Question 1170895
Equation of the ellipse in standard form: 

{{{(x-h)^2/a^2+(y-k)^2/b^2=1}}}  ...horizontal major axis ({{{a>b}}})

{{{(x-h)^2/b^2+(y-k)^2/a^2=1 }}} ...vertical major axis ({{{a>b}}})


given Co-vertices:({{{4}}},{{{1}}})->means we have {{{vertical}}} major axis

so, you need

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

given foci at ({{{2}}},{{{ 3}}})  and  ({{{2}}},{{{ -1}}}) 

then center is half way between

center: ({{{(2+2)/2}}},{{{(3-1)/2}}})=({{{2}}},{{{1}}})

distance between foci and center is {{{c}}} and -> {{{c=2}}}


Co-vertices: given one at ({{{4}}},{{{1}}})

distance from center ({{{2}}},{{{1}}}) is {{{d=sqrt((4-2)^2+(1-1)^2)=sqrt(2^2+0)=2}}}

 the other one is at ({{{x}}},{{{1}}}) same distance from center  
 ({{{2-2}}},{{{1}}})=({{{0}}},{{{1}}})

Minor axis length: {{{b=4}}}
Semiminor axis length: {{{2}}}

now we can calculate {{{a}}}

{{{a=sqrt(4^2-2^2)}}}
{{{a=sqrt(16-4)}}}
{{{a=sqrt(8)}}}



then your equation is:

{{{(2-x)^2/4+(y-1)^2/8=1 }}}


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