Question 1191089
{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}

given: 


 Vertices at  ({{{0}}},{{{3}}}) and ({{{0}}},{{{-3}}}) 
a focus at ({{{0}}},{{{5}}})
in your case the transverse axis lies on the y-axis, the vertices are located at ({{{0}}},±{{{a}}}), and the foci are located at ({{{0}}},±{{{c}}})
=>{{{a=3}}}
=> {{{c=5}}}
Center is half way between, at ({{{0}}},{{{0}}})=>{{{h=0}}}, {{{k=0}}}

in this case your equation is of the form:

{{{y^2/a^2-x^2/b^2=1}}}

 find {{{a}}} can be found using this formula:

{{{b^2 = c^2 - a^2}}}
{{{b^2 = 25 - 9}}}
{{{b^2 = 16}}}
{{{b=4 }}}



equation is:

{{{y^2/9-x^2/16=1}}}


{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(0,0,.12),locate(0.2,0.5,C(0,0)),
circle(0,5,.12),locate(0.2,5,F(0,5)),
circle(0,3,.12),locate(0,3,V(0,3)),
circle(0,-3,.12),locate(0,-3,V(0,-3)),
graph( 600, 600, -10, 10, -10, 10,-(3/4)sqrt(x^2+16),(3sqrt(x^2+16))/4)) }}}