How do you convert $y = 2 y^{2} - x^{2} + 6 x y$ $y=2y^2-x^2+6xy$ into a polar equation?

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Answer 1 · 2018-03-03T13:37:58

$r = \frac{sin θ}{2 - 3 {cos}^{2} θ + 3 sin 2 θ}$ $r=sintheta/(2-3cos^2theta+3sin2theta)$

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

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

Hence we can write $y = 2 y^{2} - x^{2} + 6 x y$ $y=2y^2-x^2+6xy$ as

$r sin θ = 2 r^{2} {sin}^{2} θ - r^{2} {cos}^{2} θ + 6 r^{2} sin θ cos θ$ $rsintheta=2r^2sin^2theta-r^2cos^2theta+6r^2sinthetacostheta$

or $sin θ = r (2 {sin}^{2} θ - {cos}^{2} θ + 6 sin θ cos θ)$ $sintheta=r(2sin^2theta-cos^2theta+6sinthetacostheta)$

or $r = \frac{sin θ}{2 {sin}^{2} θ - {cos}^{2} θ + 6 sin θ cos θ}$ $r=sintheta/(2sin^2theta-cos^2theta+6sinthetacostheta)$

= $\frac{sin θ}{2 - 3 {cos}^{2} θ + 3 sin 2 θ}$ $sintheta/(2-3cos^2theta+3sin2theta)$

How do you convert y=2y^2-x^2+6xy y=2y2−x2+6xyy=2y^2-x^2+6xy into a polar equation?