Answer
$$\eqalign{
& {\text{Vertices }}\left( { \pm 1,0} \right){\text{ and foci}}\left( { \pm 6,0} \right) \cr
& {\text{Asymptotes: }}y = 6x{\text{ and }}y = - 6x \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Vertices }}\left( { \pm 1,0} \right){\text{ }} \cr
& {\text{The equation of the hyperbola is given by: }}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1,{\text{ with}} \cr
& {\text{Vertices }}\left( { \pm a,0} \right) \cr
& {\text{Then,}} \cr
& a = 1 \cr
& \frac{{{x^2}}}{{{{\left( 1 \right)}^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& {\text{passes through}}\left( {\frac{5}{3},8} \right) \cr
& {\left( {5/3} \right)^2} - \frac{{{{\left( 8 \right)}^2}}}{{{b^2}}} = 1 \cr
& b = 6 \cr
& \cr
& \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& \frac{{{x^2}}}{{{{\left( 1 \right)}^2}}} - \frac{{{y^2}}}{{{{\left( 6 \right)}^2}}} = 1 \cr
& {x^2} - \frac{{{y^2}}}{{36}} = 1 \cr
& {\text{Asymptotes: }}y = \frac{b}{a}x{\text{ and }}y = - \frac{b}{a}x \cr
& {\text{Asymptotes: }}y = 6x{\text{ and }}y = - 6x \cr
& \cr
& {\text{Foci }}\left( { \pm c,0} \right) \cr
& c = \sqrt {{a^2} + {b^2}} = \sqrt 7 \cr
& {\text{Foci }}\left( { \pm \sqrt 7 ,0} \right) \cr
& \cr
& {\text{Graph}} \cr} $$
