Question 1187216


 the equation of the hyperbola:

given:

Vertices: ({{{0}}},{{{ 2}}}), ({{{0}}}, {{{- 2}}}) 
Focus : ({{{0}}},{{{ 6}}})

since transverse axis parallel to the y-axis, equation is

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

Center is half way between vertices: ({{{h}}},{{{k}}})=({{{(0+0)/2}}},{{{(2-2)/2}}})=({{{0}}},{{{0}}}) => {{{h=0}}}, {{{k=0}}}


the coordinates of the vertices are ({{{0}}}, ± {{{a}}})=({{{0}}}, {{{2}}})=>{{{a=2}}}

the coordinates of the foci are ({{{0}}},±{{{c}}})=({{{0}}}, {{{6}}}) =>{{{c=6}}}

find {{{b}}}

 {{{6^2=2^2+b^2}}}

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

{{{b^2=36-4}}}

{{{b^2=32}}}


and your equation is:

{{{(y-0)^2/4-(x-0)^2/32=1}}}

{{{y^2/4-x^2/32=1}}}


{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(0,6,.12), locate(0,6, F(0,6)),
circle(0,2,.12), locate(0,2, v(0,2)),
circle(0,-2,.12), locate(0,-2, v(0,-2)),
graph( 600, 600, -10, 10, -10, 10, sqrt(4(x^2/32+1)), -sqrt(4(x^2/32+1)))) }}}