Question 1204698
from the graph you see that you need equation:

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

center ({{{h}}},{{{k}}})=({{{-2}}},{{{-1}}})

the length of the transverse axis is the distance between vertices and you see 0n the graph that {{{2a=4}}} => {{{a=2}}}

so far equation is

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

{{{(y +1)^2/4 - (x+2)^2/b^2 = 1}}}


I picked point ({{{-1}}}, {{{-5}}}) which lie on one asymptote

{{{(-5 +1)^2/4 - (-1+2)^2/b^2 = 1}}}
{{{16/4 - 1/b^2 = 1}}}
{{{16/4 - 1 = 1/b^2}}}
{{{16/4 - 4/4 = 1/b^2}}}
{{{b^2 = 1/3 }}}

{{{b = 1/sqrt(3) }}}


the equations of the asymptotes are {{{y}}}= ± {{{(a/b)(x-h)+k}}}

 {{{y}}}= ± {{{(2/1/sqrt(3))(x+2)-1}}}

 {{{y}}}= ± {{{2sqrt(3)(x+2)-1}}}


{{{y= 2sqrt(3)(x+2)-1}}} => {{{y=2sqrt(3) x + 4sqrt(3) - 1}}}
or
{{{y= - 2sqrt(3)(x+2)-1}}}=> {{{y=-2sqrt(3) x - 4sqrt(3) - 1}}}


and, your equation is:


{{{(y - (-1))^2/2^2 - (x - (-2))^2/(1/sqrt(3))^2 = 1}}}


{{{(y +1)^2/4 - (x+2)^2/(1/3) = 1}}}


{{{(y +1)^2/4 - 3(x+2)^2 = 1}}}


{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(-1,-5,.12), locate(-1,-5,p(-1,-5)),
circle(-2,-1,.12), locate(-2,-1,C(-2,-1)),
circle(-2,1,.12), locate(-2,1,V(-2,1)),
circle(-2,-3,.12), locate(-2,-3,V(-2,-3)),

graph( 600, 600, -10, 10, -10, 10, -2sqrt(3x^2+12x+13)-1, 2sqrt(3x^2+12x+13)-1,2sqrt(3) x + 4sqrt(3) - 1,-2sqrt(3) x - 4sqrt(3) - 1)) }}}