2.(x+4)² + y²/4 = 1
更新1:
correction for #2 it's (x+4)²/1 + y²/4 = 1
ELLIPSE EQUATION PROBLEM!!!! I JUST NEED TO KNOW THE CENTER OR (h,k) PLS ANSWER AS SOON AS POSSIBLE?
回答 (3)
2016-10-18 11:08 am
For the equation of ellipse [(x - h)²/a²] + [(y - k)²/b²] = 1, the center is (h, k).
1.
(x²/4) + [(y - 2)²/25] = 1
[(x - 0)²/2²] + [(y - 2)²/5²] = 1
The center is (0, 2).
2.
(x + 4)² + (y²/4²) = 1
[(x - (-4))²/1²] + [(y - 0)²/2²] = 1
The center is (-4, 0).
1.
(x²/4) + [(y - 2)²/25] = 1
[(x - 0)²/2²] + [(y - 2)²/5²] = 1
The center is (0, 2).
2.
(x + 4)² + (y²/4²) = 1
[(x - (-4))²/1²] + [(y - 0)²/2²] = 1
The center is (-4, 0).
2016-10-18 12:17 pm
1) (0.2)
2) (-4,0)
2) (-4,0)
2016-10-18 11:47 am
No calculation required. You can just read off (0, 2) and (-4, 0)
You just need to pick out the h and k values of any ellipse written as
(x - h)^2/a^2 + (y - k)^2/b^2 = 1
You just need to pick out the h and k values of any ellipse written as
(x - h)^2/a^2 + (y - k)^2/b^2 = 1
收錄日期: 2021-04-18 15:48:29
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https://hk.answers.yahoo.com/question/index?qid=20161018024500AA60vBv