Reduce $F(x,y)=0$ to canonical form of parabola

Question

I know, from calculating $\delta$ and $\Delta$, that the following function represents parabola.

$$x^2-2xy+y^2-2x+27y+10=0$$

But how can I get it to its canonical form: $y^2=2px$ ?

Essentially I need to sketch it, not precisely draw it.

Answer 1

OK. Well, your first three terms are just $(x-y)^2$. That suggests changing variables to $$ u = (x+y) \\ v = (x - y) $$ which means that $$ x = (u+v)/2\\ y = (u - v)/2, $$ so that you have $$ v^2 - 2x + 2y + 25y + 10 = 0\\ v^2+2 (y-x) + 25 y + 10 = 0\\ v^2 - 2v + 25 (u-v)/2 + 10 = 0 \\ $$ That's now something for which you can do the completing-the-square to get an aligned parabola.