题目 #17137

2 (1)如图, 延长BD交x轴于D点, ∵点A是反比例函数\(y_{1}\)= _{x}图象上一点，过点A作y轴的平行线, k 1 交函数\(y_{2}\)= _{x}的图象于点B, 且k > 0所以S△OBD= _{2}k, S△OAD= 1 ∵S△AOB= 3, S△AOB= S△OBD − 1 S△OAD= _{2}k −1 = 3,解得k = 8故k的值为8； 2 1 (2)如图, 过点C作CE ⊥BA, ∵点A的横坐标为4,点A是反比例函数\(y_{1}\)= _{x}图象上一点，∴A(4, _{2}) ∵ 8 1 2 2 1 BA平行于y轴所以点B的横坐标为4,\(y_{2}\)= _{x}, ∴B(4,2) ∴\(y_{OB}\)= _{2}x ∵\(y_{1}\)= _{x}∴ _{x}= _{2}x(x > 0)解得 , x = 2 1 3 2 ∴正比例函数\(y_{OB}\)的图象与反比例函数\(y_{1}\)= _{x}图象的交点C的坐标为(2,1) ∴CE = 2 ∴BA = 2 − _{2}= _{2}∴ S△ABC= _{2}\(\times ^{1} _{2}\times 2 = _{2}\)所以S△AOC= S△AOB −S△ABC= 3 − _{2}= _{2}故△AOC的面积为 _{2}.^{3} 3 3 3 3