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

Question

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

Answer 1

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,

Answer 2

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.

Answer 3

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.

Answer 4

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 $\rho\ne{0}$ $$\sin{\varphi}=\rho \cos^2{\varphi}.$$