Question

如图，已知A，B两点是反比例函数y=2/x（x＞0）的图象上任意两点，过A，B两点分别作y轴的垂线，垂足分别为C

如图，已知A，B两点是反比例函数y=2/x（x＞0）的图象上任意两点，过A，B两点分别作y轴的垂线，垂足分别为C，D，连接AB，AO，BO，求梯形ABCD的面积与△AOB面积之比

Answer 1

设A(a, 2/a), B(b, 2/b), 则C(0, 2/a), D(0, 2/b)
CA = 2/a, DB = 2/b
CD = |a - b|
梯形ABCD的面积S1 = (1/2)(CA + DB)*CD = (1/2)(2/a + 2/b)|a-b| = (a+b)|a-b|/ab
AB的方程为：(y - 2/a)/(x - a) = (2/b - 2/a)/(b - a) = -2/ab
2x + aby -2(a+b) = 0
O到AB的距离为d = |2*0 + ab*0 - 2(a+b)|/√[2² + (ab)²] = 2(a+b)/√(4 + a²b²)
AB = √[(a - b)² + (2/a - 2/b)²]
= (|a-b|/ab)√(4 + a²b²)
△AOB面积S2 = (1/2)*AB*d = (1/2)(|a-b|/ab)√(4 + a²b²)*2(a+b)/√(4 + a²b²)
= (a+b)|a-b|/(ab)
S1 = S2