直线AB的方程为ykxm与W的方程联立消去y得
1kx
2
2
2kmxm220
故x1x2
2kmm22x1x22所以1k2k1
OAx1x2y1y2OB
x1x2kx1mkx2m1k2x1x2kmx1x2m2
1km
222
2
2
k11k2k24222k1k1
2km
2
2
2
m2
又因为x1x20所以k210从而OAOB2

f综上当AB⊥x轴时OA取得最小值2OB解法二Ⅰ同解法一Ⅱ设AB的坐标分别为x1y1x2y2

xi2yi2xiyixiyi2i12
令sixiyitixiyii12则siti2且si0ti0i12所以
OAx1x2y1y2OB11s1t1s2t2s1t1s2t24411s1t1s2t222s1t1s2t22
当且仅当s1t1s2t2即
x1x2时