Reduce conic sections to standard form

Question

I have some problems in reducing these conic sections to their standard form and i ask for help to solve them.

$x^2-4xy+4y^2+x+y=5$

$x^2-2xy+3y^2+x+y=6$

Also if I don't ask too much, I would like them to be solved this way:

Example:

$x^2+2xy+5y^2+2x+2y+2=0$

$x^2+2xy+5y^2+2x+2y+2=(x+y)^2+4y^2+2(x+y)+1+1=(x+y+1)^2+4y^2+1=X^2+Y^2+1$

With $X=x+y+1$ and $Y=2y$.

Thank you.

Answer 1

Start by completing the square for $x^2-4xy$ to get rid of the $-4xy$ term, i.e. $$x^2-4xy+4y^2+x+y=5 \iff (x-2y)^2+x+y=5$$ Now let $X=x-2y$ and $Y=3y$ to give $$X^2+X+Y=5.$$ Can you see what the curve is now?

The second one is similar, but this time choose $Y=\sqrt{2}y$ so that $x^2-2xy+3y^2$ becomes $X^2+Y^2$.