Question 1136330
Find the equation in standard form of the hyperbola that has 

foci at ({{{8}}}, {{{1}}}) , ({{{-8}}}, {{{1}}}) 
and transverse axis with length {{{14}}} 


The standard form of a hyperbola that opens sideways is 

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

For the hyperbola that opens up and down, it is 

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

 In both cases, the center of the hyperbola is given by ({{{h}}}, {{{k}}}). 
The vertices are {{{a }}}spaces away from the center.
The "foci" of an hyperbola are "inside" each branch, and each focus is located some fixed distance {{{c}}} from the center.
The endpoints of the transverse axis are called the vertices of the hyperbola. 
The distance between the vertices is {{{2a}}}. The distance between the foci is {{{2c}}}.

if foci at ({{{8}}}, {{{1}}}) , ({{{-8}}}, {{{1}}}) , the distance between is {{{16}}}, then

{{{2c=16}}} 

{{{c=8}}}


if transverse axis with length {{{14}}}, means distance between vertices is 


{{{2a=14}}}->{{{a=7}}}

and  vertices are at ({{{7}}}, {{{0}}}) , ({{{-7}}}, {{{0}}})
and its midpoint is the center of the hyperbola => which is at ({{{-7}}}, {{{0}}})

{{{(x - 0)^2 / 7^2 - (y - 0)^2 / b^2 = 1}}}

find {{{b}}}

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

{{{b^2=8^2-7^2}}}

{{{b^2=64-49}}}

{{{b^2=15}}}

{{{x^2 / 49 - y^2 /15 = 1}}}