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

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Answer 1 · 2016-12-27T14:32:50

$r = \frac{sin θ}{1 - sin θ cos θ}$ $r=sintheta/(1-sinthetacostheta)$

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 $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$

Hence, $y = {(x - y)}^{2} + x y$ $y=(x-y)^2+xy$

$\Leftrightarrow r sin θ = {(r cos θ - r sin θ)}^{2} + r cos θ \times r sin θ$ $hArrrsintheta=(rcostheta-rsintheta)^2+rcosthetaxxrsintheta$

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

or $r = 1 - sin θ cos θ$ $r=1-sinthetacostheta$

or $r = \frac{sin θ}{1 - sin θ cos θ}$ $r=sintheta/(1-sinthetacostheta)$
graph{y=(x-y)^2+xy [-2.344, 2.656, -0.43, 2.07]}

How do you convert y=(x-y)^2+xy y=(x−y)2+xyy=(x-y)^2+xy into a polar equation?