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

Question

1391 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-15T21:17:40

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

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

use formulas:
$x = r cos θ and y = r sin θ$ $x=rcostheta and y=rsintheta$

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

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