Question

过抛物线y^2=2px的焦点下的直线与抛物线交于A,B两点,求证1/丨FA丨+1/丨FB丨=2/P

Answer 1

设A(x1,y1)   B(x2,y2)

设直线  y=k(x-p/2) 

将上面式子代入y^2=2px  得

y^2=2p(y/k+p/2)

y^2-(2p/k)y-p^2=0

则y1*y2=-p^2

x1*x2=(y1^2/2p)*(y2^2/2p)

=(y1y2/2p)^2

=p^2/4

由抛物线定义

1/丨FA丨+1/丨FB丨=1/(x1+p/2)+1/(x2+p/2)

=(x1+x2+p)/[x1x2+p/2(x1+x2)+p^2/4]

=(x1+x2+p)/[p/2(x1+x2)+p^2/2]

=(x1+x2+p)/[p/2(x1+x2+p)]

=2/p