Question

抛物线y^2=4x有一弦被点M(2,1)所平分,求该弦所在直线l的方程

Answer 1

设被点M(2,1)平分的弦为AB,A(x1,y1),B(x2,y2),AB的斜率为k.

则y1²=4x1,y2²=4x2,y1+y2=2

相减得y1²-y2²=4(x1-x2)

所以(y1+y2)(y1-y2)=4(x1-x2)

k=(y1-y2)/(x1-x2)=4/(y1+y2)=2

故直线l的方程为y-1=2(x-2),即2x-y-3=0.