初三数学抛物线

Question

如题所述

Answer 1

(1)根据表格，x = -2和x = 0时，抛物线上的点的纵坐标相同，即对称轴为x = -1, h = -1
点(0, -3)和(1, 0)在抛物线上：
a + k = -3
4a + k = 0
解得a = 1, k = -4
抛物线为y = (x + 1)^2 - 4 = x^2 + 2x - 3 = (x - 1)(x + 3)
x = -3, y = m = 0
ak + h = 1*(-1)-4 = -5

(2) A(-3, 0), B(1, 0), C(0, -3), D(-1, -4)
令P(-1, p), 对称轴与x轴交于D'(-1, 0)
tan∠BCO = OB/OC = 1/3
tan∠APD = AD'/D'P = 2/p = 1/3, p = 6
P(-1, 6)
PC的斜率为k = (6 + 3)/(－1－0)＝　－9，方程为y＝-9x - 3
PC与x轴交于P'(-1/3, 0)
△APC的面积 = △APP'的面积+△AP'C的面积 = (1/2)AP'*P的纵坐标 + （1/2）AP'*(-C的纵坐标)
= (1/2)*(-1/3 + 3)*6 + (1/2)*(-1/3+3)*3
= 12

(3）此时抛物线变为y = x^2 + 2x - 3 - k
D1(-1, -4-k), C1(0, -3-k)
CD的方程为y = x - 3
显然FD与FD1相互垂直，FD的斜率为1，则FD'的斜率为-1，即F的纵坐标与DD1的中点的纵坐标相同。
将CD的方程与新抛物线联立，得其纵坐标为y = (-1/2)(7 + sqrt(4k + 1)) (sqrt: 平方根)
该值 = (-4 - 4 - k)/2 （DD中点的纵坐标)
(-1/2)(7 + sqrt(4k + 1)) = (-4 - 4- k)/2
k^2 - 2k = 0
k = 2 (舍去k = 0)