问题 解答题
已知椭圆C:
x2
a2
+
y2
b2
=1(a>b>0)的离心率为
2
2
,以原点为圆心,椭圆的短半轴长为半径的圆与直线x-y+
2
=0相切.
(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
OA
+
OB
=t
op
(O为坐标原点),当|
PA
-
PB
|<
2
5
3
时,求实数t取值范围.
答案

(Ⅰ)由题意知e=

c
a
=
2
2
,所以e2=
c2
a2
=
a2-b2
a2
=
1
2

即a2=2b2.(2分)

又因为b=

2
1+1
=1,所以a2=2,S△OBC
OA
+S△OCA
OB
+S△OBA
OC
=
0

故椭圆C的方程为

x2
2
+y2=1.(4分)

(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),

y=k(x-2)
x2
2
+y2=1.
得(1+2k2)x2-8k2x+8k2-2=0.△=64k4-4(2k2+1)(8k2-2)>0,k2
1
2
.(6分)

x1+x2=

8k2
1+2k2
x1x2=
8k2-2
1+2k2
OA
+
OB
=t
OP
∴(x1+x2,y1+y2)=t(x,y),

x=

x1+x2
t
=
8k2
t(1+2k2)
y=
yy+y2
t
=
1
t
[k(x1+x2)-4k]=
-4k
t(1+2k2)

∵点P在椭圆上,∴

(8k2)2
t2(1+2k2)2
+2
(-4k)2
t2(1+2k2)2
=2,∴16k2=t2(1+2k2).(8分)

|

PA
-
PB
|<
2
5
3
,∴
1+k2
|x1-x2|<
2
5
3
,∴(1+k2)[(x1+x2)2-4x1x2]<
20
9

(1+k2)[

64k4
(1+2k2)2
-4•
8k2-2
1+2k2
]<
20
9
,∴(4k2-1)(14k2+13)>0,∴k2
1
4
.(10分)

1
4
k2
1
2
,∵16k2=t2(1+2k2),∴t2=
16k2
1+2k2
=8-
8
1+2k2

-2<t<-

2
6
3
2
6
3
<t<2
,∴实数t取值范围为(-2,-
2
6
3
)∪(
2
6
3
,2)
.(12分)

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