Question

How do you convert `2x^2=-2y^2` to polar form?

Answer 1

r = 0 and theta is arbitrary.

Explanation:

As `(x, y)=(r cos theta, r sin theta)`, the given equation reduces to

`r^2(cos^theta+sin^2 theta)=r^2(1)=r^2=0`, for every `theta`.

So, r =0 and `theta` is arbitrary.

Interestingly, this is a proof for "0-dimensional point in space is a null

space vector#..