What the is the polar form of #sqrt(x^2+y^2) = 6x+7y-x^2y-2xy #?

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Answer 1 · 2016-10-12T13:50:03

We know the relations

#x=rcostheta and y =rsintheta#

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

where r and #theta# are the polar coordinate of a point having rectangular coordinate #(x,y)#

The given equation in rectanglar form is

#sqrt(x^2+y^2)=6x+7y-x^2y-2xy#

#=>sqrt(r^2)=6rcostheta+7rsintheta-r^3cos^2thetasintheta-2r^2costhetasintheta#

#=>r=6rcostheta+7rsintheta-r^3cos^2thetasintheta-r^2sin2theta#

#=>6costheta+7sintheta-r^2cos^2thetasintheta-rsin2theta=1#

This is the polar form of the given equation.