Question 1191119

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

given: 

Center at ({{{-2}}},{{{2}}})=>{{{h=-2}}}, {{{k=2}}}

a focus at ({{{6}}},{{{2}}})=> focus is  ({{{h - c}}}, {{{k }}}) =  ({{{6}}},{{{2}}})=>{{{-2 - c=6}}}=>{{{-2-6=c}}}=>{{{c=-8}}}

a vertex at ({{{4}}},2).=>   ({{{h + a}}}, {{{k}}}) = ({{{4}}}, {{{2}}})=>{{{-2 + a=4}}}=>{{{a=2+4}}}=>{{{a=6}}}

{{{b^2=c^2-a^2}}}
{{{b^2=(-8)^2-6^2}}}
{{{b^2=64-36}}}
{{{b^2=28}}}

equation is:

{{{(x-(-2))^2/6^2-(y-2)^2/28=1}}}

{{{(x+2)^2/36-(y-2)^2/28=1}}}



{{{ drawing( 600, 600, -15, 15, -10, 10,
circle(-2,2,.12),locate(-2,2,C(-2,2)),
circle(6,2,.12),locate(6,2,F(6,2)),
circle(4,2,.12),locate(4,2,V(4,2)),
graph( 600, 600, -15, 15, -10, 10, (1/3)(6-sqrt(7)*sqrt(x^2+4x-32)), (1/3)(sqrt(7)*sqrt(x^2+4x-32)+6)  )) }}}