Question

What the is the polar form of `y = y/x^2+(2x-2)(y-5)`?

Answer 1

`r^4cos^2thetasin2theta-3/2 r^3costhetasin2theta-10r^3cos^3theta+10r^2cos^2theta+rsintheta =0`

Explanation:

`y=y/x^2 +(2x-2)(y-5)`

`y=(y+x^2(2x-2)(y-5))/x^2`

`x^2y=y+x^2(2x-2)(y-5)`

`x^2y=y+x^2(2xy-10x-2y+10)`

`x^2y=y+2x^3y-10x^3-2x^2y+10x^2`

`0=y+2x^3y-10x^3-2x^2y+10x^2-x^2y`

`0=2x^3y-3x^2y-10x^3+10x^2+y`

Now use the formula
`x=rcostheta and y=rsin theta`

`0=2r^3cos^3thetaxxrsintheta-3r^2cos^2thetaxxrsintheta-10r^3cos^3theta+10r^2cos^2theta+rsintheta`

`0=2r^4cos^3thetasintheta-3r^3cos^2thetasintheta-10r^3cos^3theta+10r^2cos^2theta+rsintheta`

`0=r^4cos^2theta(2sinthetacostheta)-3/2 r^3costheta(2sinthetacostheta)-10r^3cos^3theta+10r^2cos^2theta+rsintheta`

`0=r^4cos^2thetasin2theta-3/2 r^3costhetasin2theta-10r^3cos^3theta+10r^2cos^2theta+rsintheta`