Question 1079428
 

Equation of a hyperbola, since the hyperbola has it axis along the x-axis

 {{{(x-h)^2/a^2 - (y-k)^2/b^2   = 1 }}} where ({{{h}}}, {{{k}}} ) is the center and 
{{{a}}} = semi-major axis
{{{b}}} = semi-minor axis

if a center at ({{{5}}}, {{{0}}}), means {{{h=5}}} and {{{k=0}}}, so we have

{{{(x-5)^2/a^2 - (y-0)^2/b^2   = 1 }}}

{{{(x-5)^2/a^2 - y^2/b^2   = 1 }}}

it has minor axis length {{{8}}}, so {{{b = 8/2=4}}}

if vertices along the major axis at ({{{-1}}}, {{{0}}}) and ({{{11}}},{{{ 0}}})=> has major axis length {{{12}}} (distance from {{{-1}}} to {{{11}}}), so major axis {{{a = 12/2=6 }}}

and, your equation is:

{{{(x-5)^2/6^2 - y^2/4^2   = 1}}}
 

{{{(x-5)^2/36 - y^2/16   = 1 }}}


{{{drawing( 600, 600, -10, 15, -10, 10,
circle(5,0,.15),circle(-1,0,.15),circle(11,0,.15),
locate(5,0.8,C(5,0)),locate(-1,0.8,v(-1,0)),locate(11,0.8,v(11,0)),
 graph( 600, 600, -10, 15, -10, 10,-sqrt((16(x-5)^2/36-1)), sqrt((16(x-5)^2/36-1)))) }}}