题目内容
以椭圆
+
=1内的点M(1,1)为中点的弦所在直线方程为______.
| x2 |
| 16 |
| y2 |
| 4 |
设点M(1,1)为中点的弦所在直线与椭圆相交于点A(x1,y1),B(x2,y2).
则
+
=1,
+
=1,
相减得
+
=0,
∵1=
,1=
,kAB=
..
∴
+
=0,解得kAB=-
.
故所求的直线方程为y-1=-
(x-1),化为x+4y-5=0.
故答案为x+4y-5=0.
则
| ||
| 16 |
| ||
| 4 |
| ||
| 16 |
| ||
| 4 |
相减得
| (x1+y1)(x1-y1) |
| 16 |
| (x2+y2)(x2-y2) |
| 4 |
∵1=
| x1+x2 |
| 2 |
| y1+y2 |
| 2 |
| y1-y2 |
| x1-x2 |
∴
| 2 |
| 16 |
| 2kAB |
| 4 |
| 1 |
| 4 |
故所求的直线方程为y-1=-
| 1 |
| 4 |
故答案为x+4y-5=0.
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