(-c+my)2 |
a2 |
y2 |
b2 |
整理得:(b2m2+a2)y2-2mcb2y-b4=0,
设A(x1,y1),B(x2,y2),则y1,y2为上述方程的两个根,
∴y1+y2=
2mcb2 |
b2m2+a2 |
b4 |
b2m2+a2 |
∵OA⊥OB,
∴(-c+my1)(-c+my2)+y1y2=0.
∴c2-mc(y1+y2)+(m2+1)y1y2=0,将①代入,整理得:
a2c2-(c2b2+b4)m2-b4=0,
∴(c2b2+b4)m2=a2c2-b4≥0,
∴a2c2≥(a2-c2)2,又e=
c |
a |
∴e4-3e2+1≤0,
∴
3-
| ||
2 |
3+
| ||
2 |
∴
3-
| ||
2 |
∴
| ||
2 |
故选D.