Eliminating the $xy$ term of an equation for a conic gives $\tan 2\beta = 2/0$

Question

So, I have the following :

$$x^2 + 2xy + y^2 + 2\sqrt{2}\;x - 2\sqrt{2}\;y+4=0$$

I know that to take out the $xy$ term from $A x^2 + B x y + C y^2 + \cdots = 0$, I must use : $$\tan 2\beta=\frac{B}{A-C}$$

But the problem is that I get $$\tan 2\beta =\frac{2}{0}$$

What should I do ?

Answer 1

Notice that $x^2+2xy+y^2=(x+y)^2$, not sure what $a$, $A$, $B$, and $C$ are in this case..

Answer 2

$\tan2\beta=\dfrac{2}{0}$ is infinity, then $\beta=\dfrac{\pi}{4}$.

Get to parabola $B^2-4AC=4-4=0$, the rotation is

$$x=\dfrac{x'}{\sqrt{2}}-\dfrac{y'}{\sqrt{2}}$$

$$y=\dfrac{x'}{\sqrt{2}}+\dfrac{y'}{\sqrt{2}}$$

With focal axis $y=-x$