The Y-intercept of the common tangent to the parabola $y^2=32 x$ and $x^2=108 y$ is

A common tangent to both parabolas must have the same slope m and y-intercept c in its equation $y = mx + c$.

For $y^2 = 32x$, which has standard form $y^2 = 4ax$ with $4a = 32$ giving $a = 8$,
the tangent with slope m is $y = mx + \frac{8}{m}$, yielding y-intercept $\frac{8}{m}$.

For $x^2 = 108y$, in standard form $x^2 = 4by$ with $4b = 108$ giving $b = 27$,
the tangent with slope m is $y = mx - 27m^2$, yielding y-intercept $-27m^2$.

Equating the y-intercepts for a common tangent:

$\frac{8}{m} = -27m^2$

Multiplying both sides by m gives $8 = -27m^3$, so $m^3 = -\frac{8}{27}$ and $m = -\frac{2}{3}$.

Substituting into the first y-intercept expression:

$c = \frac{8}{-\frac{2}{3}} = -12$

The y-intercept of the common tangent is therefore $\boxed{-12}$.