解：将y=ax+1代入双曲线方程，得：
（3-a²)x²-2ax-2=0
于是XA+XB=2a/(3-a²)
XA*XB=-2/(3-a²)
YA+YB=a(XA+XB)+2=2a²/(3-a²)+2=6/(3-a²)
YA*YB=(aXA+1)(aXB+1)=a²(XA*XB)+a(XA+XB)+1
=-2a²/(3-a²)+2a²/(3-a²)+1=1
故│AB│=√[XA-XB)²+(YA-YB)²]
=√[(XA+XB)²+(YA+YB)²-4(XA*XB+YA*YB)]
=√[4a²/(3-a²)²+36/(3-a²)²+8/(3-a²)-4]
=√[-4(a²-2)(a²-3)/(3-a²)²]=2√[(a²-2)/(3-a²)]
由此可知(a²-2)/(a²-3)<0,即2
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