Question

抛物线y²=2px（p>0)的焦点作一条直线交抛物线于A（x1,y1）B（x2,y2）两点。求证：y1y2=-p²

Answer 1

焦点(p/2,0)
若垂直x轴，是x=p/2
则y²=p²
y1=-p,y2=p
y1y2=-p²

若有斜率
y=k(x-p/2)
x=y/k+p/2
所以y²=2py/k+p²
y²-2py/k-p²=0
y1y2=-p²

综上
y1y2=-p²

Answer 2

OA向量*OB向量 =x1x1+y1y2

过抛物线y²=2px(p＞0)焦点(p/2 ,0)的直线
y=k(x- p/2)和抛物线方程联立
==> k²(x- p/2)²=2px

k²x²-(k²+2p)x+k²p²/4=0
==>x1x2 =p²/4 ,和抛物线方程联立 ==>y1y2=- p²
==>OA向量*OB向量 =x1x1+y1y2
= -3p²/4