How do you convert #3y= 3x^2-6xy+3x # into a polar equation?

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Answer 1 · 2016-06-17T12:44:05

#=>r=(tantheta-1)/(costheta-2sintheta)#

We know that if a point in X-Y plane has rectangular coordinate
#(x,y)# and its polar coordinate is #(r,theta)#,then we have the relation

#x=rcostheta and y =rsintheta#

and # r=sqrt(x^2+y^2)#

Now the given equation is

#3y= 3x^2-6xy+3x #

Dividing bothsides by 3

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

Inserting #x=rcostheta and y =rsintheta# in the given equation

#rsintheta=r^2cos^2theta-2rcostheta*rsintheta+rcostheta#

Dividing both sides by #rcostheta#

#=>tantheta=rcostheta-2rsintheta+1#

#=>r=(tantheta-1)/(costheta-2sintheta)#