PQ为过抛物线焦点F的弦,作PQ的垂直平分线交抛物线对称轴于R点,求证|FR|=1/2|PQ|

我们仅举y²=2px的情形,此处p>0焦点F（p/2,0）设PQ方程：x=my+p/2代入抛物线y²=2pxy²-2pmy-p²=0韦达定理：y1+y2=2pm,y1×y2=-p²x1+x2=y1²/2p+y2²/2p=(y1²+y2²)/2p=[(y1+y2)²-2y1y2]/2p=(4p²m²+2p²)/2p=2pm²+p所以PQ中点坐标（pm²+p/2,pm）弦长公式：｜PQ｜=√(1+m²)[(y1+y2)²-4y1y2]=√(1+m²)(4p²m²+4p²)=2p(1+m²)与直线PQ垂直的直线斜率为-m所以PQ的垂直平分线的方程：y-pm=-m(x-pm²-p/2)令y=0,则x=pm²+3p/2｜FR｜=pm²+3p/2-p/2=pm²+p=p(1+m²)由此1/2｜PQ｜=｜FR｜