Question

What the is the polar form of `y^2 = (x-3y)^2`?

Answer 1

`tan theta = 1/2 and tan theta = 1/4`, representing radial lines `theta=tan^(-1)(1/2)`. its opposite `theta=pi+tan^(-1)(1/2), theta=tan^(-1)(1/4)` and its opposite `theta=pi+tan^(-1)(1/4)`.

Explanation:

If `a^2=b^2, a=+-b`.
In rectangular form, this represents the pair of straight lines
`y = +-(3x-y)`.

Substitute `x=r cos theta and y = r sin theta`.
The result is the pair of equations
`tan theta = 1/2 and tan theta = 1/4`.

`tan theta` is positive in both the 1st and the 3rd quadrants. These bifurcate into four equations for two pairs of opposite radial lines, as given in the answer.
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