Question

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

Answer 1

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

Explanation:

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

Put `x=rcostheta` and `y=rsintheta` ; we get :-

`r^2 sin^2theta=((rcostheta-3)^2)/(rsintheta) -r^2cos^2theta`

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