How do you convert $- y = 3 y^{2} - 4 x^{2} - 2 x$ $-y=3y^2-4x^2 -2x$ into a polar equation?

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Answer 1 · 2016-04-17T19:57:16

$3 r^{2} {sin}^{2} θ + r sin θ - 4 r^{2} {cos}^{2} θ - 2 r cos θ = 0$ $3r^2sin^2theta + r sin theta -4r^2 cos^2 theta -2 r cos theta = 0$

$3 y^{2} + y - 4 x^{2} - 2 x = 0$ $3y^2+y-4x^2-2x=0$

Use formulas

$x = r cos θ, y = r sin θ$ $x=rcos theta, y= r sin theta$

$3 r^{2} {sin}^{2} θ + r sin θ - 4 r^{2} {cos}^{2} θ - 2 r cos θ = 0$ $3r^2sin^2theta + r sin theta -4r^2 cos^2 theta -2 r cos theta = 0$

How do you convert -y=3y^2-4x^2 -2x −y=3y2−4x2−2x-y=3y^2-4x^2 -2x into a polar equation?