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在讲授椭圆这部分内容时,我曾给学生出了这样一道题目:“过点P(2,1)作直线与椭圆x2/16+y2/4=1交于A、B两点,若点P平分弦AB,求弦AB所在的直线方程.”学生很快就想出了两种解法:一种是设弦AB所在的直线方程为y-1=k(x-2),然后将直线方程代入椭圆方程来解题;另一种是用两点法. 这时,有一个学生举手,说自己还有第三种解法,她的解法如下: 如图1,设A(x,y),因为点P平分弦AB,所以B点坐标为(4-x,2-y). 因为A、B两点在椭圆x2+4y2=16上,
In teaching this part of the ellipse, I gave this question to the students: “Across the point P(2,1), make a straight line and ellipse x2/16+y2/4=1 to cross at two points A and B, if P divides the string AB and finds the equation of the line where the string AB is located.” The student quickly came up with two solutions: one is to set the line AB where the equation of the line is y-1=k(x-2), and then the straight line The equation is substituted into the elliptic equation to solve the problem; the other is the two-point method. At this time, a student raises his hand and says that he still has a third solution. Her solution is as follows: As shown in Figure 1, let A (x, y) ), because the point P divides the chord AB, so the coordinates of the B point are (4-x, 2-y). Because A and B are two points on the ellipse x2+4y2=16,