How do you convert $y^2 = x^2 ((2+x) / (2-x))$ to polar form?

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Answer 1 · 2016-10-15T01:08:19

Please see the explanation for the process.
$r = (2(sin^2(theta) - cos^2(theta)))/(cos^3(theta) + cos(theta)sin^2(theta))$

Given:

$y^2 = x^2((2 + x)/(2 - x))$

Multiply both sides by $2 - x$ :

$2y^2 - xy^2 = 2x^2 + x^3$

add $xy^2 - 2x^2$ to both sides and flip:

$x^3 + xy^2 = 2y^2 - 2x^2$

Now substitute $rcos(theta$ for x and $rsin(theta)$ for y:

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

Divide both sides by $r^2$ :

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

Divide both sides by the coefficient of r:

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

How do you convert y^2 = x^2 ((2+x) / (2-x))y^2 = x^2 ((2+x) / (2-x)) to polar form?