Rotation of axes with zero inverse cotangent

[medicalphysicsguy]

We are rotating axes to eliminate the xy term in

\(\displaystyle Ax^2 + Bxy + Cy^2+Dx+Ey+F=0\)

To do this we need to choose an angle such that \(\displaystyle cot 2\phi=(A-C)/B\)

But the easiest problems like

\(\displaystyle x^2-2xy+y^2-2\sqrt{2}x-2\sqrt{2}y=0\)

I get (A-C)/B to be zero which is undefined for inverse cotangent. So I don't see how to find \(\displaystyle \phi\) for the axis translation.

What am I missing?

Thanks

 

[galactus]

You're OK. You have \(\displaystyle cot(2\theta)=0\).

Which means \(\displaystyle \theta = 45 \;\ \text{degrees}\)

\(\displaystyle cot^{-1}(0)=\frac{\pi}{2}\)