The given equation of the hyperbola is:

$\frac{(y+4)^2}{16} - \frac{(x+3)^2}{4} = 1$

This is in the standard form of a vertical hyperbola:

$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$

Here, the center of the hyperbola is $(h, k)$, and $a^2$ and (b^