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

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Answer 1 · 2018-05-10T04:33:25

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

$x = r cos θ$ $x = r cos theta$

$y = r sin θ$ $y = r sin theta$

$3 y = 2 y^{2} - x^{2}$ $3 y = 2y^2 - x^2$

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

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

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

We stop here.

How do you convert 3y=2y^2-x^2 3y=2y2−x23y=2y^2-x^2 into a polar equation?