Verified answer

Rearrange to get:

5x^2 + 10x + 9y^2 - 36y = 4

Complete the square on the x and y terms on the LHS:

=> 5[x^2 + 2x] + 9[y^2 - 4y] = 4

=> 5[(x + 1)^2 - 1] + 9[(y - 2)^2 - 4] = 4

Expand:

=> 5(x + 1)^2 - 5 + 9(y-2)^2 - 36 = 4

=> 5(x+1)^2 + 9(y-2)^2 = 4 + 5 + 36 = 45

Divide through by 45 to get:

[(x + 1)^2]/9 + [(y - 2)^2]/5 = 1

=> [(x + 1)^2]/3^2 + [(y - 2)^2]/ (sqrt(5))^2 = 1

This is now of the form of an equation of an ellipse: (x-p)^2 / a^2 + (y-q)^2/b^2 = 1

Where the origin is point (p,q) and the foci are at +/-(ae,0) and e is the eccentricity and

e = sqrt(1 - b^2/a^2

and the ellipse meets the axis at (a,0), (-a,0), (0,b) and (0,-b) *** See extra note ***

Hence we have p = -1 and q = 2

Hence centre at point (-1, 2)

Also, a = 3 and b = sqrt(5). hence vertices at (3,0), (-3,0), (0,sqrt5) and (0, -sqrt5)

e = sqrt(1 - 5/9) = sqrt(4/9) = 2/3

Hence foci at +/- (3*2/3), 0) = +/- (2,0)

Extra note: Oh, just realised this gives the foci and vertices when centre is at origin. You will need to adjust each by the translation p,q. as necessary. Haven't got much time now.

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