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))#