Answer
hyperbola, center $(0,0)$, vertices $(-5,0),(5,0)$, foci $(-\sqrt {26},0),(\sqrt {26},0)$, and asymptotes $y=\pm\frac{1}{5}x$.
Work Step by Step
1. Given $\frac{x^2}{25}-y^2=1$, we can identify it as a hyperbola with a horizontal transverse axis, $a=5, b=1$ and $c=\sqrt {a^2+b^2}=\sqrt {26}$.
2. We can find its center $(0,0)$, vertices $(-5,0),(5,0)$, foci $(-\sqrt {26},0),(\sqrt {26},0)$, and asymptotes $y=\pm\frac{b}{a}x=\pm\frac{1}{5}x$.
