Question 1149990
the hyperbola has a horizontal transverse axis and its standard form of the equation is:

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

 the hyperbola has a vertical transverse axis and its standard form of the equation is:

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


 ({{{h}}},{{{k}}}) is the center 

 {{{a }}}is the distance between {{{center }}}and {{{vertex}}}, the length of the transverse axis is {{{2a}}}

the length of the conjugate axis is {{{2b}}}

 {{{c}}} is the distance between {{{center}}} and {{{focus}}}

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

The {{{foci}}} of the hyperbola is ({{{h}}},{{{k}}} ± {{{c}}}).

The {{{vertices}}} of the hyperbola is ({{{h}}},{{{k }}}± {{{a}}}).

{{{Asymptotes}}} of the hyperbola are: {{{y-k}}}= ±{{{ (a/b)(x-h)}}}.

The formula for {{{eccentricity}}}{{{ e}}} is: {{{e=sqrt(a^2+b^2)/a}}}


given:


the hyperbola that has eccentricity {{{3}}}, =>{{{3=sqrt(a^2+b^2)/a}}}

center at ({{{0}}},{{{0}}}), =>{{{h=0}}},{{{k=0}}} and your formula becomes 

{{{y^2/a^2-x^2/b^2=1}}} (the standard equation for a hyperbola with a vertical transverse axis)

vertex at ({{{0}}},{{{8}}})=> {{{a=8}}}

then {{{3=sqrt(8^2+b^2)/8}}}........solve for {{{b^2}}}

{{{24=sqrt(8^2+b^2)}}}.......square both sides

{{{24^2=8^2+b^2}}}

{{{b^2=24^2-8^2}}}

{{{b^2=576-64}}}

{{{b^2=512}}}

and, your formula is: {{{y^2/64-x^2/512=1}}}


{{{drawing ( 600, 600, -25, 25, -25, 25,
circle(0,8,.12),locate(0,8,v(0,8)),

graph( 600, 600, -25, 25, -25, 25,-sqrt(64(1+x^2/512)) ,sqrt(64(1+x^2/512)))) }}}