过抛物线y2=4x的焦点F的直线交该抛物线于A,B两点,O为坐标原点.若|AF|=3,△AOB面积.解析：∵抛物线y^2=4x∴其焦点F(1,0)∵过F直线交该抛物线于A,B两点,|AF|=3∴|AF|=x(A)+p/2=3==>x(A)=3-1=2代入抛物线y^2=8==>y1=-2...