|a|＝2,|b|＝1,a•b＝√3×(1/2)－1×(√3/2)＝0∴a⊥b由x⊥y得[a＋(t²－3)b]•(－ka＋tb)＝0=>－k|a|²＋(t³－3t)|b|²＝0从而k＝(t³－3t)/4∴(k＋t²)/t＝1/4(t＋2)²－...