Answer
Hyperbola.
Vertices: $ \quad (5,0), \quad (-5,0)$
Foci: $ \quad (\sqrt{26},0), \quad (-\sqrt{26},0)$
Asymptotes: $ \displaystyle \quad y=\frac{1}{5}x, \quad y=-\frac{1}{5}x$
Work Step by Step
Both variables squared, a minus between terms$\Rightarrow$hyperbola
$\displaystyle \frac{x^{2}}{5^{2}}-\frac{y^{2}}{1}=1$
Table 4:
$\begin{array}{cccc}
{\text{Foci}}&{\text { Vertices }}&{\text{Equation}}&{\text{asymptotes}}\\\hline
(h\pm c,k)&{(h \pm a, k)}&{ \displaystyle \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1,}&{b^{2}=c^{2}-a^{2}\displaystyle \quad y-k=\pm\frac{b}{a}(x-h)}\end{array}$
$a=5,\quad b=1, \quad $Find $c$
$c^{2}=a^{2}+b^{2}=25+1=26$
$c=\sqrt{26}$
Vertices: $ \quad (h \pm a, k)=(0\pm 5,0),$
Foci: $ \quad (h\pm c,k)=(0\pm\sqrt{26},0),$
Asymptotes: $ \displaystyle \quad y-k=\pm\frac{b}{a}(x-h)$
$y=\displaystyle \pm\frac{1}{5}(x)$
