Question

设抛物线y^2=2px的焦点为f,经过点f的直线与抛物线交于a、b两点,又m是其准线上一点,试证:直线ma、mf、mb

Answer 1

F(p/2,0),设AB直线方程为：y=k(x-p/2),代入抛物线方程,k^2*(x-p/2)^2=2px,k^2*x^2-p(k^2+1)x+p^2/4=0,

解得：x1=[p(k^2+1)+2p√(k^2+1)]/(2k^2),

x2=[p(k^2+1)-2p√(k^2+1)]/(2k^2),

再代入直线方程求得：y1=p(1+√(k^2+1))/k, y2=p(1-√(k^2+1))/k

即A(x1,y1),B(x2,y2)

设M(-p/2,y3),则MF的斜率kmf=-y3/p,

MA的斜率kma=(y3-y1)/(-p/2-x1)=(kp(1+√(k^2+1)-k^2*y3)/[p(k^2+1+√(k^2+1))],

MB的斜率kmb=(y3-y2)/(-p/2-x2)= (kp(1-√(k^2+1)-k^2*y3)/[p(k^2+1-√(k^2+1))],

kma+kmb=-2*y3/p=2*kmf,

所以,MA,MF,MB斜率成等差数列