Question 97784
The standard form of the equation of an ellipse is 

{{{ (x-h)^2/a^2  + (y-k)^2/b^2 =1 }}} where (h,k) is the center of the ellipse

we first try and put the given equation in the standard form 

{{{x^2+9y^2-24x+18y+9=0}}}

{{{ x^2 -24x + 144  + 9y^2 + 18y + 9 -144 =0}}}

{{{ x^2 -24x + 144  + 9(y^2 + 2y + 1) =144}}}
{{{(x-12)^2 + 9(y + 1)^2 =144 }}}

{{{((x-12)^2)/144 + (9(y + 1)^2)/144 =1 }}}

{{{((x-12)^2)/144 + (y + 1)^2/16 =1 }}}

{{{((x-12)^2)/12^2 + (y - (-1))^2/4^2 =1 }}}

so we have the center (h,k) as (12,-1)

also a=12 and b=4 
Since a > b the major axis is horizontal

vertices are given by (h-a,k) and (h+a,k)
so we have ( 12-12,-1) and (12+12,-1)
vertices are (0,-1) and (24,-1)

focii are at points  given by (h-c,k) and (h+c,k) where c = {{{sqrt(a^2 -b^2)}}}
(12- 8*sqrt2,-1)  and (12+ 8*sqrt2,-1)