panda716
幼苗
共回答了9个问题采纳率:88.9% 举报
1. OB=OC, C(0, -3)
tan﹤ACO=1/3 = AO/OC = AO/3
AO = 1
A(-1, 0)
代入A, B, C三点的坐标:
a - b + c = 0
9a + 3b + c = 0
c = -3
解得a = 1, b = -2, c = -3
y = x² - 2x - 3
2. D(2 -3)
设F(m, 0), E(n, n² - 2n - 3)
使以A,D,E,F为顶点的四边形是平行四边形, AE与DF平行且长度相等.
AE的斜率p = (n² - 2n - 3)/(n+1) = n-3
DF的斜率q = 3/(m-2)
p = q, m-2 = 3/(n-3) (1)
AE² = (n+1)² + (n² - 2n - 3)²
DF² = (m-2)² + (0+3)² = (m-2)² + 9
(n+1)² + (n² - 2n - 3)² = (m-2)² + 9 (2)
(1)代入(2)并简化: (n+1)²(n-3)² = 9
(n+1)(n-3) = ±3
(i) (n+1)(n-3) = -3
n(n-2) = 0
n = 0或n = 2
n = 0时, m = 1, E(0, -3), F(1, 0)
n = 2时, E(2 -3), 与D重合舍去.
(ii)(n+1)(n-3) = 3
n² - 2n - 6 = 0
n = 1±√7
n = 1+√7时, m = 4+√7, F(4+√7, 0)
n = 1-√7时, m = 4-√7, F(4-√7, 0)
要完整的话还需考虑 AF与DE平行且长度相等.
3. 现在没时间完成,容后再做.
1年前
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