so hmm check the picture below
the hyperbola is more or less like so, since [tex]\bf \cfrac{(x-{{ h}})^2}{{{ a}}^2}-\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1
\qquad center\ ({{ h}},{{ k}})\qquad
vertices\ ({{ h}}\pm a, {{ k}})\\\\
-----------------------------\\\\[/tex]
[tex]\bf \cfrac{x^2}{5}-\cfrac{y^2}{22}=1\implies \cfrac{(x-0)^2}{(\sqrt{5})^2}-\cfrac{(y-0)^2}{(\sqrt{22})^2}=1\quad
\begin{cases}
a=\sqrt{5}\\
b=\sqrt{22}\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
c=\sqrt{a^2+b^2}\\
c=\sqrt{5+22}\\
c=\sqrt{27}
\end{cases}
\\\\\\
\textit{distance from right-focus point to vertex}\qquad \sqrt{27}-\sqrt{5}[/tex]
notice, the vertex is "a" distance from the center, the center is at the origin, thus the vertices is at ±√(5), 0
so, the distance in the picture, from the vertex to the right-focus point, or the other focus point for that matter, will then be, the distance "c" minus the distance of the vertex from the center
