Question 1177460
given:

Center ({{{3}}},{{{0}}})
one focus at ({{{3}}},{{{2}}})
length of minor axis is {{{4}}}


Equation of the ellipse: 

since center and foci have same {{{x}}} coordinate, major axis is vertical, so you need equation

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

Center ({{{3}}},{{{0}}}) =>{{{ h=3}}}, {{{k=0}}}

focus at ({{{3}}},{{{2}}})-> {{{c=2}}}

 minor axis is {{{2b=4}}} ->{{{b=2}}}

Use the equation {{{c^2=a^2-b^2}}} along with the given coordinates of the vertices and foci, to solve for {{{b^2}}}.

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

{{{a^2=2^2+2^2}}}

{{{a^2=8}}}

{{{a=sqrt(8)}}}

{{{a=sqrt(4*2)}}}

{{{a=2sqrt(2)}}}


{{{(x-3)^2/4+y^2/8 =1}}}


{{{drawing ( 600, 600, -5, 5, -5, 5, 
circle(3,0,.012), locate(3,0.3,C),
circle(3,2,.012), locate(3,2,F),
graph( 600, 600, -5, 5, -5, 5, -sqrt(2)*sqrt(-x^2+6x-5),sqrt(2)*sqrt(-x^2+6x -5))) }}}