Question 915212
If the pool table is longest at {{{60in}}}, then your {{{a=60/2=30}}}, and if it is widest at {{{40in}}}, then {{{b=40/2=20}}} 

The equation for an ellipse at the origin is: 

{{{x^2/a^2+y^2/b^2=1 }}}

And plugging in: 

{{{x^2/(30)^2+y^2/(20)^2=1 }}}

{{{x^2/900+y^2/400=1 }}}

That is your equation. 

The foci are given as the points of distance {{{c}}} from the center, where: 

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

So solving for "{{{c}}" yields:

{{{30^2-20^2=c^2 }}}

{{{900-400=c^2 }}}

{{{500=c^2}}} 

{{{sqrt(500)=c }}}


{{{22.36=c }}} or  {{{-22.36=c }}}

The foci are at ({{{-22.36}}},{{{0}}}) and ({{{22.36}}},{{{0}}})
vertices: ({{{-30}}},{{{0}}}) and ({{{30}}},{{{0}}})

center: ({{{0}}},{{{0}}})


{{{drawing( 600, 600, -45, 45, -40, 40,circle(22.36,0,.7), circle(-22.36,0,.7), circle(30,0,.7), circle(-30,0,.7), graph( 600, 600, -45, 45, -40, 40, sqrt(400-4x^2/9),-sqrt(400-4x^2/9) )) }}}