∵OD⊥AB,kOD=1/2
∴kAB=-2
设直线AB为：y=-2x+b
A(x1,y1),B(x2,y2)
则y1²=2px1,y2²=2px2,
∴(y1y2)²=4p²x1x2,
∵OA⊥OB,kOA•kOB=-1
∴y1/x1• y2/x2= y1y2/ x1x2
=y1y2•y1y2/x1x2•y1y2=
(y1y2)²/ x1x2•y1y2
=4p²x1x2/ x1x2•y1y2
=4p²/ y1y2=-1
∴y1y2=-4p²,x1x2=4p²
将y=-2x+b代入y²=2px
得(-2x+b)²=2px
即4x²-(4b+2p)x+b²=0
∴x1x2= b²/4
∴b²/4=4p²,b²=16p²
∴b=4p(取正值)
直线AB为：y=-2x+4p
将点D(2,1)代入
1=-2×2+4p
∴p=5/4