论文部分内容阅读
一引例对双曲线方程x~2/a~2-y~2/b~2=1 (1)我们同时给出一个与它有关的直线方程: b(t~2+a~2)x-a(t~2-a~2)y-2ta~2b=0 (2)这里t是参数。我们先介绍一下这个直线方程在求方程(1)所表示双曲线切线中的作用。引例问:过点P_1(4,0),P_2(6,2(3~(1/2)))P_3(3,2),P_4(0,2)能否作双曲线x~2/9-y~2/4=1的切线,若能,求出切线方程。解:已知a=2,b=3代入(2)得: 2(t~2+9)x-3(t~2-9)y-36t=0 (3) ①将P_1点坐标代入(3),得 8(t~2+9)-36t=0 2t~2-9t+18=0,t无实数解,这时我们说过P_1点的切线不存在。
An example of the hyperbolic equation x~2/a~2-y~2/b~2=1 (1) We also give a linear equation related to it: b(t~2+a~2)xa( t~2-a~2)y-2ta~2b=0 (2) where t is a parameter. Let us first introduce the role of this linear equation in finding the hyperbolic tangent represented by equation (1). Quoting Q: P_1(4,0), P_2(6,2(3~(1/2)))P_3(3,2), P_4(0,2) can be hyperbolic x~2/9 The tangent of -y ~ 2/4 = 1, if possible, find the tangent equation. Solution: It is known that a=2 and b=3 are substituted into (2): 2(t~2+9)x-3(t~2-9)y-36t=0 (3) 1 Substituting P_1 point coordinates ( 3), get 8(t~2+9)-36t=0 2t~2-9t+18=0. There is no real solution to t. At this time, we said that the tangent of P_1 does not exist.