How do you convert 2y=y^2-4x^2 -2x 2y=y24x22x into a polar equation?

1 Answer
Oct 16, 2016

Please see the explanation for the process.

r = 2(sin(theta) - cos(theta))/(sin^2(theta) - 4cos^2(theta))r=2sin(θ)cos(θ)sin2(θ)4cos2(θ)

Explanation:

Write as:

y^2 - 4x^2 - 2y - 2x = 0y24x22y2x=0

Substitute rsin(theta)rsin(θ) for every y and rcos(theta)rcos(θ) for every x:

(rsin(theta))^2 - 4(rcos(theta))^2 - 2(rsin(theta)) - 2(rcos(theta)) = 0(rsin(θ))24(rcos(θ))22(rsin(θ))2(rcos(θ))=0

Put common factors outside of the ()s:

(sin^2(theta) - 4cos^2(theta))r^2 - 2r(sin(theta) - cos(theta)) = 0(sin2(θ)4cos2(θ))r22r(sin(θ)cos(θ))=0

Most the second term to the right:

(sin^2(theta) - 4cos^2(theta))r^2 = 2r(sin(theta) - cos(theta))(sin2(θ)4cos2(θ))r2=2r(sin(θ)cos(θ))

Divide both sides by (sin^2(theta) - 4cos^2(theta))r(sin2(θ)4cos2(θ))r

r = 2(sin(theta) - cos(theta))/(sin^2(theta) - 4cos^2(theta))r=2sin(θ)cos(θ)sin2(θ)4cos2(θ)