3x^2+4y^2-36x+24y+132=0

Looking at the equation we can tell it will be an ellipse because of the coefficients of x

^{2}and y^{2}are not the same but they are the same sign (so not a hyperbola). It's easiest to find the critical points like the center foci and vertices if it's written in the form: (x-h)^{2}/(a^{2})+(y-k)^{2}/(b^{2}) = 1First we reorganize a little:

3x

^{2}-36x+4y^{2}+24y = -132Then we try to complete the square for x and y. We'll start with y:

3x

^{2}-36x + 4(y^{2}+6x +9) = -132+36 (since we factored by 4)3x

^{2}-36x + (y+3)^{2}= -96now for x:

3(x

^{2}-12x+36)+4(y+3)^{2}=-96+108 (since I factored out a 3, we added 108 not 36)3(x-6)

^{2}+4(y+3)^{2}=12Finally, we divide by 12 to get the right hand side equal to 1:

(x-6)

^{2}/4 + (y+3)^{2}/3 = 1Now we can read off the points you need:

The center is shifted from (0,0) to (6,-3) (this is from the h, k).

The major axis is√4 = 2 left and right from the center (since the larger number is under the x term we can tell it will be longer in the horizontal direction). So the major vertices are: (4,-3)) and (8,-3). The minor vertices are: (6,-3-√3) (6,3+√3)).
The foci distance "c" is found by the ellipse relation: c

^{2}=a^{2}-b^{2}. 4-3 = 1. so c = √1 = 1. Finally the foci points are a distance c from the center in the major axis direction: F1 = (5,-3) and F2 = (7,-3)