How do you convert $2 y = y^{2} - x^{2} - 4 x$ $2y=y^2-x^2 -4x$ into a polar equation?

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Answer 1 · 2016-07-31T09:51:28

$r = - (\frac{2 sin (θ) + 4 cos (θ)}{cos (2 θ)})$ $r=-((2sin(theta)+4cos(theta))/cos(2theta))$

$2 y = y^{2} - x^{2} - 4 x$ $2y=y^2-x^2-4x$

$x = r cos (θ)$ $x=rcos(theta)$
$y = r sin (θ)$ $y=rsin(theta)$

Plug these values in the given equation

$2 r sin (θ) = r^{2} {sin}^{2} (θ) - r^{2} {cos}^{2} (θ) - 4 r cos (θ)$ $2rsin(theta)=r^2sin^2(theta)-r^2cos^2(theta)-4rcos(theta)$

$2 r sin (θ) + 4 r cos (θ) = - r^{2} ({cos}^{2} (θ) - {sin}^{2} (θ))$ $2rsin(theta)+4rcos(theta)=-r^2(cos^2(theta)-sin^2(theta))$

$r (2 sin (θ) + 4 cos (θ)) = - r^{2} (cos (2 θ))$ $r(2sin(theta)+4cos(theta))=-r^2 (cos(2theta))$

Used the identity $cos (2 θ) = {cos}^{2} (θ) - {sin}^{2} (θ)$ $cos(2theta)=cos^2(theta)-sin^2(theta)$

$r = - (\frac{2 sin (θ) + 4 cos (θ)}{cos (2 θ)})$ $r=-((2sin(theta)+4cos(theta))/cos(2theta))$

How do you convert 2y=y^2-x^2 -4x 2y=y2−x2−4x2y=y^2-x^2 -4x into a polar equation?