The technique we want to use is called completing the square. We shall use it on the #x# terms first and then the #y#.

Rearrange to

#9x^2 + 4y^2 - 36x + 8y = -31#

Focussing on #x#, divide through by the #x^2# coefficient and add the square of half the coefficient of the #x^1# term to both sides:

#x^2 + 4/9y^2 - 4x + 8/9y +(-2)^2 = -31/9 + (-2)^2#

#(x-2)^2 + 4/9y^2 + 8/9y = 5/9#

Divide through by #y^2# coefficient and add square of half the coefficient of the #y^1# term to both sides:

#9/4(x-2)^2 + y^2 + 2y + (1)^2= 5/4+ (1)^2#

#9/4(x-2)^2 + (y+1)^2 = 9/4#

Divide by #9/4# to simplify:

#(x-2)^2 + 4/9(y+1)^2 = 1#

#(x-2)^2/1 + ((y+1)^2)/(9/4) = 1#

General equation is

#(x-a)^2/h^2 + (y-b)^2/k^2 = 1#

where #(a,b)# is the centre and #h, k# are the semi-minor/major axis.

Reading off the centre gives #(2, -1)#.

In this case, the #y# direction has a bigger value than the #x#, so the ellipse will be stretched in the #y# direction. #k^2 > h^2#

The vertices are obtained by moving up the major axis from the centre. Ie #+-sqrt(k)# added to the y coordinate of the centre.

This gives #(2, 1/2) and (2, -5/2)#.

The co-vertices lie on the minor axis. We add #+-sqrt(h)# to the centre's x coordinate to find these.

#(1,-1) and (3,-1)#

Now, to find the foci:

#c^2 = k^2 - h^2#

#c^2 = 9/4 - 1#

#c^2 = 5/4 implies c = +-sqrt(5)/2#

Foci will be situated along the line #x = 2# at #+-sqrt(5)/2# from #y = -1#.

#therefore# foci at #(2, (-2+sqrt(5))/2) and (2,(-2-sqrt(5))/2)#

Finally the eccentricity is found using

#e=sqrt(1-h^2/k^2)#

#e=sqrt(1-1/(9/4)) = sqrt(1-4/9) = sqrt(5)/3#