设|PA|=a,
∠APB=2∠APO,
sin∠APO=|OA|/|PA|=1/√(a²+1)
cos∠APB=1-2sin^2∠APO=(a²-1)/(a²+1)
|PA×PB|=|PA|·|PB|cos∠APB=a²(a²-1)/(a²+1)
=(a²+1)+2/(a²+1)-3
》-3+2√2
选D
设|PA|=a,
∠APB=2∠APO,
sin∠APO=|OA|/|PA|=1/√(a²+1)
cos∠APB=1-2sin^2∠APO=(a²-1)/(a²+1)
|PA×PB|=|PA|·|PB|cos∠APB=a²(a²-1)/(a²+1)
=(a²+1)+2/(a²+1)-3
》-3+2√2
选D