A comet follows the hyperbolic path described by x^2/ 5 - y^2/22 = 1, where x and y are in millions of miles. If the sun is the focus of the path, how close to the sun is the vertex of the path?
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
