Question

如图，抛物线y=ax2+bx-2经过A（4，0），B（1，0）两点

(1)求出抛物线的解析式；（2)若P是抛物线上x轴上方的一动点，过P作PM⊥x轴，垂足为M，是否存在P点，使得以A，P，M为顶点的三角形与△OAC相似？若存在，请求出符合条件的点P的坐标；若不存在，请说明理由；（3）在直线AC上方的抛物线上有一点D，使得△DCA的面积最大，求出点D的坐标．

Answer 1

(1) y = a(x - 4)(x - 1) = ax² - 5ax + 4a
4a = -2, a = -1/2
b = -5a = 5/2
y = -x²/2 + 5x/2 - 2

(2)
OA = 4, OC = 2, OA : OC = 2 : 1
要使以A，P，M为顶点的三角形与△OAC相似, 须PM : MA = 2 : 1或MA : PM = 2 : 1
M(m, 0), P(m, -m²/2 + 5m/2 - 2)
MA = 4 - m, MP = -m²/2 + 5m/2 - 2
(i) PM : MA = 2 : 1
-m²/2 + 5m/2 - 2 : 4 - m = 2 : 1
8 - 2m = -m²/2 + 5m/2 - 2
m² - 9m + 20 = (m - 4)(m - 5) = 0
m = 4, 点A, 舍去
m = 5, x轴下方

(ii)MA : PM = 2 : 1
-m²/2 + 5m/2 - 2 : 4 - m = 1 : 2
m² - 6m + 8 = (m - 2)(m - 4) = 0
m = 2, P(2, 1)
m = 4, 点A, 舍去

(3)
AC定长，显然抛物线过D的切线与AC平行时, AC上的高最大，即△DCA的面积最大。
AC斜率k = (0 + 2)/(4 - 0) = 1/2
切线y = x/2 + d
-x²/2 + 5x/2 - 2 = x/2 + d
x² - 4x + 4 + 2d = 0
∆ = 16 - 4(4 + 2d) = -8d = 0
d = 0
y = x/2
x² - 4x + 4 = (x - 2)² = 0
x = 2
D(2, 1)

Answer 2

好好学习 别在网上找答案了！