连接PD,设DQ⊥AP垂足为Q
∵ABCD为矩形,∴∠B=90°,AD=BC,CD=AB
∵AB=2,AP=x,∴BP=√(AP²-AB²)=√(x²-4),CD=2
∵AD=3,∴BC=3,∴PC=BC-BP=3-√(x²-4)
∵∠B=90°,∴S△ABP=1/2·AB·BP=√(x²-4)
∵∠C=90°,∴S△DPC=1/2·CD·CP=3-√(x²-4)
∵DQ⊥AP,∴S△APD=1/2·AP·DQ=1/2xy
∵ABCD面积=S△ABP+S△DPC+S△APD=AD·AB
∴√(x²-4)+3-√(x²-4)+1/2xy=6 化简后得到：y=6/x