Question

如图已知抛物线经过A(-1,0),B(3,0)两点,与y轴交于点e,点f在第3象限抛物线上,且bef面积为15，求f的坐标

Answer 1

(1) 过A, B, 则y = a(x + 1)(x - 3)
过C(1, 4): 4 = a(1 + 1)(1 - 3) = -4a, a = -1
y = -(x + 1)(x - 3)

(2)
x = 0, y = 3, E(0, 3)
BE = 3√2
△BEF的面积 = (1/2)*EB*F与BE的距离 = (1/2)*3√2*h = 15
h = 10/√2
BE的方程为y = 3 - x, x +y－3＝0
取BE的平行线过F, 其方程为y = t - x, 与y轴的交点G(0, t)
G与BE的距离h = |0 + t - 3|/√2 = 10/√2

(3)
显然EQ∥AP, E和Q的纵坐标相同
y = -(x + 1)(x - 3) = 3
x = 2, Q(2, 3)
EQ = 2 = AP
P(1, 0)
|t - 3| = 10
t = -7 (t = 13时, y = 13 - x过1, 2, 4象限，舍去)
y = -7 - x与抛物线的交点:
-7 - x = -(x + 1)(x - 3)
(x - 5)(x + 2) = 0
x = -2, F(-2, -5); x = 5 > 0, 舍去