高中数学 高三 一轮复习 椭圆 教学设计

高中数学高三一轮复习椭圆教学设计一、教学内容分析（一）【核心概念】课标要求与教材定位  椭圆是高中数学平面解析几何的核心内容，是圆锥曲线学习的开端，也是坐标法思想应用的典范。根据《普通高中数学课程标准（2017年版2020年修订）》的要求，学生需经历从具体情境中抽象出椭圆模型的过程，掌握椭圆的定义、标准方程及简单几何性质，并能运用这些知识解决与椭圆相关的位置关系及简单实际问题。本节课是北师大版高中数学一轮复习“主题3第8章第5节椭圆”的内容，其地位在于：一方面，它是对必修课程中解析几何思想的深化与拓展；另一方面，它为后续学习双曲线、抛物线提供了研究范式和方法基础。（二）【知识网络】教学内容的核心地位与作用  本节课的教学内容聚焦于椭圆的核心概念、标准方程及其基本量的内在联系。具体包括：  1.椭圆的定义（包括第一定义和统一定义，本阶段重点复习第一定义及其在解题中的应用）。  2.椭圆的标准方程的两种形式（焦点在x轴、焦点在y轴）及其推导过程所蕴含的坐标法思想。  3.椭圆的基本几何性质：范围、对称性、顶点、长轴、短轴、焦距以及离心率。  4.a,b,c三个基本量的几何意义和数量关系（a2=b2+c2a^2=b^2+c^2a2=b2+c2），以及它们与椭圆扁平程度（离心率e）的关联。  5.直线与椭圆的位置关系及其判定方法，为后续综合问题埋下伏笔。二、学情与目标分析（一）【重要】学情分析：已知与未知的衔接  学生已在高二阶段完成了椭圆的新课学习，对椭圆的基础知识有初步印象。进入高三一轮复习，他们面临的主要挑战和已有基础如下：  1.知识基础：学生已掌握椭圆的定义、标准方程和基本几何性质，但部分学生对a,b,c的几何意义理解不够深刻，容易混淆；对离心率反映椭圆形状的实质缺乏本质理解；对于用坐标法解决几何问题的思想已有初步体会，但熟练度和规范性有待提高。  2.能力基础：学生具备一定的代数变形能力和数形结合意识，但在面对复杂情境时，往往不能主动、恰当地运用坐标法思想，运算求解能力和逻辑推理能力仍需系统强化。  3.心理特征：高三学生目标明确，意志力增强，但面对一轮复习中知识点的整合与提升，可能存在畏难情绪。因此，教学设计需注重知识的系统化、网络化构建，通过典型例题的深入剖析，帮助学生克服难点，建立信心。（二）【教学目标】核心素养导向下的多维目标设定  根据上述分析，结合数学学科核心素养，确立本节课的教学目标如下：  1.知识与技能：    （1）【基础】准确理解并掌握椭圆的定义、标准方程及其推导过程，能根据条件熟练求解椭圆的标准方程。    （2）【重要】熟练掌握椭圆的范围、对称性、顶点、离心率等简单几何性质，并能灵活运用这些性质解决相关问题。    （3）理解a,b,c的几何意义及其内在联系（a2=b2+c2a^2=b^2+c^2a2=b2+c2），能利用它们进行基本量的计算和判断。  2.过程与方法：    （1）通过复习椭圆定义的生成过程，进一步体会数形结合、等价转化的数学思想。    （2）通过对椭圆标准方程的推导回顾，强化坐标法在解析几何中的应用，提升运算求解能力。    （3）通过探究离心率与椭圆扁平程度的关系，感悟用代数形式刻画几何特征的精髓，提升直观想象与逻辑推理素养。  3.情感、态度与价值观：    （1）在知识网络的构建中，体会数学知识的系统性与内在统一性，培养严谨求实的科学态度。    （2）通过对椭圆美感的欣赏（对称性、简洁的方程），激发学习数学的兴趣和探索精神。（三）【难点】教学重、难点定位  1.教学重点：椭圆的定义、标准方程及其简单几何性质的理解与应用。  2.教学难点：灵活运用椭圆的定义和性质解决与焦点三角形、离心率相关的综合问题，特别是建立代数方程与几何特征之间的关联。三、教学策略与方法（一）【核心理念】“以生为本，问题驱动”的教学策略  本节课贯彻“学生为主体，教师为主导”的新课程理念。采用“问题驱动”和“变式探究”的教学模式。通过精心设计的问题链，引导学生主动回顾、积极思考、合作交流。教师的作用在于适时点拨、总结提炼、深化拓展，帮助学生将零散的知识点串联成线、编织成网。（二）教学方法  1.启发式讲授法：用于核心概念（如定义、方程、离心率）的精准辨析与深化理解。  2.讲练结合法：通过典型例题的示范讲解与学生的即时练习，巩固基础知识，提升解题技能。  3.变式探究法：围绕核心考点，设置变式问题，引导学生在变化中抓不变，在联系中促迁移，培养思维的深刻性和灵活性。  4.数形结合法：贯穿始终，引导学生借助图形理解代数表达，运用代数运算解决几何问题。四、教学实施过程（核心环节，详细展开）（一）【基础】知识梳理，温故知新（约10分钟）  1.问题驱动，唤醒记忆：    教师首先展示一个生活中的椭圆实例（如行星运行轨道、油罐车横截面等），并提出问题链：    （1）【基础】“请同学们回顾，椭圆的第一定义是什么？用数学语言如何表述？”    （2）“在平面内，到两个定点的距离之和等于常数（大于两定点间距离）的点的轨迹叫椭圆。这里的‘常数’和‘距离’如何用代数符号表示？”    学生思考并回答，教师引导提炼出关键词：“平面内”、“两个定点（焦点）”、“距离之和”、“常数（2a2a2a）”、“2a>∣F1F2∣2a>|F_1F_2|2a>∣F1​F2​∣”。    （3）“有了定义，我们如何将这种几何关系转化为代数方程？”    引导学生回忆起“建系设点列式化简”的坐标法步骤。以焦点在x轴为例，建立适当坐标系，设P(x,y)P(x,y)P(x,y)为椭圆上任意一点，焦点F1(−c,0)F_1(c,0)F1​(−c,0),F2(c,0)F_2(c,0)F2​(c,0)，根据定义列出方程∣PF1∣+∣PF2∣=2a|PF_1|+|PF_2|=2a∣PF1​∣+∣PF2​∣=2a。    2.方程推导，重温坐标法：    教师带领学生简要回顾方程的化简过程（两次平方法），重点关注化简的关键步骤和令b2=a2−c2b^2=a^2c^2b2=a2−c2的合理性（使方程形式简洁，且b具有明显的几何意义）。    最终得出焦点在x轴上的标准方程：    x2a2+y2b2=1(a>b>0)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\quad(a>b>0)a2x2​+b2y2​=1(a>b>0)    类比提问：“如果焦点在y轴上，标准方程又是什么？”    引导学生得出：    y2a2+x2b2=1(a>b>0)\frac{y^2}{a^2}+\frac{x^2}{b^2}=1\quad(a>b>0)a2y2​+b2x2​=1(a>b>0)    3.几何性质，数形互译：    教师展示一个标准椭圆的图形（焦点在x轴），引导学生从方程和图形两个角度观察、归纳其几何性质：    （1）范围：由方程可得x2a2≤1\frac{x^2}{a^2}\le1a2x2​≤1且y2b2≤1\frac{y^2}{b^2}\le1b2y2​≤1，即−a≤x≤aa\lex\lea−a≤x≤a，−b≤y≤bb\ley\leb−b≤y≤b。图形直观体现为椭圆位于直线x=±ax=\pmax=±a和y=±by=\pmby=±b所围成的矩形内。    （2）对称性：以−xx−x代xxx，方程不变，说明图形关于y轴对称；以−yy−y代yyy，方程不变，说明图形关于x轴对称；同时以−xx−x、−yy−y代xxx、yyy，方程不变，说明图形关于原点中心对称。坐标轴是椭圆的对称轴，原点是椭圆的对称中心（即中心）。    （3）顶点：曲线与对称轴的交点称为顶点。令y=0y=0y=0，得x=±ax=\pmax=±a，对应顶点A1(−a,0)A_1(a,0)A1​(−a,0)、A2(a,0)A_2(a,0)A2​(a,0)；令x=0x=0x=0，得y=±by=\pmby=±b，对应顶点B1(0,−b)B_1(0,b)B1​(0,−b)、B2(0,b)B_2(0,b)B2​(0,b)。同时指出长轴A1A2A_1A_2A1​A2​长为2a2a2a，短轴B1B2B_1B_2B1​B2​长为2b2b2b。    （4）离心率：定义e=cae=\frac{c}{a}e=ac​。教师提问：“e的大小反映了椭圆的什么几何特征？”引导学生观察，当e越接近0时，ccc越接近0，由a2=b2+c2a^2=b^2+c^2a2=b2+c2知bbb越接近aaa，椭圆越圆；当e越接近1时，ccc越接近aaa，bbb越接近0，椭圆越扁。强调离心率是刻画椭圆扁平程度的重要量。    【重要】此时，教师需在黑板上板书出核心关系图：定义→坐标法\xrightarrow{坐标法}坐标法<pathd="M0241v40hc47.335.3847811012816.73263..63..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​标准方程→代数研究\xrightarrow{代数研究}代数研究<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​几何性质→相互印证\xrightarrow{相互印证}相互印证<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​深化定义，形成一个完整的知识闭环。（二）【高频考点】典例精析，深化理解（约25分钟）  本环节通过精选例题，对核心考点进行多层次、多角度的剖析。  【高频考点1】椭圆定义与标准方程的求解  例1：【基础】求适合下列条件的椭圆的标准方程：    （1）两个焦点的坐标分别是(−4,0)(4,0)(−4,0),(4,0)(4,0)(4,0)，椭圆上一点P到两焦点距离之和为10。    （2）两个焦点的坐标分别是(0,−2)(0,2)(0,−2),(0,2)(0,2)(0,2)，并且椭圆经过点(−32,52)(\frac{3}{2},\frac{5}{2})(−23​,25​)。    （3）长轴长是短轴长的2倍，且经过点(2,−6)(2,6)(2,−6)。  分析：    （1）【基础】学生快速判断焦点在x轴，由已知c=4c=4c=4，2a=10⇒a=52a=10\Rightarrowa=52a=10⇒a=5，则b2=a2−c2=25−16=9b^2=a^2c^2=2516=9b2=a2−c2=25−16=9，得方程为x225+y29=1\frac{x^2}{25}+\frac{y^2}{9}=125x2​+9y2​=1。    （2）【重要】判断焦点在y轴，设方程为y2a2+x2b2=1\frac{y^2}{a^2}+\frac{x^2}{b^2}=1a2y2​+b2x2​=1(a>b>0a>b>0a>b>0)。由焦点坐标知c=2c=2c=2，有a2−b2=4a^2b^2=4a2−b2=4①。又椭圆过点(−32,52)(\frac{3}{2},\frac{5}{2})(−23​,25​)，代入方程得(52)2a2+(−32)2b2=1\frac{(\frac{5}{2})^2}{a^2}+\frac{(\frac{3}{2})^2}{b^2}=1a2(25​)2​+b2(−23​)2​=1，即254a2+94b2=1\frac{25}{4a^2}+\frac{9}{4b^2}=14a225​+4b29​=1②。联立①②解方程组求a2,b2a^2,b^2a2,b2。教师需引导学生规范解方程组的过程（可令m=a2,n=b2m=a^2,n=b^2m=a2,n=b2），得a2=10,b2=6a^2=10,b^2=6a2=10,b2=6。强调“代入验证”的步骤。    （3）【难点】条件“长轴长是短轴长的2倍”并未指明焦点位置，需分类讨论。设标准方程。      情形一：焦点在x轴，设x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1a2x2​+b2y2​=1，则2a=2×2b⇒a=2b2a=2\times2b\Rightarrowa=2b2a=2×2b⇒a=2b。      情形二：焦点在y轴，设y2a2+x2b2=1\frac{y^2}{a^2}+\frac{x^2}{b^2}=1a2y2​+b2x2​=1，则a=2ba=2ba=2b。    分别将点(2,−6)(2,6)(2,−6)代入两种情形下的方程，求出a2,b2a^2,b^2a2,b2的值，并检验是否满足a>b>0a>b>0a>b>0。最终得出两解：x237+y2374=1\frac{x^2}{37}+\frac{y^2}{\frac{37}{4}}=137x2​+437​y2​=1和y2148+x237=1\frac{y^2}{148}+\frac{x^2}{37}=1148y2​+37x2​=1。此题旨在训练学生思维的严密性，注意焦点位置不明时的分类讨论。  【教学意图】例1覆盖了直接法、待定系数法和分类讨论法求解椭圆方程的基本题型，巩固了a,b,c的关系，是解决所有椭圆问题的基础。  【高频考点2】椭圆的几何性质及其应用  例2：【重要】已知椭圆x225+y29=1\frac{x^2}{25}+\frac{y^2}{9}=125x2​+9y2​=1的左、右焦点分别为F1,F2F_1,F_2F1​,F2​，点P在椭圆上。    （1）若∣PF1∣=2|PF_1|=2∣PF1​∣=2，求∣PF2∣|PF_2|∣PF2​∣及△PF1F2\trianglePF_1F_2△PF1​F2​的面积。    （2）若∠F1PF2=60∘\angleF_1PF_2=60^\circ∠F1​PF2​=60∘，求△PF1F2\trianglePF_1F_2△PF1​F2​的面积。  分析：    （1）【基础】由椭圆定义知∣PF1∣+∣PF2∣=2a=10|PF_1|+|PF_2|=2a=10∣PF1​∣+∣PF2​∣=2a=10，∴∣PF2∣=8\therefore|PF_2|=8∴∣PF2​∣=8。又c2=a2−b2=25−9=16c^2=a^2b^2=259=16c2=a2−b2=25−9=16，∴∣F1F2∣=2c=8\therefore|F_1F_2|=2c=8∴∣F1​F2​∣=2c=8。在△PF1F2\trianglePF_1F_2△PF1​F2​中，三边为2,8,8，为等腰三角形。可先求底边上的高，或用海伦公式，或由余弦定理先求一个角再求面积。教师引导学生用最简洁的方法：取PF2PF_2PF2​中点D，连接F1DF_1DF1​D，则F1D⊥PF2F_1D\perpPF_2F1​D⊥PF2​，且∣F1D∣=∣F1P∣2−(∣PF2∣/2)2=22−42|F_1D|=\sqrt{|F_1P|^2(|PF_2|/2)^2}=\sqrt{2^24^2}∣F1​D∣=∣F1​P∣2−(∣PF2​∣/2)2<pathd="M263,681c0.7,0,18,39.7,52,119c34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,704.7,510.7,1060.3,512,1067l00c4.7,7.3,11,11,19,11H40000v40H1012.3s271.3,567,271.3,567c38.7,80.7,84,175,136,283c52,108,89.167,185.3,111.5,232c22.3,46.7,33.8,70.3,34.5,71c4.7,4.7,12.3,7,23,7s12,1,12,1s109,253,109,253c72.7,168,109.3,252,110,252c10.7,8,22,16.7,34,26c22,17.3,33.3,26,34,26s26,26,26,26s76,59,76,59s76,60,76,60zMHv40hz">​=22−42<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​发现此法不成立，因22<422^2<4^222<42，不能构成直角三角形。故回到余弦定理求∠F1PF2\angleF_1PF_2∠F1​PF2​或∠PF1F2\anglePF_1F_2∠PF1​F2​。引导学生发现PF2PF_2PF2​是底边，PF1PF_1PF1​和F1F2F_1F_2F1​F2​是腰，用等腰三角形性质求解更优。计算cos⁡∠PF1F2=∣PF1∣2+∣F1F2∣2−∣PF2∣22∣PF1∣∣F1F2∣=4+64−642×2×8=432=18\cos\anglePF_1F_2=\frac{|PF_1|^2+|F_1F_2|^2|PF_2|^2}{2|PF_1||F_1F_2|}=\frac{4+6464}{2\times2\times8}=\frac{4}{32}=\frac{1}{8}cos∠PF1​F2​=2∣PF1​∣∣F1​F2​∣∣PF1​∣2+∣F1​F2​∣2−∣PF2​∣2​=2×2×84+64−64​=324​=81​，则sin⁡∠PF1F2=378\sin\anglePF_1F_2=\frac{3\sqrt{7}}{8}sin∠PF1​F2​=837<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​，面积S=12∣PF1∣∣F1F2∣sin⁡∠PF1F2=12×2×8×378=37S=\frac{1}{2}|PF_1||F_1F_2|\sin\anglePF_1F_2=\frac{1}{2}\times2\times8\times\frac{3\sqrt{7}}{8}=3\sqrt{7}S=21​∣PF1​∣∣F1​F2​∣sin∠PF1​F2​=21​×2×8×837<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​=37<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​。    （2）【热点】焦点三角形问题。设∣PF1∣=m,∣PF2∣=n|PF_1|=m,|PF_2|=n∣PF1​∣=m,∣PF2​∣=n。由定义：m+n=10m+n=10m+n=10①。由余弦定理：∣F1F2∣2=m2+n2−2mncos⁡60∘|F_1F_2|^2=m^2+n^22mn\cos60^\circ∣F1​F2​∣2=m2+n2−2mncos60∘，即64=m2+n2−mn64=m^2+n^2mn64=m2+n2−mn②。将①式平方得m2+n2+2mn=100m^2+n^2+2mn=100m2+n2+2mn=100③。联立②③，可解得mn=12mn=12mn=12。于是S△PF1F2=12mnsin⁡60∘=12×12×32=33S_{\trianglePF_1F_2}=\frac{1}{2}mn\sin60^\circ=\frac{1}{2}\times12\times\frac{\sqrt{3}}{2}=3\sqrt{3}S△PF1​F2​​=21​mnsin60∘=21​×12×23<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​=33<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​。    【变式与拓展】若将条件改为“∠F1PF2=90∘\angleF_1PF_2=90^\circ∠F1​PF2​=90∘”，则面积如何？引导学生发现此时m2+n2=64m^2+n^2=64m2+n2=64，结合m+n=10m+n=10m+n=10可求得mn=18mn=18mn=18，面积S=9S=9S=9。  【教学意图】例2以焦点三角形为载体，深入考查椭圆的定义、余弦定理、三角形面积公式等知识的综合运用，体现了数形结合思想和方程思想。其中(2)问的解法（整体代换求mnmnmn）是处理此类问题的典型技巧，应重点强调。  【高频考点3】离心率的求值或范围问题  例3：【难点】设椭圆x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1a2x2​+b2y2​=1(a>b>0)的左、右焦点分别为F1,F2F_1,F_2F1​,F2​，点P是椭圆上异于长轴端点的任意一点，若△PF1F2\trianglePF_1F_2△PF1​F2​的内心为I，且满足PI→=λIF1→+μIF2→\overrightarrow{PI}=\lambda\overrightarrow{IF_1}+\mu\overrightarrow{IF_2}PI<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">=λIF1​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​+μIF2​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​，其中λ,μ\lambda,\muλ,μ为非零实数，求离心率e的取值范围。  分析：此题综合性较强，需引导学生逐步分析。  第一步：理解条件转化。内心I是内切圆圆心，到三边距离相等。向量关系PI→=λIF1→+μIF2→\overrightarrow{PI}=\lambda\overrightarrow{IF_1}+\mu\overrightarrow{IF_2}PI<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">=λIF1​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​+μIF2​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​暗示了点I与P,F1,F2的某种线性关系，但直接处理较困难。需联想到内切圆的性质和面积法。  第二步：寻找等量关系。设△PF1F2\trianglePF_1F_2△PF1​F2​的内切圆半径为r，三边长为∣PF1∣=m|PF_1|=m∣PF1​∣=m,∣PF2∣=n|PF_2|=n∣PF2​∣=n,∣F1F2∣=2c|F_1F_2|=2c∣F1​F2​∣=2c，周长为m+n+2c=2a+2cm+n+2c=2a+2cm+n+2c=2a+2c。三角形面积S可表示为S=12r(m+n+2c)=12r(2a+2c)=r(a+c)S=\frac{1}{2}r(m+n+2c)=\frac{1}{2}r(2a+2c)=r(a+c)S=21​r(m+n+2c)=21​r(2a+2c)=r(a+c)。  另一方面，S也可用坐标法表示，设P(x0,y0)P(x_0,y_0)P(x0​,y0​)，则S=12∣F1F2∣⋅∣y0∣=c∣y0∣S=\frac{1}{2}|F_1F_2|\cdot|y_0|=c|y_0|S=21​∣F1​F2​∣⋅∣y0​∣=c∣y0​∣。  第三步：建立联系与转化。由c∣y0∣=r(a+c)c|y_0|=r(a+c)c∣y0​∣=r(a+c)得r∣y0∣=ca+c\frac{r}{|y_0|}=\frac{c}{a+c}∣y0​∣r​=a+cc​。由于I在△PF1F2\trianglePF_1F_2△PF1​F2​内部，几何上可得到某种范围。但直接处理向量条件，可以考虑内心I的坐标公式（用三边坐标表示），但较为复杂。  【另辟蹊径】回顾内心性质：内心I将△PF1F2\trianglePF_1F_2△PF1​F2​分成三个小三角形，它们的面积之和等于大三角形面积。S△IF1F2:S△IPF2:S△IPF1=∣F1F2∣:∣PF2∣:∣PF1∣=2c:n:mS_{\triangleIF_1F_2}:S_{\triangleIPF_2}:S_{\triangleIPF_1}=|F_1F_2|:|PF_2|:|PF_1|=2c:n:mS△IF1​F2​​:S△IPF2​​:S△IPF1​​=∣F1​F2​∣:∣PF2​∣:∣PF1​∣=2c:n:m。而题目中PI→=λIF1→+μIF2→\overrightarrow{PI}=\lambda\overrightarrow{IF_1}+\mu\overrightarrow{IF_2}PI<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">=λIF1​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​+μIF2​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​等价于以I为起点，向量PI→\overrightarrow{PI}PI<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">可以用IF1→\overrightarrow{IF_1}IF1​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​和IF2→\overrightarrow{IF_2}IF2​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​线性表示，这意味着P,I,F1,F2四点共面（自然成立），且系数λ,μ\lambda,\muλ,μ与P相对于边F1F2的位置有关。一个常见的结论是，在三角形中，对于内心I，有PI→=∣PF2∣周长PF1→+∣PF1∣周长PF2→\overrightarrow{PI}=\frac{|PF_2|}{周长}\overrightarrow{PF_1}+\frac{|PF_1|}{周长}\overrightarrow{PF_2}PI<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">=周长∣PF2​∣​PF1​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​+周长∣PF1​∣​PF2​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​？或类似的加权形式。更直接地，可以利用坐标法。但考虑到高三一轮复习的层次，此题可作如下简化处理：引导学生从特例入手，寻找离心率的边界。  教师引导：“非零实数λ,μ\lambda,\muλ,μ的存在性意味着什么？能否从极端位置考虑？”考虑P从短轴端点向长轴端点移动时，内心的位置变化。当P在短轴端点(0,b)时，三角形为等腰三角形，内心I在y轴上，此时PI→\overrightarrow{PI}PI<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">与IF1→+IF2→\overrightarrow{IF_1}+\overrightarrow{IF_2}IF1​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​+IF2​<pathd="M0241v