How do you convert #(x^2 + y^2)^2 = 4(x^2-y^2)# into polar form?

Question

4090 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-31T13:48:20

#r^2=cos(2theta)#

The pass transformations

# {(x = r cos(theta)), (y = r sin(theta)):} #

give

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

or equivalently

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

See below the graphic representation.

enter image source here