解:(1)解析:∵D为函数y=k/x(k>0)上一点,O(0,0),C(4,0),△ODC是以CO为斜边的等腰直角形设D(x,y)|OD|=√(x^2+y^2),|OC|=√[(x-4)^2+y^2)∴x^2+y^2= (x-4)^2+y^2==>x=2|OD|=4/√2=2√2==>y=2==>y=k/2=2==>k=4 (==>是推得的意思)∴反比例函数y=4/x (2)∵反比例函数y=4/x,AB⊥X轴,BE⊥Y轴,∴B(1,4),A(1,0),E(0,4)∵⊿ABO与⊿A’BO关于直线OB对称∴⊿ABO≌⊿A’BO==>∠OBA=∠OBA’, ∠BOA=∠BOA’延长BA’交X轴于G,设OG=x∴∠O...
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