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

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Answer 1 · 2016-09-16T15:56:13

$r(cos^2theta-3sin^2theta+sin2theta)+sinthetacos^2theta=0$

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

$x=rcostheta$ , $y=rsintheta$ i.e. $x=r/costheta$ , $y=r/sintheta$ and $r^2=x^2+y^2$ ,

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

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

Now multiplying it by $sin^2thetacos^2theta$ , we get

$rsinthetacos^2theta=-r^2cos^2theta+3r^2sin^2theta-2r^2sinthetacostheta$ or

$sinthetacos^2theta=-r(cos^2theta-3sin^2theta+2sinthetacostheta)$ or

$r(cos^2theta-3sin^2theta+sin2theta)+sinthetacos^2theta=0$

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