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

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Answer 1 · 2016-07-31T09:51:28

#r=-((2sin(theta)+4cos(theta))/cos(2theta))#

#2y=y^2-x^2-4x#

#x=rcos(theta)#
#y=rsin(theta)#

Plug these values in the given equation

#2rsin(theta)=r^2sin^2(theta)-r^2cos^2(theta)-4rcos(theta)#

#2rsin(theta)+4rcos(theta)=-r^2(cos^2(theta)-sin^2(theta))#

#r(2sin(theta)+4cos(theta))=-r^2 (cos(2theta))#

Used the identity #cos(2theta)=cos^2(theta)-sin^2(theta)#

#r=-((2sin(theta)+4cos(theta))/cos(2theta))#