Question 1187217
 the equation of the hyperbola:

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

the length of the transverse axis is {{{ 2a}}}
the coordinates of the vertices are  ({{{h}}} ± {{{a}}}, {{{k}}})
the length of the conjugate axis is  {{{2b}}}
the coordinates of the co-vertices are  ({{{h}}} , {{{k}}} ±{{{ b}}} )
the distance between the foci is {{{2c}}}, where  {{{c^2=a^2+b^2}}}
the coordinates of the foci are ({{{h}}} ±{{{c}}}, {{{k}}})

given:

Co -Vertices: ({{{1}}}, {{{5}}}) , ({{{1}}}, {{{1}}}) 

we know that the center is half way between: ({{{(1+1)/2}}},{{{(5+1)/2}}})=({{{1}}},{{{3}}}) => so, {{{ h=1}}} and {{{ k=3}}}


use Co -Vertices: ({{{1}}}, {{{5}}}), {{{ h=1}}} and {{{ k=3}}} to find {{{b}}}

({{{1}}} , {{{3}}}±{{{b}}} )=({{{1}}}, {{{5}}}) 

({{{1}}} , {{{3}}}±{{{b}}} )=({{{1}}}, {{{5}}})=>{{{3}}}±{{{b=5}}} =>±{{{b=2}}}


the coordinates of the foci are ({{{h }}}±{{{c}}}, {{{k}}})= ({{{4}}},{{{ 3}}}) =>({{{1}}} ±{{{c}}},{{{ 3}}})= ({{{4}}}, {{{3}}}) =>{{{1}}} ±{{{c=4}}}=>±{{{c=4-1}}}=>±{{{c=3}}}


then {{{c^2=a^2+b^2}}}=>{{{3^2=a^2+2^2}}}=>{{{a^2=9-4=5}}}


and your equation is:

{{{(x-1)^2/5-(y-3)^2/4=1}}}


{{{ drawing( 600, 600, -10, 10, -10, 10, 
circle(4,3,.12), locate(4,3,F(4,3)),
circle(1,1,.12), locate(1,1,co-v(1,1)),
circle(1,5,.12), locate(1,5,co-v(1,5)),
circle(1,3,.12), locate(1,3,C(1,3)),
graph( 600, 600, -10, 10, -10, 10, sqrt((x-1)^2/5-1)+3, -sqrt((x-1)^2/5-1)+3) )}}}