Answer
Vertices: $(\pm 10,0)$
Foci: $(\pm 10\sqrt 2, 0)$
Asymptotes: $y = \pm x$
Work Step by Step
Recall: The hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ has the foci $(\pm c,0)$ where $c^2=a^2+b^2$, the vertices $(\pm a,0)$, and the asymptotes $y=\pm(b/a)x$.
We have $x^2-y^2=100$.
This equation can be written as $\frac{x^2}{100}-\frac{y^2}{100}=1$.
Then,
$a^2=100\to a=10$
$b^2=100\to b=10$
$c=\sqrt{a^2+b^2}=\sqrt{10^2+10^2}=\sqrt{10^2\cdot 2}=10\sqrt{2}$
$b/a=10/10\to b/a=1$
So, the foci are $(\pm 10\sqrt{2},0)$, the vertices are $(\pm 10,0)$, and the asymptotes are $y=\pm x$.
