Question

设抛物线y2=16x的焦点为F,过点Q(-4,0)的直线l与抛物线交于AB两点,若QA=2QB,求直线l斜率

Answer 1

抛物线y^2=16x的焦点为(4,0),准线为x=-4.

作AA'、BB'垂直准线,垂足为A'、B'.

则题意知,AA'=2BB',QA'=2QB'.

即yA=2yB,yA^2/16+4=2(yB^2/16+4),解得：yA=-8、yB=-4或yA=8、yB=4.

而直线l的斜率=(yB-yA)/(yB^2/16-yA^2/16)=16/(yA+yB)=-4/3或4/3.