Converting $x^2 + 6y - 9 = 0$ to polar Hi I'm trying to solve this problem but am having difficulty removing the remaining r. I have tried in here  but cannot get an answer.
Any help is appreciated 
 A: As wrote Patrick Da Silva,
$$y = -\frac{1}{6} x^2 + \frac{3}{2}. \tag{*}$$
Put $$\begin {gather}
x=\rho \cos{\varphi}, \\
y-\frac{3}{2}=\rho \sin{\varphi},
\end{gather}$$
translating pole to the point $\left(0, \,\frac{3}{2}\right).$
Then $(*)$ becomes
$$\rho \sin{\varphi}=\rho^2 \cos^2{\varphi},$$
or, for $\rho\ne{0}$
$$\sin{\varphi}=\rho \cos^2{\varphi}.$$
A: Your curve is a polynomial. You can look at it this way. 
$$
y = -\frac 16 x^2 + \frac 32.
$$
If you want to switch into polar coordinates, it'll give you the same equation, in polar coordinates. You can't expect to "find $x$ and $y$" or to "find the value of $r$ and $\theta$". Are you asked to do something in particular?
Hope that helps,
A: Direct translation gives $r^2\cos^2\theta +6r\sin\theta-9=0$. If we use one of the common conventions, that $r\ge 0$, we can solve explicitly for $r$ using the Quadratic Formula, getting
$$r=-3\cos\theta+3\sqrt{\sin^2\theta +1}.$$
But the second form is perhaps less attractive than the first implicit form.
There are various other ways to manipulate, for example by rewriting $\cos^2\theta$ as $1-\sin^2\theta$. All the answers we get by correct manipulations are correct. 
A: I'll assume you want to find a function $r(\theta)$ for the curve. 
After substituting for $x$ and $y$, we should anticipate solving a quadratic formula. This is why you are having trouble solving for $r$. Try solving for $x$ in a general quadratic equation, and you will be faced with a similar obstruction. 
$$r^2\cos^2\theta+6r\sin\theta-9=0.$$
We now apply the quadratic formula: 
\begin{align*}
r&={-b\pm\sqrt{b^2-4ac}\over 2a}\\
&={-6\sin\theta\pm\sqrt{36\sin^2\theta+36\cos^2\theta}\over 2\cos^2\theta}\\
&={-3\sin\theta\pm3\over \cos^2\theta}.
\end{align*}
Some care will be needed in choosing which of $\pm$ are chosen. 
