k=√2∵向量AF=3向量FB∴│AF│=3│BF│分别过点A,B作AC,BD垂直于准线设│BF│=a,∴│AF│=3a∴│BD│=a/e,│AC│=3a/e过点B作BG垂直于AC∴AG=3a/e-a/e=2a/e∴cos∠GAB=│AG│/│AB│=2a/e/4a=1/2e=√3/3∴tan∠GAB=...请问"过点B作BG垂直于AC∴AG=3a/e-a/e=2a/e"是如何得出的？AG=AC-GC=AC-BD=3a/e-a/e