Question

已知椭圆x^2/a^2+y^2/b^2=1(a>b>0)的左右焦点分别为F1,F2,P为椭圆M上任意一点,

Answer 1

设P(x,y),则x^2/a^2+y^2/b^2=1(a>b>0)

F1(-c,0),F2(c,0)

PF1=(-c-x,-y),PF2(c-x,-y)

PF1●PF2

=x^2-c^2+y^2

当x^2+y^2=b^2时,PF1●PF2取得最小值b^2-c^2

∴2c^2≤b^2-c^2≤3c^2

即3c^2≤a^2-c^2≤4c^2

4c^2≤a^2≤5c^2

∴1/5≤e^2≤1/4

√5/5≤e≤1/2