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

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Answer 1 · 2016-05-03T08:52:00

$r = \frac{2 cos θ - sin θ}{3 {sin}^{2} θ - {cos}^{2} θ}$ $r=(2costheta-sintheta)/(3sin^2theta-cos^2theta)$

If $(r, θ)$ $(r,theta)$ is in polar form and $(x, y)$ $(x,y)$ in Cartesian form the relation between them is as follows:

$x = r cos θ$ $x=rcostheta$ , $y = r sin θ$ $y=rsintheta$ , $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$ and $tan θ = \frac{y}{x}$ $tantheta=y/x$

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

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

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

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

$r = \frac{2 cos θ - sin θ}{3 {sin}^{2} θ - {cos}^{2} θ}$ $r=(2costheta-sintheta)/(3sin^2theta-cos^2theta)$

How do you convert -y=3y^2-x^2 -2x −y=3y2−x2−2x-y=3y^2-x^2 -2x into a polar equation?