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#
