Parabolas and axis of symmetry?

Question

I have the parabola $$(x+y)^2 = 8(x−y)$$ and know that the axis of symmetry is $$x+y=0$$ but I know when this is the case the left hand side equals 0 but apart from that I can't see how this equation was found. Can someone please help and explain a $(x+y)^2 = 8(x−y)$ way to find it (if there is one)?

Answer 1

Try a change of variable $X=x+y$ and $Y=x-y$. Then you have $$Y=\frac{1}{8}X^2\ .$$ Its axis of symmetry is $X=0$, i.e. $x+y=0$.

Answer 2

Think of "regular" parabolas, with equation $y = ax^2 + bx + c$. These always have an axis of symmetry given by $x = -b/2a$. We could replace the standard coordinates of the plane to the coordinate $w,z$, with $w = x + y$ and $z = x - y$ (turning the coordinate axes of the plane 45 degrees counterclockwise). Then the equation of $(x+y)^2 = 8(x-y)$ becomes $w^2 = 8z$, so we have the parabola $z = 1/8 \cdot w^2$ with axis of symmetry $w = 0$, which is precisely $x + y = 0$.