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