首先考虑斜率存在的情况。设直线方程为y=kx+b设A(x1,y1)B(x2,y2)C(x3,y3)D(x4,y4)联立直线椭圆双曲线方程,由韦达定理x1+x2=-50bk/(16+25k^2)x3+x4=2bk/(1-k^2)由AC=BD,得到x1+x2=x3+x4-50bk/(16+25k^2)=2bk/(1-k^2)bk=0,k=0或b=0若k=0,x1,2=±5sqrt(16-b^2)/4,x3,4=±sqrt(b^2+1)由AB=3CD,x2-x1=3(x4-x3),代入后得到b=±16/13,l:y=±16/13若b=0,x1,2=±20/sqrt(16+25k^2),x3,4=1/sqrt(1-k^2)由AB=3CD,x2-x1=3(x4-...
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