Question

已知：斜率为1的直线l过抛物线y^2=2px(p>0)的焦点F,且与抛物线交于A,B两点

Answer 1

直线为为 y=x-p/2

直接用抛物线第一定义,准线为 x=-p/2

AB = AF+BF = x1+p/2+x2+p/2 = x1+x2+p

AB=4,所以 x1+x2+p=4

x=y+p/2 带入 y^2=2px,有 y^2=2py+p^2

x1+x2+p = y1+y2+2p = 2p+2p=4p=4; p=1

即 y^2 = 2x