Question

求学霸帮忙解

Answer 1

(1)
过(-2, 0), (4, 0), 则可表达为y = a(x + 2)(x - 4)
过(0, -8), 则-8 = -8a, a = 1
y = (x + 2)(x - 4)
D(-1, -5)

(2)
设想将y = x - 4向右下方平移，直至与抛物线相切，P即为所求的点(底BD为定值，高最大)
令切线为y = x + t
y = x + t = (x + 2)(x - 4)
x² - 3x - 8 - t = 0
其判别式=9 + 4(8 + t) = 0, t = -41/4时, 二者相切, 此时上述方程变为(x - 3/2)² = 0
P(3/2, -35/4)
它与y = x - 4, 即x - y - 4 = 0的距离h = |3/2 + 35/4 - 4|/√2 = 25√2/8
BD = 5√2
三角形BDP的最大面积为S = (1/2)*BD*h = 125/8

(3)
Q(q, q - 4), G(q, (q + 2)(q - 4))
显然∠DQG不可能为直角
BD的斜率为1, DG的斜率k = [(q + 2)(q - 4) - (-5)]/[q - (-1)] = q - 3
(i) ∠DGQ为直角
此时DG∥x轴， 二者的纵坐标相等, 即(q + 2)(q - 4) = -5, (q + 1)(q - 3) = 0
q = 3 (舍弃q = -1, 此为点D)
Q(3, -1)

(ii) ∠GDQ为直角
DG和BD的斜率之积为-1, 即k = -1 = q - 3
q = 2, Q(2, -2)