Converting $x^2 + 6y - 9 = 0$ to polar.

Question

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}

Answer 1

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 $$ r = \frac{3}{1+\sin(\theta)}$$

Answer 2

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 '$+$': $$ r=\frac3{1+\sin(\phi)} $$

Answer 3

$$x^2+6y=9$$

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 $$r=\frac{3}{\sin \theta - 1} \lor r=\frac{3}{1+ \sin \theta} $$