As the equation #(x+11)^2/144+(y-5)^2/121=1# is of form
#(x-h)^2/a^2+(y-k)^2/b^2=1#, whose centre is #(h,k)# and major and minor axis are #2a# and #2b# respectively.
As such here centre is #(-11,5)# and major axis is #24# and minor axis is #22#.
For focus we should calculate eccentricity of ellipse and as #b^2=a^2(1-e^2)#, #e=sqrt(1-b^2/a^2)=sqrt(1-121/144)=sqrt23/12#
and foci are given by #(h+-ae,k)# i.e. they are
#(-11+-sqrt23,5)# i.e. #(-11-sqrt23,5)# and #(-11+sqrt23,5)#
graph{(x+11)^2/144+(y-5)^2/121=1 [-30, 10, -8, 18]}
Note - Here we had #a > b#, hence major axis is parallel to #x#-axis. But if we have #b >a#, then major axis is parallel to #y#-axis and #e=sqrt(1-a^2/b^2)# and then foci are #(h,k+-ae)#
