There is no #xy# term here, so tha ellipse has its major and minor axis parallel to #x# and #y# axes and hence we can write the equation in form
#(x-h)^2/a^2+(y-k)^2/b^2=1#, whose center is #(h,k)# and major and minor axes are #2a# and #2b# (if #a>b#, otherwise reverse i.e. minor axis is #2a# and major axis is#2b#). Focii are always along major axis and here at a distance of #ae# (or #be# if #2b# is major axis) on either side of center, where Let us convert the equation accordingly
#9x^2+25y^2-18x+200y+184=0# can be written as
#9(x^2-2x)+25(y^2+8y)+184=0#
or #9(x^2-2x+1)+25(y^2+8y+16)-9-400+184=0#
or #9(x-1)^2+25(y+4)^2=225#
or #(x-1)^2/25+(y+4)^2/9=1#
or #(x-1)^2/5^2+(y+4)^2/3^2=1#
Hence, center of ellipse is #(1,-4)# and major axis is #10# and minor axis is #6#.
Verices will be #5# units on either side of center parallel to #x#-axis and #3# units on either side of center parallel to #y#-axis
i.e. vertices are #(6,-4)#, #(-4,-4)#, #(1,-7)# and #(1,-1)#.
Eccentricity of ellipse is #sqrt(1-b^2/a^2)=sqrt(1-9/25)=4/5#
and hence focii are #5xx4/5=4# units on either side of #(1,-4)# i.e. focii are at #(-3,-4)# and #(5,-4)#.
graph{((x-1)^2+(y+4)^2-0.03)((x-6)^2+(y+4)^2-0.03)((x+4)^2+(y+4)^2-0.03)((x-1)^2+(y+7)^2-0.03)((x-1)^2+(y+1)^2-0.03)((x+3)^2+(y+4)^2-0.03)((x-5)^2+(y+4)^2-0.03)(9x^2+25y^2-18x+200y+184)=0 [-8.83, 11.17, -8.72, 1.28]}
