解:由对称性,不妨设点A(a,0),P(acost,bsint).∵AP⊥OP,即向量AP⊥向量OP.∴向量AP·向量OP=0。∴(acost-a,bsint)·(acost,bsinb)=0.===>(acost)²-a²cost+(bsint)²=0.====>c²cos²t-a²cost+a²-c²=0.===>e²cos²t-cost+1-e²=0.===>(cost-1)[e²cost-(1-e²)]=0.===>cost=(1-e²)/e²<1.===>2e²>1.===>e>√2/2.∴√2/2<e<1.