设过焦点的直线是x=1/2+ky,设S(x1,y1),B(x2,y2),则OA·OB=x1x2+y1y2=(1/2+ky1)(1/2+ky2)+y1y2=1/4+k/2×(y1+y2)+(k^2+1)(y1y2)
联立y^2=2x与x=1/2+ky,得y^2-2ky-1=0,所以y1+y2=2k,y1y2=-1
所以,OA·OB=1/4+k/2×(y1+y2)+(k^2+1)(y1y2)=-3/4
设过焦点的直线是x=1/2+ky,设S(x1,y1),B(x2,y2),则OA·OB=x1x2+y1y2=(1/2+ky1)(1/2+ky2)+y1y2=1/4+k/2×(y1+y2)+(k^2+1)(y1y2)
联立y^2=2x与x=1/2+ky,得y^2-2ky-1=0,所以y1+y2=2k,y1y2=-1
所以,OA·OB=1/4+k/2×(y1+y2)+(k^2+1)(y1y2)=-3/4