现行高二《平面解析几何》第101页有一道典型习题:过抛物线y2=2px的焦点的一条直线和抛物线相交,两交点的纵坐标分别为y1,y2,求证:y1y2=-p2. 分析设过抛物线的焦点F(p/2,0)的直线方程为x=my+p/2.将上式代入方程y2=2px,化简y2=2pmy-p2=0.因为y1,y2是该方程的两根,所以y1y2=-p2,这里之所以将过F(p/2,0)的直线方程设为x=mx+p/2,而不 The current high school “flat analytic geometry” on page 101 has a typical problem: a line passing through the focus of the parabola y2 = 2px intersects with a parabola. The ordinates of the two intersections are y1, y2, respectively. Proof: y1y2 = -p2. The linear equation of the parabolic focus F(p/2,0) is x=my+p/2. Substitute the above equation into the equation y2=2px, simplify y2=2pmy-p2=0. Since y1,y2 is the equation Two, so y1y2=-p2, the reason why the line equation passing F(p/2,0) is x=mx+p/2 instead of