How do you convert y=(x-2y)^2-2x^2y -2y^2 y=(x−2y)2−2x2y−2y2 into a polar equation?

1 Answer
Καδήρ Κ.
Jul 21, 2017

rcos^2theta-2rsin2theta+2rsin^2theta-2r^2cos^2thetasintheta-sintheta=0rcos2θ−2rsin2θ+2rsin2θ−2r2cos2θsinθ−sinθ=0

Explanation:

Let;s expand this first :

y=(x-2y)^2-2x^2y-2y^2=y=(x−2y)2−2x2y−2y2=

x^2-4xy+4y^2-2x^2y-2y^2=x^2-4xy+2y^2-2x^2y=>x2−4xy+4y2−2x2y−2y2=x2−4xy+2y2−2x2y⇒

y=x^2-4xy+2y^2-2x^2yy=x2−4xy+2y2−2x2y

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

y=rsinthetay=rsinθ
x=rcosthetax=rcosθ

rsintheta=r^2cos^2theta-4r^2sinthetacostheta+2r^2sin^2theta-2r^3cos^2thetasinthetarsinθ=r2cos2θ−4r2sinθcosθ+2r2sin2θ−2r3cos2θsinθ

=>sintheta=rcos^2theta-2rsin2theta+2rsin^2theta-2r^2cos^2thetasintheta=>⇒sinθ=rcos2θ−2rsin2θ+2rsin2θ−2r2cos2θsinθ⇒

rcos^2theta-2rsin2theta+2rsin^2theta-2r^2cos^2thetasintheta-sintheta=0rcos2θ−2rsin2θ+2rsin2θ−2r2cos2θsinθ−sinθ=0

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