题目内容
过抛物线y2=2x的焦点F作直线交抛物线于A,B两点,若|AB|=,|AF|<|BF|,则|AF|为( )
A. | B. | C. | D. |
B
设过抛物线焦点的直线为,联立得整理得k2x2-(k2+2)x+k2=0,
x1+x2=,x1x2=.
|AB|=x1+x2+1=+1=,得k2=24,
代入k2x2-(k2+2)x+k2=0得12x2-13x+3=0,
解得x1=,x2=.又|AF|<|BF|,
故|AF|=x1+=.选B
x1+x2=,x1x2=.
|AB|=x1+x2+1=+1=,得k2=24,
代入k2x2-(k2+2)x+k2=0得12x2-13x+3=0,
解得x1=,x2=.又|AF|<|BF|,
故|AF|=x1+=.选B
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