问题
解答题
设F1(-1,0),F2(1,0),动点M满足|MF1|+|MF2|=2
(1)求M的轨迹C的方程; (2)设直线l:y=
|
答案
(1)设动点M(x,y),
∵F1(-1,0),F2(1,0),∴|MF1|+|MF2|=2
>2=|F1F2|,2
则M的轨迹为以F1,F2为焦点,以2
为长轴的椭圆,2
则a=
,c=1,b2=a2-c2=1.2
方程为:
+y2=1;x2 2
(2)联立
,得9x2-4x-12=0.y=
(x-1)7 7
+y2=1x2 2
设A(x1,y1),B(x2,y2),
则x1+x2=
,x1x2=-4 9
.12 9
=(x1+1,y1),F1A
=(x2+1,y2),F1B
∴
•F1A
=(x1+1,y1)•(x2+1,y2)F1B
=(x1+1)(x2+1)+y1y2=
x1x2+8 7
(x1+x2)+6 7 8 7
=
×(-8 7
)+12 9
×6 7
+4 9
=0.8 7