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 · 2016-02-02T11:57:35

here is how,

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

$or, y = x^{2} - 2 x y + y^{2} - x y$ $or,y=x^2-2xy+y^2-xy$

$or, r sin θ = (x^{2} + y^{2}) - 3 x y$ $or,rsintheta=(x^2+y^2)-3xy$

$or, r sin θ = r^{2} - 3 r cos θ r sin θ$ $or,rsintheta=r^2-3rcosthetarsintheta$

$or, 2 r sin θ = 2 r^{2} - 3 \cdot r^{2} \cdot 2 cos θ sin θ$ $or,2rsintheta=2r^2-3*r^2*2costhetasintheta$

$or, 2 sin θ = 2 r - 3 r sin 2 θ$ $or,2sintheta=2r-3rsin2theta$

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