[tex]\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)
\end{cases}\\\\
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\cfrac{(y+11)^2}{15^2}-\cfrac{(x-7)^2}{8^2}=1\implies \cfrac{(y-(-11))^2}{15^2}-\cfrac{(x-7)^2}{8^2}=1\\\\
-------------------------------\\\\
c=\textit{distance from the center to either foci}\\\\
c=\sqrt{a^2+b^2}\implies c=\sqrt{15^2+8^2}\implies \boxed{c=17}[/tex]
now, if you notice, the positive fraction, is the one with the "y" variable, and that simply means, the hyperbola traverse axis is over the y-axis, so it more or less looks like the picture below.
now, from the hyperbola form, we can see the center is at (7, -11), and that the "c" distance is 17.
so, from -11 over the y-axis, we move Up and Down 17 units to get the foci, which will put them at (7, -11-17) or (7, -28) and (7 , -11+17) or (7, 6)
