Convert hyperbola in rectangular form to polar form

Step 1
Staring with
${\displaystyle 3y^2-16y-x^2+16=0}$
Put it first in the standard format. Compute the square in y
${\displaystyle 3{(y-d\frac{8}{3})}^2-d\frac{64}{3}-x^2+16=0}$
${\displaystyle 3{(y-d\frac{8}{3})}^2-x^2=d\frac{16}{3}}$
${\displaystyle d\frac{9}{16}{(y-d\frac{8}{3})}^2-d\frac{3}{16}{x}^2=1}$
Thus the center is at
${\displaystyle (0,\frac{8}{3}),a=\frac{4}{3}}$ and ${\displaystyle b=\frac{4}{\sqrt{3}}}$
The focal distance
${\displaystyle c=\sqrt{a^2+b^2}=4\sqrt{d\frac{1}{9}+d\frac{1}{3}}=d\frac{8}{3}}$
Hence, the foci are at $(0,0)$ and $(0,\frac{16}{3})$
Taking the focus that is at the origin, then ${\displaystyle x=r\cos \theta ,y=r\sin \theta }$
Back to the original equation
${\displaystyle 3y^2-16y-x^2+16=0}$
Substituting the polar expressions,
${\displaystyle 3{\left(r\sin \theta \right)}^2-16r\sin \theta -r^2{\cos }^2\theta +16=0}$
Using ${\displaystyle {\cos }^2\theta =1-{\sin }^2\theta }$ and collecting terms
${\displaystyle r^2(4{\sin }^2\theta -1)-16r\sin \theta +16=0}$
From the quadratic formula, and taking the positive root
$r=d\frac{1}{2(4{\sin }^2\theta -1)} (16\sin \theta -\sqrt{256{\sin }^2\theta -(256{\sin }^2\theta -64)})$
And this simplifies to
${\displaystyle r=d\frac{1}{2(4{\sin }^2\theta -1)}(16\sin \theta -8)}$
Factoring the denominator and cancelling equal terms
${\displaystyle r=d\frac{4}{2\sin \theta +1}}$