Question

一道解析几何求解

一个椭圆x^2/2+y^2=1，如图，设A,B是椭圆上位于x轴上方的两点，且直线AF1与直线BF2平行，AF2与BF1交于点P
求证：PF1+PF2是定值

Answer 1

设A点横坐标为x1,B点横坐标为x2
过A作AM垂直x轴于M,过B作BN垂直x轴于N
AF1/BF2=F1M/F2N=(x1+c)/(x2-c)
PF1/BF1=AF1/(AF1+BF2)=(x1+c)/[(x1+c)+(x2-c)]=(x1+c)/(x1+x2)
PF1=[(x1+c)/(x1+x2)]*BF1
PF2/AF2=BF2/(AF1+BF2)=(x2-c)/[(x1+c)+(x2-c)]=(x2-c)/(x1+x2)
PF2=[(x2-c)/(x1+x2)]*AF2
PF1+PF2=[(x1+c)/(x1+x2)]*BF1+[(x2-c)/(x1+x2)]*AF2
=[(x1+c)/(x1+x2)]*(a+ex2)+[(x2-c)/(x1+x2)]*(a-ex1)
=[(x1+c)*(a+ex2)+(x2-c)(a-ex1)]/(x1+x2)
=(ax1+cex2+ax2+cex1)/(x1+x2)
=a+ce
=根2+1*(根2)/2
=(3根2)/2

Answer 2

某个高考题，注意用平行得到的比例线段把PF1和PF2转换成焦半径，然后设一个斜率k，把A和B的坐标表示出来，套焦半径公式就可以得到一个定值，记得是3√3/2