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

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Answer 1 · 2018-05-26T10:14:01

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

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

Substitute in:
$x = r cos θ$ $x=rcostheta$
$y = r sin θ$ $y=rsintheta$

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

$r^{2} {(sin θ)}^{2} + 3 r^{2} {(cos θ)}^{2} + r^{2} (cos θ sin θ) = - r sin θ$ $r^2(sintheta)^2+3r^2(costheta)^2+r^2(costhetasintheta)=-rsintheta$

Divide both sides by $r$ $r$

$r {(sin θ)}^{2} + 3 r {(cos θ)}^{2} + r (cos θ sin θ) = - sin θ$ $r(sintheta)^2+3r(costheta)^2+r(costhetasintheta)=-sintheta$

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

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

How do you convert y=-y^2-3x^2-xy y=−y2−3x2−xyy=-y^2-3x^2-xy into a polar equation?