Question 952225
given:

Center is ({{{3}}},{{{2}}}); 
{{{a=3c}}}; 
foci:({{{1}}},{{{2}}}),({{{5}}},{{{2}}})


 the equation of the ellipse is:

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

so, for your ellipse it is:

   {{{(x-3)^2/a^2 + (y-2)^2/b^2 = 1}}}     (center at ({{{3}}},{{{2}}}), means {{{h=3}}} and {{{h=2}}})


    for an ellipse,{{{ a^2 -b^2 = c^2}}}

because the foci have the same {{{y}}}-coordinate, you know that it is a {{{horizontal}}} ellipse (major axis is horizontal)

distance from ({{{1}}},{{{2}}}) to ({{{5}}},{{{2}}}) ={{{ 2c}}} , where {{{c}}} is the distance from the center to each focus


    {{{2c = 4 }}}==> {{{c = 2}}} (which is also the distance from ({{{3}}},{{{2}}}) to either ({{{1}}},{{{2}}}) or ({{{5}}},{{{2}}}))
    
since {{{a = 3c}}}, that means that {{{a = 3*2 = 6}}}
  
 
then, {{{b^2= 6^2 - 2^2}}} => {{{b^2=36 - 4}}}}}} =>{{{ b^2=32}}}

  and your equation is:

    {{{(x - 3)^2 / 36 + (y - 2)^2 / 32 = 1 }}}


{{{drawing(600, 600, -10, 10, -10, 10,
circle(3,2,.12),locate(3,2,C(3,2)),
circle(1,2,.12),locate(1,2,F(1,2)),
circle(5,2,.12),locate(5,2,F(5,2)),
 graph(600, 600, -10, 10, -10, 10,-sqrt((1-(x - 3)^2 / 36)*32)+2 ,sqrt((1-(x - 3)^2 / 36)*32)+2)) }}}