抛物线Y^2=2PX中过焦点F的直线与抛物线交于A,B两点,求AF分之一加BF分之一的值答案貌似和P有关.

问题描述:

抛物线Y^2=2PX中过焦点F的直线与抛物线交于A,B两点,求AF分之一加BF分之一的值
答案貌似和P有关.

可设点A(2pa²,2pa),B(2pb²,2pb).(a≠b)又F(p/2,0).由A,F,B三点共线,可得ab=-1/4.又由抛物线定义知,|AF|=2pa²+(p/2).|BF|=2pb²+(p/2).∴(1/|AF|)+(1/|BF|)=[1/(2pa²+(p/2))]+[1/(2pb²+(p/2))]=(2/p)[1/(4a²+1)+1/(4b²+1)]=(2/p)[(4a²+4b²+2)/(16a²b²+4a²+4b²+1)]=(2/p)[(4a²+4b²+2)/(4a²+4b²+2)]=2/p.

设抛物线y²=2px(p>0),焦点坐标为F(p/2,0),A(x1,y1),B(x2,y2),
过点F的直线方程为x=my+(p/2),
代入y²=2px,得y²=2pmy-p²=0,∴y1y2= -p²,
x1x2=(y1²/2p) (y2²/2p)=p²/4.
由抛物线的定义可知,AF=x1+(p/2),BF=x2+(p/2),
∴1/AF+1/BF
=1/[ x1+(p/2)]+1/[ x2+(p/2)]
=(x1+x2+p)/[x1x2+p(x1+x2)/2+(p²/4)] (通分化简)
将x1x2= p²/4,x1+x2=AB-p,代入上式,得
1/AF+1/BF=AB/[(p²/4)+p(AB-p)/2+(p²/4)]=2/p,
即1/AF+1/BF=2/p.