What the is the polar form of sqrt(x^2+y^2) = 6x+7y-x^2y-2xy √x2+y2=6x+7y−x2y−2xy?

1 Answer
P dilip_k
Oct 12, 2016

We know the relations

x=rcostheta and y =rsinthetax=rcosθandy=rsinθ

Again x^2+y^2=r^2x2+y2=r2

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

The given equation in rectanglar form is

sqrt(x^2+y^2)=6x+7y-x^2y-2xy√x2+y2=6x+7y−x2y−2xy

=>sqrt(r^2)=6rcostheta+7rsintheta-r^3cos^2thetasintheta-2r^2costhetasintheta⇒√r2=6rcosθ+7rsinθ−r3cos2θsinθ−2r2cosθsinθ

=>r=6rcostheta+7rsintheta-r^3cos^2thetasintheta-r^2sin2theta⇒r=6rcosθ+7rsinθ−r3cos2θsinθ−r2sin2θ

=>6costheta+7sintheta-r^2cos^2thetasintheta-rsin2theta=1⇒6cosθ+7sinθ−r2cos2θsinθ−rsin2θ=1

This is the polar form of the given equation.

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