Given $4$ polar coordinates $(3, -\pi/6)$, $(1, \pi/3)$, $(3, 5\pi/6)$, $(-3, 4\pi/3)$, graph and find the polar equation.
I know that the general polar equation is $r = ep / 1+- e \cos (\theta)$. After plotting the points, I am not sure how the graph looks like, but I assume it looks like this : 
Since it's hyperbola, the eccentricity will be $> 1$. Then I am kind of stuck. Not sure how can I determine the sign and cos/sin for the polar equation though I think I can find $e$ and $p$ by plugging-in 2 points for $\theta$ and $r$ into the general polar equation and then solving it as a system of equations. Am I on the right way? Thanks!
The polar form of the conic section you quote,
$$r = \frac{p}{1+\epsilon \cos{\theta}}$$
is very restrictive and does not cover a hyperbola that has an offset and rotation such as that in your graph. If you convert back to cartesian form, the polar equation is equivalent to
$$\frac{\left ( x + \frac{\epsilon p}{1-\epsilon^2}\right)^2}{p^2/(1-\epsilon^2)^2} + \frac{y^2}{p^2/(1-\epsilon^2)} = 1$$
Thus, the conic, whether an ellipse or hyperbola, is centered on the $x$-axis and has axes of symmetry parallel to the $x$ and $y$ axes. Thus, the conic having the given polar form is completely determined using only two points.
Obviously, you need a more general parameteric form; I do not what it is off the top of my head. That said, you can start with the cartesian form:
$$a x^2+ 2 b x y+c y^2 + d x + e y + f=0$$
Thus, you need six points for the general conic. Perhaps you can eliminate two parameters, say $e$ and $f$, but remember you are imposing a symmetry that may or may not be valid for your plot.