问题
解答题
已知椭圆C:
(Ⅰ)求椭圆C的方程; (Ⅱ)若过点M(2,0)的直线与椭圆C相交于A,B两点,设P为椭圆上一点,且满足
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答案
(Ⅰ)由题意知e=
=c a
,所以e2=2 2
=c2 a2
=a2-b2 a2
.1 2
即a2=2b2.(2分)
又因为b=
=1,所以a2=2,S△OBC•2 1+1
+S△OCA•OA
+S△OBA•OB
=OC
.0
故椭圆C的方程为
+y2=1.(4分)x2 2
(Ⅱ)由题意知直线AB的斜率存在.设AB:y=k(x-2),A(x1,y1),B(x2,y2),P(x,y),
由
得(1+2k2)x2-8k2x+8k2-2=0.△=64k4-4(2k2+1)(8k2-2)>0,k2<y=k(x-2)
+y2=1.x2 2
.(6分)1 2
x1+x2=
,x1•x2=8k2 1+2k2
∵8k2-2 1+2k2
+OA
=tOB
∴(x1+x2,y1+y2)=t(x,y),OP
∴x=
=x1+x2 t
,y=8k2 t(1+2k2)
=yy+y2 t
[k(x1+x2)-4k]=1 t -4k t(1+2k2)
∵点P在椭圆上,∴
+2(8k2)2 t2(1+2k2)2
=2,∴16k2=t2(1+2k2).(8分)(-4k)2 t2(1+2k2)2
∵|
-PA
|<PB
,∴2 5 3
|x1-x2|<1+k2
,∴(1+k2)[(x1+x2)2-4x1•x2]<2 5 3 20 9
∴(1+k2)[
-4•64k4 (1+2k2)2
]<8k2-2 1+2k2
,∴(4k2-1)(14k2+13)>0,∴k2>20 9
.(10分)1 4
∴
<k2<1 4
,∵16k2=t2(1+2k2),∴t2=1 2
=8-16k2 1+2k2
,8 1+2k2
∴-2<t<-
或2 6 3
<t<2,∴实数t取值范围为(-2,-2 6 3
)∪(2 6 3
,2).(12分)2 6 3