Question

过点M（3,1）做抛物线y^2=4x的弦的中点的轨迹方程是什么?

Answer 1

设A(x1,y1),B(x2,y2)中点N(xo,yo)得x1+x2=2xo,y1+y2=2yo,A,B点在曲线上有y1^2=4x1,y2^2=4x2,相减得y1^2-y2^2=4(x1-x2),于是直线斜率：K=(y1-y2)/(x1-x2)=4/(y1+y2)=4/(2yo)=2/yo,又MN,AB共线又得K=(yo-1)/(xo-3)=2/yo,进而得中点轨迹方程为yo(yo-1)=2(xo-3),当y1+y2=2=2yo时,得yo=1,xo=3.即中点N(3,1).（显然y1+y2=2是不恒成立的,如：过点M且垂直于x轴的直线易知y1+y2=2yo=0）