Question 696743
{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}

you have   {{{x^2/25-y^2/9=1}}}

The hyperbola is centered on a point ({{{h}}}, {{{k}}}), which is the "center" of the hyperbola. In your case {{{h=0}}} and {{{k=0}}}; so the center is at ({{{0}}}, {{{0}}})

{{{a^2=25}}}...=>...{{{a=5}}}....the vertices are {{{a = 5}}} units to either side

{{{b^2=9}}}...=>...{{{b=3}}} 

vertices: ({{{0-a}}},{{{ 0}}}) and ({{{0+a}}}, {{{0}}})......=>....({{{0-5}}}, {{{0}}}) and =>....({{{-5}}}, {{{0}}}) and ({{{5}}}, {{{0}}})

The equation {{{c^2 = a^2 + b^2}}} tells us that {{{c^2 = 25 + 9 =34}}}, so {{{c = sqrt(34)}}} which is {{{c = 5.8}}}, ..=>...and the {{{foci}}} being {{{5.8}}} units to either side of the center.


{{{ graph( 600, 600, -10, 10, -10, 10, sqrt(9x^2/25-9),-(sqrt(9x^2/25-9))) }}}