设抛物线方程为y^2=2px(p>0),过(p/2,0)引直线L交抛物线于A(x1,y1),B(x2,y2),求证x1x2,y1y2为定值

Question

如题所述

Answer 1

当L⊥x轴时
x1=x2=p/2
x1*x2=p^2/4
y1=p,y2=-p
y1*y2=-p^2
当L不垂直x轴时
设y=k(x-p/2)
代入y^2=2px
得
k^2x^2-(pk^2+2p)x+k^2p^2/4=0
韦达定理
x1x2=k^2p^2/4/k^2=p^2/4
y1y2
=k^2(x1x2-(x1+x2)*p/2+p^2/4)
=-p^2

x1x2,y1y2为定值
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Answer 2

设过(P/2，0)的直线为：y-0=k(x-p/2) y=kx-kp/2 代入y^2=2px得 (kx)^2-k^2px+(pk/2)^2=2px
(kx)^2-(pk^2+2p)x+(pk/2)^2=0
x1x2=(pk/2)^2/k^2=p^2/4 x1+x2=(pk^2+2p)/k^2
y1y2=k(x1-p/2)k(x2-p/2)=k^2[x1x2-(x1+x2)p/2+p^2/4]=-p^2