convert hyperbola in rectangular form to polar form i am trying to convert the rectangular equation of a conic (hyperbola) to a polar form.
the rectangular equation is:
$$3y^2 - 16y -x^2 + 16 = 0.$$
i substituted $r\sin\theta$ for y and $r\cos\theta$ for x, and tried to simplify, but I am stuck.
i tried substituting $1-\sin^2\theta$ for the resulting $\cos^2\theta$ term, and ended up with the following expression after some manipulation:
$$r^2(2\sin\theta-1)(2\sin\theta+1)-16r\sin\theta + 16 = 0.$$
the solution to the question is supposed to be:
$$r = \frac{4}{1 + 2\sin\theta}.$$
I just can't figure out how to get from where I am to the expected solution.  Perhaps I made the wrong substitution? I just need to figure out the intermediate steps to get from the original rectangular equation to the final equation in polar form given above.
Hope someone can help me -- thanks!
 A: Staring with $3 y^2 - 16 y - x^2 + 16 = 0 $
Put it first in the standard format.  Complete the square in $y$
$ 3 \left(y - \dfrac{8}{3} \right)^2 - \dfrac{64}{3} - x^2 + 16 = 0 $
$ 3 \left(y - \dfrac{8}{3} \right)^2 - x^2 = \dfrac{16}{3} $
$ \dfrac{9}{16} \left( y - \dfrac{8}{3} \right)^2 - \dfrac{3}{16} x^2 = 1 $
Thus the center is at $\left(0, \dfrac{8}{3} \right) $ , $ a = \dfrac{4}{3} $ and $b = \dfrac{4}{\sqrt{3}} $
The focal distance $ c = \sqrt{a^2 + b^2} = 4 \sqrt{ \dfrac{1}{9} + \dfrac{1}{3} } = \dfrac{8}{3} $
Hence, the foci are at $ (0, 0) $ and $ \left(0, \dfrac{16}{3} \right) $
Taking the focus that is at the origin, then $ x = r \cos \theta, y = r \sin \theta $
Back to the original equation
$3 y^2 - 16 y - x^2 + 16 = 0 $
Substituting the polar expressions,
$ 3 (r \sin \theta)^2 - 16 r \sin \theta - r^2 \cos^2 \theta + 16 = 0 $
Using $\cos^2 \theta = 1 - \sin^2 \theta$ and collecting terms
$ r^2 ( 4 \sin^2 \theta - 1 ) - 16 r \sin \theta + 16 = 0$
From the quadratic formula, and taking the positive root
$ r = \dfrac{1}{ 2(4 \sin^2 \theta - 1 ) } \left ( 16 \sin \theta - \sqrt{ 256 \sin^2 \theta - (256 \sin^2 \theta - 64 ) } \right) $
And this simplifies to
$ r = \dfrac{1}{2 (4 \sin^2 \theta - 1) } \left( 16 \sin \theta - 8 \right) $
Factoring the denominator and cancelling equal terms
$ r = \dfrac{ 4 }{ 2 \sin \theta + 1 } $
