Question

已知过抛物y^2=2px的焦点F的直线交抛物线于A(X1,Y1)B（X2,Y2)（1）求证X1X2为定值（2）1/FA+

Answer 1

1)

AB直线方程为：y=k(x-p/2)

代人：y^2=2px得：

k^2(x-p/2)^2=2px

k^2x^2-(pk^2-2p)x+k^2p^2/4=0

x1x2=p^2/4

是定值

2）

FA^2=(x1-p/2)^2+y1^2

=x1^2-px1+p^2/4+2px1

=x1^2+px1+p^2/4

=(x1+p/2)^2

FA=x1+p/2

同样可得：FB=x2+p/2

1/FA+1/FB

=1/(x1+p/2)+1/(x2+p/2)

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

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

=(x1+x2+p)/(p^2/4+p(x1+x2)/2+p^2/4)

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

=2/p

为定值