How do you convert $y = 3 x^{2} - 2 x - x y^{2}$ $y= 3x^2 -2x-xy^2$ into a polar equation?

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Answer 1 · 2017-03-10T16:15:53

$r^{2} {sin}^{2} θ - 3 r cos θ + tan θ + 2 = 0$ $r^2sin^2theta-3rcostheta+tantheta+2=0$

The relation between rectangular coordinates $(x, y)$ $(x,y)$ and polar coordinates $(r, θ)$ $(r,theta)$ is given by

$x = r cos θ$ $x=rcostheta$ , $y = r sin θ$ $y=rsintheta$ , $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$

Hence $y = 3 x^{2} - 2 x - x y^{2}$ $y=3x^2-2x-xy^2$ can be written as

$r sin θ = 3 r^{2} {cos}^{2} θ - 2 r cos θ - r^{3} cos θ {sin}^{2} θ$ $rsintheta=3r^2cos^2theta-2rcostheta-r^3costhetasin^2theta$

or $sin θ = 3 r {cos}^{2} θ - 2 cos θ - r^{2} cos θ {sin}^{2} θ$ $sintheta=3rcos^2theta-2costheta-r^2costhetasin^2theta$

or $r^{2} cos θ {sin}^{2} θ - 3 r {cos}^{2} θ + sin θ + 2 cos θ = 0$ $r^2costhetasin^2theta-3rcos^2theta+sintheta+2costheta=0$

or $r^{2} {sin}^{2} θ - 3 r cos θ + tan θ + 2 = 0$ $r^2sin^2theta-3rcostheta+tantheta+2=0$

How do you convert y= 3x^2 -2x-xy^2 y=3x2−2x−xy2y= 3x^2 -2x-xy^2 into a polar equation?