设抛物线y2=2px(p>0)的准线为l:x=- p 2 .如图所示,分别过点A,B作AM⊥l,BN⊥l,垂足为M,N.过点B作BC⊥AM交于点C.则|AM|=|AF|,|BN|=|BF|.∵AM∥x轴,∴∠BAC=∠AFx=60°.在Rt△ABC中,|AC|= 1 2 |AB|.又|AM|-|BN|=|AC|,∴|AF|?|BF|= 1 2 (|AF|+|BF|),化为 |AF| |BF| =3.故选:C.