Please help-What is the equation of the ellipse with foci (0,6), (0,-6) and co-vertices (2,0),(-2,0)

My answer and logic is below but it can't be correct

I thought the answer would be x^2/4 + y^2/40 = 1 but that can't be because the choices are:
x^2/1 + y^2/40 = 1 or x^2/1 + y^2/36
I'm really confused because I think b^2=4, c^2=36 and a^2 = 40

c^2 = 36

b^2 = 4
a^2 - b^2 = c^2, so
a^2 = b^2 + c^2 = 40
OK so far.
The major axis is the horizontal x = 0 line and the equation is
x^2/40 + y^2/4 = 1

I agree with your answer.

Thank you-there must just be a mistake on the homework sheet for the selection of answers

To find the equation of an ellipse with foci and co-vertices given, you can follow these steps:

1. Determine the distance between the foci. In this case, the foci are located at (0,6) and (0,-6), so the distance is 2c = 12.

2. Determine the distance between the co-vertices. The co-vertices are located at (2,0) and (-2,0), so the distance is 2a = 4.

3. Use the formulas for an ellipse:

- The formula for the distance between co-vertices and foci is given by:
c^2 = a^2 - b^2, where a is half the distance between co-vertices, b is half the distance between minor axes, and c is half the distance between the foci.

- The equation of an ellipse centered at the origin is given by:
x^2/a^2 + y^2/b^2 = 1, where a is the length of the semi-major axis, and b is the length of the semi-minor axis.

4. Substitute the values into the formula and solve for b^2:

c^2 = a^2 - b^2
36 = 4 - b^2
b^2 = -32

5. Since b^2 is negative, this means there are no real solutions for b^2. Therefore, the ellipse with the given foci and co-vertices does not exist.

Based on your explanation, it appears that you incorrectly calculated the values of b^2 = 4, c^2 = 36, and a^2 = 40. Therefore, the equation x^2/4 + y^2/40 = 1 cannot be correct. The correct equation would be x^2/1 + y^2/36 = 1, as one of the given choices.