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

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Answer 1 · 2016-07-31T09:51:28

$r=-((2sin(theta)+4cos(theta))/cos(2theta))$

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

$x=rcos(theta)$
$y=rsin(theta)$

Plug these values in the given equation

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

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

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

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

$r=-((2sin(theta)+4cos(theta))/cos(2theta))$

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