging $x$ and $y$ doesn't change the equation ; thus parabola (P) is symmetrical with respect to the $y=x$ straight line, thus is its axis.


*point $S=(2,2)$ belongs to (P).


*(P) does not intersect the $x$ axis, because, if we set $y=0$ in the equation of (P), we get the quadratic equation $x^2 - 8x +32 = 0$ that has no real roots.  Thus, by symmetry, (P) does not intersect the $y$ axis.
Conclusion  : (P) lies entirely in the first quadrant.
Remark: As point S is situated on the axis of symmetry of (P), it qualifies it as the apex of (P).
A: From $x^2+y^2-2xy$ you should directly see that you should let $u=\dfrac1{\sqrt2}(x-y)$ (the $\sqrt2$ is for normalization) and $v=\dfrac1{\sqrt2}(x+y)$.
Then you obtain $2u^2 -8\sqrt2v+32=0$, which becomes $v=\dfrac{\sqrt2}8u^2+2\sqrt2$.
A: Well, if you rearrange the equation $x^2 + y^2 - 2xy - 8x - 8y +32 = 0 $, you will get $$(x-y)^2=4\cdot 2\cdot (x+y-4)$$ which is clearly of the form $Y^2=4aX$, where $Y=x-y \,$ and $X=x+y-4$.
So, this is a parabola with axis $y=x$ and vertex at $(2,2)$. 
And $x+y=4$ is the tangent to the parabola at the vertex.
Hope, this helps. 
A: Set $u=x+y$ and $v=x-y$. Then $v^2-8u+32=0$ so $v^2= 8(u-4)$. Thus in coordinate system made by $u,v$ focus is at $G(6,0)$ and vertex at $V(4,0)$. The matrix which takes $(x,y)$ to $(u,v)$ is 
$$M=\left(%
\begin{array}{cc}
  1 & 1 \\
  1 & -1 \\
\end{array}%
\right)$$
So $$M^{-1}= {1\over 2}\left(%
\begin{array}{cc}
  1 & 1 \\
  1 & -1 \\
\end{array}%
\right)$$
So $G$ goes to $F(3,3)$ and vertex $V$ to $O(2,2)$. And since equations $x^2 +  8x  +32 = 0 $ and $y^2- 8y +32 = 0 $ don't have the solution, a parabola does not intersect the coordinate lines and so it lies entirely in 1. quadrant. 
Hope it is OK because my linear algebra is a little rusty. :( 
