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

1 Answer
Feb 23, 2018

rectangular form given:

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

x=rcostheta

y=rsintheta

Substituting for the terms

rsintheta=(rsintheta)^2/(rcostheta)+(rcostheta-3)(rsintheta-2)

rsinthetarcostheta=(rsintheta)^2+rcostheta(rcostheta-3)(rsintheta-2)

r^2sinthetacostheta=r^2sin^2theta+rcostheta(r^2costhetasintheta-2rcostheta-3rsintheta+6)

r^2sinthetacostheta-r^2sin^2theta-rcostheta(r^2costhetasintheta-2rcostheta-3rsintheta+6)=0

r^2sinthetacostheta-r^2sin^2theta-r^3cos^2thetasintheta+2r^2cos^2theta+3r^2costhetasintheta-6rcostheta=0

-r^3cos^2thetasintheta+r^2(sinthetacostheta-sin^2theta+2cos^2theta+3costhetasintheta)-6rcostheta=0

-r^3cos^2thetasintheta+r^2(4costhetasintheta-sin^2theta+2cos^2theta)-6rcostheta=0

4costhetasintheta-sin^2theta+2cos^2theta=2cos^2theta+4costhetasintheta-sin^2theta
Thus,

r^3cos^2thetasintheta-r^2(4costhetasintheta-sin^2theta+2cos^2theta)+6rcostheta=0
Dividing throughout by cos^3theta
r^3tantheta-r^2(4secthetatantheta-secthetatan^2theta+2sectheta)+6rsec^2theta=0

r(r^2tantheta-r(4secthetatantheta-secthetatan^2theta+2sectheta)+6sec^2theta)=0