Question

抛物线Y^2=2PX中,一直线过（P,0）点,与抛物线交于A B两点.能推出什么结论?

Answer 1

设直线为 y=k(x-p)

代入k^2(x-p)^2=2px

即k^2x^2-(2k^2p+2p)x+k^2p^2=0

韦达定理得x1+x2=(2k^2p+2p)/k^2,x1x2=p^2

y1y2=-2p^2

向量OA·向量OB=x1x2+y1y2=p^2-2p^2=-p^2.(为定值）