抛物线y=x²与过点P(-1,-1)的直线l交于P1、P2两点.求在线段P1P2上满足条件1/PP1+1/PP2=2/PQ的点Q的轨迹方程.看不懂题目的见图http://hiphotos.baidu.com/%D1%A9%C9%BD%BD%A3%BF%CDty/pic/item/14ba2dea3d6d55fb6eeef7936d224

抛物线y=x²与过点P(-1,-1)的直线l交于P1、P2两点.求在线段P1P2上满足条件1/PP1+1/PP2=2/PQ的点Q的轨迹方程.看不懂题目的见图http://hiphotos.baidu.com/%D1%A9%C9%BD%BD%A3%BF%CDty/pic/item/14ba2dea3d6d55fb6eeef7936d224
抛物线y=x²与过点P(-1,-1)的直线l交于P1、P2两点.求在线段P1P2上满足条件1/PP1+1/PP2=2/PQ的点Q的轨迹方程.
看不懂题目的见图http://hiphotos.baidu.com/%D1%A9%C9%BD%BD%A3%BF%CDty/pic/item/14ba2dea3d6d55fb6eeef7936d224f4a20a4dd09.jpg

抛物线y=x²与过点P(-1,-1)的直线l交于P1、P2两点.求在线段P1P2上满足条件1/PP1+1/PP2=2/PQ的点Q的轨迹方程.看不懂题目的见图http://hiphotos.baidu.com/%D1%A9%C9%BD%BD%A3%BF%CDty/pic/item/14ba2dea3d6d55fb6eeef7936d224
设直线l方程为y=k(x+1)-1,代入y=x²得
x²-kx-(k-1)=0
显然需判别式△=k²+4(k-1)≥0,得k≥-2+2√2或k≤-2-2√2
设P1(x1,x1²),P2(x2,x2²),Q(x,y)
根据韦达定理有：
x1+x2=k ①
x1*x2=1-k ②
根据条件1/PP1+1/PP2=2/PQ可得：
1/(x1+1)+1/(x2+1)=2/(x+1) ③
1/(x1²+1)+1/(x2²+1)=2/(y+1) ④
将①②代入③④得：
2/(x+1) =1/(x1+1)+1/(x2+1)=(x1+x2+2)/(x1x2+x1+x2+1)=(k+2)/(1-k+k+1)
解得x=4/(2+k)-1
于是k=4/(x+1)-2 ⑤
2/(y+1) =1/(x1²+1)+1/(x2²+1)=[(x1+x2)²-2x1x2+2]/[(x1x2)²+(x1+x2)²-2x1x2+1]
=[k²-2+2k+2]/[(1-k)²+k²-2+2k+1]=(k²+2k)/(2k²)=(k+2)/(2k)
解得y=4k/(k+2)-1 ⑥
将⑤代入⑥,化简得
2x+y-1=0(-√2-1