Question

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?

Answer 1

So hmm check the picture below

the hyperbola is more or less like so, since $\bf \cfrac{(x-{{ h}})^2}{{{ a}}^2}-\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1 \qquad center\ ({{ h}},{{ k}})\qquad vertices\ ({{ h}}\pm a, {{ k}})\\\\ -----------------------------$

$\bf \cfrac{x^2}{5}-\cfrac{y^2}{22}=1\implies \cfrac{(x-0)^2}{(\sqrt{5})^2}-\cfrac{(y-0)^2}{(\sqrt{22})^2}=1\quad \begin{cases} a=\sqrt{5}\\ b=\sqrt{22}\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ c=\sqrt{a^2+b^2}\\ c=\sqrt{5+22}\\ c=\sqrt{27} \end{cases} \\\\\\ \textit{distance from right-focus point to vertex}\qquad \sqrt{27}-\sqrt{5}$

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