设AB:x=ty+2
y^2=4x 与x=ty+2联立消去x得：
y^2=4ty+8==>y^2-4ty-8=0
设A(x1,y1),B(x2,y2)
则y1+y2=4t,y1y2=-8
∴ (y1-y2)^2=(y1+y2)^2-4y1y2
=16t^2+32 ,|y1-y2|=√(16t^2+32)
∴SΔAOB=SΔAOM+SΔBOM
=1/2*|0M|*(|y1|+|y2|)
=1/2*2*|y1-y2|
=|y1-y2|
=√(16t^2+32)≥4√2
t=0时,取等号,此时AB⊥x轴
∴三角形OAB面积的最小值为4√2