What is the center, foci, and major and minor vertices of the ellipse (x+32)^2+9(y+3)^2=1104

Steps would be appreciated.

The standard form would be

(x+32)^2/(1104) + (y+3)^2/(1104/9) = 1
showing that the center is located at x= -32, y = -3,
i.e., the point (-32,-3), and that the
horizontal semi-axis has a length sqrt(1104) ...about 33.2
while the vertical semi-axis has a length sqrt(1104/9) ...about 11.1
so that the vertices are roughly at
(-65.2,-3),(1.2,-3),(-32,-14.1),(-32,8…
[major,major,minor,minor]

The foci will be on the major (horizontal) axis,
and the distance from the center to either focus is
sqrt(a^2 - b^2) = sqrt(1104 - (1104/9))
= about 31.32,
so the foci are around
(-63.32,-3) and (-0.68,-3).