Question

点A,B在反比例函数y =k/x的图象上，且A.B的横坐标分别为a,2a(a>o),AC垂直x轴于c，且三角形AOC的面积为2

求△AOB的面积

Answer 1

解：点A(a，k/a) ，点B(2a，k/2a)，AC⊥ x轴于点C，所以C(a，0)，SAOC = (1/2) * a * |(k/a)| = |k|/2 = 2 => |k| = 4 => k = ±4

当k = 4时，点A(a，4/a) ，点B(2a，2/a)，过点B作BD垂直于x轴，D是垂足，所以SAOB = SOABD – SOBD = SAOC + SABDC – SOBD = 2 + (1/2)*(4/a + 2/a)*a – (1/2)*2a*(2/a) = 2 + 3 – 2 = 3 。(当k = -4 时也一样)