Question

已知AB是抛物线y2＝2px的任意一条焦点弦，且A（x1，y1）B（x2，y2）求证：X1X2=P²／4 y1y2＝﹣P

Answer 1

解：焦点坐标为(p/2，0)
∵有两个焦点
∴弦不平行于x轴
则设焦点弦所在直线为x=ky+m
则p/2=0+m，即m=p/2
交x=ky+m代入抛物线方程
得：y²=2p(ky+m)
即y²-2pky-2pm=0
则y1y2=-2pm=-2p×p/2=-p²
y1+y2=2pk
x1x2=(ky1+m)(ky2+m)
=k²y1y2+m²+km(y1+y2)
=-k²p²+p²/4+kp/2×2kp
=-k²p²+p²/4+k²p²
=p²/4
证毕

Answer 2

焦点F(p/2,0) 设斜率为k
y=k(x-p/2) y^2=k^2(x^2-px+p^2/4)
y^2=2px
x^2-(p+2p/k^2)x+p^2/4=0
x1x2=p^2/4
y=k(x-p/2) x=y/k+p/2
y^2=2px
y^2=2py/k+p^2
y^2-2py/k-p^2=0
y1y2=-p^2