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

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Answer 1 · 2017-07-21T10:12:01

$rcos^2theta-2rsin2theta+2rsin^2theta-2r^2cos^2thetasintheta-sintheta=0$

Let;s expand this first :

$y=(x-2y)^2-2x^2y-2y^2=$

$x^2-4xy+4y^2-2x^2y-2y^2=x^2-4xy+2y^2-2x^2y=>$

$y=x^2-4xy+2y^2-2x^2y$

Now to swith to polar coordinates we do the following substitutions :

$y=rsintheta$
$x=rcostheta$

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

$=>sintheta=rcos^2theta-2rsin2theta+2rsin^2theta-2r^2cos^2thetasintheta=>$

$rcos^2theta-2rsin2theta+2rsin^2theta-2r^2cos^2thetasintheta-sintheta=0$

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