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

1 Answer
Oct 11, 2016

Please see the explanation for "How".

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

Explanation:

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

#rsin(theta) = (rsin(theta))^2 + 3(rcos(theta))^2 - rcos(theta)#

Add #rcos(theta)# to both sides:

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

Factor out #r# and #r^2#:

#r(sin(theta) + cos(theta)) = r^2(sin^2(theta) + 3cos^2(theta))#

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

#r(sin(theta) + cos(theta)) = r^2(sin^2(theta) + cos^2(theta) + 2cos^2(theta))#

Use the identity #sin^2(theta) + cos^2(theta) = 1#

#r(sin(theta) + cos(theta)) = r^2(1 + 2cos^2(theta))#

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

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