如果椭圆x236+y29=1的弦被点(4,2)平分,则这条弦所在的直线方程是(  ) A.x-2y=0 B.x+2y-4=0 C.2x+3y-12=0 D.x+2y-8=0

问题描述:

如果椭圆

x2
36
+
y2
9
=1的弦被点(4,2)平分,则这条弦所在的直线方程是(  )
A. x-2y=0
B. x+2y-4=0
C. 2x+3y-12=0
D. x+2y-8=0

设这条弦的两端点为A(x1,y1),B(x2,y2),斜率为k,

x12
36
+
y12
9
=1
x22
36
+
y22
9
=1

两式相减再变形得
x1+x2
36
+k
y1+y2
9
=0

又弦中点为(4,2),故k=
1
2

故这条弦所在的直线方程y-2=
1
2
(x-4),整理得x+2y-8=0;
故选D.