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