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
如果椭圆
+=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
数学人气:559 ℃时间:2019-10-05 13:57:52
优质解答
设这条弦的两端点为A(x
1,y
1),B(x
2,y
2),斜率为k,
则
,
两式相减再变形得
+k=0又弦中点为(4,2),故k=
−,
故这条弦所在的直线方程y-2=
−(x-4),整理得x+2y-8=0;
故选D.
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