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x2
9
+
y2
4
=1内一定点(1,0)作弦,则弦中点的轨迹方程为______.
人气:447 ℃ 时间:2019-10-17 04:20:51
解答
设弦两端点坐标为(x1,y1),(x2.y2),诸弦中点坐标为(x,y).弦所在直线斜率为k
x21
9
+
y21
4
=1

x22
9
+
y22
4
=1

两式相减得;
1
9
(x1+x2)(x1-x2)+
1
4
(y1+y2)(y1-y2)=0
2x
9
+
2y
4
k= 0

又∵k=
y
x−1
,代入上式得
2x/9+2y^2/4(x-1)=0
2x
9
+
2y2
4(x−1)
=0

整理得诸弦中点的轨迹方程:4x2+9y2-4x=0
故答案为4x2+9y2-4x=0
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