How do you convert $y= 3x^2 -2xy+4xy^2$ into a polar equation?

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Answer 1 · 2016-11-08T16:44:57

$4r^2costhetasin^2theta+r(3cos^2theta-2costhetasintheta)-sintheta=0$

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

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

Hence, $y=3x^2-2xy+4xy^2$

$hArrrsintheta=3r^2cos^2theta-2r^2costhetasintheta+4r^3costhetasin^2theta$

or $sintheta=3rcos^2theta-2rcosthetasintheta+4r^2costhetasin^2theta$

$4r^2costhetasin^2theta+r(3cos^2theta-2costhetasintheta)-sintheta=0$

How do you convert y= 3x^2 -2xy+4xy^2 y= 3x^2 -2xy+4xy^2 into a polar equation?