Polar to Cartesian relations:
#x=Rsintheta#
#y=R cos theta#
Our equation:
#3x^2+4y^2-6x=9#
Substitute:
#3R^2sin^2theta+4R^2cos^2theta-6Rsintheta=9#
Simplify:
#3R^2(sin^2theta+cos^2theta)+R^2cos^2theta-6Rsintheta=9#
#3R^2(1+cos^2theta)-6Rsintheta=9#
#R^2(1+cos^2theta)-2Rsintheta-3=0#
We can if we wish rewrite this in the form #R=f(theta)# by solving the quadratic here in #R#. As the original equation wasn't given in the form #y=g(x)# we shouldn't feel obliged to, but it might be a useful formulation.
#R=(2sintheta+-sqrt(4sin^2theta+12(1+cos^2theta)))/(2(1+cos^2theta))#
#R=(sintheta+-sqrt(sin^2theta+3(1+cos^2theta)))/(1+cos^2theta)#
#R=(sintheta+-sqrt(4+2cos^2theta))/(1+cos^2theta)#
