Question

How do you convert `y=(x-y)^2+xy` into a polar equation?

Answer 1

`r=sintheta/(1-sinthetacostheta)`

Explanation:

The relation between Cartesian coordinates `(x,y)` and polar coordinates `(r,theta)` is given by

`x=rcostheta`, `y=rsintheta` and `r^2=x^2+y^2`

Hence, `y=(x-y)^2+xy`

`hArrrsintheta=(rcostheta-rsintheta)^2+rcosthetaxxrsintheta`

or `sintheta/r=cos^2theta+sin^2theta-2sinthetacostheta+sinthetacostheta`

or `r=1-sinthetacostheta`

or `r=sintheta/(1-sinthetacostheta)`
graph{y=(x-y)^2+xy [-2.344, 2.656, -0.43, 2.07]}