Answer
Vertices: $(\pm 10,0)$
Foci: $(\pm 10\sqrt 2, 0)$
Asymptotes: $y = \pm x$
Work Step by Step
First, divide both sides by 100 to get the equation in standard form.
So we have, $\frac{x^{2}}{100}-\frac{y^{2}}{100}=1$
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.
$a^{2} = 100,$ so $a = 10$, and because the hyperbola is centered at the origin, then the verticles are $(-10,0)$ and $(10,0)$
The distance from the center to the foci is equal to $c$ where $c^{2} = a^{2} +b^{2}$
Plugging in values, we get $c^{2} = 100 + 100$ and $c = \sqrt 200$ which simplifies to $10\sqrt 2$
Thus, the foci are $(-10\sqrt 2,0)$ and $(10\sqrt 2,0)$.
The slope of the asymptotes passing through the center of the hyperbola is $\pm \frac{b}{a}$
Plugging in values, we get $m = \frac{10}{10}$ which simplifies to $m = 1$ so the equations of the asymptotes are $y = \pm x$
