Answer
center $(0,0)$, transverse axis $x-axis$, vertices $(-5,0),(5,0)$, foci $(-\sqrt {34},0),(\sqrt {34},0)$, and asymptotes $y =\pm\frac{3}{5}x$.
See graph.
Work Step by Step
1. Given $\frac{x1. Given $\frac{x^2}{25}-\frac{y^2}{9}=1$, we can find the center $(0,0)$, transverse axis $x-axis$, $a=5,b=3$, vertices $(-5,0),(5,0)$, $c=\sqrt {a^2+b^2}=\sqrt {25+9}=\sqrt {34}$, foci $(-\sqrt {34},0),(\sqrt {34},0)$, and asymptotes $y=\pm\frac{b}{a}x=\pm\frac{3}{5}x$.
2. See graph.^2}{25}-\frac{y^2}{9}=1$, we can find the center $(0,0)$, transverse axis $x-axis$, $a=5,b=3$, vertices $(-5,0),(5,0)$, $c=\sqrt {a^2+b^2}=\sqrt {25+9}=\sqrt {34}$, foci $(-\sqrt {34},0),(\sqrt {34},0)$, and asymptotes $y=\pm\frac{b}{a}x=\pm\frac{3}{5}x$.
2. See graph.
