(1) 梯形顶点的坐标：O(0,0),A(6,0),B(3,4),C(0,4)
OA = 6
AB = √[(3-6)² + (4-0)²] = 5
设M(t,0),t∈[0,6]时MN与y轴平行,此时OM=AN = t
M,N的横坐标相同,设N(t,n)
AB的方程为:(y - 0)/(x - 6) = (4 - 0)/(3 - 6) = -4/3
y = -4(x-6)/3
n = -4(t - 6)/3
OM² = AN²
t² = (t - 6)² + (n - 0)² = (t - 6)² + 16(t - 6)²/9 = 25(6-t)²/9
t = ±5(6-t)/3
t = 15 (> 6,舍去)
t = 15/4
(2)
AB的斜率为k = (4 - 0)/(3 - 6) = -4/3
MN⊥AB时,MN的斜率为-1/k = 3/4
设M(t,0),t∈[0,6],此时N(p,4(6-p)/3)
OM² = AN²
t² = (p - 6)² + 16(6 - p)²/9 = 25(6 - p)²/9 (a)
[4(6-p)/3 - 0]/(p - t) = 3/4 (b)
由(a)(b):t = 9/4 （另一值t = 9>6,舍去)