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.
Write an equation of the ellipse with foci at (0, ±10), and vertices at (0, ±11).
5 answers
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
b^2 = (11^2 - 10^2) = 21 since the axis is vertical, we have
x^2/21 + y^2/121 = 1
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
x^2/21 + y^2/121 = 1
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
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
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