Question

设直线l与抛物线y^2=16x相交所得的弦ab被点m(3,2)所平分,则直线l的方程是

Answer 1

yA+yB=2yM=2*2=4 yA^2=16xA.(1) yB^2=16xB.(2) (1)-(2):yA^2-yB^2=16xA-16xB (yA+yB)*(yA-yB)=16(xA-xB) (yA-yB)/(xA-xB)=16/(yA+yB) k(AB)=16/4=4 直线l的方程是:y-2=4*(x-3) y=4x-10