Cartesian to polar eq. is this correct

given the cartesian equation x^2 -12y - 36 = 0 find a polar equation with the same graph.

r(cos^2(Θ)) - r(12sin(Θ)) = 36
r(cos^2(Θ) - 12sin(Θ)) = 36
r = (36)/(cos^2(Θ) - 12sin(Θ))

am i correct?

No.
x = r cosѲ, y = r sinѲ
x^2 – 12y – 36 = 0
r^2 (cos^2)(Ѳ) – 12r sinѲ – 36 = 0
r^2 - r^2 (sin^2)(Ѳ) – 12r sinѲ – 36 = 0
r^2 – (r sinѲ + 6)^2 = 0
(r + r sinѲ + 6)(r – r sinѲ – 6) = 0
r(1+sinѲ) + 6 = 0 or r(1-sinѲ) – 6 = 0
r > 0
therefore
r = 6/(1 – sinѲ)