Question

What the is the polar form of `x^2-y^2 = 6x+-x^2y-y`?

Answer 1

I get `r(\cos^2(\theta)-\sin^2\theta))=6\cos(\theta)\pm \sin(\theta)+r^2\cos^2(\theta)\sin(\theta)`.

Explanation:

Put in `x=r\cos(\theta), y=r\sin(\theta)`. Then

`r^2(\cos^2(\theta)-\sin^2\theta))=6r\cos(\theta)\pm r\sin(\theta)+r^2\cos^2(\theta)\sin(\theta)`

Since every term has a factor of `r`, either `r=0` or you can divide by `r`. The latter gives

`r(\cos^2(\theta)-\sin^2\theta))=6\cos(\theta)\pm \sin(\theta)+r^2\cos^2(\theta)\sin(\theta)`

This already contains `r=0` when `\tan(\theta)=\pm 6` and so it's the complete solution.