Manipulating a polar equation I started with this:
$\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$
And via substitution, got this far:
$r=\left(\frac{144}{9-25\sin^{2}\theta}\right)^{.5}$
For the fact that Desmos plots these the same, I assume I'm right so far.
The goal in this section is to end with the form
$r=\frac{ep}{1-e\sin\theta}$
and I am at a loss what the next manipulation might be.
 A: The right focus is $(5, 0)$.  Shifting the hyperbola to the left by $5$ units results in the new equation:
$ \dfrac{(x + 5)^2}{16} - \dfrac{y^2}{9} = 1 $
Now using polar coordinates, but measuring the angle $\theta $ from the positive $y$ axis direction, then   $x = -r \sin \theta , y = r \cos \theta $
Hence,
$ \dfrac{ (-r \sin \theta + 5 )^2 }{16} - \dfrac{ (r \cos \theta)^2 }{9} = 1 $
Multiplying through by $144$,
$ 9 ( r^2 \sin^2 \theta - 10 r \sin \theta + 25 ) - 16 r^2 \cos^2 \theta = 144 $
Simplifiying,
$ r^2 ( 25 \sin^2 \theta - 16 ) - 90 r \sin \theta + 81 = 0 $
Using the quadratic formula,
$ r = \dfrac{1}{ 2(25 \sin^2 \theta - 16) } \left( 90 \sin \theta - \sqrt{ 8100 \sin^2 \theta - (8100 \sin^2 \theta -  5184 ) } \right)$
And this simplifies to,
$ r = \dfrac{ 90 \sin \theta -72 }{2(5 \sin \theta - 4)(5 \sin \theta + 4) } $
which simplifies further to
$ r = 9 \dfrac{ 5 \sin \theta - 4 }{ (5 \sin \theta - 4)(5 \sin \theta + 4 ) } =  \dfrac{ 9 }{ 5 \sin \theta + 4 } $
$e = \sqrt{1 + \dfrac{3}{4}^2 } = \dfrac{5}{4}$
Therefore,
$ r = \dfrac{ 9/4 }{ 1 + e \sin \theta } $
