Find the center, vertices, and foci of the ellipse 9x^2 + 4y^2 + 36x -8y + 4 =0

We start with writing the terms containing x and y together. We
get 9x^2 + 36x + 4y^2 – 8y + 4 =0

=> 9(x^2 + 4x) + 4(y^2 –
2y) + 4 =0

Now complete the
squares

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

divide both the sides by 36

=>
(x^2 + 4x +4)/4 + (y^2 – 2y +1)/9 = 1

=> (x + 2) ^2 / 4 + (y
– 1)/9 = 1

This is in the form (x – h) ^2/a^2 + (y – k) ^2/b^2 =
1

a = 3, b= 2, h = -2 and k=1.

The
center of the ellipse is (h, k) or (-2, 1).

The vertices of the
ellipse are (h, k-a) = (-2, -2) and (h, k + a) = (-2, 4)

The foci
are the points (h, k + sqrt (a^2 + b^2) = (-2, 1 + sqrt 5) and (h, k- sqrt (a^2 + b^2) = (-2, 1-
sqrt 5)