这个简单,
设直线OA为y=kx,(k>0)；设直线OB为y=(-1/k)x.（∵OA⊥OB）
直线OA与抛物线y^2=2x交于O,A两点,将y=kx代入y^2=2x得x=0或x=2/(k^2),由直线OA方程得A点为（2/(k^2),2/k）.
同上,直线OB与抛物线交于O,B两点,将y=(-1/k)x代入y^2=2x得x=0或x=2k^2,由直线OB方程得B点为（2k^2,-2k）.
|OA|=√{[2/(k^2)]^2+[2/k]^2}=(2/k)·√[(1/k^2)+1],
|OB|=√{[2k^2]^2+[-2k]^2}=(2k)·√[k^2+1],
于是
S△OAB=(1/2)·|OA|·|OB|
=2·√{[(1/k^2)+1]·[k^2+1]}
=2·√[k^2+1/(k^2)+2]
=2·√{[k+(1/k)]^2}
=2·[k+(1/k)]≥4,当且仅当k=1/k时,即k=1时取等号.
综上,S△OAB最小值为4.