本刊93年第2期《一个有用的截距不等式》一文,用一个不等式解决了一类涉及圆锥曲线上两点成轴对称的高考题,确实使人耳目一新。本文试图从此两点及它的对称轴所在的直线方程出发,来解决此类问题,同样显得简捷明快。设椭圆C:x~2/a~2+y~2/b~2=1上存在不同两点A、B,若AB中点为M(m,n),则C关于M对称的曲线C′的方程为:(x-2m)~2/a~2+(y-2n)~2/b~2=1。显然,AB是C与C′的公共弦,C-C′得AB所在直线方程为: b~2mx+a~2ny-b~2m~2-a~2n~2=0 (Ⅰ)而线段AB的垂直平分线,即A、B两点的对称轴方程: The article “A Useful Intercept Inequality” in the second issue of the journal, 93th Anniversary, uses an inequality to solve a class of college entrance examination questions involving axisymmetric symmetry on two points on a conic curve, which is truly refreshing. This article attempts to solve this problem from the two points and the equation of the line where its axis of symmetry is located. It also looks simple and clear. Let there be two different points A and B on the ellipse C:x~2/a~2+y~2/b~2=1. If the AB midpoint is M(m,n), then the curve C about C symmetry The equation for ’ is: (x-2m)~2/a~2+(y-2n)~2/b~2=1. Obviously, AB is the common chord of C and C’. The linear equation for CC’s AB is: b~2mx+a~2ny-b~2m~2-a~2n~2=0 (I) and the vertical line AB The bisector, that is, the symmetry axis equations at points A and B: