Find center, vertices, foci.

(x+4)^2/36 + (y-3)^2/100 =1

center is (-4,3)

major axis is vertical,since 100 > 36

a = 10
b = 6
c^2 = a^2-b^2

foci at (0,±c)

To find the center, vertices, and foci of the ellipse given by the equation:

(x+4)^2/36 + (y-3)^2/100 =1

We need to compare it with the standard form of an ellipse:

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

Here, we can see that the x-coordinates are squared over 36, while the y-coordinates are squared over 100. To find the center, we use the values of h and k.

The center of the ellipse is given by the coordinates (h, k), which in our case is (-4, 3).

So, the center of the ellipse is at (-4, 3).

To find the vertices, we need the lengths of the major and minor axes (2a and 2b). In our equation, a^2 = 36, so a = 6, and b^2 = 100, so b = 10.

The vertices are located on the major axis. Since the major axis is along the x-axis, we add or subtract 6 horizontally from the center to find the vertices.

The vertices are at (-4 + 6, 3) = (2, 3) and (-4 - 6, 3) = (-10, 3).

So, the vertices of the ellipse are (2, 3) and (-10, 3).

To find the foci, we need the distance between the center and the foci. This distance, denoted as c, can be found using the formula c = sqrt(a^2 - b^2).

In our case, c = sqrt(36 - 100) = sqrt(-64). Since the square root of a negative number is not defined in the real number system, we can infer that the given ellipse does not have real foci.

Therefore, the ellipse does not have any foci.

To summarize:
- Center: (-4, 3)
- Vertices: (2, 3) and (-10, 3)
- Foci: None