抛物线y2=2px与直线ax+y-4=0交于两点A、B,且点A的坐标是(1,2),设抛物线的焦点为F,则|FA|+|FB|等于( )
A.7 B.3
C.6 D.5
抛物线y2=2px与直线ax+y-4=0交于两点A、B,且点A的坐标是(1,2),设抛物线的焦点为F,则|FA|+|FB|等于( )
A.7 B.3 C.6 D.5
A
由题意可求得抛物线方程为y2=4x,直线方程为2x+y-4=0.
由
得x2-5x+4=0.
设此方程的两根为x1、x2,则x1+x2=5,|FA|+|FB|=(x1+1)+(x2+1)=(x1+x2)+2=5+2=7.