解(1)抛物线Y=1/2X²的焦点F坐标为(0,1/2)准线y=-1/2
由题意知,直线L的斜率存在,故设其方程为y=kx+1/2代入抛物线Y=1/2X²得
X²-2kx-1=0
⊿=4k²+4>0恒成立,设A(x1,1/2 x1²),B(x2,1/2 x2²),则x1+x2=2k,x1x2=-1
由y′=x得k1=x1,k2=x2,所以k1k2=x1x2=-1,即PA与PB垂直.
又PA、PB的方程分别为y-1/2 x1²=x1(x-x1),y-1/2 x2²=x2(x-x2)两式解得x=k=(x1+x2)/2
代入PA方程得y-1/2 x1²=x1[(x1+x2)/2-x1]=-1/2 x1x2=-1/2,所以点P在准线L上
(2)存在m=1满足条件.
FA●FB=mFP得,
(x1,1/2 x1²-1/2)●(x2,1/2 x2²-1/2)=m(K,-1/4)
又k=(x1+x2)/2,x1x2=-1,化简得1/m²=m,∴m=1