论文部分内容阅读
众所周知,过一点的双曲线最多只有两条切线。但是,笔者却可以求出四条。题:求通过点p(O,-1)的双曲线x~2-4y~2=1的切线。解:设过点P的切线方程为y=kx-1,下面由切线与双曲线有唯一交点来确定k。把y=kx-1代入x~2-4y~2=1并整理,得 (1-4k~2)x2+8kx-5=0 (＊) 当1-4k~2=0,即k=±1/2时,方程(＊) 有唯一解,从而直线与双曲线有唯一交点。当1-4k≠0时,令△=16k~2+20(1-4k~2)=0得k=±5~(1/2)/4,即k=±5~(1/2)/4也为所
As we all know, there is only a maximum of two tangents for a hyperbolic curve. However, the author can find four. Question: Find the tangent of the hyperbolic x~2-4y~2=1 passing the point p(O,-1). Solution: Set the tangent equation of the over point P to y=kx-1. The following is determined by the unique intersection of the tangent and hyperbola. Substituting y=kx-1 into x~2-4y~2=1 and finishing, we get (1-4k~2)x2+8kx-5=0 (*) When 1-4k~2=0, that is k=± At 1/2, the equation (*) has a unique solution so that the straight line has the unique intersection with the hyperbola. When 1-4k≠0, let △=16k~2+20(1-4k~2)=0 get k=±5~(1/2)/4, that is, k=±5~(1/2) /4 is also