设直线AB解析式为y=kx+b, 将点A(0,6),B(6,0)代入,得

解得

则直线AB解析式为y=-x+6, 设Pt,-t2

+2t+6其中0

∴PN=PM-MN=-t2+2t+6-(-t+6)=-t2+2t+6+t-6=-t2+3t, ∴S△PAB=S△PAN+S△PBN =PN·AG+PN·BM =PN·(AG+BM) =PN·OB =×-t2+3t×6 =-t2+9t =-(t-3)2+,

∴当t=3时,△PAB的面积有最大值;

(3)如图2,

图2

∵PH⊥OB于点H, ∴∠DHB=∠AOB=90°, ∴DH∥AO, ∵OA=OB=6,

∴∠BDH=∠BAO=45°,

13

∵PE∥x轴、PD⊥x轴, ∴∠DPE=90°,

若△PDE为等腰直角三角形, 则∠EDP=45°,

∴∠EDP与∠BDH互为对顶角,即点E与点A重合,

则当y=6时,-x2

+2x+6=6, 解得x=0(舍去)或x=4, 即点P(4,6).

14