Question

如图，抛物线y=ax2+bx﹣4与x轴交于A(4，0)、B(﹣2，0)两点，与y轴交于点C，点P是线段AB上一动点(端点除外)

过点P作PD∥AC，交BC于点D，连接CP．
(1)求该抛物线的解析式；
(2)当动点P运动到何处时，BP2=BD•BC；
(3)当△PCD的面积最大时，求点P的坐标．

Answer 1

(1) y = a(x - 4)(x + 2)
x = 0, y = -8a = -4, a = 1/2
y = (x - 4)(x + 2)/2 = x²/2 - x - 4

(2)C(0, -4)
BC = 2√5
P(p, 0), p > -2
BP² = (p + 2)²
AC和PD的斜率均为1, PD的方程: y = x - p
BC的方程: x/(-2) + y/(-4) = 1
联立, D((p-4)/3, (2p + 4)/3)
BD² = [(p - 4)/3 + 2]² + [(2p + 4)/3]² = 5(p+2)²/9
BD = (√5)(p+2)/3
BP² = BD•BC
(p + 2)² = 2√5* (√5)(p+2)/3
p = 4/3
P(4/3, 0)

(3)PD² = [(p - 4)/3 -p]² + [(2p + 4)/3]² = 8(p+2)²/9
PD = 2(√2)(p+2)/3
AC = 4√2
PD的方程: y = x - p, x - y - p = 0
A与PD的距离h = |4 - 0 - p|/√2 = (4 - p)/√2
△PCD的面积S = 梯形ACDP的面积 - △APC的面积
= (1/2)(PD + AC)*h - (1/2)AP*|C的纵坐标|
= (1/2)[2(√2)(p+2)/3 + 4√2]*(4 - p)/√2 - (1/2)(4 - p)*4
= -p²/3 + 2p/3 + 8/3
= -(p - 1)²/3 + 3
p = 1时, S最大
P(1, 0)