過拋物線y∧2＝2px(p＞0)的焦點(diǎn)F的直線與拋物線相交于P,Q兩點(diǎn),線段PQ的中垂線交拋物線對(duì)稱軸于R,求‖PQ‖＝2‖F(xiàn)R‖

數(shù)學(xué)人氣：381 ℃時(shí)間：2019-11-04 23:12:17

優(yōu)質(zhì)解答

設(shè)P點(diǎn)坐標(biāo)（x1,y1）Q（x2,y2）
由拋物線且PQ過焦點(diǎn)F得‖PQ‖＝‖PF‖+‖QF‖=x1+p/2+x2+p/2=x1+x2+p
PQ的斜率為k=（y2-y1）/(x2-x1)=(y2-y1)/[(y2^2-y1^2)/2p]=2p/(y1+y2)
∴PQ中垂線斜率為-1/k=-(y1+y2)/2p
PQ中垂線方程為y-（y1+y2)/2=-(y1+y2)/2p[x-(x1+x2)/2]
交對(duì)稱軸x軸的交點(diǎn)坐標(biāo)R為令y=0 解得x=（x1+x2）/2+p
‖F(xiàn)R‖=（x1+x2）/2+p-p/2=(x1+x2+p)/2
故‖PQ‖=2‖F(xiàn)R‖