Answer
$${x^2} - \frac{{{y^2}}}{{25}} = 1$$
Work Step by Step
$$\eqalign{
& {\text{Vertices }}\left( { \pm 1,0} \right) \cr
& {\text{Asymptotes }}y = \pm 5x \cr
& {\text{Vertices }}\left( {\underbrace { - 1}_x,\underbrace 0_y} \right){\text{ and }}\left( {\underbrace 1_x,\underbrace 0_y} \right) \cr
& y = y{\text{ so the hyperbola has the standard form}} \cr
& \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& {\text{Characteristics of the hyperbola}} \cr
& {\text{Vertices }}\left( { - a,0} \right){\text{ and }}\left( {a,0} \right) \cr
& {\text{Vertices }}\left( {\underbrace { - 1}_{ - a},\underbrace 0_0} \right){\text{ and }}\left( {\underbrace 1_a,\underbrace 0_0} \right) \Rightarrow a = 1 \cr
& \cr
& {\text{Asymptotes:}} \pm \frac{b}{a}x \cr
& \pm \frac{b}{a}x = \pm 5x \cr
& \frac{b}{a} = 5 \to b = 5 \cr
& \cr
& \underbrace {\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1}_ \Downarrow \cr
& \frac{{{x^2}}}{{{{\left( 1 \right)}^2}}} - \frac{{{y^2}}}{{{{\left( 5 \right)}^2}}} = 1 \cr
& {x^2} - \frac{{{y^2}}}{{25}} = 1 \cr} $$
