What the is the polar form of #y^2 = (x-3y)^2-x^2 #?

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Answer 1 · 2016-10-27T07:29:27

We know the relations

#x=rcostheta and y =rsintheta#

Again #x^2+y^2=r^2#

where #(r,theta)# are the polar coordinates of a point corresponding to the rectangular coordinates #(x,y)#

The given equation in rectanglar form is

#y^2=(x-3y)^2-x^2#

#=>y^2=x^2-6xy+9y^2-x^2#

#=>8y^2=6xy#

#=>8r^2sin^2theta=6xxrcosthetaxxrsintheta#

#=>8sin^2theta=6costhetasintheta#

#=>4(1-cos2theta)=3sin2theta#

#=>3sin2theta+4cos2theta=4#

This is the polar form of the given equation.