How do you convert y^2 = x^2 ((2+x) / (2-x))y2=x2(2+x2x) to polar form?

1 Answer
Oct 15, 2016

Please see the explanation for the process.
r = (2(sin^2(theta) - cos^2(theta)))/(cos^3(theta) + cos(theta)sin^2(theta))r=2(sin2(θ)cos2(θ))cos3(θ)+cos(θ)sin2(θ)

Explanation:

Given:

y^2 = x^2((2 + x)/(2 - x))y2=x2(2+x2x)

Multiply both sides by 2 - x2x:

2y^2 - xy^2 = 2x^2 + x^32y2xy2=2x2+x3

add xy^2 - 2x^2xy22x2 to both sides and flip:

x^3 + xy^2 = 2y^2 - 2x^2x3+xy2=2y22x2

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

r^3cos^3(theta) + r^3cos(theta)sin^2(theta) = 2r^2sin^2(theta) - 2r^2cos^2(theta)r3cos3(θ)+r3cos(θ)sin2(θ)=2r2sin2(θ)2r2cos2(θ)

Divide both sides by r^2r2:

r(cos^3(theta) + cos(theta)sin^2(theta)) = 2(sin^2(theta) - cos^2(theta))r(cos3(θ)+cos(θ)sin2(θ))=2(sin2(θ)cos2(θ))

Divide both sides by the coefficient of r:

r = (2(sin^2(theta) - cos^2(theta)))/(cos^3(theta) + cos(theta)sin^2(theta))r=2(sin2(θ)cos2(θ))cos3(θ)+cos(θ)sin2(θ)