2023版考前三个月冲刺专题练　第28练　定点、定值问题【无答案版】

第28练　定点、定值问题

[考情分析]　解析几何是数形结合的典范，是高中数学的主要知识模块，定点和定值问题是高考考查的重点知识，在解答题中一般会综合考查直线、圆、圆锥曲线等，试题难度较大，多次以压轴题出现．

一、定点问题

例1　(2022·全国乙卷)已知椭圆E的中心为坐标原点，对称轴为x轴、y轴，且过A(0，－2)，B两点．

(1)求E的方程；

(2)设过点P(1，－2)的直线交E于M，N两点，过M且平行于x轴的直线与线段AB交于点T，点H满足＝.证明：直线HN过定点．

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规律方法　求解定点问题常用的方法

(1)“特殊探路，一般证明”，即先通过特殊情况确定定点，再转化为有方向、有目标的一般性证明．

(2)“一般推理，特殊求解”，即先由题设条件得出曲线的方程，再根据参数的任意性得到定点坐标．

(3)求证直线过定点(x0，y0)，常利用直线的点斜式方程y－y0＝k(x－x0)来证明．

跟踪训练1　(2022·上海模拟)已知F1，F2分别为椭圆E：＋＝1的左、右焦点，过F1的直线l交椭圆E于A，B两点．

(1)当直线l垂直于x轴时，求弦长|AB|；

(2)当·＝－2时，求直线l的方程；

(3)记椭圆的右顶点为T，直线AT，BT分别交直线x＝6于C，D两点，求证：以CD为直径的圆恒过定点，并求出定点坐标．

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二、定值问题

例2　(2020·新高考全国Ⅰ)已知椭圆C：＋＝1(a>b>0)的离心率为，且过点A(2,1)．

(1)求C的方程；

(2)点M，N在C上，且AM⊥AN，AD⊥MN，D为垂足．证明：存在定点Q，使得|DQ|为定值．

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规律方法　求圆锥曲线中定值问题常用的方法

(1)引出变量法：其解题流程为

→

↓

→

↓

→

(2)特例法：从特殊入手，求出定值，再证明这个值与变量无关．

跟踪训练2　(2022·南通模拟)已知F1(－，0)，F2(，0)为双曲线C的焦点，点P(2，－1)在C上．

(1)求C的方程；

(2)点A，B在C上，直线PA，PB与y轴分别交于点M，N，点Q在直线AB上，若＋＝0，·＝0，证明：存在定点T，使得|QT|为定值．

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