Completing square in hyperbolic equation

Question

I tried completing the square for the following hyperbola: $x^2+y^2-3xy+2x+3y+2=0$ by sending $3xy+2$ to the other side and finally stuck here: $(x+1)^2 +2(y+\frac{3}{4})^2 = \frac{1}{8} -3xy$ I am not sure how to deal with the $-3xy$ term, should I divide the whole equation by terms on RHS, would that enable me to find the centre of hyperbola by comparing it to standard equation?

Answer 1

You can find the center by solving $\frac{\partial}{\partial x}(x^2+y^2-3xy+2x+3y+2)=\frac{\partial}{\partial y}(x^2+y^2-3xy+2x+3y+2)=0$ to get $x=\frac{13}{5},y=\frac{12}{5}$.

Then $(x+\frac{13}{5})^2+(y+\frac{12}{5})^2-(3(x+\frac{13}{5}))(y+\frac{12}{5})+2(x+\frac{13}{5})+3(y+\frac{12}{5})+2=x^2+y^2-3xy+\frac{41}{5}$ is the equation for the hyperbola centered at the origin.

Edit: Now let $\theta$ be the angle the hyperbola is rotated from the standard form then $\tan{2\theta}=\frac{-3}{1-1}$ gives us $2\theta=\frac{\pi}{2}$ or $\theta=\frac{\pi}{4}$ so we can use $x=x'\cos(\frac{\pi}{4})-y'\sin(\frac{\pi}{4})$ and $y=x'\cos(\frac{\pi}{4})+y'\sin(\frac{\pi}{4})$. Expanding this you get $-\frac12 x'^2+\frac52 y'^2+\frac{41}{5}$. Now the inverse transformation is $x'=x\cos(\frac{\pi}{4})+y\sin(\frac{\pi}{4})$ and $y'=-x\sin(\frac{\pi}{4})+y\cos(\frac{\pi}{4})$

Putting it all together we get my comment.