Question 1180953
You have the distance between the foci is 

{{{8sqrt(2) = 2c}}} -> {{{c =4sqrt(2)}}}

A property of hyperbolas is that the absolute value of the difference of the distances from the foci is equal to the length of the transverse axis: 

{{{8 = 2a}}} -> {{{a = 4}}}

Another property is that 

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

from which you can find {{{b}}} (half the conjugate axis), {{{b = 4}}}. 

Since the foci are on a horizontal line, your hyperbola is oriented {{{horizontally}}}. 

The {{{center}}} halfway between the foci, in this case at the {{{origin}}}. 

So the equation is 

{{{x^2/a^2 - y^2/b^2 = 1 }}}

with {{{a = b = 4}}} in this case. 

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

{{{x^2/16 - y^2/16 = 1 }}}



{{{drawing ( 600, 600, -15, 15, -15, 15,
circle(4sqrt(2), 0,.15),locate(4sqrt(2), 0.7,F),
circle(-4sqrt(2), 0,.15),locate(-4sqrt(2), 0.7,F),
graph( 600, 600, -15, 15, -15, 15,-sqrt(x^2 - 16),sqrt(x^2 - 16) )) }}}