Question

在平面直角坐标系中 抛物线与x轴交于点A(3.0)B(-1,0)

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Answer 1

(1)

BO = 1, tan∠OBC = OC/BO = OC/1 = 3, OC = 3, C(0, 3)

于是可以表达为y = a(x - 3)(x + 1)

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

y = -(x - 3)(x + 1) = -x² + 2x + 3

(2)

对称轴x = (-1 + 3)/2 = 1

A关于对称轴的对称点为B(-1, 0)

过BC的直线为x/(-1) + y/3 = 1

取x = 1, y = 6

P(1. 6)

(3)

抛物线定点为(1, 4)

令此直线为y = t, t < 4

圆与x轴相切, 则|t|为半径；令EF的中点为M(1, t), 则ME=MF = |t|

不妨取E(t + 1, t)

t = -(t+1)² + 2(t+1) + 3

t = (-1 ± √17)/2