How do you convert y=y^2+3x^2 -x y=y2+3x2−x into a polar equation?

1 Answer
Douglas K.
Oct 11, 2016

Please see the explanation for "How".

r = (sin(theta) + cos(theta))/(1 + 2cos^2(theta))r=sin(θ)+cos(θ)1+2cos2(θ)

Explanation:

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

rsin(theta) = (rsin(theta))^2 + 3(rcos(theta))^2 - rcos(theta)rsin(θ)=(rsin(θ))2+3(rcos(θ))2−rcos(θ)

Add rcos(theta)rcos(θ) to both sides:

rsin(theta) + rcos(theta) = (rsin(theta))^2 + 3(rcos(theta))^2rsin(θ)+rcos(θ)=(rsin(θ))2+3(rcos(θ))2

Factor out rr and r^2r2:

r(sin(theta) + cos(theta)) = r^2(sin^2(theta) + 3cos^2(theta))r(sin(θ)+cos(θ))=r2(sin2(θ)+3cos2(θ))

Pull out a cos^2(theta)cos2(θ) from 3cos^2(theta)3cos2(θ):

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

Use the identity sin^2(theta) + cos^2(theta) = 1sin2(θ)+cos2(θ)=1

r(sin(theta) + cos(theta)) = r^2(1 + 2cos^2(theta))r(sin(θ)+cos(θ))=r2(1+2cos2(θ))

Divide both sides by r(1 + 2cos^2(theta))r(1+2cos2(θ)):

r = (sin(theta) + cos(theta))/(1 + 2cos^2(theta))r=sin(θ)+cos(θ)1+2cos2(θ)

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