Question

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

Answer 1

`r^2cos^2theta-r^3sin^3theta-6rcostheta+9=0`

Explanation:

`y^3=(x-3)^2`

`y^3=x^2-6x+9`

`0=x^2-y^3-6x+9`

Use the formulas:
`x=rcostheta and y=rsin theta`

`0=r^2cos^2theta-r^3sin^3theta-6rcostheta+9`