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

1 Answer
Oct 10, 2016

Please see the explanation for the process.

#r = -((3cos(theta))/(3(cos(theta))(sin(theta)) - 2))#

Explanation:

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

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

Factor out r whenever possible:

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

Swap the last two terms:

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

Combine the #r^2# terms on the right:

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

Substitute 1 for #cos^2(theta) + sin^2(theta):#

#3r^2(cos(theta))(sin(theta)) = 2r^2 - 3rcos(theta) #

Combine like terms:

#r^2(3cos(theta))(sin(theta)) - 2) = - 3rcos(theta) #

Move everything to the left side:

#r^2(3cos(theta))(sin(theta)) - 2) + 3rcos(theta) = 0 #

Divide both sides by #(3cos(theta))(sin(theta)) - 2)#

#r^2 + r((3cos(theta))/(3(cos(theta))(sin(theta)) - 2)) = 0 #

Divide both sides by r:

#r + ((3cos(theta))/(3(cos(theta))(sin(theta)) - 2)) = 0 #

Solve for r:

#r = -((3cos(theta))/(3(cos(theta))(sin(theta)) - 2))#