check the picture below. So with those points, ther hyperbola looks more or less like so.
now, notice the distance from vertex to vertex, half-way is the center, at 30, -15.
now, from the center to the given focus point, at -11, - 15, is 41 units, bear in mind the distance from the center to either foci is "c". Thus c = 41
since the traverse axis for the hyperbola is the x-axis, thus the fraction with the "x" variable is the positive fraction. thus
[tex]\bf \textit{hyperbolas, horizontal traverse axis }\\\\
\cfrac{(x-{{ h}})^2}{{{ a}}^2}-\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1
\qquad
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}\pm a, {{ k}})\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{{{ a }}^2+{{ b }}^2}
\end{cases}[/tex]
[tex]\bf \begin{cases}
h=30\\
k=-15\\
a=40
\end{cases}\implies \cfrac{(x-30)^2}{30^2}-\cfrac{[y-(-15)]^2}{b^2}=1
\\\\\\
\cfrac{(x-30)^2}{30^2}-\cfrac{(y+15)^2}{b^2}=1\\\\
-------------------------------\\\\
\textit{now, we know }c=41\implies c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2
\\\\\\
\sqrt{c^2-a^2}=b\implies \sqrt{41^2-30^2}=b\implies \sqrt{781}=b\\\\
-------------------------------\\\\
\cfrac{(x-30)^2}{30^2}-\cfrac{(y+15)^2}{(\sqrt{781})^2}=1\implies \cfrac{(x-30)^2}{900}-\cfrac{(y+15)^2}{781}=1[/tex]
