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

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Answer 1 · 2018-05-26T10:14:01

$r=-(sintheta)/(sin^2theta+3cos^2theta+costhetasintheta)$

Rewrite as:
$y^2+3x^2+xy=-y$

Substitute in:
$x=rcostheta$
$y=rsintheta$

$(rsintheta)^2+3(rcostheta)^2+(rcostheta)(rsintheta)=-rsintheta$

$r^2(sintheta)^2+3r^2(costheta)^2+r^2(costhetasintheta)=-rsintheta$

Divide both sides by $r$

$r(sintheta)^2+3r(costheta)^2+r(costhetasintheta)=-sintheta$

Factorise out $r$ :
$r(sin^2theta+3cos^2theta+costhetasintheta)=-sintheta$

Make $r$ the subject:
$r=-(sintheta)/(sin^2theta+3cos^2theta+costhetasintheta)$

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