How do you convert $y=(3x-2y)^2-x^2y -5y^2$ into a polar equation?

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Answer 1 · 2016-07-02T05:03:01

$r^2cos^2thetasintheta+r(6sin2theta+10sin^2theta-9)+sintheta=0$

For converting a equation with Cartesian coordinates into a polar equation, one can use the relations $x=rcostheta$ and $y=rsintheta$

Hence, $y=(3x-2y)^2-x^2y-5y^2$ can be written as

$rsintheta=(3rcostheta-2rsintheta)^2-(rcostheta)^2(rsintheta)-5(rsintheta)^2$ or

$rsintheta=9r^2cos^2theta+4r^2sin^2theta-12r^2sinthetacostheta-r^3cos^2thetasintheta-5r^2sin^2theta$

or $rsintheta=9r^2(1-sin^2theta)-r^2sin^2theta-6r^2sin2theta-r^3cos^2thetasintheta$

or $sintheta=9r-10rsin^2theta-6rsin2theta-r^2cos^2thetasintheta$

or $r^2cos^2thetasintheta+r(6sin2theta+10sin^2theta-9)+sintheta=0$

How do you convert y=(3x-2y)^2-x^2y -5y^2 y=(3x-2y)^2-x^2y -5y^2 into a polar equation?