过抛物线y∧2=2px(p>0)的焦点F的直线与抛物线相交于P,Q两点,线段PQ的中垂线交抛物线对称轴于R,求‖PQ‖=2‖FR‖
人气:227 ℃ 时间:2019-11-04 23:12:17
解答
设P点坐标(x1,y1)Q(x2,y2)
由抛物线且PQ过焦点F得‖PQ‖=‖PF‖+‖QF‖=x1+p/2+x2+p/2=x1+x2+p
PQ的斜率为k=(y2-y1)/(x2-x1)=(y2-y1)/[(y2^2-y1^2)/2p]=2p/(y1+y2)
∴PQ中垂线斜率为-1/k=-(y1+y2)/2p
PQ中垂线方程为y-(y1+y2)/2=-(y1+y2)/2p[x-(x1+x2)/2]
交对称轴x轴的交点坐标R为 令y=0 解得x=(x1+x2)/2+p
‖FR‖=(x1+x2)/2+p-p/2=(x1+x2+p)/2
故‖PQ‖=2‖FR‖
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