Question 1144012
Find the equation of the hyperbola with the following properties. 
Foci at ({{{-8}},{{{-7}}}) and ({{{-8}}},{{{-13}}})
Vertices at ({{{-8}}},{{{-9}}}) and ({{{-8}}},{{{-11}}})

recall: ({{{h}}},{{{ k}}}), which is the "center" of the hyperbola.
The point on each branch closest to the center is that branch's "vertex". The vertices are some fixed distance {{{a }}}from the center. 

The "foci" of an hyperbola are "inside" each branch, and each focus is located some fixed distance {{{c}}} from the center.

Take note that ALL of the points given to you (both vertices and foci) all have a x-coordinate of {{{-8}}}. So this tells us that the hyperbola opens {{{up }}}and {{{down}}} and has a form


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



So we need to find the values of {{{h}}}, {{{k}}}, {{{a}}}, and {{{b}}}.

Now let's find the midpoint of the line connecting the vertices. This midpoint is the center of the hyperbola.

{{{x}}} mid=Average the {{{x}}}-coordinates of the vertices
{{{y}}} mid=Average the {{{y}}}-coordinates of the vertices

so, ({{{h}}}, {{{k}}})=({{{(-8-8)/2}}}, {{{(-7-13)/2}}})=({{{-8}}},{{{-10}}})


=>{{{h=-8}}}, {{{k=-10}}}

and center is at ({{{h}}}, {{{k}}})=({{{-8}}}, {{{-10}}})


If the {{{x}}} term has the minus sign then the hyperbola will open up and down.


if vertices at ({{{-8}}},{{{-9}}}) and ({{{-8}}},{{{-11}}}) on same vertical line, hyperbola opens up and down 

and 
we know that vertices are at: 
({{{h}}},{{{k+b}}})=({{{-8}}},{{{-9}}})....since {{{h=-8}}}, {{{k=-10}}}, solve for {{{b}}}

({{{-8}}},{{{-10+b}}})=({{{-8}}},{{{-9}}})=>{{{-10+b=-9}}}=>{{{b=-9+10}}}=>{{{b=1}}}

 and same for other vertices
 
({{{h}}},{{{k-b}}})=({{{-8}}},{{{-11}}}) 

({{{-8}}},{{{-10-b}}})=({{{-8,-11}}}) =>{{{-10-b=-11}}}=>{{{-10+11=b}}}=>{{{b=1}}}

so far we know that:

{{{b= 1}}} (distance from vertex to center) 
{{{c = 3}}} (distance from focus to center)

now find {{{a}}}

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

{{{3^2 -1^a^2 = 1^2 }}}

{{{9 -a^2 = 1}}}

{{{8 = a^2 }}}


so, your hyperbola open up and down and it is:

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

{{{(y+10)^2/1-(x+8)^2/8=1}}}



{{{drawing( 600, 600, -15, 15, -15, 15,
circle(-8,-10,.12),locate(-8,-10,C(-8,-10)),
circle(-8,-9,.12),locate(-8,-9,V(-8,-9)),
circle(-8,-11,.12),locate(-8,-11,V(-8,-11)),
circle(-8,-7,.12),locate(-8,-7,F(-8,-7)),
circle(-8,-13,.12),locate(-8,-13,F(-8,-13)),
 graph( 600, 600, -15, 15, -15, 15,- sqrt((x+8)^2/8+1)-10, sqrt((x+8)^2/8+1)-10)) }}}