Given the equation of an ellipse is 4x^2 + y^2 + 24x - 6y + 9 = 0

(i) Find the centre,foci and vertices
(ii)calculate the lengths of the major and minor axes

complete the square ....

4(x^2 + 6x + ...) + *y^2 - 6y + .....) = -9

4(x^2 + 6x + 9) + *y^2 - 6y + 9) = -9 + 36 + 9
4(x+3)^2 + (y+3)^2 = 36
divide by 36

(x+3)^2 /9 + (y+3)^2 /36 = 1

take it from here ...

To find the center, focus, and vertices of an ellipse, we first need to rewrite the equation of the ellipse in a standard form. The standard form of the equation of an ellipse is given by:

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

where (h, k) represents the center of the ellipse, and "a" and "b" are half the lengths of the major and minor axes, respectively.

Let's begin by rearranging the given equation:

4x^2 + y^2 + 24x - 6y + 9 = 0

Group the x-terms and the y-terms:

(4x^2 + 24x) + (y^2 - 6y) + 9 = 0

Complete the square for both the x-terms and the y-terms:

4(x^2 + 6x) + (y^2 - 6y) + 9 = 0
4(x^2 + 6x + 9) + (y^2 - 6y + 9) - 4(9) - 9 = 0
4(x + 3)^2 + (y - 3)^2 - 36 - 9 = 0
4(x + 3)^2 + (y - 3)^2 = 45.

Now, we have the equation in standard form. Comparing it with the standard form, we can see that (h, k) = (-3, 3). Therefore, the center of the ellipse is located at (-3, 3).

To find the lengths of the major and minor axes, we need to determine the values of "a" and "b." These are given by:

a = sqrt(45)
b = sqrt(45)

So, both the major and minor axes have lengths equal to sqrt(45).

Next, we can find the foci and vertices. For an ellipse, the distance between the center and each focus is given by c. The relationship between a, b, and c is given by the equation:

c^2 = a^2 - b^2.

Therefore, to find c, we can use the equation:

c = sqrt(a^2 - b^2).

Substituting the values of a and b, we get:

c = sqrt(45 - 45) = sqrt(0) = 0.

Since c = 0, the foci coincide with the center (-3, 3).

Finally, the vertices lie on the major axis of the ellipse. For an ellipse centered at (h, k), the vertices are located at (h ± a, k). Substituting the values of h, k, and a, we get:

Vertex 1: (-3 + sqrt(45), 3)
Vertex 2: (-3 - sqrt(45), 3)

Therefore, the vertices of the ellipse are approximately (-0.472, 3) and (-5.528, 3).