Question 1191092
the equation of the hyperbola:

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

eccentricity of hyperbola {{{e = c/a}}}=> {{{c=3}}} and {{{a=2}}}

{{{b^2=3^2-2^2=5}}}

focus is  ({{{h - c}}},{{{ k }}}) =  ({{{4}}}, {{{0}}} ) 

=>{{{h-c=4}}} =>{{{h-3=4}}}=>{{{h=7
and {{{k=0}}}


then, center is at ({{{7}}},{{{ 0 }}}) 


and your equation is:

{{{(x-7)^2/4-(y-0)^2/5=1 }}}

{{{(x-7)^2/4-y^2/5=1 }}}


{{{drawing ( 600, 600, -10, 15, -10, 10,
circle(4,0,.12), locate(4,0.5,F(4,0)),

graph( 600, 600, -10, 15, -10, 10,- sqrt(5((x-7)^2/4-1)) ,sqrt(5((x-7)^2/4-1))) )}}}