Answer
Vertices: $(-6,0)$ and $(6,0)$
Foci: $(-10,0)$ and $(10,0)$
Asymptotes: $y = \pm \frac{4}{3} x$
Work Step by Step
The following is a horizontal hyperbola because the $x^{2}$ term is first, a vertical hyperbola has the $y^{2}$ term first.
$\frac{x^{2}}{36}-\frac{y^{2}}{64}=1$
General equation of a horizontal hyperbola:
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ where $a$ is the distance from the center to the vertices.
$a^{2} = 36,$ so $a = 6$, and because the hyperbola is centered at the origin, then the vertices are $(-6,0)$ and $(6,0)$
The distance from the center to the foci is equal to $c$ where $c^{2} = a^{2} +b^{2}$
Plugging in values, we get $c^{2} = 36 +64$ and $c = 10$
Thus, the foci are $(-10,0)$ and $(10,0)$.
The slope of the asymptotes passing through the center of the hyperbola is $\pm \frac{b}{a}$
Plugging in values, we get $m = \frac{8}{6}$ which simplifies to $m = \frac{4}{3}$ so the equations of the asymptotes are $y = \pm \frac{4}{3} x$
