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?