Please help-I'm totally lost-

What is the equation of the ellipse with foci(-10,0), (10,0) and co-vertices (0,-3),(0,3)
x^2/109 + y^2/9 = 1
X62/100 + y^2/9 = 1
Please show me the steps because I have others to do and I'm really stuck

To find the equation of the ellipse, we need to determine the distance between the foci and the distance between the co-vertices.

1. Finding the distance between the foci:
- The distance between the foci of an ellipse is given by the formula: c = √(a^2 - b^2), where:
- 'c' represents the distance between the foci,
- 'a' represents the semi-major axis of the ellipse,
- 'b' represents the semi-minor axis of the ellipse.

- In this case, we can observe that the x-coordinates of the foci are -10 and 10. So, the distance between the foci is 2a = 20.
Therefore, 'a' = 20/2 = 10.

- Similarly, we can observe that the y-coordinates of the co-vertices are -3 and 3. So, the distance between the co-vertices is 2b = 6.
Therefore, 'b' = 6/2 = 3.

- Now, we can find 'c' using the formula c = √(a^2 - b^2):
c = √(10^2 - 3^2)
= √(100 - 9)
= √91.

2. Writing the equation of the ellipse:
- The standard equation of an ellipse with center (h, k) is: ((x - h)^2 / a^2) + ((y - k)^2 / b^2) = 1.

- In this case, the center is the origin (0, 0) because the foci and co-vertices are symmetric about the origin.

- Using the values we found earlier, the equation of the ellipse becomes:
(x^2 / 10^2) + (y^2 / 3^2) = 1

- Simplifying, we get:
x^2/100 + y^2/9 = 1.

Therefore, the equation of the ellipse is x^2/100 + y^2/9 = 1.