How do you convert $-y=3y^2-x^2 -2x$ into a polar equation?

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Answer 1 · 2016-05-03T08:52:00

$r=(2costheta-sintheta)/(3sin^2theta-cos^2theta)$

If $(r,theta)$ is in polar form and $(x,y)$ in Cartesian form the relation between them is as follows:

$x=rcostheta$ , $y=rsintheta$ , $r^2=x^2+y^2$ and $tantheta=y/x$

Hence, $-y=3y^2-x^2-2x$ can be written as

$-rsintheta=3r^2sin^2theta-r^2cos^2theta-2rcostheta$ or

$-sintheta=3rsin^2theta-rcos^2theta-2costheta$ or

$r(3sin^2theta-cos^2theta)=2costheta-sintheta$ or

$r=(2costheta-sintheta)/(3sin^2theta-cos^2theta)$

How do you convert -y=3y^2-x^2 -2x -y=3y^2-x^2 -2x into a polar equation?