Question

二次函数图象经过点A(-3,0),B(-1,8),C(0,6),直线 与y轴交于点D,

Answer 1

抛物线：y = a(x + 3)(x - 1)

x = 0,y = -3a = 6,a = -2

y = -2(x + 3)(x - 1)

tan∠DAO = OD/AO = 2/3

AD的倾斜角为∠DAO,

(1) P在直线AD的上方

∠PAO = ∠DAO + ∠PAD=∠DAO+45°

或

(2) P在直线AD的下方

∠PAO = ∠PAD - ∠DAO =45° - ∠DAO

(1)

tan∠PAO = tan(∠DAO+45°) = (tan∠DAO+ tan45°)/(1 - tan∠DAOtan45°) = 5

AP的斜率为5,方程为y = 5(x + 3)

与抛物线联立：-2(x + 3)(x - 1) = 5(x + 3),x = -3/2

P(-3/2,15/2)

(2)

tan∠PAO = tan(45° - ∠DAO ) = ( tan45° - tan∠DAO)/(1 + tan∠DAOtan45°) = -1/5

AP的斜率为-1/5,方程为y = (-1/5)(x + 3)

与抛物线联立：-2(x + 3)(x - 1) = (-1/5)(x + 3),x = 11/10

P(11/10,-41/50)