由双曲线第二定义AF=e(x1-a²/c) BF=e(x2-a²/c),然后利用倾斜角为60°,分别过A、B作x轴的垂线那么可得AF=2(x1-c) , BF=2(c-x2),又题目中说AF=3FB,则x1+3x2=4c由AF=e(x1-a²/c)=2(x1-c)可得x1=a(2c-a)/(2a-c),同理x2=a(a+2c)/(2a+c)然后将其代入x1+3x2=4c中有(2c-a)(c+2a)+3(2c+a)(2a-c)=4c(4a²-c²)/a,整理可得4a²-4c²+12ac=4c(4a²-c²)/a,将其两边同时除以a²可得4-4e²+12e=4e(4-e²)即e³-e²-e+1=0 则(e²-1)(e-1)=0, e=1 不对 换种方法:设BF=x则AF=3xAD=3x/e BC=x/e则AE=2x/e由倾斜角为60°知AB=2AE故有AB=AF+BF=4x=2AE=2*2x/e则e=1是不是题目有问题我怎么2种方法算出来都是1????????? 图:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/f703738da97739120f16fc78f8198618367ae2f3?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)