How do you convert #y=2y^2-x^2+6xy # into a polar equation?

Question

1925 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-03T13:37:58

#r=sintheta/(2-3cos^2theta+3sin2theta)#

The relation between Cartesian coordinates #(x,y)# and polar coordinates #(r,theta)# is given by

#x=rcostheta#, #y=rsintheta# and #x^2+y^2=r^2#

Hence we can write #y=2y^2-x^2+6xy# as

#rsintheta=2r^2sin^2theta-r^2cos^2theta+6r^2sinthetacostheta#

or #sintheta=r(2sin^2theta-cos^2theta+6sinthetacostheta)#

or #r=sintheta/(2sin^2theta-cos^2theta+6sinthetacostheta)#

= #sintheta/(2-3cos^2theta+3sin2theta)#