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近年来,在各省、市高考模拟考试或全国高考、会考中.出现了一个命题的新方向——探求多重参数互相制约条件的问题,这类问题形式多样、方法灵活、多变,技巧性强。学生普遍感到束手无策,本文试图通过一些具体的例子来阐述这类题型的解法与技巧. 一、巧妙转化.方程架桥 例1 已知圆x~2+y~2=1及双曲线.x~2-y~2=1.而直线y=kx+b交二曲线于四个不同的点A、B、C、D(如图),为使|AB|=|CD|,求k、b应满足的条件. 分析:将|AB|=|CD|转化为线段,AD的中点M与线段BC的中点N重合,然后用韦达定理列方程架桥铺路,问题可迎刃而解. 解:将y=kx+b代入x~2+y~2=.1.得 x~2+k~2x~2+2h·bx+b~2-1=0. 故 x_Nx_1+x_2/2=-kb/1+k~2; 同理,将y=kx+b代入x~2-y~2=1,得 x_M=kb/1-k~2由x_M=x_N,可得k=0或b=0.又因A、B、C、D是四个不同的点.故K,b应满足的条件是:k=0,0< |b|<1 或b=0,0<|k|<1.
In recent years, a new direction of propositions has emerged in the provincial and municipal college entrance examination mock examinations or national college entrance examinations and examinations. This is a question of exploring multiple constraints on each other’s constraints. These types of problems have various forms, are flexible, changeable, and have strong skills. . Students generally feel helpless, this article attempts to illustrate the solution and skills of this type of question through some specific examples. First, the ingenious transformation. Equation bridge example 1 known circle x~2+y~2=1 and hyperbola.x ~2-y~2=1. And the straight line y=kx+b intersects the two curves at four different points A, B, C, and D (as shown in the figure). For |AB|=|CD|, find k, b. Conditions to be satisfied. Analysis: Convert the |AB|=|CD| to a line segment. The midpoint M of AD coincides with the midpoint N of the line segment BC. Then the bridge is paved with the Vedic theorem equation, and the problem can be solved. : Substitute y=kx+b into x~2+y~2=.1. Get x~2+k~2x~2+2h•bx+b~2-1=0. So x_Nx_1+x_2/2=- Kb/1+k~2; Similarly, substituting y=kx+b into x~2-y~2=1 yields x_M=kb/1-k~2 from x_M=x_N, where k=0 or b is available. =0. Also because A, B, C, D are four different points. So K, b should meet the following conditions: k = 0, 0 < |b| <1 or b = 0, 0 <|k| <1.