Converting $x^2 + 6y - 9 = 0$ to polar. So far I got here
\begin{align}
(r\cos\phi)^2 & + 6 r \sin\phi- 9 = 0\\
(r\cos\phi)^2 & = 9 - 6r \sin\phi
\end{align}
 A: Since $r^2\cos^2(\phi)+6r\sin(\phi)-9=0$,
$$
\begin{align}
r
&=\frac{-6\sin(\phi)\pm\sqrt{36\sin^2(\phi)+36\sin^2(\phi)}}{2\cos^2(\phi)}\\
&=\frac{-3\sin(\phi)\pm3}{\cos^2(\phi)}\\
&=\frac3{\sin(\phi)\pm1}
\end{align}
$$
Note that both '$+$' and '$-$' yield the same curve. '$-$' simply gives the curve $180^\circ$ around with negative the radius. So we can simply pick '$+$':
$$
r=\frac3{1+\sin(\phi)}
$$
A: You could also solve for $r$ as a function of $\theta$: write $\cos^2(\theta)$ in terms of $\sin(\theta)$, put all terms on the left, and factor.  You should end up with
$$ r = \frac{3}{1+\sin(\theta)}$$
A: $$x^2+6y=9$$
As always we choose $x=r \cos \theta$ and $y=r \sin \theta$
Thus $$(r \cos \theta)^2 +6r \sin \theta -9=0$$
With the identity $\cos^2 \theta = 1- \sin^2 \theta$ we find
$$r^2 - (r \sin \theta)^2 + 6r \sin \theta -9=0$$
Which equals $$(r \sin \theta -3)^2=r^2$$
Thus $r=r \sin \theta -3$ or $r= 3-r \sin \theta$ and subsequently we see $$r=\frac{3}{\sin \theta - 1} \lor r=\frac{3}{1+ \sin \theta} $$
