一道易错的圆锥曲线题_高中圆锥曲线易错题
时间：2019-01-27 03:22:43 来源：雅意学习网 本文已影响 人 
【摘要】文章例举了一道易错的圆锥曲线题,此题系直线与曲线交点的中点问题,是解析几何的重头戏,有韦达定理和点差法两种解法。点差法的前提条件是两个交点的存在性。 【关键词】圆锥曲线题 韦达定理 点差法
One problem about the conical curve in which students often make mistakes
Shuai Wanping
【Abstract】In this paper, the writer has explained one problem about the conical curve in which students often make mistakes. The problem is one which is relative to the midpoint of the beeline and the curve and is most important in the part analytic geometry. For solving that, we can use the Weida theorem and the method of difference of dots. The precondition of the method of difference of dots is that we must have two dots.
【Keywords】Problem about conical curve Weida theorem Method of difference of dots
已知双曲线 ,过点P(1,1)能否作一条直线l与双曲线交于A、B两点,且P为AB的中点;若存在,求AB的弦长。
法一、解:假设能作出这样的直线l,与双曲线交点为A(x1,y1)、B(x2,y2)。
①当直线的斜率不存在时,直线方程为x=1与双曲线相切,不合题意;
②当直线的斜率存在时,可设直线方程为y-1=k(x-1)。
此时联立两方程可得:
消去y得(2-k2)x2-2k(1-k)x-(1-k)2-2=0 (1)
由韦达定理得
∵P为AB的中点,∴x1+x2=2即
∴k=2
而k=2时(1)式为2x2-4x+3=0,其△ 本文为全文原貌 未安装PDF浏览器用户请先下载安装 原版全文
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