(1)y = ax² + bx - 3a 中,x=-1时,y=0;x=0时,y=-3.
分别代入解得:a = 1 b = - 2
故所求解析式为y = x² - 2x - 3
(2)y = x² - 2x - 3中,x = m时,y = - m - 1.代入得:
m² - m - 2 = 0
(m - 2)(m+1)= 0
解得:m = 2 或 m = - 1(不合题意,舍去)
则D(2,- 3)
y = x² - 2x - 3中令y=0得x = - 1或x = 3,故B(3,0)
由B(3,0),C(0,-3)易得直线BC的解析式为:y = x - 3
由 DD‘⊥BC,D(2,- 3)易得直线DD’的解析式为:y = - x - 1,其与y轴交于(0,- 1).
(0,- 1)与D(2,- 3)关于BC对称
∴ D‘(0,- 1)
(3)存在符合条件的点P.
易得BD的解析式为:y = 3x - 9
作CE∥BD交x轴于点P,则∠PCB=∠CBD.此时,BP = CD = 2,故P(1,0)
作∠BCP =∠CBD,CP交x轴于点P,(在点B右侧)
设直线BD交y轴于点M,易得M(0,- 9) 显然有△OCP≌△OBM,则P(9,0)
故符合条件的点P有(1,0)和(9,0)