find the center and vertices of the ellipse given by the equation?

Question

4x^2+y^2-24x-4y+36=0

Answer 1

center (3,2)
verticies:
(4,2) (2,2)
(3,4) (3,0)

4x^2+y^2-24x-4y+36=0
x^2 + y^2/4 - 6x - y + 9 = 0
[x^2 - 6x + 9] + [y^2 - 4y + 4]/4 = 1
(x-3)^2 /1 + (y-2)^2 / 4 = 1

So if the ellipse equation is:
(x-x1)^2/a^2 + (y-y1)^2/b^2 = 1
Then center is at (x1, y1)

So, center at (3,2)
a = 1, b = 2
Verticies at (3+1,2) (3-1,2) or (4,2) & (2,2)
and (3, 2+2) (3,2-2) or (3,4) & (3,0)

Answer 2

First, subtract by 36, giving you 4x^2+y^2-24x-4y=-36.
Rearrange, gives you (4x^2-24x)+(y^2-4y)=-36, or (4(x^2-6x))+(y^2-4y)=-36.
Add, (4*(6/2)^2) and (4/2)^2 to both sides, giving you (4(x^2-6x+9))+(y^2-4y+4)=-36+36+4, or (4*(x-3)^2)+(y-2)^2=4.
Divide by 4, gives you ((x-3)^2/1^2)+((y-2)^2/2^2)=1.

So, the center is (3,2).
And the vertices are (3+1,2), (3-1,2), (3,2+2), (3,2-2),
or (4,2), (2,2), (3,4), (3,0).