如图 已知抛物线y=-x2+4x+5的顶点为D,与x轴交于A、B两点(A左B右),与y轴 交于C点,

2025-02-25 00:29:44
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回答1:

(1) y = -(x + 1)(x - 5)
A(-1, 0), B(5, 0), C(0, 5)
对称轴x = (-1 + 5)/2 = 2, E与对称轴的距离也是2 - 0 = 2, 即E的横坐标为2+2 = 4, E(4, 5)
AE: (y - 0)/(5 - 0) = (x + 1)/(4 + 1), y = x + 1

(2) AE关于x翻折,横坐标不变,纵坐标变号,即新方程为-y = x + 1, y = -x - 1
与抛物线联立:-(x+1)(x - 5) = -(x +1)
x + 1 = 0, x = -1 (点A,舍去)
x - 5 = 1, x = 6, F(6, -7)

(3)两个三角形的底GE相同,只需考虑其上的高即可。
令P(m, -(m + 1)(m - 5)), G(m, m + 1)
B与GE(x - y + 1 = 0)的距离为H = |5 - 0 + 1|/√2 = 6/√2

P与GE(x - y + 1 = 0)的距离为h = |m + (m +1)(m - 5) + 1|/√2 = |(m + 1)(m - 4)|/√2
h : H = |(m + 1)(m - 4)| : 6 = 2 : 3
|(m + 1)(m - 4)| = 4
(m + 1)(m - 4) = 4, m = 2±√11

(m + 1)(m - 4) = -4, m = 2 ±√3