I need Help finishing this problem with an ellipse. The question says to find the vertices and foci?

Question

x^2 + 9y^2 - 24x + 18y + 9 = 0
x^2 - 24 x + 9y^2 + 18y = -9 then I factored out 9 & completed the square
x^2 - 24 x + 144 + 9(y^2 + 2y +1) = -9 + 144 + 9
(x - 12)^2 + 9(y+1)^2 = 144 then when I divided by 144 I got
(x-1)^2/12 + (y+1)^2/16 = 1
I feel like this is not right can someone help

Answer 1

x^2 + 9y^2 - 24x + 18y + 9 = 0
or (x^2 - 2*x*12 + 12^2) + 9(y^2 + 2*y*1 +1) + 9 - 12^2 - 9 = 0
or (x - 12)^2 + 9(y + 1)^2 = 12^2 = 144
or (x - 12)^2/144 + 9(y + 1)^2/144 = 1
or ((x - 12)^2)/12^2 + ((y + 1)^2)/4^2 = 1

Transform the equation to new coordinate system
X = x - 12 and Y = y + 1

Standard equation of ellipse in the new coordinate system is
X^2/a^2 + Y^2/b^2 = 1

For this equation, the centre is at (X=0,Y=0) in new coordinate system.
i.e. (x - 12 = 0, y + 1 = 0) i.e. (12,-1) in old coordinate system

The coordinates of foci are (X = a*e, Y = 0) and (X = -a*e, Y = 0) in new coordinate system.

Question: What is 'e'?
Answer: 'e' is the eccentricity of the ellipse and e = sqrt[1 - (b/a)^2]
Here a = 12, b = 4
So e = sqrt[1 - (4/12)^2] = (2/3)*sqrt(2)

So in new coordinate system foci are (X = 8*sqrt(2), Y = 0) and (X = -8*sqrt(2), Y = 0)

In old coordinate system foci are (x = X + 12 = 12 + 8*sqrt(2), y = Y - 1 = 0 - 1 = -1) and (12 - 8*sqrt(2), -1)

Vertices are points where the ellipse cuts the major axis.
These are (X = a = 12, Y = 0) and (X = -a = -12, Y = 0) in new coordinate system.
So in old coordinate system the vertices are (24,-1) and (0,-1).

I wrote vertex instead of centre in the previous solution. But I have corrected it now.

Answer 2

(((x - h)^2)/(a^2)) + (((y - k)^2)/(b^2)) = 1

x^2 + 9y^2 - 24x + 18y + 9 = 0
x^2 - 24x + 9y^2 + 18y + 9 = 0
(x^2 - 24x) + (9y^2 + 18y) + 9 = 0
(x^2 - 24x + 144 - 144) + 9(y^2 + 2y) + 9 = 0
((x - 12)^2 - 144) + 9(y^2 + 2y + 1 - 1) + 9 = 0
(x - 12)^2 - 144 + 9((y + 1)^2 - 1) + 9 = 0
(x - 12)^2 - 144 + 9(y + 1)^2 - 9 + 9 = 0
(x - 12)^2 - 144 + 9(y + 1)^2 = 0
(x - 12)^2 + 9(y + 1)^2 = 144
(((x - 12)^2)/144) + (((y + 1)^2)/16) = 1

Here's the reason, since the coefficient of (x - 12)^2 is 1, you just divide 144 by 1, and you get 144, same goes for the 9(y + 1)^2, since the coefficient is 9, divide 144 by 9, and you get 16

a^2 - b^2 = c^2
144 - 16 = c^2
c^2 = 128
c = sqrt(128)
c = sqrt(64 * 2)
c = 8sqrt(2)


Center : (h,k) : (12,-1)
Foci : (h±c, k),
Foci = (12±8sqrt(2),-1)
Foci = ((12 + 8sqrt(2)),-1) and ((12 - 8sqrt(2)),-1)

a^2 = 144
a = 12

Vertice ((h ± a),k)
Vertice ((12 ± 12),-1)
Vertice (0,-1) and (24,-1)

So

Vertice : (0,-1) and (24,-1)
Foci : ((12 - 8sqrt(2)),-1) and ((12 + 8sqrt(2)),-1)
Center : (12,-1)

Answer 3

There is a mistake on the very last line.
x^2 + 9y^2 - 24x + 18y + 9 = 0
x^2 - 24 x + 9y^2 + 18y = -9
x^2 - 24 x + (3y)^2 + 18y = -9+9+144
(x-12)^2+(3y+3)^2=144
(x-12)^2+9(y+1)^2=144
(x-12)^2/144+(y+1)^2/16=1

I believe this is the correct answer. Hope this helps!

Answer 4

Your denomenator in the first term should be 12^2 (twelve squared) or 144. The second term denomenator is correct and can be written as 4^2.