题目内容
已知椭圆
+
=1(a>b>0)的离心率为
,直线y=
x+1与椭圆相交于A、B两点,点M在椭圆上,且满足
=
+
.则b=______.
| x2 |
| a2 |
| y2 |
| b2 |
| ||
| 2 |
| 1 |
| 2 |
| OM |
| 1 |
| 2 |
| OA |
| ||
| 2 |
| OB |
∵由e=
∴a=2b;
设椭圆方程为
+y2=b2
将直线方程与椭圆方程联立得
消去y得:x2+2x+2-2b2=0
则x1=-1+
,x2=-1-
,
=
+
=(
+
,
+
)
∴xM=
+
=-
+(1-
)
yM=
+
=
+
∵M在椭圆上,
代入椭圆方程得xM2+(1+
)xM+1+
-2b2=0
求得b2=1,b=1
故答案为:1
| ||
| 2 |
∴a=2b;
设椭圆方程为
| x2 |
| 4 |
将直线方程与椭圆方程联立得
消去y得:x2+2x+2-2b2=0
则x1=-1+
| 2b2-1 |
| 2b2-1 |
| OM |
| 1 |
| 2 |
| OA |
| ||
| 2 |
| OB |
| x1 |
| 2 |
| ||
| 2 |
| y1 |
| 2 |
| ||
| 2 |
∴xM=
| x1 |
| 2 |
| ||
| 2 |
1+
| ||
| 2 |
| 3 |
| ||
| 2 |
yM=
| ||
| 2 |
| ||||
| 2 |
1+
| ||
| 2 |
| xM |
| 2 |
∵M在椭圆上,
代入椭圆方程得xM2+(1+
| 3 |
| ||
| 2 |
求得b2=1,b=1
故答案为:1
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