Question 462331
((x-2)^2)/36+((y-8)^2)/144=1

From the general equation
[((x-h)^2)/a^2]+[((y-k)^2)/b^2]=1 we know that the center of the ellipse is at  (h,k). So for our ellipse equation the center of the ellipse is at (2,8).

c is the distance from the center to a focus point. The expression for c (when b > a) is

{{{c = sqrt( b^2-a^2 ) }}}
So for our ellipse
{{{c = sqrt( 12^2-6^2 ) }}}
{{{c = sqrt( 144-36 ) }}}
{{{c = sqrt( 108 ) }}}
{{{c = 10.3923 }}}
The expression for the foci of a general ellipse is
Foci: (c+h, k) (-c+h, k)
So for our ellipse the foci are (10.3923+2, 8) (-10.3923+2, 8)
(12.3923, 8) (-8.3923, 8)