第1个回答 2020-02-10
F(p/2,0),
设AB:x=my+p/2,
代入y^2=2px,得y^2-2mpy-p^2=0,
△=4p^2(m^2+1),
设A(x1,y1),B(x2,y2),,
由|AB|=3|FB|,得|AF|=2|FB|,|y1|=2|y2|,
不妨设y1<0<y2,则y1=-2y2,于是
y1+y2=-y2=2mp,①
y1y2=-2y2^2=-p^2②
②/①,得2y2=-p/(2m),y2=-p/(4m),
代入①,约去p,得1/(4m)=2m,m^2=1/8,m=-√2/4,y2=p/√2,
|AB|=√[△(1+m^2)=2p(1+m^2)=9p/4,
S=(1/2)(p/2)(y2-y1)=3py2/4=3p^2/(4√2),
由|AB|=3S/√2得9p/4=9p^2/8,p>0,
所以p=2.选D.本回答被提问者和网友采纳