椭圆M:
|
由题意可知F1(-c,0),F2(c,0),设点P为(x,y)
∵
+x2 a2
=1∴x2=y2 b2 a2 (b2-y2) b2
∴
=(-c-x,-y),PF1
=(c-x,-y)PF2
∴PF1
=x2-c2+y2=•PF2
-c2+y2a2 (b2-y2) b2
=a2-c2-c2y2 b2
当y=0时PF1
取到最大值3c2,即a2-c2=3c2,•PF2
∴a2=4c2∴e=
=c a 1 2
故答案为:1 2
椭圆M:
|
由题意可知F1(-c,0),F2(c,0),设点P为(x,y)
∵
+x2 a2
=1∴x2=y2 b2 a2 (b2-y2) b2
∴
=(-c-x,-y),PF1
=(c-x,-y)PF2
∴PF1
=x2-c2+y2=•PF2
-c2+y2a2 (b2-y2) b2
=a2-c2-c2y2 b2
当y=0时PF1
取到最大值3c2,即a2-c2=3c2,•PF2
∴a2=4c2∴e=
=c a 1 2
故答案为:1 2