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

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Answer 1 · 2018-01-05T08:26:21

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

Given $2 y = - x^{2} + 4 x y$ $2y=-x^2 + 4xy$

and

$x = r cos (θ)$ $x=rcos(theta)$ and $y = r sin (θ)$ $y=rsin(theta)$ we can substitute:

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

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

Since $r \neq 0$ $r ne 0$ for all $θ$ $theta$ we can divide through by $r$ $r$ :

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

Now we can factor $r$ $r$ from the right side:

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

Now we can solve for $r$ $r$ :

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

and for no very good reason, I'd rather rewrite the denominator:

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

How do you convert 2y= -x^2+4xy 2y=−x2+4xy2y= -x^2+4xy into a polar equation?