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

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Answer 1 · 2016-05-19T10:43:15

Polar equation is $r = \frac{sin θ - 3 cos θ}{{cos}^{2} θ + 2 cos 2 θ}$ $r=(sintheta-3costheta)/(cos^2theta+2cos2theta)$

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

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

Using them we can convert $y = 3 x^{2} + 3 x - 2 y^{2}$ $y=3x^2+3x-2y^2$ as follows.

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

or $r sin θ = 3 {(r cos θ)}^{2} + 3 r cos θ - 2 {(r sin θ)}^{2}$ $rsintheta=3(rcostheta)^2+3rcostheta-2(rsintheta)^2$

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

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

or $sin θ = r {cos}^{2} θ + 3 cos θ + 2 r cos 2 θ$ $sintheta=rcos^2theta+3costheta+2rcos2theta$

or $r = \frac{sin θ - 3 cos θ}{{cos}^{2} θ + 2 cos 2 θ}$ $r=(sintheta-3costheta)/(cos^2theta+2cos2theta)$

How do you convert y= 3x^2+3x-2y^2 y=3x2+3x−2y2y= 3x^2+3x-2y^2 into a polar equation?