Question 1177017
Standard form: 

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

given:

foci: ({{{-6}}},{{{5}}}), ({{{6}}},{{{5}}}) 

since foci is at ({{{c}}},{{{5}}}) =>{{{c=6}}}

vertices:({{{-5}}},{{{5}}}), ({{{5}}},{{{5}}})

since vertices is at ({{{a}}},{{{5}}}) =>{{{a=5}}}

then 

{{{b^2=c^2-a^2}}}

{{{b^2=6^2-5^2}}}

{{{b^2=36-25}}}

{{{b^2=11}}}

so, your equation is:

{{{x^2/25-(y-5)^2/11=1}}}


{{{drawing( 600,600, -10, 10, -10, 10, 
circle(-6,5,.12),locate(-6,5.5,F(-6,5)),
circle(6,5,.12),locate(6,5.5,F(6,5)),
circle(-5,5,.12),locate(-5,5,V(-5,5)),
circle(5,5,.12),locate(5,5,V(5,5)),
 graph( 600,600, -10, 10, -10, 10, (1/5)(25-sqrt(11)*sqrt(x^2-25)), (1/5)(sqrt(11)*sqrt(x^2 - 25) + 25))) }}}