Have you heard of the cycloid? It is parametrized by
$x = a\dfrac{\theta + \sin \theta}{2}$
$y = a\dfrac{1-\cos \theta}{2}$
and is the solution to
$\frac{{dx}}{{dy}} = \sqrt {\frac{{a - y}}{y}} $
Since your equation is
$\frac{{dy}}{{dt}} = \sqrt {\frac{{ay - b}}{y}} $
we can go like this:
Put
$\frac{{dt}}{{dy}} = \sqrt {\frac{y}{{ay - b}}} $
Now let
$y = \frac{b}{a}{\cosh ^2}\theta $
We get
$dt = \frac{{2b}}{{{a^{3/2}}}}{\cosh ^2}\theta d\theta $
So
$dt = \frac{{2b}}{{{a^{3/2}}}}\left( {\frac{1}{2} + \frac{{\cosh 2\theta }}{2}} \right)d\theta $
and integrating gives
$t = \frac{{2b}}{{{a^{3/2}}}}\left( {\frac{{2\theta }}{4} + \frac{{\sinh 2\theta }}{4}} \right)+C$
So your solution is parametrized by ($\phi = 2\theta$, suppose initial conditions make $C=0$)
$\eqalign{ & y = \frac{b}{{2a}}\left( {1 + \cosh \phi } \right) \cr & t = \frac{b}{{2{a^{3/2}}}}\left( {\phi + \sinh \phi } \right) \cr} $
In the same way the cycloid is a "deformed" circumeference your solution is a "deformed" hiperbola. The cycloid has a closed form for $(x,y)$ coordinates so you might be able to find one for the above curve.