高中数学的奇技淫巧 - 知乎专栏
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{"database":{"Post":{"":{"title":"小技巧
圆锥曲线中的四点共圆问题","author":"ying-zi-hhh-92","content":"先来看一个例题已知椭圆Γ:x?/4+y?=1与直线l:y=2x+1交于点A，B，以AB为直径的圆交Γ于C，D两点，则直线CD的斜率是?A.1
D.-2(这个图花了我30min，手机党的悲哀)图画出来了题目答案一清二楚了，就是D项了。等等，这不是艺术课......然而常规方法应付这个题是悲剧的，学渣无从下手，学霸也不敢算.....。。。。。。。进入正题。。。。。解答这个问题之前先来看看4-4课本的一道习题:(找这两张图，又花了30min)可以看出，若两条倾斜角互补的直线L1,L2交椭圆与A，B，C，D 4点，那么PA·PB=PC·PD(P是两直线的焦点)再由相交弦定理的逆定理得到ABCD四点共圆.总结 若A，B，C，D是椭圆上的四点，且K(AB)+K(CD)=0那么A，B，C，D四点共圆反之亦然。当然前提是斜率存在，其实斜率不存在时即AD，BC是两条垂直X轴的直线时，ABCD依旧共圆。下面给出证明看不懂自习补充二次曲线系(￢_￢)设两条直线即椭圆(看图那么过A B C D的曲线方程(λ是参数):要使ABCD四点共圆，则必须有:解释一下，第一个式子表示x?，y?项的系数相等，第二个是xy的系数是零才能表示圆，于是由第二个式子证明了结论，即k1+k2=0，第一个式子可以帮助求出圆的方程。如果K不存在的话四边形ACBD是等腰梯形就不多说了。......然而事情还没有结束！这个结论置于双曲线和抛物线依旧成立。技术有限，双曲线画不出来了╮( o?ωo? )╭证明的话和椭圆一样的方法小技巧大作用，希望能在考试中帮你节约哪怕一分钟的时间以上工作全部在我RAM只有0.5GB的手机上完成，已累趴！T^T【高考之后】思考 给定一个椭圆，如何利用尺规作图找出它的中心和长短轴？（以下内容纯属兴趣）利用上面的定理可以轻松做到，读者不妨试试。1.利用圆规画一个圆，和已知椭圆交于四点。2.连对角线3.作出该对角线所夹角的一条角平分线，交椭圆LH两点4.做LH的垂直平分线交椭圆于IJ两点，IJ就是椭圆的长轴或短轴，后面的就略过了","updated":"T12:53:21.000Z","canComment":false,"commentPermission":"anyone","commentCount":23,"likeCount":151,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T20:53:21+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":23,"likesCount":151},"":{"title":"椭圆的准圆（圆与椭圆“切线”和“两直线垂直”的引论）","author":"zhou-zhi-peng-68-2","content":"貌似木有人发过这个……第一次发好紧张QAQ椭圆准圆的概念准圆是圆锥曲线两条互相垂直的切线的交点的轨迹，由法国数学家Gaspard Monge首先发现，所以又被叫做“蒙日圆”。对于椭圆E：\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1，它的准圆方程为C：x^2+y^2=a^2+b^2相信大部分同学已经在做题中遇到了类似的情形，只是题目不提“准圆”这个名词罢了。这里给大家简介下椭圆准圆的两个简单性质，过多的奇妙性质，留给各位慢慢咀嚼。而写下这篇短短的论述，更多是希望各位同学能够对于椭圆与其有密切关系的圆，在“切线问题”、“两直线垂直”等问题上有所收获和突破，掌握其中的思路和方法。性质①：准圆引切线互相垂直（椭圆的外切矩形）对于椭圆E：\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1，在圆C：x^2+y^2=a^2+b^2上任意一点P(m,n)引椭圆E的两条切线l_{1}和l_{2}，则有l_{1}⊥l_{2}，证明如下：∵P(m,n)在圆C上，则有m^2+n^2=a^2+b^2，设切线的方程为y-n=k(x-m)，通过联立椭圆方程：\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1
以及切线方程：y-n=k(x-m)
②得到二次方程：(a^{2} k^{2}+b^{2})x^{2} +2ka^{2} (n-mk)x+a^{2}(km-n)^{2} -a^{2}b^{2} =0让\\Delta =4k^{2} a^{4} (km-n)^{2}-4(a^{2} k^{2}+b^{2} )a^{2} [(km-n)^{2} -b^{2} ]等于0，列出关于k的方程化简整理得到：(m^{2} -a^{2} )k^{2} -2mnk+n^{2} -b^{2} =0，由题设，l_{1} 和l_{2} 的斜率k_{1} 和k_{2} 满足方程于是，由韦达定理，k_{1}\\cdot k_{2} =\\frac{n^{2} -b^{2} }{m^{2} -a^{2} } ，又因为P(m,n)在圆C上，则有m^2+n^2=a^2+b^2于是，k_{1}\\cdot k_{2} =\\frac{n^{2} -b^{2} }{m^{2} -a^{2} } =\\frac{a^{2}+b^{2}-m^{2} -b^{2} }{m^{2} -a^{2} } =-1，∴得到l_{1}⊥l_{2}相关例题已知椭圆E：\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1（a&b&0）离心率为\\frac{\\sqrt{6} }{3} ，椭圆上的点到右焦点的最短距离为\\sqrt{3} -\\sqrt{2} （1）求椭圆E的标准方程（2）过圆x^2+y^2=4上的点P向椭圆E引两条切线，，切点分别为M,N，切线分别交圆于另两点A,B，求\\triangle PAB面积的最大值题中的问题（2），根据以上的推导，我们得到了一个重要的性质——AB为圆的直径（直径所对的圆周角为直角），因此存在\\left|PA \\right|^2+\\left| PB\\right| ^2=\\left|AB \\right| ^2=R^2（半径），由此，有关三角形的周长，面积的性质，可以通过这个定值使用基本不等式，柯西不等式等等来解决最值问题。性质②：过原点的两条互相垂直的直线交椭圆所得圆的弦长为定值对于椭圆E：\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1，和圆C：x^2+y^2=a^2+b^2，过原点的两条直线l_{1}和l_{2}，有l_{1}⊥l_{2}，若l_{1}和l_{2}分别交E于A,B，直线AB交C于M,N，则\\left|MN \\right| 为定值，证明如下：设直线AB的方程为y=kx+m，∵l_{1}⊥l_{2}，∴l_{1}和l_{2}对应的斜率k_{1} 和k_{2} 满足k_{1}\\cdot k_{2} =-1于是，联立椭圆方程和直线方程，得到\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1=(\\frac{y-kx}{m}) ^{2} x^{2} 的系数为：\\frac{1}{a^{2} } -\\frac{k^{2} }{m^{2} } ，y^{2} 的系数为：\\frac{1}{b^{2} } -\\frac{1 }{m^{2} } ，于是k_{1}\\cdot k_{2} =\\frac{(\\frac{1}{a^{2} } -\\frac{k^{2} }{m^{2} })}{( \\frac{1}{b^{2} } -\\frac{1 }{m^{2} }) } =-1，解出得到\\frac{m^{2} }{k^{2}+1 } =\\frac{a^2b^2}{a^2+b^2} ，计算原点到直线AB的距离d=\\frac{\\left| m \\right| }{\\sqrt{k^{2}+1 } } ，于是d^{2} =\\frac{a^2b^2}{a^2+b^2} （定值）又因为圆C的半径已经确定为\\sqrt{a^2+b^2} ，因此所得的弦长\\left|MN \\right| 也为定值。当然，上述的证明也可以通过设出A,B两点的坐标(x_{1},y_{1} )和(x_{2},y_{2} )，通过向量OA,OB的点乘为0得到m和k的关系，但由于计算繁琐，这里给出的是二次曲线系的证明方法。相关例题已知椭圆E：\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1（a&b&0）离心率为\\frac{\\sqrt{6} }{3} ，定义椭圆E的准圆C为：x^2+y^2=a^2+b^2。\n（1）若椭圆E的准圆的半径为2，求椭圆E的标准方程（2）在（1）的条件下，过原点的两条直线l_{1}和l_{2}，满足l_{1}⊥l_{2}，若l_{1}和l_{2}分别交E于A,B，直线AB交C于M，N两点，\\left|MN \\right| 是否为定值？若是，求出定值，不是，请说明理由。这个问题在上面的证明过程中已经给的很详细了:-D\n引论例题已知l是圆O：x^2+y^2=2的切线，l 交椭圆C：\\frac{x^2}{6}+\\frac{y^2}{3}=1于A,B两点，求\\triangle AOB面积的取值范围。这就是在简介部分为大家提到的了：希望各位同学能够对于椭圆与其有密切关系的圆，在“切线问题”、“两直线垂直”等问题上有所收获和突破，掌握其中的思路和方法。那么针对于切线的问题，最常思考就是——切线是否与相关直线有着垂直的关系。并且大部分的题目，都离不开这个性质，虽然本题中的圆并非是前面所提到的椭圆的准圆，但这个思想仍然适用。以下给出解题：不妨设切线l的方程为：y=kx+m，由于l与圆相切，由点到直线的距离得到d=\\frac{\\left| m \\right| }{\\sqrt{k^{2}+1 } } =\\sqrt{2} ，由此得到\\frac{m^{2} }{{k^{2}+1 } } =2，之后探究由切线得到的两个特殊的点A,B一般探究直线的斜率问题，如果能够得到OA\\bot OB，那么对于之后求解面积就会有很大帮助于是，计算k_{OA}\\cdot k_{OB} ，联立直线方程和椭圆方程，同样，根据上面提到的二次曲线系有\\frac{x^2}{6}+\\frac{y^2}{3}=1=(\\frac{y-kx}{m}) ^{2} ，∴x^{2} 的系数为：\\frac{1}{6 } -\\frac{k^{2} }{m^{2} } ，y^{2} 的系数为：\\frac{1}{3} -\\frac{1 }{m^{2} } k_{OA}\\cdot k_{OB} =\\frac{(\\frac{1}{6 } -\\frac{k^{2} }{m^{2} })}{( \\frac{1}{3 } -\\frac{1 }{m^{2} }) } ，分子分母同时去分母，并且由\\frac{m^{2} }{{k^{2}+1 } } =2将m^2整体代换得到k_{OA}\\cdot k_{OB} =\\frac{2-4k^2}{4k^2-2}=-1，于是得到我们想要的结论：OA\\bot OB再然后就容易多了，由于涉及直角三角形的面积需要两直角边的边长，不妨设出椭圆的极坐标形式，将x^2和y^2用\\rho ^2cos^2\\theta ，\\rho ^2sin^2\\theta 整体代换得到：\\rho ^2(1+sin^2\\theta )=6，于是OA,OB的长度（\\rho 的值）满足这个方程，又∵两直线的垂直关系，不妨设：\\left| OA \\right|^2 =\\frac{6}{1+sin^2\\theta } ，\\left| OB \\right|^2 =\\frac{6}{1+cos^2\\theta } ，∴(S_{\\triangle AOB})^2=\\frac{1}{4} \\times \\frac{6}{1+sin^2\\theta } \\cdot \\frac{6}{1+cos^2\\theta } (S_{\\triangle AOB})^2=\\frac{1}{4} \\times \\frac{6}{1+sin^2\\theta } \\cdot \\frac{6}{1+1-sin^2\\theta } =(S_{\\triangle AOB})^2=\\frac{1}{4} \\times \\frac{6}{1+sin^2\\theta } \\cdot \\frac{6}{2-sin^2\\theta } 又∵sin^{2} \\theta的有界性，介于[0,1]，通过二次函数的单调性，可得S_{\\triangle AOB}的取值范围为[2,\\frac{3\\sqrt{2} }{2} ]通过以上的证明，我们不难猜测，当圆的半径满足与椭圆有关的数值（a^2,b^2）的一个表达式的时候，这条结论应该都成立，于是，列举了三个椭圆并进行计算探究后，得到了以下结论：对于椭圆E：\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1，当存在圆C：x^2+y^2=\\frac{a^2b^2}{a^2+b^2} 时，与圆C相切的直线与椭圆E相交于A,B两点，则有OA\\bot OB恒成立。证明如下：对于椭圆E：\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1，和圆C：x^2+y^2=\\frac{a^2b^2}{a^2+b^2} ，直线l与圆C相切，不难得到m和k的关系满足：\\frac{m^2}{k^2+1} =\\frac{a^2b^2}{a^2+b^2} ，之后联立直线方程和椭圆方程，运用二次曲线系，得到：有\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1=(\\frac{y-kx}{m}) ^{2} ，∴x^{2} 的系数为：\\frac{1}{a^2 } -\\frac{k^{2} }{m^{2} } ，y^{2} 的系数为：\\frac{1}{b^2} -\\frac{1 }{m^{2} } ，∴k_{OA}\\cdot k_{OB} =\\frac{(\\frac{1}{a^2 } -\\frac{k^{2} }{m^{2} })}{( \\frac{1}{b^2 } -\\frac{1 }{m^{2} }) } ，将式子中的k^2替换为\\frac{m^2(a^2+b^2)}{a^2b^2} -1，经过化简整理后得到k_{OA}\\cdot k_{OB}=\\frac{\\frac{1}{m^2}-\\frac{1}{b^2} }{\\frac{1}{b^2}-\\frac{1}{m^2}} =-1（定值），∴满足我们所求的条件OA\\bot OB，即证。总结根据以上三个题目的详细证明和解答，我们得到了椭圆和圆密切相关的两组特殊数据①圆半径的平方R^2=a^2+b^2，即我们所说的准圆方程②圆半径的平方R^2=\\frac{a^2b^2}{a^2+b^2} 以上两个圆与已知椭圆，在“切线”和“两直线垂直”方面都有着及其特殊的性质，因此，当下次在题目中遇到圆和椭圆的问题时，不妨先看看圆的半径是否满足椭圆的这些特殊关系，一旦满足，将会瞬间“柳暗花明又一村”，大大降低解题的难度，节省解题的时间。\n完结撒花~O(∩_∩)O~~","updated":"T15:38:46.000Z","canComment":false,"commentPermission":"anyone","commentCount":38,"likeCount":168,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T23:38:46+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"/v2-b35bed50ab17a604c6b39_r.png","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":38,"likesCount":168},"":{"title":"并不高深，关于压轴题的谐星想法（四）。。","author":"zhou-jia-le-61","content":"好，大家就当我说的话是放屁吧，今年考的是数列，还只有两问，还贼jier简单。。。。由4道一模新题略窥20题的解题思路。\n\n北京历年高考20题的出题思路都是考察数集（我并认为叫数集比叫数列更合适）中各元素间的性质，抑或是给出一个新定义，考察对新定义的理解与运用。所以，对于20题，重点不是在于学习各种数列求和、求通项的技巧，而是在于学习推理与证明的本真思路。\n\n通览4道题，第一问都是十分简单，且大多不用写过程。但这里我想说一点，如果能随随便便就看出答案，千万不要把数一写就跑到下一问去了。第一问的作用，往往要么是一个“小引理”，要么是一个易于上手的对新定义的实践。所以还是要对第一问进行推导再往下做，这是对之后两问的负责。\n\n再说说第二和第三问。这4道题很好地把高中常见的数学思想及推理方法给覆盖到了。下面逐个简单提点一下。\n\n朝阳：分类讨论，循环递降（归一化），奇偶性，从“1，2，3”做起，枚举。\n\n丰台：反证，分类讨论，一般与特殊的辩证统一。\n\n海淀：必要与充分，引理化。（诶，它这个第三问我是真不能驾驭）\n\n丰台：一般与特殊的辩证统一（不止一次用到），奇偶性，分类讨论。\n\n此外还有一些：数学归纳法，符号化，特值，倒排（反序）思想，抽屉原理，分析法……\n\n很多时候，沿着题目所指的路走下去，就能大概看到证明的方向了，之后就是一些小思路了。\n\n当然，北京高考考过的模型是绝不会再次进入高考的，但对于对20题有梦想的各位老哥而言，每一次模拟的20题都是不可多得的好机会，应在学习了答案的证法后，自己推导一遍乃至数遍，最好能对答案进行改进与创新（当然不是所有题都能这样），最后不妨像我这样把出现的数学思想及推理方法列出并整理出来，到考场上也不至于蒙圈。祝各位老哥高考顺利。ps：一般与特殊的辩证统一是指对一般成立的对特殊也一定成立，以及添加适当条件使一般问题特殊化而所加条件不干扰命题本身。附：4道20题","updated":"T14:22:32.000Z","canComment":false,"commentPermission":"anyone","commentCount":8,"likeCount":18,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T22:22:32+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":8,"likesCount":18},"":{"title":"神奇的圆锥曲线公式③","author":"erin-mi","content":"定理：过已知双曲线上的任意一点,分别作两条渐近线的平行线,则其平行线与渐进线围城的平行四边形面积是定值（图中四边形OMPN的面积），该定值为\\frac{ab}{2} 。","updated":"T10:42:37.000Z","canComment":false,"commentPermission":"anyone","commentCount":11,"likeCount":58,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T18:42:37+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"/v2-507b010ce66c8c2f78d9a7c_r.png","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":11,"likesCount":58},"":{"title":"排列组合中的分组分配问题","author":"qian-qian-87-4-27","content":"好多同学对分组问题都有个疑惑，到底什么时候考虑平均分组？平均分组问题该如何解决？为什么要用平均分组？考虑一种简单的情况：将4个元素(a，b，c，d)平均分为2组，则每组应有2个元素。第一步：从4个元素中取出2个为第1组，有C_{4}^{2} 种取法.再从剩下的2个元素中取2个为第2组，有C_{2}^{2} 种取法，则按乘法原理，到目前为止，有C_{4}^{2} \\ast C_{2}^{2} 种分法；第二步：假设第一步中先取了（a，b）为第1组,则（c，d）自动为第2组。因为此时是不涉及分给谁的，所以这种分法和先取（c，d）再取（a，b）实际上算同一种分法。也就是说第一步得到的分法是有序的，2个组就有A_{2}^{2} 种顺序。要消除这种有序，只需除以顺序的数目。第三步：最后平均分的分法就是\\frac{C_{4}^{2} \\ast C_{2}^{2} }{A_{2}^{2} } 。所谓平均分组是指将所有的元素分成所有组元素个数相等或部分组元素个数相等。平均分成的组，不管它们的顺序如何，都是一种情况，所以分组后要除以A_{m}^{m} ，即m!，其中m表示组数。举个栗子1、全部均匀分组：将12本书按4:4:4平均分成三堆，有多少种不同的分法？\\frac{C_{12}^{4} \\ast C_{8}^{4}\\ast C_{4}^{4} }{A_{3}^{3} } =57752、部分均匀分组：将12本书按2:2:2:6分成四堆有多少种不同的分法？\\frac{C_{12}^{2} \\ast C_{10}^{2}\\ast C_{8}^{2} \\ast C_{6}^{6}}{A_{3}^{3} } 看出来了吗？套公式嘛，谁不会啊！客官别急，再来个栗子变1：将12本书按4:4:4平均分给小明、小红、小华三个人，有多少种分法？PS：上述方法处理有分配对象和无分配对象的问题都可以噢~今天520，没啥好表示的，留一道有对象的作业送给你吧~祝你早点找到对象！！！明天继续更多的可以关注Inner peace~","updated":"T05:42:24.000Z","canComment":false,"commentPermission":"anyone","commentCount":7,"likeCount":90,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T13:42:24+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"/v2-efd27b80b6ffa61_r.jpg","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":7,"likesCount":90},"":{"title":"神奇的圆锥曲线公式④","author":"erin-mi","content":"定理：①椭圆中，记焦点三角形的底角分别为\\alpha 和\\beta ，则该椭圆的离心率为e=\\frac{sin\\left( \\alpha +\\beta
\\right) }{sin\\left( \\alpha
\\right)+sin\\left( \\beta
} ②双曲线中，记焦点三角形的底角分别为\\alpha 和\\beta 则该双曲线的离心率为e=\\frac{sin\\left( \\alpha +\\beta
\\right) }{sin\\left( \\alpha \\right) -sin\\left( \\beta
\\right) } ，且（sin\\alpha -sin\\beta ＞0）例题：标准答案：知道公式后的解答：第一步：判断相关量的正负（为了方便记∠PF1F2为A，记∠PF2F1为B）显然
cosB＜0第二步：通过公式计算离心率e=\\frac{\\frac{15}{25} }{\\frac{\\sqrt{5} }{5} } =\\frac{3\\sqrt{5} }{5} ","updated":"T09:59:23.000Z","canComment":false,"commentPermission":"anyone","commentCount":13,"likeCount":126,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T17:59:23+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"/v2-39b261dab6bebf6b42cc6_r.png","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":13,"likesCount":126},"":{"title":"导数压轴题的套路总结","author":"wu-han-85-32","content":" （多图预警）共分为三大类问题：1.双变量问题2.单变量问题 3.证明n项求和的不等式一点点微小的见解，难免有纰漏之处，欢迎指正与补充。一、二、三、字丑见谅。高考狗没有时间编辑公式了QAQ也算为高考攒个人品（大雾）最后预祝高考党们高考顺利owo!","updated":"T15:05:43.000Z","canComment":false,"commentPermission":"anyone","commentCount":97,"likeCount":911,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T23:05:43+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":97,"likesCount":911},"":{"title":"圆锥曲线极坐标系的特殊应用","author":"mo-luo-tuo-fu-14","content":"记一圆锥曲线的焦点为F（-c，0），P为椭圆上一点，过P作PN与其准线（x=-a^{2} /c）垂直，垂足为N，记角PFO=\\theta ,O为原点坐标，图大概长这样：由椭圆的第二定义(平面上到定点F的距离与到定直线的距离之比为常数e(即椭圆的离心率，e=c/a)的点的集合)我们有（不会用知乎的编辑器，原谅我字丑）值得一提的是如果在右焦点建极坐标系只需把c*cos\\theta 的系数变为+1以下是其应用过程是很炫酷，但考试的时候还是要推一遍，但也不麻烦，我的语言不甚严谨，具体还需各位看官自己整理这种建极坐标系极点不在原点的方法给了我一些启发，让我想到了一道高考题题目具体记不清了好像是突然想起来题目了，别吐槽字丑，a=2，b=根号3到这里就完了点个赞吧","updated":"T09:59:19.000Z","canComment":false,"commentPermission":"anyone","commentCount":10,"likeCount":59,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T17:59:19+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":10,"likesCount":59},"":{"title":"神奇的圆锥曲线公式⑤","author":"erin-mi","content":"公式:1.向量OA点乘向量OB=-\\frac{3}{4} p_{}^{2} 2.三角形AOB面积=\\frac{p^{2} }{4} \\sqrt{cot\\Theta ^{2}+1 } ","updated":"T01:23:49.000Z","canComment":false,"commentPermission":"anyone","commentCount":16,"likeCount":34,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T09:23:49+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"/v2-8cd39cca7c39_r.png","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":16,"likesCount":34},"":{"title":"阿尔卑斯圆","author":"qian-qian-87-4-27","content":"有一个姑娘，她有一些任性她还有一些嚣张…唔，今天说的不是小燕子。说的是数学里的一个圆，它叫阿波罗尼斯圆。（上课的时候我喜欢叫它阿尔卑斯圆，听起来像阿尔卑斯糖，就变成了阿尔卑斯圆本篇重点不是如何证明结论哈，而是怎么使用结论。但还是简单提供证明过程：引例：设A、B是平面内的两个定点，平面内的动点P到点A的距离与到点B的距离的比为定值\\lambda （\\lambda &0），求点P的轨迹。解：可设定线段AB的长为2a（a&0），以线段AB所在直线为x轴，以线段AB的中垂线为y轴，建立直角坐标系，则A（-a，0），B（a，0），P（x，y），由\\frac{PA}{PB} =\\lambda ,得\\frac{\\sqrt{\\left( x+a \\right)^{2}+y^{2}
} }{\\sqrt{\\left( x-a \\right)^{2}+y^{2}
} } =\\lambda ，化简得：\\left( 1-\\lambda^{2}
\\right) x^{2} +\\left( 1-\\lambda^{2}
\\right) y^{2} +2a\\left( 1-\\lambda^{2}
\\right)x+a^{2} \\left( 1-\\lambda^{2}
\\right)=0当\\lambda =1时，方程即为x=0，此时轨迹是y轴，也就是线段AB的中垂线；当\\lambda &0且\\lambda \\ne 1时，配方整理得：\\left( x-\\frac{a\\left( \\lambda ^{2}+1
\\right) }{ \\lambda ^{2}-1} \\right) ^{2} +y^{2} =\\left( \\frac{2a\\lambda }{\\lambda ^{2}-1 }
\\right) ^{2} 此时轨迹是一个圆，就是阿波罗尼斯圆。对话的方式简单点，阿波罗尼斯圆就是：动点P到两个定点之间的距离之比为一个常数（大于0且不等于1），这个点的轨迹为阿波罗尼斯圆。此类型解题步骤：第一步：设动点坐标（x,y）；第二步：根据条件列等式或不等式；第三步：化简得证，○( ＾皿＾)っHiahia…上例题：【例】已知圆O：x^{2} +y^{2} =1，圆M：\\left( x-4 \\right) ^{2} +y^{2} =4,动点P在直线x+\\sqrt{3} y-b=0上，过点P分别作圆O、圆M的切线，切点分别为A、B。若A、B满足PB=2PA的点有且只有两个，则b的范围——分析：将动点A、B转化为定点即圆心，再利用切线长公式列式，化简即可得点P的轨迹方程。图片里的答案写错了，-20/3＜b＜4。留个家庭作业给你们~变式1：已知A（-2,0），B（1,0），直线l:y=x+m，若直线上有且只有一个点P，满足PA=\\sqrt{2} PB，求m的值变式2：条件改为“PA\\geq \\sqrt{2} PB”，求m的值再举个栗子，08年江苏高考题满足条件AB=2，AC=\\sqrt{2} BC的\\Delta ABC的面积的最大值——对于本题，较常见的解法是通过设变量，建立目标函数来解决最大值问题。粗略的解题过程放在最后。现在我们用阿氏圆来解：比较两种解法可知，采用解法二，借助阿氏圆，利用数形结合的思想来处理，显然要便捷得多！我可以无愧的说自己是一枚合格的吃货了，最近有人邀请我回答，戳进去瞧瞧，相信你会失望的附上上题解答更多的可以关注Inner peace~","updated":"T11:56:25.000Z","canComment":false,"commentPermission":"anyone","commentCount":13,"likeCount":69,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T19:56:25+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"/v2-8e3cd3e90ebe7_r.jpg","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":13,"likesCount":69},"":{"title":"极值点：我也不想偏移，我也很无奈啊","author":"zsgs2008","content":"一直以来，很多小朋友问我为什么不愿意分享点干货给大家。拖了这么久，放一次大招。编辑公式太麻烦，直接上图咯，图片可能不容易看。我的QQ群里有PDF格式，有需要的同学可以加群，群文件里有，自行下载咯。极值点偏移基本被玩废了，但伪极值点偏移很多同学还没摸清脉络，所谓伪极值点偏移就是指形式看起来和极值点很相像，但不等号迷の反向。我在本文里给大家提供一些思路，欢迎参考。另：本文可复制，可粘贴，可共享。你可以拿去用，说你也发现了这些。但不能去吹嘘”我最先发明“”神级结论“”秒杀“之类的，了解我的人都知道，我反感这些词！我是唐老师，专注初高中数学高分培训，提倡扎实稳健搞定高中数学，不喜浮夸，不爽秒杀，我为高中数学高分代言。","updated":"T09:37:06.000Z","canComment":false,"commentPermission":"anyone","commentCount":58,"likeCount":270,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T17:37:06+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"/v2-1babbf68d3d5a5ab2946_r.png","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":58,"likesCount":270},"":{"title":"蛋白质相关计算","author":"Kushim-Jiang","content":"马上就要高考了，同为高三的我来临考攒人品，讲一个看似无用实则无用的破点子。一、知识储备蛋白质的单体称为氨基酸（简记为 AA），其结构简式为\\mathrm{NH_2CH(R)COOH}，其中 R 基可能含 N、S 等。两分子氨基酸脱去一分子水形成酰胺键（肽键），形成二肽，二肽亦能和其他氨基酸脱水缩合形成多肽，当多肽中首末两个氨基酸脱水缩合后便形成环肽。人体所需氨基酸约 20 种，其中脯氨酸、色氨酸、谷氨酰胺、天冬酰胺、赖氨酸中 R 基含 1 个 N，精氨酸、组氨酸 R 基中含 2 个 N，半胱氨酸、甲硫氨酸 R 基中含 1 个 S，丝氨酸、谷氨酰胺、苏氨酸、天冬酰胺、酪氨酸 R 基中含 1 个 O，天冬氨酸、天冬氨酸 R 基中含 2 个 O（无需背记）。二、不含 R 基的蛋白质各类原子数计算不少教辅书在介绍该考点时往往会给出一系列公式，其中绝大多数书籍仅给出前两条：N 原子数 = 氨基酸数
= 肽键数 + 肽链数O 原子数 = 氨基酸数 + 肽链数
= 肽键数 + 2 × 肽链数C 原子数 = 2 × 氨基酸数
= 2 × 肽键数 + 2 × 肽链数H 原子数 = 2 × 氨基酸数 + 2 × 肽链数 = 2 × 肽键数 + 4 × 肽链数这么简洁的公式，简直太好记了！李华哭着对我说。我摸摸李华的狗头说，嗯我也觉得蛮好记的那你滚吧。以下引入「环肽法」。首先假设所有氨基酸均相互脱水缩合形成若干个环肽，此时可将其表记为\\mathrm{(NHCH(R)CO)_n}，考察每一个链结内各种原子的个数，我们可以得到：N 原子数 = 1 × 氨O 原子数 = 1 × 氨C 原子数 = 2 × 氨H 原子数 = 2 × 氨然后加入若干个水分子，让其形成若干条肽链。此时 H、O 原子数随肽链形成而增多，于是我们可以得到：N 原子数 = 氨O 原子数 = 氨 + 1 × 链C 原子数 = 2 × 氨H 原子数 = 2 × 氨 + 2 × 链同理，形成若干个环肽时，每个链结连接两个肽键，每个肽键连接两个链结（绕口令），因此此时氨基酸数等于肽键数，即有：N 原子数 = 1 × 键O 原子数 = 1 × 键C 原子数 = 2 × 键H 原子数 = 2 × 键之后，每加 1 分子水，便断开 1 个肽键形成 1 条肽链，若看作将 1 个链结两侧添上 -H 和 -OH，则等价于从多肽中脱出 1 个氨基酸分子，而原有的环肽仍为环肽结构。因此各种原子的个数包含两个部分：环肽上的原子和脱下来的诸\\mathrm{NH_2CH(R)COOH}内的原子，即有：N 原子数 = 键 + 1 × 链O 原子数 = 键 + 2 × 链C 原子数 = 2 × 键 + 2 × 链H 原子数 = 2 × 键 + 4 × 链整理便是：N = 氨 = 键 + 链O = 氨 + 链 = 键 + 2链C = 2氨 = 2键 + 2链H = 2氨 + 2链 = 2键 + 4链简记为「氨链单体补氢氧，键链单体补 AA」，即「原子个数表记为氨基酸数和肽链数的和时，在单体中各种原子个数乘氨基酸数后补上水中氢氧原子个数乘肽链数；原子个数表记为肽键数和肽链数的和时，在单体中各种原子个数乘肽键数后补上 AA 中各种原子个数乘肽链数」。另外，上式中的「氨 = 键 + 链」亦为较重要的结论。稍做几题。【例 1】结晶牛胰岛素为双链五十一肽，求不含 R 基的各种原子个数。〖解〗根据「氨链单体补氢氧」显然可得，N 原子数 = 1×51 + 0 = 51，O 原子数 = 1×51 + 2×1 = 53，C 原子数 = 2×51 + 0 = 102，H 原子数 = 2×51 + 2×2 = 106。【例 2】已知一种蛋白质不含 R 基后剩余部分可表示为\\mathrm{C}_x\\mathrm{H}_{y}\\mathrm{O}_8\\mathrm{N}_4，求 x，y。〖解〗若以「氨链单体补氢氧」为据则有：O = 氨 + 链 = 8，N = 氨 = 4，解得氨 = 4，链 = 4，于是 x = 2氨 = 8，y = 2氨 + 2链 = 16。【例 3】有一种蛋白质由 3 条肽链和 4 条环肽凭 R 基连接而成，每条肽链或环肽上均有 15 个肽键，求不含 R 基的各种原子个数。〖解〗根据「键链单体补 AA」显然可得，N 原子数 = 1×105 + 1×3 = 108，O 原子数 = 1×105 + 2×3 = 111，C 原子数 = 2×105 + 2×3 = 216，H 原子数 = 2×105 + 4×3 = 222。三、含 R 基的蛋白质各类原子数计算当原题条件未明确表示不含 R 基时皆默认含 R 基，此时仅靠氨基酸数、肽键数、肽链数无法得知具体的各类原子数或氨基羧基数，均只能称作「至少若干个」。（1）若原题中所示氨基酸均不含 O（或 N），可根据 O（或 N）的个数确定氨基酸数。【例 4】今有一化合物为一条肽链，其分子式为\\mathrm{C_{55}H_{70}O_{19}N_{10}}，已知将它完全水解后只得到下列四种氨基酸：甘（\\mathrm{-H}）、丙（\\mathrm{-CH_3}）、苯丙（\\mathrm{-CH_2C_6H_5}）、谷（\\mathrm{-(CH_2)_2COOH}）。则该多肽是______肽，该多肽进行水解后，需______个水分子，得到______个谷氨酸分子。〖解〗由于四种氨基酸均不含 N，则有 N = 氨 = 10，故为十肽。水解需 9 个水分子。由于谷氨酸 R 基含 O，则 O = 氨 + 链 + R = 10 + 1 + R = 19，即 R = 8。故得到 4 个谷氨酸分子。【例 5】有一条多肽链，分子式为\\mathrm{C}_x\\mathrm{H}_y\\mathrm{O}_p\\mathrm{N}_q\\mathrm{S}，将它彻底水解后，只得到下列四种氨基酸：丙（\\mathrm{-CH_3}）、苯丙（\\mathrm{-CH_2C_6H_5}）、赖（\\mathrm{-(CH_2)_3NH_2}）、半胱（\\mathrm{-CH_2HS}），分析推算可知，水解得到的氨基酸个数为______。〖解〗由于四种氨基酸均不含 O，则有 O = 氨 + 链 = 氨 + 1 = p，故氨 = p - 1。（2）若所给氨基酸 R 基中同时含 S、N、O 等元素，则需根据特征元素计算。【例 6】若某一多肽链由 201 个氨基酸组成，其分子式为\\mathrm{C}_x\\mathrm{H}_y\\mathrm{O}_a\\mathrm{N}_b\\mathrm{S}_2（a＞201，b＞202），并且是由下列 5 种氨基酸组成的：半胱（\\mathrm{-CH_2HS}）、丙（\\mathrm{-CH_3}）、苯丙（\\mathrm{-CH_2C_6H_5}）、赖（\\mathrm{-(CH_2)_3NH_2}）、天冬（\\mathrm{-CH_2COOH}），那么该多肽链彻底水解后将会得到赖氨酸______个、天冬氨酸______个。〖解〗赖氨酸仅含 N，N = 氨 + R = 201 + R = b，R = b - 201，故得到赖氨酸 (b - 201) 个。天冬氨酸仅含 O，O = 氨 + 链 + R = 201 + 1 + R = a，R = a - 202，故得到天冬氨酸 (a - 202)/2 个。（3）二硫键（2 -SH → -S-S- + 2H）需要添 H。【例 7】经测定，某多肽链分子式是\\mathrm{C}_{21}\\mathrm{H}_x\\mathrm{O}_y\\mathrm{N}_4\\mathrm{S}_2，其中含有一个二硫键。已知该多肽是由下列氨基酸中的几种作为原料合成的：苯丙（\\mathrm{-CH_2C_6H_5}）、天冬（\\mathrm{-CH_2COOH}）、丙（\\mathrm{-CH_3}）、亮（\\mathrm{-CH_2CH(CH_3)_2}）、半胱（\\mathrm{-CH_2HS}）。下列有关该多肽的叙述，正确的是______。A. 该多肽水解后产生的氨基酸分别是半胱氨酸、天冬氨酸、丙氨酸和亮氨酸；B. 该多肽中 H 原子数和 O 原子数分别是 32 和 5；C. 该多肽形成过程中至少需要 4 种 tRNA；D. 该多肽在核糖体上形成，形成过程中相对分子质量减少了 56。〖解〗均不含 N，N = 氨 = 4，为四肽。仅半胱含 S，S = 半胱 = 2，AC 错。苯丙共 9C，天冬 4C，丙 3C，亮 6C，半胱 3C，由于仅有 2×3 + 9 + 6 = 21 成立，故为两个半胱、苯丙、亮。氢 + 2 = 2氨 + 2链 + R = 8 + 2 + 22，H = 30，B 错。故选 D。四、作业【练 1】已知天冬酰胺（\\mathrm{-CH_2CONH_2}），现有分子式为\\mathrm{C_{63}H_{103}O_{18}N_{17}S_2}的多肽，其中含 2 个天冬酰胺。在上述多肽中最多有肽键______个。A. 14；B. 15；C. 16；D. 17。【练 2】某多肽的分子式为\\mathrm{C_{42}H_{65}O_9N_{11}}，它彻底水解后只得到以下 3 种氨基酸：甘（\\mathrm{-H}）、苯丙（\\mathrm{-CH_2C_6H_5}）、赖（\\mathrm{-(CH_2)_3NH_2}），则此多肽中含有的赖氨酸个数为______个。A. 2；B. 3；C. 5；D. 8。【练 3】现有氨基酸 100 个，其中氨基总数为 102 个，羧基总数为 101 个，则由这些氨基酸合成的含有 2 条肽链的蛋白质中，含有氧原子的数目是______个。A. 101；B. 102；C. 103；D. 104。【练 4】现有一种十三肽，分子式为\\mathrm{C_{54}H_{95}O_{20}N_{17}}，已知将它彻底水解后只得到下列氨基酸：丙（\\mathrm{-CH_3}）、赖（\\mathrm{-(CH_2)_3NH_2}）、天冬（\\mathrm{-CH_2COOH}），将该十三肽彻底水解后有______个丙氨酸，______个赖氨酸，______个天冬氨酸。和未水解的化合物相比，水解后______元素数目增加，______元素数目减少。答案见评论区。敲完字，夜已深，不由得题句一首，以献出卷组：你们考卷尽管出，考到这些算我输。","updated":"T17:13:21.000Z","canComment":false,"commentPermission":"anyone","commentCount":26,"likeCount":72,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T01:13:21+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":26,"likesCount":72},"":{"title":"关于曲线系的一点补充","author":"liu-ying-kun-51","content":"(知乎小白，才疏学浅，见谅。如有错误恳请批评与指正。)0.0有关曲线系的知识，之前已经有文章提供了详细的介绍，包括直线系和曲线系，以及它们的简单应用。这里再提一下，当一条直线与二次曲线相交的情况：设直线AB：y=kx+m ①
二次曲线：ax?+by?+cxy+dx+ey+f=0 ②则由①可得：\\frac{y-kx}{m}=1代入②中，使所有项均为二次，得ax^2+by^2+cxy+dx\\cdot \\frac{y-kx}{m}+dy\\cdot \\frac{y-kx}{m}=f\\cdot (\\frac{y-kx}{m})^2=0③设OA·OB：(y-k_1x)(y-k_2x)=0，则OA·OB一定能分解成③。对比两式系数，我们发现：k_{OA}\\cdot k_{OB}=x?系数/y?系数k_{OA}+ k_{OB}=-xy项系数/y?项系数从而，我们得出过原点的两条直线OA与OB的性质。下面看一道题，题很简单，结论各位应该已经耳熟能详了。例：已知抛物线y?=2px，过原点的两条垂直直线OA、OB交抛物线于A、B。\n
试证：直线AB过x轴上一定点。解：设AB:x=my+n
则OA·OB：y^2-2px\\cdot\\frac{x-my}{n}=0
OA⊥OB\\Rightarrow k_{OA}\\cdot k_{OB}=\\frac{-2p}{n}=-1
即n=2p，故AB过定点(-2p,0)。利用它，2017年全国Ⅰ卷理数的第20题第二问也可以做。（2017全国Ⅰ，理20）已知椭圆C:\\frac{x^2}{4}+y^2=1,P(0,1)。（2）设直线l不经过P点且与C交于A，B两点。若直线PA与PB斜率之和为-1，证明：l过定点。解：(2) 平移坐标系，使P坐标为(0,0)。
则C':\\frac{x^2}{4}+y^2+2y=0
设l_{ab}:y=kx+b. \\Rightarrow \\frac{y-kx}{b}=1
代入C'，我们得到C':\\frac{x^2}{4}+y^2+2\\frac{y^2-kxy}{b}=0①
设PA：y=m_1x ，PB：y=m_2x
则PA·PB：(m_1x-y)(m_2x-y)=0
即m_1m_2x^2-(m_1+m_2)xy+y^2=0②，其中，m_1+m_2=-1 .
①与②表示同一曲线.
易知b≠ -2.由①得:
\\frac{x^2}{4(1+\\frac{2}{b})}+y^2-\\frac{2k}{b+2}xy=0③
与②对比系数得：\\frac{2k}{b+2}=-1.即b= -2k-2.
故直线恒过(2,-2).
故在原坐标系中，l过定点(2,-1).(凌晨一点多，编不下去了。。。所以未完待续)
参考文献：数学那玩意 韩旭","updated":"T17:36:37.000Z","canComment":false,"commentPermission":"anyone","commentCount":25,"likeCount":52,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":true,"rating":"none","sourceUrl":"","publishedTime":"T01:36:37+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"/v2-bfdf23c59fb9a695717fe_r.jpg","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":25,"likesCount":52},"":{"title":"关于抽象函数周期性的二三规律","author":"lin-an-46-47","content":"
抽象函数相信大家都不陌生，关于抽象函数的做法也有十分多种，各有千秋。这里就先介绍些普遍的抽象函数周期性的解法。
两图预警！！----------------------------不正经的分割线------------------------------另：1.若f（x）关于两点A（a，0），B（b，0）（a＞b），对称则有f（x）周期为2（a-b）
2.若f（x）关于两直线x=a，x=b对称（a＞b），则f（x）是以2（a-b）为周期的函数
这里就不给上证明了类比上面的方法可以自己证，若考试的时候想不起来可以画一个sin函数，可以立马想起来。
如：f(x)=sinx，其关于点（0,0）对称，关于点（π，0）对称，则函数f（x）的周期T＝2（π-0）=2π。函数f（x）＝sinx，其关于直线x＝π/2对称，关于直线x＝3π/2对称，则函数h（x）的周期T＝2（3π/2-π/2）＝2πPS：有没有好心人教我怎么用知乎自带的公式编辑器QAQ，写文章要在WPS里写还复制不到知乎里来。参考资料：1《课堂实录》
------作者的数学老师邓献宝","updated":"T10:16:53.000Z","canComment":false,"commentPermission":"anyone","commentCount":17,"likeCount":45,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T18:16:53+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"/v2-db0e89ad74a0c27f8a8eaff5bc5375dc_r.jpg","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":17,"likesCount":45},"":{"title":"圆锥曲线中焦点三角形的一些性质（1）","author":"liu-ying-kun-51","content":"(知乎小白，才疏学浅，见谅。如有错误恳请批评与指正。)0-0在高考中，有许多与圆锥曲线相关的选择填空题。其中，相当一部分都是围绕着一个名词——焦点三角形展开。而如果熟悉了焦点三角形的性质再做题，许多题就可以快速排除答案，从而提升做题速度，为后面的大题留足时间。今天，我们先来谈一谈，椭圆与双曲线中的焦点三角形所具备的性质。如图，\\Delta PF_1F_2 是椭圆\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1的焦点三角形。其中，P(x_0,y_0),c?=a?-b?，F_1,F_2为焦点，则：C_{\\Delta PF_1F_2}=2(a+c).\n若\\angle{F_1PF_2}=\\theta ,则S_{\\Delta PF_1F_2}=b^2tan{\\frac{\\theta }{2}}=c\\left| y_0 \\right| .Max\\left[ S_{\\Delta PF_1F_2} \\right]=bc ,当且仅当P与短轴两端点重合时取得.\n\\angle{F_1PF_2}有最大值，当且仅当P与短轴两端点重合时取得.\na-c\\leq \\left| PF_1 \\right|\\leq a+c ,P与左顶点重合时取最小值，与右顶点重合时取最大值.\n若\\angle{PF_1F_2}=\\alpha ,\\angle{PF_2F_1}=\\beta ,则e=\\frac{sin{(\\alpha +\\beta )}}{sin{\\alpha }+sin{\\beta }}.记\\Delta PF_1F_2的内切圆圆心为I，延长PI与x轴交于M，则I点分\\overrightarrow{PM}的比为\\frac{1}{e}.以焦半径为直径的圆与以长轴为直径的圆内切.将PF_1延长至交椭圆于Q，以PQ为直径的圆与左准线相离。\n简单提一下其中几条的证明。（2）利用余弦定理和椭圆的第一定义，联立得出\\left| PF_1 \\right| \\cdot \\left| PF_2 \\right|即可。（7）如图。（8）画图，利用圆心距等于两圆半径之差即可。S_{\\Delta PF_1F_2}=b^2cot{\\frac{\\theta }{2}}=c\\left| y_0 \\right| .\\Delta PF_1F_2内切圆圆心坐标为±a.\n以焦半径为直径的圆和以实轴为直径的圆内切.e=\\frac{sin{(\\alpha +\\beta )}}{sin{\\alpha }-sin{\\beta }}.\nPT为\\angle{F_1PF_2}的角平分线，则焦点在PT上的投影为以实轴为直径的圆（除实轴端点）.以焦点弦为直径的圆与对应准线相交.\n应该还是比较好想明白的。本文技术含量不高，还请各位谅解。下一次，我会总结在抛物线中焦点弦的性质。（之所以单独拿出来，是因为抛物线的结论比较多）参考资料：我的笔记本","updated":"T16:46:48.000Z","canComment":false,"commentPermission":"anyone","commentCount":10,"likeCount":32,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T00:46:48+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"/v2-6aea8a529cd6565_r.jpg","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":10,"likesCount":32},"":{"title":"高中数学干货之圆锥曲线（—）","author":"hzqsns-78","content":"前言（introduction）：本人今年高考完，闲来无事便心血来潮想为大家整理下我高中三年所积累的一点知识，平时要是没事会继续为大家提供更多有用的知识，这是我第一次在知乎上写文章，如有错误还望各位能及时指出QvQ。阅读须知（Instructions）：1.编辑不易，收藏同时不要忘记点赞、点赞、点赞。2.本文干货很多，请仔细阅读。3.本文技巧的东西比较少，记忆的东西比较多。正文（Main body）：圆锥曲线历年都是高考中的热点也是难点，但其中真正难倒我们的往往不是思路而是复杂的运算，所以我们平时应该要去积累一些简化运算的技巧，所以在这里为大家提供一些简化运算的东西。1.圆锥曲线之硬解定理（1）椭圆（2）双曲线（3）抛物线上述结构都有相似的地方，建议重点去记椭圆的地方，因为在大题中考的概率比较大，双曲线主要在小题里面出现，不要说什么记不住什么的。。我掌握了这些解圆锥曲线的速度真的很快，为导数又提供了时间，从而让数学又提升了一个层次。今天主要是大题的干货，过几天有时间会给你们选填的技巧知识好了QwQ。","updated":"T15:50:21.000Z","canComment":false,"commentPermission":"anyone","commentCount":40,"likeCount":296,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T23:50:21+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"/v2-e8bdac26b4dd599c299cde_r.jpg","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":40,"likesCount":296},"":{"title":"圆锥曲线中的优化算法","author":"feng-qing-yang-28-19","content":"
此文和本人发的圆锥曲线秒杀法是姊妹篇，前者侧重于道，本文侧重于术，建议将二者结合起来看。
本文主要内容包括:如何利用化齐次、双根、隐函数求导和放缩、柯西不等式简化圆锥曲线中的计算。
此外也欢迎大家来我知乎专栏读书.作文,
微信公众号: 有酒可以留客谈、
做客，上面我会定期发布和高考作文相关内容北京地区有数学、语文补习需求的可以私信本人，价格绝对比机构划算。定场文，本人写的高考作文纽带","updated":"T13:04:34.000Z","canComment":false,"commentPermission":"anyone","commentCount":14,"likeCount":125,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T21:04:34+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":14,"likesCount":125},"":{"title":"三角形里的两类三角恒等变换（上）","author":"Aundy","content":"更多三角公式可以参见（日语版，但是问题不大）。（下）的部分估计等几天再更。一定要叫我填坑哟！非商业目的随意转载。","updated":"T06:14:59.000Z","canComment":false,"commentPermission":"anyone","commentCount":7,"likeCount":37,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T14:14:59+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":7,"likesCount":37},"":{"title":"三角形里的两类三角恒等变化（全）","author":"Aundy","content":"以上是正文。下面是我从摘抄的一些可能有用的结论（关于反三角函数/逆三角函数，实在用不了别的方法的时候偷偷摸摸用一下……最好解答题里不要出现）基本关系（商数关系）、平方关系的a)、加法定理、三角函数的对称性与周期性（诱导公式）、降幂公式、辅助角公式需要掌握。完结撒花！非商业转载随意。","updated":"T05:50:39.000Z","canComment":false,"commentPermission":"anyone","commentCount":10,"likeCount":50,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T13:50:39+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":10,"likesCount":50},"":{"title":"极点极线解一类圆锥曲线题","author":"lin-an-46-47","content":"写在前面
作者很赞同作者老师的一句话：圆锥曲线有千万个结论。那么对于圆锥曲线的解题，对于普通生而言，就应该是基本方法为主，技巧为辅。而且对于技巧应该有选择，笔者认为最好的选择标准有两个，1.容易记，2.用得上。若是一个方法十分厉害能通杀所有圆锥曲线，然而几页纸的结论记不住也毫无用处，同样若是记了好方法高考不考，对于目标高考的学生还是没用。
今天要写的极点极线是我认为好记而用处也不小的，话不多说，上图。PS：非常感谢上篇文章教我用知乎编辑器的那位小可爱，我想告诉你，我还是没有学会...
综上，极点极线应该适用于以下几类题
1.题中给出了形如|PA|·|PB|=|QA||QB|等一类四个线段的乘积，比值类，就可以考虑是否能用
2.题目要求的是要求某点恒在哪条条直线上或者是某条直线恒过某点此类题即可考虑。
3.速求切线。若是点在曲线上题目中要用到有关于切线的，可以用公式直接求出切线方程，这是辅助功能了。","updated":"T05:48:52.000Z","canComment":false,"commentPermission":"anyone","commentCount":4,"likeCount":12,"state":"published","isLiked":false,"slug":"","isTitleImageFullScreen":false,"rating":"none","sourceUrl":"","publishedTime":"T13:48:52+08:00","links":{"comments":"/api/posts//comments"},"url":"/p/","titleImage":"/v2-4f9edb05077_r.jpg","summary":"","href":"/api/posts/","meta":{"previous":null,"next":null},"snapshotUrl":"","commentsCount":4,"likesCount":12}},"User":{"lin-an-46-47":{"isFollowed":false,"name":"林安","headline":"北隗","avatarUrl":"/v2-ffa8a2adfd_s.jpg","isFollowing":false,"type":"people","slug":"lin-an-46-47","bio":null,"hash":"fc9a4da30f982d755ad8959","uid":105200,"isOrg":false,"description":"北隗","profileUrl":"/people/lin-an-46-47","avatar":{"id":"v2-ffa8a2adfd","template":"/{id}_{size}.jpg"},"isOrgWhiteList":false},"Aundy":{"isFollowed":false,"name":"Aundy-MikuFan","headline":"相信我，外表很man少女心，二次元狂热，喜欢语言学、物理学、数学。巨蟹座。","avatarUrl":"/744d2e652f467a40cc5855cdb6305cdb_s.jpeg","isFollowing":false,"type":"people","slug":"Aundy","bio":"游荡者","hash":"a50b4dfecd","uid":64,"isOrg":false,"description":"相信我，外表很man少女心，二次元狂热，喜欢语言学、物理学、数学。巨蟹座。","profileUrl":"/people/Aundy","avatar":{"id":"744d2e652f467a40cc5855cdb6305cdb","template":"/{id}_{size}.jpeg"},"isOrgWhiteList":false},"qian-qian-87-4-27":{"isFollowed":false,"name":"数学老师心里苦","headline":"微信公主号：数学老师的羊圈","avatarUrl":"/v2-dbc20e84b6fe0_s.jpg","isFollowing":false,"type":"people","slug":"qian-qian-87-4-27","bio":"高中数学老师","hash":"58cf221d2f17dc5ff04f55","uid":691600,"isOrg":false,"description":"微信公主号：数学老师的羊圈","profileUrl":"/people/qian-qian-87-4-27","avatar":{"id":"v2-dbc20e84b6fe0","template":"/{id}_{size}.jpg"},"isOrgWhiteList":false},"feng-qing-yang-28-19":{"isFollowed":false,"name":"吴磊","headline":"
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