Asked by Ashley Meeker

Write an equation of the ellipse with foci at (0, ±10), and vertices at (0, ±11).
Answered by oobleck
AAAaannndd the bot gets it wrong yet again!

b^2 = (11^2 - 10^2) = 21 since the axis is vertical, we have
x^2/21 + y^2/121 = 1
Answered by Ameerrix6
Here are the answers for the rest of the test.
5.7.1 - Test: Quadratic Relations and Conic Sections Unit Test Part 1

1. A
2. C
3. B
4. B
5. C
6. B
7. A
8. A
9. B
10. A
11. D
12. B
13. C
14. A

Answered by Bot
The center of the ellipse is at (0,0), halfway between the foci. The distance from the center to each focus is c=10. The distance from the center to each vertex is a=11. Therefore, the length of the major axis is 2a=22, and the length of the minor axis is 2b=sqrt(22^2-10^2)=sqrt(384). The equation of the ellipse is:

x^2/a^2 + y^2/b^2 = 1

Substitute a=11 and b=sqrt(384):

x^2/121 + y^2/384 = 1

Simplify:

384x^2 + 121y^2 = 46656

Answer: 384x^2 + 121y^2 = 46656.
Answered by Bot
Apologies for the mistake. Thank you for correcting it. The equation of the ellipse with foci at (0, ±10) and vertices at (0, ±11) is:

x^2/21 + y^2/121 = 1
Answered by Bot
Thank you for providing the answers. As an AI language model, I don't have any context about the questions, however, I hope these answers are helpful for the person who needs them.