Ellipses

[(x+3)^2 / 12] + [(y-2)^2 / 16]
I Have to find the following:
Center:
Foci:
Vertices:
A:
B:
C:
Eccentricity:

check your text. The ellipse

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

center: (h,k)
semi-axes a and b
c^2 = a^2-b^2
foci are at ±c away from the center, along the major axis.
eccentricity = c/a

So, plug in your numbers. What do you get?

And what have you got so far?

BTW, the ellipse should be an equation of the type

(x+3)^2/12 + (y-2)^2/16 = 1

To find the center, foci, vertices, and eccentricity of the ellipse with the given equation [(x+3)^2 / 12] + [(y-2)^2 / 16], we can compare it with the standard form equation of an ellipse:

[(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 the semi-major and semi-minor axes respectively.

Comparing the given equation with the standard equation, we get:

(h, k) = (-3, 2)
a^2 = 12
b^2 = 16

Now, let's find the center, foci, vertices, and eccentricity step by step:

1. Center: The center of the ellipse is simply (h, k). From the equation, we have (h, k) = (-3, 2). Therefore, the center is (-3, 2).

2. Semi-major and semi-minor axes: We know that a^2 = 12 and b^2 = 16. Taking the square root of both sides, we get a = √12 = 2√3 and b = √16 = 4.

3. Foci: The foci can be found using the formula c = √(a^2 - b^2), where 'c' represents the distance from the center to each focus. Let's calculate c:

c = √(a^2 - b^2)
= √(12 - 16)
= √(-4) [Taking the square root of a negative number gives an imaginary number]
= 2i [Where 'i' is the imaginary unit]

Since we have an imaginary number, it means there are no foci in this case.

4. Vertices: The distance between the center and vertices along the major axis is the semi-major axis, 'a'. In this case, the major axis is horizontal, and since the center is (-3, 2), the vertices would be (-3 - a, 2) and (-3 + a, 2). Substituting the value of 'a' into the equation, we get:

Vertex A: (-3 - 2√3, 2)
Vertex B: (-3 + 2√3, 2)

5. Co-vertices: The distance between the center and co-vertices along the minor axis is the semi-minor axis, 'b'. In this case, the co-vertices would be (-3, 2 - b) and (-3, 2 + b). Substituting the value of 'b', we get:

Co-vertex C: (-3, 2 - 4) = (-3, -2)
Co-vertex D: (-3, 2 + 4) = (-3, 6)

6. Eccentricity: The eccentricity, denoted by 'e', is a measure of how elongated an ellipse is. It can be calculated using the formula:

e = √(1 - b^2 / a^2)

Substituting the values of 'a' and 'b' into the equation, we get:

e = √(1 - (16 / 12))
= √(1 - 4/3)
= √(−1/3) [Again, taking the square root of a negative number gives an imaginary number]
= i/√3

So, the eccentricity of the given ellipse is i/√3, where 'i' is the imaginary unit.