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Additional Product Information
Features and Benefits
"Focus on Modeling" sections illustrate modeling techniques as well as how mathematics can be applied to model real-life situations. These sections, as well as others, are devoted to teaching students how to create their own mathematical models, rather than using prefabricated formulas.
Real-world applications from engineering, physics, chemistry, business, biology, environmental studies, and other fields demonstrate how mathematics is used to model real-life situations.
The authors introduce the graphing calculator early, so it can be used to graph and solve equations prior to the introduction of functions. Material that deals with graphing devices is labeled with an icon so that those who prefer not to use the graphing calculator can easily skip this material.
The chapters on trigonometry have been written so professors can begin with either the right triangle approach or the unit circle approach.
Each approach to trigonometry is accompanied by the applications appropriate for that approach, clarifying the reason for different approaches to trigonometry.
"Mathematics in The Modern World" vignettes show that mathematics is a living science crucial to the scientific and technological progress of recent times, as well as to the social, behavioral, and life sciences.
Discovery�Discussion problems at the end of every section encourage conceptual, critical thinking, and writing skills.
Special projects are included in each chapter to engage students and make them active learners by having them work (perhaps in groups) on extended projects that give a feeling of substantial accomplishment when completed.
Review Sections and Chapter Tests at the end of each chapter help students gauge their learning progress. Brief answers to the odd-numbered exercises in each section and to all questions in the Chapter Tests are given at the back of the book.
This edition is fully supported with iLrn™ (part of the CengageNOW suite of technology products).
What's New
All of the chapters (other than Chapter "P" Prerequisites) are now followed by a "Focus on Modeling" section. Chapter "P" Prerequisites is now the only chapter be followed by a "Problem Solving" section.
There are six new "Focus on Modeling" sections, containing high-interest applications ranging from surveying to architecture.
Each exercise section now contains a special group heading for "Applications" exercises, to give these more prominence and to make them easier to assign. Each such exercise has a title to clarify its purpose at a glance.
One of the very long chapters (Chapter 9, Analytic Trigonometry) has been shortened, and the extra material from it, plus some related sections from the Analytic Geometry chapter (previously Chapter 11), have been combined into a new, more logically structured Chapter 9: Polar Coordinates and Vectors.
This edition includes several new Discovery Projects, for example the Discovery Project in 3.2 on "Relations and Functions," which was added to satisfy those users who asked for some material on relations.
More than 20% of the exercises are new, including many more application exercises. Many others have been updated and revised.
Each chapter now begins with a one-page "Chapter Overview" to introduce the ideas of the chapter and indicate how they can be applied to the real world.
Alternate Formats
Choose the format that best fits your student's budget and course goals | 677.169 | 1 |
Problem Solving Strategy
Grade: 6.5
The files below contain a problem solving strategy, including a lot of hints, tips and trick. It consists of nine seperate chapters.
Solving a problem in 7 steps (53kb)
Hints of how to solve any basic Statics problem. Of course there isn't one method to solve all Statics problems, but this chapter might be able to help anyway.
How to handle 3D problems (40kb)
3D problems are often difficult, especially because they're hard to draw. Also the rules are slightly different in 2D than in 3D. It is all explained in this chapter.
Truss structures and frames (42kb)
Things that are wise to do, and things that you absolutely shouldn't do when solving a truss problem are all mentioned in this chapter.
Using signs: always a difficult thing (49kb)
There are only 2 options, + or -. Yet which one to choose is often a riddle even the best students often fear. A method which might just eliminate those fears is described in this chapter.
Distributed Loads (41kb)
Distributed loads can be very easy, as long as you know how to handle them. And that's exactly what this chapter explains.
Sections (25kb)
How do you find the normal force, shear force and bending moment at a given point? It's not that hard, once you find a method that suits you. And it might just be the method described in this chapter.
Diagrams (33kb)
A bit similar to the previous chapter, but this time we're going to find the normal force, shear force and bending moment of the entire structure. Once more a method is explained how to find the diagrams.
Virtual Work (38kb)
Some students find it useless, but it's still often a part of the exam. This chapter contains a step by step guide through the process of deriving the virtual work equation, and solving it.
Cables and pressure vessels (37kb)
And eventually comes cables and pressure vessels, possibly one of the most difficult subjects of Statics. What tricks can you use to solve the problems, and what laws apply? It can all be found in this last chapter.
Full Version
There have been a few complains about the correctness of the data presented in these files, especially since I sometimes use different conventions than the book does. I therefore want to emphasize that these files are here to help you find a method with which you can solve problems. If you find you already have a good method of solving certain types of Statics problems, I urge you to use your own method.
Definitions
Grade: 7
There are a few things in statics which aren't pure logical. Those 'facts' are noted below. Read them carefully. | 677.169 | 1 |
Graphics with Mathematica
In this book we generate graphic images using the software Mathematica thus providing a gentle and enjoyable introduction to this rather technical software and its graphic capabilities.
The programs we use for generating these graphics are easily adaptable to many variations.
These graphic images are enhanced by introducing a variety of different coloring techniques.
Detailed instructions are given for the construction of some interesting 2D and 3D fractals using iterated functions systems as well as the construction of many different types of Julia sets and parameter sets such as the Mandelbrot set.
The mathematics underlying the theory of Iterated function systems and Julia sets is given an intuitive explanation, and references are provided for more detailed study | 677.169 | 1 |
ability theory is a rapidly expanding field and is used in many areas of science and technology. Beginning from a basis of abstract analysis, this mathematics book develops the knowledge needed for advanced students to develop a complex understanding of probability. The first part of the book systematically presents concepts and results from analysis before embarking on the study of probability theory. The initial section will also be useful for those interested in topology, measure theory, real analysis and functional analysis. The second part of the book presents the concepts, methodology and fundamental results of probability theory. Exercises are included throughout the text, not just at the end, to teach each concept fully as it is explained, including presentations of interesting extensions of the theory. The complete and detailed nature of the book makes it ideal as a reference book or for self-study in probability and related fields. Covers a wide range of subjects including f-expansions, Fuk-Nagaev inequalities and Markov triples.Provides multiple clearly worked exercises with complete proofs.Guides readers through examples so they can understand and write research papers independently | 677.169 | 1 |
'A Survey of Modern Algebra' by Garrett Birkhoff and Saunders Mac Lane has a really good treatment of linear algebra. Anything particular that you didn't find clear in Boas' treatment, or the whole topic of linear algebra?
She uses vectors to find distances between planes and such. Example given a plane 3x + 4y + 5z she immediately says the vector perpendicular to this plane is such and such. I don't know how she gets that.
I suggest, if those are the difficulties, you get Stewart's calculus book from the library (DO NOT buy it!) and reading the relevant chapter(s) on vectors. For the matrices part, try finding a book on matrix algebra, which isn't quite the same as linear algebra. Perhaps something like these books (btw I have not read them): | 677.169 | 1 |
About
Overview
MATHEMATICS FOR INFORMATION TECHNOLOGY 1E delivers easy-to-understand and balanced mathematical instruction. Each chapter begins with an application, goes on to present the material with examples, and closes with a summary of the relevant concepts and practice exercises. With numerous illustrations included as well, readers can understand the content from a number of different angles. Whether used in a classroom or an online distance-learning format, readers will find Mathematics for Information Technology an invaluable resource.
Features and Benefits
Chapter Openers introduce each chapter with interesting and motivational applications, illustrating the real-world nature of the chapter topics.
Chapter Objectives outline the knowledge and skills students will master in each chapter.
Key terms are highlighted and carefully defined to help students increase technological literacy.
Detailed examples in every chapter help students master a step-by-step approach to problem-solving. Geared toward IT majors, examples provide both technical and every-day scenarios to help students see how math concepts are applied in the real world.
Meet the Author
Author Bio
Alfred BastaStephan DeLong
Stephan DeLong is a professor of mathematics at Kaplan University. He received a Master of Science degree in Pure Mathematics from Lehigh University, in Pennsylvania.
Stephan DeLong is a professor of mathematics at Kaplan University. He received a Master of Science degree in Pure Mathematics from Lehigh University, in Pennsylvania.
Nadine Basta | 677.169 | 1 |
Fallacies in Mathematics by E. A. Maxwell
Book Description
As Dr Maxwell writes in his preface to this book, his aim has been to instruct through entertainment. 'The general theory is that a wrong idea may often be exposed more convincingly by following it to its absurd conclusion than by merely announcing the error and starting again. Thus a number of by-ways appear which, it is hoped, may amuse the professional, and help to tempt back to the subject those who thought they were losing interest.' The standard of knowledge expected is fairly elementary. In most cases a straightforward statement of the fallacious argument is followed by an exposure in which the error is traced to the most elementary source, and this process often leads to an analysis which is often of unexpected depth. Many students will discover just how mathematically minded they are when they read this book; nor is that the only discovery they will make. Teachers of mathematics in schools and technical schools, colleges and universities will also be sure to find something here to please them.
Other Editions...
Books By Author E. A. Maxwell
Concise review examines geometry of the straight line, circle, plane, and sphere as well as associated configurations, including the triangle and the cylinder. Suitable for university undergraduates and advanced high school students. 1962 | 677.169 | 1 |
Algebraic Expressions, Equations, Inequalities Flashcards
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
0.21 MB | 3 pages
PRODUCT DESCRIPTION
This product contains flashcards for common vocabulary words to be used prior to introducing a unit on algebraic concepts such as solving expression, equations, and inequalities. Use as review or at a math center throughout the year to continue supporting these concepts. Recommended for upper elementary and middle school students in both the general and exceptional education classrooms | 677.169 | 1 |
This programed mathematics textbook (Volume I) is for student use in vocational education courses. It was developed as part of a programed series covering 21 mathematical competencies which were identified by university researchers through task analysis of several occupational clusters. The development of a sequential content structure was also based on these mathematics competencies. After completion of this program the student should be able to: (1) know that "sum" indicates the operation of addition, (2) add two or three numeric fractions of the form a/b where 0 is less than a/b and when a/b is less than 100, (3) add two or three fractions of the form k/y, where 0 is less than k when k is less than 100 and y is the same literal denominator for all fractions, (4) add two or three literal fractions with the same denominators, and (5) add mixed fractions of the form Xa/b, where 0 is less than (X,a, and b) and these are less than 100. The material is to be used by individual students under teacher supervision. Twenty-six other programed texts and an introductory volume are available as VT 006 882-VT 006 909, and VT 006 975. (EM) | 677.169 | 1 |
Activity-Based Statistics, Student Guide, 2nd Edition
Activity-Based Statistics helps build real statistical understanding through a set of innovative hands-on activities that can be used each day in conjunction with other texts. The second edition continues to emphasize discovery by motivating students to apply the skills they have learned to discover the everyday relevance of statistics. There are over 45 activities and five long-term projects that have been updated and extended to encourage students to experience statistics in context. While the second edition includes updated technology extensions for Fathom, technology is used throughout the book to extend activities and is not required to complete them | 677.169 | 1 |
The objective of this book is to illustrate how the use of visualization can be a powerful tool for better understanding of some basic mathematical inequalities. Drawing pictures is a well-known method for problem solving, and the authors will convince you that the same is true when working with inequalities. They show how to produce figures in a systematic way for the illustration of inequalities and open new avenues to creative ways of thinking and teaching. In addition, a geometric argument cannot only show two things unequal, but also help the observer see just how unequal they are.
About the Authors
Claudi Alsina was born on 30 January 1952 in Barcelona, Spain. He received his BA and PhD in mathematics from the University of Barcelona. His post-doctoral studies were at the University of Massachusetts, Amherst. Claudi, Professor of Mathematics at the Technical University of Catalonia, has developed a wide range of international activities, research papers, publications and hundreds of lectures on mathematics and mathematics education. His latest books include Associative Functions: Triangular Norms and Copulas with M.J. Frank and B. Schweizer, WSP, 2006; Math Made Visual: Creating Images for Understanding Mathematics (with Roger B. Nelsen), MAA, 2006; Vitaminas Matematicas and El Club de la Hipotenusa, Ariel, 2008. | 677.169 | 1 |
Calculus
Calculus allows us to construct quantitative models of change. When we shoot an arrow, how fast does it travel? The fundamental idea is to study how things change instantaneously (over very small intervals of time), and each small interval is part of the whole puzzle. This foundation is what makes Calculus the universal tool for modeling continuous systems from physics to economics. | 677.169 | 1 |
Mathematica(r) For Physics
Mathematica(r) for Physics
Mathematica is a mathematical software system for researchers, students and anyone seeking an effective tool for mathematical analysis. This text aims to help readers learn the software in the context of solving physics problems. The graphical capabilities of Mathematica are emphasized and the readers are encouraged to use their intuition for the physics behind the problem. Each chapter has a short overview of the major physics and mathematics topics that are emphasized in the chapter. Examples in the form of solved problems cover a broad range of topics and mathematics procedures. | 677.169 | 1 |
Compiled and edited by J. Douglas Faires and David Wells
For more than 50 years, the Mathematical Association of America has been engaged in the construction and administration of challenging contests for students in American and Canadian high schools. The problems on these contests are constructed in the hope that all high school students interested in mathematics will have the opportunity to participate in the contests and will find the experience mathematically enriching. These contests are intended for students at all levels, from the average student at a typical school who enjoys mathematics to the very best students at the most special school.
In the year 2000, the Mathematical Association of America initiated the American Mathematics Competitions 10 (AMC 10) for students up to grade 10. The Contest Problem Book VIII is the first collection of problems from that competition covering the years 2001–2007. J. Douglas Faires and David Wells were the joint directors of the AMC 10 and AMC 12 during that period, and have assembled this book of problems and solutions.
There are 350 problems from the first 14 contests included in this collection. A Problem Index at the back of the book classifies the problems into the following major subject areas: Algebra and Arithmetic, Sequences and Series, Triangle Geometry, Circle Geometry, Quadrilateral Geometry, Polygon Geometry, Counting Coordinate Geometry, Solid Geometry, Discrete Probability, Statistics, Number Theory, and Logic. The major subject areas are then broken down into subcategories for ease of reference. The Problems are cross-referenced when they represent several subject areas.
About the Editors
J. Douglas Faires received his BS in mathematics from Youngstown State University in 1963, and earned his PhD in mathematics from the University of South Carolina in 1970. He has been a Professor at Youngstown State University since 1980, and has been actively involved in the MAA for many years. For example, he was Chair of the Ohio Section in 1981-1982 and Governor from 1997-2000. He is currently a member of the MAA's Undergraduate Research and Undergraduate Student Activities Committees and he was the National Director of the AMC 10 Competition of the American Mathematics Competitions until 2007.
Faires is a past President of Pi Mu Epsilon and he was a member of their Council for many years. Faires has been the recipient of many awards and honors. He was named the Outstanding College-University Teacher of Mathematics by the Ohio Section of the MAA in 1996; he has also received five Distinguished Professorship awards from Youngstown State University, and an honorary Doctorate of Science Degree in May 2006. Faires has authored and coauthored numerous books including Numerical Analysis (now in its eight edition!), Numerical Methods (third edition), PreCalculus (fourth edition), and First Steps for Math Olympians, a book in the MAA Problem Book Series.
David Wells received a PhD in mathematics from the University of Pittsburgh in 1973. He taught at Ohio Dominican University for six years and has been at the New Kensington campus of Penn State University since 1979. During his career at Penn State he has received both the Teresa Cohen Service Award from the Department of Mathematics and the Excellence in Teaching Award from the New Kensington Campus. He has been a member of the MAA since 1974, holding several sectional and national positions during that time. He served as Governor of the Allegheny Mountain Section from 1996 to 1999 and has recently been appointed as Chair of the MAA Committee on the American Mathematics Competitions. He has chaired the AMC 12 Contest Committee since 2001. Dr. Wells received the Meritorious Service Award from the Allegheny Mountain Section of the MAA in 2005. His non-mathematical interests include hiking and songwriting. | 677.169 | 1 |
BEGINNING ALGEBRA: CONNECTING CONCEPTS THROUGH APPLICATIONS shows students how to apply traditional mathematical skills in real-world contexts. The emphasis on skill building and applications engages students as they master algebraic concepts, problem solving, and communication skills. Students learn how to solve problems generated from realistic applications, instead of learning techniques without conceptual understanding. The authors have developed several key ideas to make concepts real and vivid for students. First, they emphasize strong algebra skills. These skills support the applications and enhance student comprehension. Second, the authors integrate applications, drawing on realistic data to show students why they need to know and how to apply math. The applications help students develop the skills needed to explain the meaning of answers in the context of the application. Third, the authors develop key concepts as students progress through the course. For example, the distributive property is introduced in real numbers, covered when students are learning how to multiply a polynomial by a constant, and finally when students learn how to multiply a polynomial by a monomial. These concepts are reinforced through applications in the text. Last, the authors' approach prepares students for intermediate algebra by including an introduction to material such as functions and interval notation as well as the last chapter that covers linear and quadratic modeling.
Biografía del autor:
Mark Clark graduated from California State University, Long Beach, with a Bachelor's and Master's in Mathematics. He is a full-time Associate Professor at Palomar College and has taught there for the past 13 years. He is committed to teaching his students through applications and using technology to help them both understand the mathematics in context and communicate their results clearly. Intermediate algebra is one of his favorite courses to teach, and he continues to teach several sections of this course each year. He has collaborated with his colleague Cynthia Anfinson to write a new intermediate and beginning algebra text published by Cengage Learning--Brooks/Cole. It is an applications-first approach to algebra; applications and concepts drive the material, supported by traditional skills and techniques.
Cynthia (Cindy) Anfinson graduated from UCSD's Revelle College in 1985, summa cum laude, with a Bachelor of Arts Degree in Mathematics and is a member of Phi Beta Kappa. She went to graduate school at Cornell University under the Army Science and Technology Graduate Fellowship. She graduated from Cornell in 1989 with a Master of Science Degree in Applied Mathematics. She is currently an Associate Professor of Mathematics at Palomar College and has been teaching there since 1995. Cindy Anfinson was a finalist in Palomar College's 2002 Distinguished Faculty Award | 677.169 | 1 |
Linear Algebra and Differential Equ has been written for a one-semester combined linear algebra and differential equations course, yet it contains enough material for a two-term sequence in linear algebra and differential equations. By introducing matrices, determinants, and vector spaces early in the course, the authors are able to fully develop the connections between linear algebra and differential equations. The book is flexible enough to be easily adapted to fit most syllabi, including courses that cover differential equations first. Technology is fully integrated where appropriate, and the text offers fresh and relevant applications to motivate student interest. Matrices and Determinants; Vector Spaces; First Order Ordinary Differential Equations; Linear Differential Equations; Linear Transformations and Eigenvalues and Eigenvectors; Systems of Differential Equations; The Laplace Transform; Power Series Solutions to Linear Differential Equations; Inner Product Spaces For all readers interested in linear algebra and differential equations.
Matrices and Determinants
1
(64)
Systems of Linear Equations
2
(15)
Matrices and Matrix Operations
17
(11)
Inverses of Matrices
28
(9)
Special Matrices and Additional Properties of Matrices
37
(6)
Determinants
43
(8)
Further Properties of Determinants
51
(7)
Proofs of Theorems on Determinants
58
(7)
Vector Spaces
65
(46)
Vector Spaces
66
(8)
Subspaces and Spanning Sets
74
(9)
Linear Independence and Bases
83
(12)
Dimension; Nullspace, Row Space, and Column Space
95
(11)
Wronskians
106
(5)
First Order Ordinary Differential Equations
111
(68)
Introduction to Differential Equations
112
(8)
Separable Differential Equations
120
(4)
Exact Differential Equations
124
(6)
Linear Differential Equations
130
(6)
More Techniques for Solving First Order Differential Equations
136
(8)
Modeling with Differential Equations
144
(9)
Reduction of Order
153
(4)
The Theory of First Order Differential Equations
157
(11)
Numerical Solutions of Ordinary Differential Equations
168
(11)
Linear Differential Equations
179
(52)
The Theory of Higher Order Linear Differential Equations
179
(10)
Homogeneous Constant Coefficient Linear Differential Equations
189
(14)
The Method of Undetermined Coefficients
203
(8)
The Method of Variation of Parameters
211
(6)
Some Applications of Higher Order Differential Equations
217
(14)
Linear Transformations and Eigenvalues and Eigenvectors
231
(62)
Linear Transformations
231
(14)
The Algebra of Linear Transformations; Differential Operators and Differential Equations | 677.169 | 1 |
Advanced Business Mathematics Accounting Essay
Published: 23, March 2015
Advanced Business Mathematics covers various different types of topic areas. However, the assignment only allows showing all the calculation/workings/formula for introducing to Management Statistics, Regression, Time Series, and Inferential Statistics, etc.
This research paper aims at achieving significant understanding of each topic area as we go along while providing an outline of notes from NCC education for indulgence as the mission and responsibilities is to accomplish the knowledge and understanding of the project.
Advanced Business Mathematics typically uses in commerce which includes more of probability, statistics, linear programming. Calculus is a more effective topic use in some cases of Advanced Business Mathematics, but not being asked to research on in the given assignment.
Further on in the project you will read and get the more knowledge and understanding of each particular calculation headings. The references provide opportunity for further knowledge on the concepts of the different types of calculation in the paper.
Advanced Business Mathematics typically used in commerce which includes;
Arithmetic
Algebra
Statistic
Probability
However, Advanced Business Mathematics can be more effective at times in some cases by using more of calculus, matrix algebra and linear programming. Most business organizations use mathematics in accounting, inventory management, marketing, and as well as financial analysis, etc. But mainly used for commercial enterprise purposes by recording the correct calculations in order to manage business operations.
In effect, Advanced Business Mathematics in this field is an important subject taught, and the knowledge of it will enhance ones' reasoning: the process of taking in information and making inferences based on what an individual knows to be true. Thus, a gauge of one's intellectual abilities, reasoning enables people to understand ideas and concepts better, and arrive at logical conclusions, problem-solving: try looking at existing solutions to common problems, e.g. design patterns. Maybe something similar will pop up that at least partially resemble the given problem, and the liability to think effective: the ability to apply the reasoning and logical to new or unfamiliar ideas, opinions, and solutions. In effect, try to see things in an open-minded way examining an idea or concept from as many angles possible. Hence, the mathematic tools are needed such as the; calculator, understanding, knowledge, time management, etc. in order to perform a good task.
Literature Review
Researcher argued that to understand the use of Advanced Business Mathematics, one needs to know little bit about how to conduct an investigation. In other words, since statistic are not just numbers that just appear out of nowhere, one should know the understanding of the topic in order to complete the assignment.
Dantzig, who turned 80 on November 8, is generally regarded as one of the founders of linear programming, along with von Neumann and Kntovich. Through his research in mathematical theory, computation, economic analysis, and applications to industrial problems, he has contributed more than any other researcher to the remarkable development of linear programming. In 1993 he said that linear programming is the capacity to solve the problems which originated. Linear programming and its offspring have come to of age and have demonstrably passed the test, and they are fundamentally affecting the economic practice of organization and management.
In addition, Advanced Business Mathematics is way different in coursework from a basic or regular mathematics coursework. In other words, in a regular mathematics coursework, there would be more concentrating on trigonometry rather than probability and so on for Advanced Business Mathematics. However, a more effective mathematics coursework in this level requires more Advanced Business Mathematics level such as, linear programming, regression, time series, and inferential statistic, etc.
Objectives
To familiarize ones' with the concepts and tools of Advanced Business Mathematics as applicable to decision in a business environment.
To provide quantitative skills which can be used in the assignment in order for one's to formulate, use and interpret mathematical models within a business context.
Understand applications of the learned mathematical principle.
Methodology
Advanced Business Mathematics is now used in most aspects of daily life. Thus, the researcher employed quantitative method to carry out the research; the quantitative research methods which includes both primary and secondary sources in order to gear towards the correct calculations of each heading, as well as for one's to formulate, use and interpret mathematical models within a business context.
Primary sources such as the student which have direct knowledge and the understanding of how to go about on calculating the given problems of the particular headings with the aid of getting information from the NCC education notes and lecturer notes as well. This type of source was chosen because without ones' getting the understanding and knowledge on the concept of a particular heading, there won't be a completion of the assignment. In effect, the key to motivation is understanding, and this will state whether or not the power to change both expectations of self and the value placed on attempting the task given.
Secondary sources included textbooks, references, and the internet which were used to attempt to the problems on how to go about calculating the workings given in the research paper. These secondary sources are very effective to the assignment given. Thus, it involved generalization, analysis, interpretation, or evaluation of the original information. These secondary sources are contrasts with the primary sources. In other words, they provide opportunity for further knowledge on the concepts of preparing the calculations.
Question 1
Introducing to Management Statistics
Observations (x)
Frequency (f)
FX (f)(x)
10
4
40
15
5
75
20
9
180
40
16
640
60
8
480
80
5
400
100
3
300
50
2115
Calculate;
Mean
Mode
Mean
∑ƒáµª
ƒ(n)
= 2115
50
= 42.3
Mode
Counting: 40
10
15
20
15
20
40
40
60
40
15
10
15
40
40
20
60
80
60
20
40
10
20
60
60
80
40
100
40
20
40
10
20
40
40
60
40
40
40
20
15
40
20
60
40
60
80
100
80
100
80
Question 1.2
Regression
Days of the month
Total sale of the product A (X)
Total sale of the product B (Y)
XY
X2
Y2
1st day
15
16
240
225
256
4th day
14
13
182
196
169
7th day
10
11
110
100
121
10th day
18
19
342
324
361
15th day
15
13
195
255
169
18th day
20
22
440
400
484
22nd day
16
15
240
256
225
25th day
5
117
255
225
289
27th day
12
16
192
144
256
30th day
13
17
221
169
289
TOTAL
148
159
2417
2264
2619
MEAN
14.8
15.9
241.7
Calculating the r (correlation) between x and y;
R= r = n∑x;y; - ∑x;∑y;
(n∑x;-(∑x;(2 n)∑y;)2-(∑y;)2(
= 10(2417)-(148*159)
10(2264-(148)2 (10(2619-159)2)
= 24170-23,532
(22,640-21904)(26190-25281)
= 638
(736)(909)
= 638
669,024
= 638
817.938873
= 0.78
Question 2.1
Calculate the Laspeyres and Paache price indices for the following data. Take from 2005 as the base year.
Litre of beer
Litre of Whiskey
Litre of wine
Year
Price
Qty
Price
Qty
Price
Qty
2005
0.95
200
19.80
10
10.50
36
2006
0.99
150
20.39
12
11.15
48
2007
1.05
120
20.99
11
12.35
60
Laspeyres formula: LPI = ∑q0 Pn
∑q0 P0
LPI05 = ∑q05 P05 * 100
∑q05 P05
= (200*0.95) + (10*19.80) + (36.10.50) *100
(200*0.95) + (10*19.80) + (36.10.50)
= 190 + 198 + 378 *100
190 + 198 + 378
= 766 *100
766
= 100%
LPI06 = ∑q05 P06 * 100
∑q05 P05
= (200*0.99) + (10*20.39) + (36*11.15) *100
(200*0.95) + (10*19.80) + (36.10.50)
= 198 + 203.9 + 401.4 *100
190 + 198 + 378
= 803.3 *100
766
= 104.8%
LPI07 = ∑q05 P07 * 100
∑q05 P05
= (200*1.05) + (10*20.99) + (36*12.35) *100
(200*0.95) + (10*19.80) + (36.10.50)
= 210 + 209.9 + 44.6 *100
190 + 198 + 378
= 864.5 *100
766
= 112.8%
Paache formula: PPI= ∑qn Pn *100
∑pn Po
PPI= ∑q05 P05*100
∑p05 P05
= (200*0.95) + (10*19.80) + (36*10.50)*100
(200*0.95) + (10*19.80) + (36*10.50)
= 190 + 198 + 378 * 100
190 + 198 + 378
= 766 *100
766
= 100%
PPI= ∑q05 P05*100
∑p05 P05
= (150*0.99) + (12*20.39) + (48*11.15) *100
(150*0.95) + (12*19.80) + (48*10.50)
= 148.5 + 244.68 + 535.2 *100
142.5 + 237.6 + 504
= 928.38 * 100
884.1
= 105%
PPI= ∑q05 P05*100
∑p05 P05
= (120*1.05) + (11*20.99) + (60*12.35) *100
(120*0.95) + (11*19.80) + (60*10.50)
= 126 + 230.89 + 741 *100
114 + 217.8 + 630
= 1097.89 *100
961.8
= 114%
Question 2.2
Time series
The number of rats captured in a grain store is summarized below. Use simple exponential smoothing with alpha = 0.2 and alpha = 0.7 to forecast the number of rats that will be caught in week 7.
= 8.15
Formula: X±Zαz(s n)
= 36.269
= 28.591
99% confident that the estimate of the mean breakfast time is between 28.591 and 36.269.
Question 2.4
Linear Programming
Minimize; cost = 9x + 3y
Subject to the following constraints;
Constraints 1: Y ≥ 5
Constraints 2: 6X + 7Y ≥ 210
Constraints 3: 7X + 15Y ≤ 525
Constraints 4: 5X + 28Y ≤ 700
Constraints 5: X ≥ 0, Y ≥ 0
Which of the following constraints are binding and non-binding?
Solving minimum cost
Corners
Expression to find minimum cost using 9x + 3y
(30,5)
9(30) + 3(5) =285
(65,5)
9(65) + 3(5) = 600
(10,23)
9(10) + 3(23) = 165
(40,18)
9(40) + 3(18) = 414
MINIMUM COST = 165
Question 2.5
Decision Tree
What is decision tree?
According to Advanced Business Mathematics lecture notes, based on my understanding, a decision tree is a tool that uses a tree-like graph shape. It is also a model of decision and their possible consequences which includes;
Notes which are decision, or choices made, and events.
Braches which re line that shows the sequences of levels and decision made, and routs into the future.
Decision tree is another way to display on algorithm and company used in operations research, specifically in decision analysis to help identify a strategy that must likely to reach goal oriented, for e.g. if a company plans to launch a new product version on the market, the firm would want to know what to be done, and how it should be done in order to geared towards goal oriented.
Using example of a typical decision tree discuss the TWO uses of the decision tree.
Since a company decided to launch a new product version or maintain its existing product version, then the cash flows will depend on the products success or failing ( not known with certainty), but if the old product is maintain by future cash flows, it can then be used to predicted with accuracy, for e.g.
Failure -100
Success 100
Old product 35
In addition, the two uses of the decision tree are to find out;
whether the firm will either have a maximum profit, revenues, values or a minimum for costs and losses to launched the product version or not.
To know if future returns are being described by a probability distribution and if he average value for the future is calculated as well.
Question 2.6
Differentiation
Some people believe that differentiation is just mathematics that involves random calculation. Explain TWO ways in which differentiation can be applied to economic analysis.
In such sense, differentiation states the way through which quality of goods is improved over time. In other words, launching new goods effectively and efficiently is a way by which an increase in radical changes occurring; often lead to changes in market shares, as well as the industry structures.
In effect, applying economic analysis using differentiation, vertical differentiation will be used. However, vertical differentiation occurs in market where several goods that are present can ordered according to the objectives quality.
Findings
The following interpretation was obtained from NCC notes that were sent to email, and from notes that was given in class.
Advanced Business Mathematics covers numerous and various different types of topics. However, the assignment only allows showing all calculation/workings/formula for introducing to Management Statistics, Regression, Time Series, and Inferential Statistics, etc. it is an important function use in business organization these days.
It is designed to ensure that effectiveness and efficiency of operations reliability of proper workings and compliance with applicable laws and regulations.
Advanced Business Mathematics typically uses in commerce which includes more of probability, statistics, linear programming. Calculus is a more effective topic uses in some cases of Advanced Business Mathematics.
Recommendation
The assignment that was done on Advanced Business Mathematics was a challenging act; still, valid and reliable information was gathered. The researcher has nothing but positive comments towards the assignment. However, since Advanced Business Mathematics cover difficult areas the researcher would like to recommend ones' to continue on the same path by doing extremely well; try not getting too comfortable and always searching for improvement. The researcher recommendation to the ones has to relation with another is for them to continue being has responsible and effective when doing their work.
The assignment has accomplished success base on the time management that was put in when attending the assignment.
Conclusion
The following information was interpreted from NCC education ( notes and notebook. The information conveyed is based on the judgment of management with an appropriate consideration to materiality. In this regard, this project will takes time out for understanding each topics and to ensure that the calculations are properly authorized and recorded. In effect, time management is effective tools that were used in order to complete the assignment.
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This syllabus presents a fused course in plane, solid, and coordinate geometry for secondary school students. Elementary set theory, logic, and the principles of separation provide unifying threads throughout this approach to geometry. There are actually two curriculum guides included; one for each of two different texts--Henderson, Pingry, and Robinson's "Modern Geometry" and Jurgensen, Donnelly, and Dolciani's "Modern Geometry." This curriculum guide is one of several prepared for secondary school mathematics instruction by Baltimore County Public Schools. (JG) | 677.169 | 1 |
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Optimization : theory and practice
Gives a detailed mathematical exposition to various optimization techniques. This book includes topics such as: Single and multi-dimensional optimization, Linear programming, Nonlinear constrained optimization and Evolutionary algorithms.Read more... | 677.169 | 1 |
Algebra for College Students, 4th Edition, is designed to provide students with the algebra background needed for further college-level mathematics courses. The unifying theme of this text is the development of the skills necessary for solving equations and inequalities, followed by the application of those skills to solving applied problems. This text contains 2 chapters, Polynomial & Rational Functions, and Counting & Probability, in addition to those found in Dugopolski's Intermediate Algebra.
"synopsis" may belong to another edition of this title.
From the Publisher:
An increased emphasis on real-data applications that involve graphs is a focus for the third edition. Some exercises have been updated throughout the text to help demonstrate concepts, motivate students, and to give students practice using new skills. Many of the real data exercises contain data obtained from the Internet. Internet addresses are provided as a resource for both students and teachers. Because internet addresses frequently change, a list of addresses will also be available on the website so that they may be more effectively maintained. An Index of Applications listing applications by subject matter is included at the front of the text. The third edition contains three new margin features that appear throughout the text: Calculator Close-Ups give students an idea of how and when to use a graphing calculator. Some Calculator Close-Ups simply introduce the features of a graphing calculator, but many are intended to enhance understanding of algebraic concepts. For this reason, many of the Calculator Close-Ups will benefit even those students who do not use a graphing calculator. Study Tips are included in the margins throughout the text. These short tips are meant to continually reinforce good study habits and keep reminding students that it is never too late to make improvements in the manner in which they study. Helpful Hints are short comments that enhance the material in the text, provide another way of approaching a problem, or clear up misconceptions. Every section in the third edition generally begins with six simple writing exercises; these exercises appear in the exercise sets. These exercises are designed to get students to review the definitions and rules of the section before doing more traditional exercises. For example, the student might be simply asked what properties of equality were discussed in this section. Each chapter begins with a Chapter Opener that discusses a real application of algebra. The discussion is accompanied by a photograph and, in most cases by a real-data application graph that helps students visualize algebra and more fully understand the concepts discussed in the chapter. In addition, each chapter contains a Math at Work feature, which profiles a real person and the mathematics that he or she uses on the job. These two features have corresponding real data exercises. Every section begins with In This Section, a list of topics that shows the student what will be covered. Because the topics correspond to the headings within each section, students will find it easy to locate and study specific concepts. Important ideas, such as definitions, rules, summaries, and strategies, are set apart in boxes for quick reference. Color is used to highlight these boxes as well as other important points in the text. At the end of every section are Warm-up exercises, a set of ten simple statements that are to be answered true or false. These exercises are designed to provide a smooth transition between the ideas and the exercise sets. They help students understand that every statement in mathematics is either true or false. They are also good for discussion or group work. The end-of-section Exercises follow the same order as the textual material and contain exercises that are keyed to examples, as well as numerous exercises that are not keyed to examples. This organization allows the instructor to cover only part of a section if necessary and easily determine which exercises are appropriate to assign. The keyed exercises give the student a place to start practicing and building confidence, whereas the non-keyed exercises are designed to wean the student from following examples in a step-by-step manner. Getting More Involved exercises are designed to encourage writing, discussion, exploration, and cooperative learning. Graphing Calculator Exercises require a graphing calculator and are identified with a graphing calculator logo. Exercises for which a scientific calculator would be helpful are identified with a scientific calculator logo. Every chapter ends with a four-part Wrap-up, which includes the following: The Chapter Summary lists important concepts along with brief illustrative examples. Enriching Your Mathematical Word Power NEW! appears at the end of each chapter and consists of multiple choice questions in which the important terms are to be matched with their meanings. This feature emphasizes the importance of proper terminology. The Review Exercises contain problems that are keyed to the sections of the chapter as well as numerous miscellaneous exercises.
The Chapter Test is designed to help the student assess his or her readiness for a test. The Chapter Test has no keyed exercises, thus enabling the student to work independently of the sections and examples. At the end of each chapter is a Collaborative Activities feature which is designed to encourage interaction and learning in groups. Instructions and suggestions for using these activities and answers to all problems can be found in the Instructor's Solutions Manual. The Making Connections exercises at the end of each chapter are designed to help your students review and synthesize the new material with ideas from previous chapters, and in some cases, review material necessary for success in the upcoming chapter. Every Making Connections exercise set includes at least one applied exercise that requires ideas from one or more of the previous chapters. NetTutor is a revolutionary new web-based learning environment for the live dissemination of mathematical content. NetTutor offers your students live, personalized tutoring via the internet. Using NetTutor's powerful WWWhiteboard software, students can post a question and receive prompt feedback from an expert in their subject. The WWWhiteboard allows students and tutors to use proper mathematical notation as well as other highlighting features - truly making this a unique learning experience. Students may also post questions to the Q&A Center and receive a reply within 24 hours. Visiting the Message Center allows students to discuss difficult concepts among themselves, while the Archive Center provides a browseable list of questions and answers maintained by the subject tutor. NetTutor is FREE and an invaluable aid for all students; the study partner who always has the answer. | 677.169 | 1 |
Algebra
Algebra is the mathematical language for taking what is known and using it to solve for what is unknown. Given that goal, the tools you'll learn in this topic are as diverse as the range of situations they apply to: from using simple equations to estimate your cab fare to using complex numbers and trigonometric functions to model the behavior of light and sound waves. | 677.169 | 1 |
Find a LawrenceThorough understanding of the theoretical underpinnings of this powerful tool can be left to the math majors. Those who ask for help in a calculus course are most often taking it as a requirement for a technical field. Here, the practical application of derivatives and integrals are what is important | 677.169 | 1 |
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This useful guide helps both new students and those who need a refresher course to acquire practical skills in calculus through a series of 20 lesson plans that require a minimal time commitment.
All key calculus topics are covered, from common functions and their graphs to differentiation, integration, and infinite series. The book contains hundreds of practice exercises without a lot of unnecessary theory or math jargon | 677.169 | 1 |
Elementary Mathmatics from an Advan (Paperback) - TaschenbuchFelix Klein:This text begins with the simplest geometric manifolds, the Grassmann determinant principle for the plane and the Grassmann principle for space; and more. Also explores affine and projective transformations; higher point transformations; transformations with change of space element; and the theory of the imaginary. Concludes with a systematic discussion of geometry and its foundations. 1939 edition. 141 figures. Elementary Mathematics from an Advanced Standpoint: Geometry Klein, Felix / Hedrick, E. R. / Noble, C. A., Dover Publications
Klein, Felix
Titel:
Elementary Mathematics from an Advanced Standpoint: Geometry
ISBN-Nummer:
0486434818
This comprehensive treatment features analytic formulas, enabling precise formulation of geometric facts, and it covers geometric manifolds and transformations, concluding with a systematic discussion of fundamentals. 1939 edition. Includes 141 figures. | 677.169 | 1 |
4.2 Applications of Eigenvalues/Eigenvectors Markov Chains A Markov chain is a discrete-time stochastic process used to model certain systems in physics and statistics, for example. With a Markov Chain, a transition matrix is used to predict future states
Chapter 1
Linear Equations vs. Nonlinear Equations
Definition In general, a linear equation with n unknowns is an equation that can be placed in the form a1x1 + a2x2 + anxn. = b In the equation above, a have known values and are called coefficients; b als
Math 1114 Practice Questions for Tests 2 & 3
Note: This is not a practice test. It is, rather, a set of practice questions designed to assist you in preparing for the test. While this set of questions is based upon past tests, we cannot guarantee that it
Math 1114 Practice Questions for Test 1
Note: This is not a practice test. It is, rather, a set of practice questions designed to assist you in preparing for the test. While this set of questions is based upon past tests, we cannot guarantee that it repre
Section 4.1 Eigenvalues and Eigenvectors Definition: Let A be an n n matrix, u be a nonzero n 1 vector, and be a constant. If then is called an eigenvalue for the matrix A and u is called the eigenvector corresponding to .
Au = u
Example: Let
3 5 A= 1 -1
Section 3.7 Cross Products
In this section, we'll be looking for a vector that is orthogonal to two given vectors. There are an infinite number of answers! One specific one is called the cross product. THE CROSS PRODUCT OF TWO THREE-DIMENSIONAL VECTORS
LE
SECTION 3.3
SCALAR MULTIPLICATION:
3 6 2 = -1 -2
Let u be a vector; let c be a scalar; and set v = cu. (1) If c > 0, then v is in the same direction as u. (2) If c < 0, then v is in the opposite direction from u.
The length or magnitude of a
u1 2 2 u= u2
Three representations Of a Vector
1. Physical-real life event such as a force or velocity. It has both magnitude and direction. 2. Geometric-picture in the coordinate plane with initial point and terminal point. It's a directed line segment which represen
Section 4.1
Eigenvalues and
Eigenvectors
Definition: Let A be an
n n matrix, u be a
nonzero n 1 vector, and
be a constant. If
Au = u
then is called an
eigenvalue for the matrix A
and u is called the
eigenvector corresponding
to .
Example: Let
3 5
A=
1 1
Section 3.7
Cross Products
In this section, well be looking for
a vector that is orthogonal to two
given vectors. There are an
infinite number of answers! One
specific one is called the cross
product.
THE CROSS PRODUCT OF TWO
THREE-DIMENSIONAL VECTORS
LET4.2 Applications of Eigenvalues/Eigenvectors Markov Chains
A Markov chain is a discrete-time stochastic process used to model certain
systems in physics and statistics, for example. With a Markov Chain, a transition
matrix is used to predict future states | 677.169 | 1 |
Elementary Theory of Numbers: Second English Edition (edited by A. Schinzel)
Kobo ebook | February 1, 1988
Since the publication of the first edition of this work, considerable progress has been made in many of the questions examined. This edition has been updated and enlarged, and the bibliography has been revised.
The variety of topics covered here includes divisibility, diophantine equations, prime numbers (especially Mersenne and Fermat primes), the basic arithmetic functions, congruences, the quadratic reciprocity law, expansion of real numbers into decimal fractions, decomposition of integers into sums of powers, some other problems of the additive theory of numbers and the theory of Gaussian integers.
Pricing and Purchase Info
$73.79 online
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From the Publisher
Since the publication of the first edition of this work, considerable progress has been made in many of the questions examined. This edition has been updated and enlarged, and the bibliography has been revised.The variety of topics covered here includes divisibility, diophantine equations, prime numbers (especially Mersenne and Fermat ... | 677.169 | 1 |
Master Math: Trigonometry: Trigonometry
Overview procedures and solutions, along with examples and applications. A complete table of contents and a comprehensive index enable you to quickly find specific topics, and the approachable style and format facilitate an understanding of what can be intimidating and tricky skills. Perfect for both students who need some extra help or rusty professionals who want to brush up, Master Math: Calculus will help you master everything from series and approximations to partial derivatives.
Related Subjects
Meet the Author
Debra Anne Ross has a double BA in Chemistry and Biology from the University of California, Santa Cruz, and and MS in Chemical Engineering from Stanford University. Debra's career encompasses biology, chemistry, biochemistry, engineering, biosensors, pharmaceutical drug discovery, and intellectual property. She is the author of the popular Master Math books, The 3:00 PM Secret: Live Slim and Strong Live Your Dreams, The 3:00 PM Secret: Ten Day Dream Diet (2009), and Arrows Through Time: A Time Travel Tale of Adventure, Courage, and Faith (2009).
Customer Reviews
Most Helpful Customer Reviews
This is the best book on learning basic algebra. It is thorough yet concise. The information is presented very clearly. The author has obviously tried to explain the concepts so that they `make sense¿ to students - and their parents. I use the book to explain algebra to my students. Like the other Master Math books by Ross, the topics flow logically and build in difficulty. What a breath of fresh air after the often confusing text books students are given in school. This book is helpful for students struggling with algebra and the parents who are tutoring them. This book is also extremely useful for older students who did not adequately learn algebra, yet find they need to know it later. Topics can easily be looked up and reviewed or learned. I highy recommend this book!
Log-IC
More than 1 year ago
Splendid clarity and progression.
A cherry on top of this fine sundae,
maybe a single page of exponential,
radical,and complex properties for
a quick reference,( but that's nitpicking).
Guest
More than 1 year ago
This is the best book out there on learning trigonometry. I especially appreciate the visually-oriented focus. Each concept is described in all its forms, such as sine. Do you know each of the different ways sine can be described? Like the other Master Math books by Ross, the topics flow logically and are in context with what precedes and follows. It is thorough yet concise, and packed full of everything you, as tutor, or your kids need to know. The real world and fun applications are wonderful! The information is explained clearly and in a way that makes sense, so that a given concept is explained in such a way you understand what is being discussed rather than just memorizing formulas. What a breath of fresh air after the often confusing text books I was and my children are given in school. I really feel I can explain trigonometry to young people using this book! if I were going back to school, and taking math or science, this book would be in my backpack.
Anonymous
More than 1 year ago
Anonymous
More than 1 year ago
i am not sure if i am missing something but on page 14 the book says x times $1.00 per glass will equal 20. then shows this equation
X + ($1.00 per glass)=$20.00 terrible. There are many books that are better. I found websites more helpful. | 677.169 | 1 |
Performs basic scientific and trigonometric functions.
Ideal for general math, pre-algebra, algebra 1 and 2, trigonometry and biology.
Features conversions and fraction calculations.
Performs basic scientific and trigonometric functions. Ideal for general math, pre-algebra, algebra 1 and 2, trigonometry and biology. Features conversions and fraction calculations. | 677.169 | 1 |
Statistics for the Utterly Confused is your user-friendly introduction to elementary statistics, designed especially for non-math majors.Required courses in statistics are cause for alarm among more than 500,000 undergraduates in such disciplines as nursing, allied health, pre-law, pre-medicine…
When it comes to understanding one of your most intimidating courses–English Grammar–even good students can be confused. This guide is a must-have for everyone, from students taking the GED to professionals writing business plans, as it explores the structures of English grammar and how to use them easily and proficiently.
When it comes to understanding one of your most intimidating courses – calculus – even good students can be confused. Intended primarily for the non-engineering calculus student (though the more serious calculus student will also benefit), Calculus for the Utterly Confused is your ticket to success. Calculus concepts are explained and applied in such diverse fields as business, medicine, finance, economics, chemistry, sociology, physics, and health and environmental sciences. The message of Calculus for the Utterly Confused is simple: You don t have to be confused anymore. With the wealth of expert advice from the authors who have taught many, many confused students, you ll discover a newer, fresher, clearer way to look at calculus. Don t wait another minute – get on the road to higher grades and greater confidence, and go from utterly confused to totally prepared in no time!
Play Ball! Everything baseball—from the popular Utterly Confused Series What's a foul ball? Or a swinging third strike? New fans, parents, and first-time coaches need no longer be Utterly Confused about baseball, as the popular series introduces the basics of the sport in a fun and easy-to-follow guide. | 677.169 | 1 |
MATH1131 Calculus
2.1: Title
MATH1131 Calculus
2.2: Introduction
Limits as x . We are interested in what happens to
a function f (x) as x becomes very large. For example, in
mathematical studies of ecology we may wish to know whether
the populat
2012 HSC Mathematics Extension 2
Sample Answers
When examination committees develop questions for the examination, they may
write sample answers or, in the case of some questions, answers could include.
The committees do this to ensure that the questions
2011 Mathematics Extension 2
HSC Examination
Sample Answers
When examination committees develop questions for the examination, they may
write sample answers or, in the case of some questions, answers could include.
The committees do this to ensure that th
2002200120042005Recent Documents
Intro to Complex Numbers
Chris Tisdell
Chris Tisdell
Complex Numbers
1 / 57
Where are we going?
We will learn about a new kind of number known as a complex
number.
We will discover the basic properties of complex numbers and
investigate some of their math
MATH1131 Calculus
Chapter 8
Integration
1
Integration is concerned with the geometric problem of finding
the area of a given region.
We start by trying to calculate the area between the x axis and
the curve y = f (x) for a x b, where
f (x) < M, i.e. f is
MATH1131 Calculus
Chapter 1
Sets, inequalities and
functions
1
Sets and elements
A set is a collection of distinct objects.
The objects in a set are called the elements or members of
the set.
Example 1.1
MATH1131 is an element of the set of all mathematic
MATH1131 Calculus
Chapter 5
The Mean Value Theorem and
its applications
1
We are interested in questions like:
If f exists on Dom(f ), a subset of R,
can we relate values of f to values of f ?
can we compare values of f to values of another
differentiabl
Showing 1 to 2 of 2
Very thorough introductory course for Maths at University level. Certainly, recommend to students who have completed Extension 1 or outgoing Mathematics students since this course is essentially a recap of Extension 1/Mathematics topics such as integration and differentiation. It i split into 2 sections - Algebra and Calculus.
Keep up to date with the online tutorials and computer quizzes. There are no reminders for them so it is your responsibility to meet the deadlines.
Attempt as many tutorial questions as possible and keep your notes compact so you can refer to them later on in the course/exam time.
Get the best marks possible for your class tests - this will save you having to outperform in the final exam. | 677.169 | 1 |
Communication is considered an important component of mathematics education in order for d
Communication is considered an important component of mathematics education in order for d
Jan 18th, 2016
SoccerBoss
Category:
Mathematics
Price: $5 USD
Question description
Communication is considered an important component of mathematics education in order for deep learning to occur. Choose a specific algebra concept that you think you understand well, and explain it in your own words. Then, present a math problem for your classmates to solve, making use of this concept. | 677.169 | 1 |
Mathematics for PhysicMore...
This University--and the response has been enthusiastic from students and instructors alike. Because physics students are often uncomfortable using the mathematical tools that they learned in their undergraduate courses, MATHEMATICS FOR PHYSICISTS provides students with the necessary tools to hone those skills. Lea designed the text specifically for physics students by using physics problems to teach mathematical concepts.
Preface
Describing the Universe
A Universal Language
Scalar and Vector Fields
Curvilinear Coordinates
The Helmholtz Theorem
Vector Spaces
Matrices
Problems
Complex Variables
All About Numbers
Functions of Complex Variables
Complex Series
Complex Numbers and Laplace's Equation
Poles and Zeros
The Residue Theorem
Using the Residue Theorem
Conformal Mapping
The Gamma Function
Problems
Differential Equations
Some Definitions
Common Differential Equations Arising in Physics
Solution of Linear, Ordinary Differential Equations
Numerical Methods
Partial Differential Equations: Separation of Variables
Problems
Fourier Series
Fourier's Theorem
Finding the Coefficients
Fourier Sine and Cosine Series
Use of Fourier Series to Solve Differential Equations
Convergence of Fourier Series
Problems
Laplace Transforms
Definition of the Laplace Transform
Some Basic Properties of the Transform
Use of the Laplace Transform to Solve a Differential Equation
Some Additional Useful Tricks
Convolution
The General Inversion Procedure
Some More Physics
Problems
Generalized Functions in Physics
The Delta Function
Developing a Theory of Distributions
Properties of Distributions
Sequences and Series
Distributions in N Dimensions
Describing Physical Quantities Using Delta Functions
The Green's Function
Problems
Fourier Transforms
Definition of the Fourier Transform
Some Examples
Properties of the Fourier Transform
Causality
Use of Fourier Transforms in the Solution of Partial Differential | 677.169 | 1 |
Description: For a one- or two-term introductory course in discrete mathematics. Focused on helping students understand and construct proofs and expanding their mathematical maturity, this best-selling text is an accessible introduction to discrete mathematics. Johnsonbaugh's algorithmic approach emphasizes problem-solving techniques. The Seventh Edition reflects user and reviewer feedback on both content and organization | 677.169 | 1 |
Summary and Info
Bignum math is the backbone of modern computer security algorithms. It is the ability to work with hundred-digit numbers efficiently using techniques that are both elegant and occasionally bizarre. This book introduces the reader to the concept of bignum algorithms and proceeds to build an entire library of functionality from the ground up. Through the use of theory, pseudo-code and actual fielded C source code the book explains each and every algorithm that goes into a modern bignum library. Excellent for the student as a learning tool and practitioner as a reference alike BigNum Math is for anyone with a background in computer science who has taken introductory level mathematic courses. The text is for students learning mathematics and cryptography as well as the practioner who needs a reference for any of the algorithms documented within. | 677.169 | 1 |
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This title is In Stock in the Booktopia Distribution Centre. We will send you a confirmation email with a Tracking Code to follow the progress of your parcel when it ships.
Maths Quest 9 for the Australian Curriculum & eBookPLUS + Maths Quest 9 for the Australian Curriculum Homework Book Value Pack Author: Jacaranda
ISBN: 9780730302360 Format: Multi-Copy Pack Number Of Pages: 0 Published: 27 June 2013 Description: Maths Quest 9 for the Australian Curriculum provides students with essential mathematical skills and knowledge through the content strands of Number and Algebra, Measurement and Geometry, and Statistics and Probability. The Curriculum focuses on students becoming proficient in mathematical understanding, fluency, reasoning and problem solving. Maths Quest 9 for the Australian Curriculum is specifically written and designed to meet the requirements and aspirations of the Australian Mathematics Curriculum. The student textbook contains the following features: clear and engaging design judicious use of ICT resources a numeracy chapter two chapters on problem solving Individual pathways activities for every exercise a Hungry brain class activity for each chapter two new ProjectsPLUS activities interactivities eLessons references to the content and proficiency strands of the new Austral | 677.169 | 1 |
If you want to improve your Algebra word problem-solving skills, this book is filled with what you need the most: Practice
"400 Practice Algebra Word Problems (With Help and Solutions)" will make a great standalone or supplemental practice guide for you if you're serious about developing your math word problem-solving skills or raising your grades in school.
It contains 400 practice word problems that will sharpen your skills at solving problems involving addition, subtraction, multiplication, division, mixed-operations, systems of equations, mixtures, rates and time, work, and even more
It starts simple and will gradually build your skills from the ground up by presenting word problems from basic to more difficult. And in case you come upon any word problem that gives you trouble, it provides sample equations for each word problem to give you a hint or a nudge in the right direction. Solutions are also given to ensure that you will arrive at the correct answers.
But that's not all. "400 Practice Algebra Word Problems (With Help and Solutions)" also contains an entire section dedicated to giving you hints, tips, and useful tricks that they don't teach you in school to help you master the hardest part about solving word problems--translating the written words into mathematical equations.
And unlike other books, it won't lock you into a rigid, step-by-step solving process or force you to solve word problems in any particular way. It gives you the opportunity to practice and learn in the way that suits you best | 677.169 | 1 |
Summary and Info
This is the second of three volumes that originated from a series of lectures in mathematics given to high school students by professors of Kyoto University in Japan. The main purpose of the lectures was to show the beauty of mathematics using material that is accessible to those with little preliminary knowledge. Discussion of the theory of trigonometric and elliptic functions and examination of various aspects of the Poncelet closure theorem illustrates the use of algebraic geometry as a method of studying geometric properties of figures using algebra as a tool. The book was originally published in Japanese in 1995 by Iwanami Shoten, Publishers, Tokyo, Japan. Author and translator information is not given | 677.169 | 1 |
A Treatise On the Science & Practical Detail of
Focus by Grade Level (K–8) – Achieve the Core – a collection of PDFs detailing the mathematical content emphasized in the standards by grade level. AskIITians offers you all IIT study material, question bank, mock test papers or free test series and that too in perfect alignment with the IIT JEE format. Such readymade books are only skimmed down version of original topicbooks. Fooled by Randomness: The Hien Role of Chance in the Markets and in Life, 2nd ed.
The student uses the process skills to understand and apply relationships in right triangles. The student is expected to: (A) determine the lengths of sides and measures of angles in a right triangle by applying the trigonometric ratios sine, cosine, and tangent to solve problems; and (B) apply the relationships in special right triangles 30�-60�-90� and 45�-45�-90� and the Pythagorean theorem, including Pythagorean triples, to solve problems. (10) Two-dimensional and three-dimensional figures Instructor's Calculator Supplement to Accompany COLLEGE ALGEBRA AND TRIGONOMETRY. Many of these come out quite easily if you express everything on the most complex side in terms of sine and cosine only. In most examples where you see power 2 (that is, 2), it will involve using the identity sin2 θ + cos2 θ = 1 (or one of the other 2 formulas that we derived above) Elements of Trigonometry, and Trigonometrical Analysis, Preliminary to the Differential Calculus. He gains his students' interest when he shows how to use probability math to gain an advantage in gambling (which gets him in a bit of trouble when his superiors find out). And while drug kingpin Russel Bell puts his evening lessons in economics to good use during the day, his enthusiasm fails to rub off to his underlings or his partner-in-crime, Avon Barksdale download A Treatise On the Science & Practical Detail of Trigonometrical Surveying, with Their Applications to Surveying in General, (Besides a Minute ... by Figures, of the Repeating Circle ...) pdf. In MATH 13300, subjects include more applications of the definite integral, an introduction to infinite sequences and series and Taylor expansions. MATH 13300 also includes an introduction to multivariable calculus, such as functions of several real variables, partial derivatives, gradients, and the total derivative, and integration of functions of several variables Trigonometry : an analytic approach. In Fig. 10 AB, CD represent two parallel straight lines. Transversal A straight line such as PQ which cuts them is called a transvenal read A Treatise On the Science & Practical Detail of Trigonometrical Surveying, with Their Applications to Surveying in General, (Besides a Minute ... by Figures, of the Repeating Circle ...) online. See also: E = MC Hammer, in which writers who cannot do math try to represent it onscreen, and Mad Mathematician, in which people who do enjoy math are portrayed as utterly deranged. Note: Americans say "math" on the basis that it's an uncountable noun, but in the UK, they say "maths" on the basis that it's short for "mathematics," which, while still an uncountable noun, ends with "s."
To find out more about Aberdeen's course and what it entails, go to High School Diploma. Guidance for K–8 and High School in determining the major areas of focus. Claim Distribution (K–8, HS, and EOC) – WA Assessment System – guidance on the clusters that should be a major focus, an additional/supporting focus, and how the math practices are applied across the content of the standards Logorithmic and Trigonometric Tables: Revised Edition. This note contains the following subtopics: Exponential and Logarithmic Functions, Trigonometric Functions, Circular Functions and Their Graphs, Trigometric Identities and Equations. This note explains the following topics: Annual Temperature Cycles, Trigonometric Functions, Trigonometric Models: Vertical Shift and Amplitude, Frequency and Period, Phase Shift, Examples, Phase Shift of Half a Period, Equivalent Sine and Cosine Models Algebra and Trigonometry: Graphs and Models with Graphing Calculator Manual (2nd Edition). | 677.169 | 1 |
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Read More use various graphing calculators to solve problems more quickly. Perhaps most important-this book effectively prepares readers for further courses in mathematicsCustomer Reviews
Algebra and Trigonometry with Analytic Geometry
by Earl William Swokowski
An ordinary text
This is a common secondary level text. Like most current texts of this type, it fails to give the reader any understanding of the breadth of this subject as developed over the last 150 years. There are a number of useful tricks of interest to engineers, computer graphics specialists, and other 3D modelers that this text does not even point to for additional reading. However, for its intended audience, it delivers adequately but still does not entice the imagination much | 677.169 | 1 |
AND HEAD OF THE MATHEMATICAL DEPARTMENT OF BEDFORD COLLEGE FORMERLY FELLOW OF MAGDALEN COLLEGE, OXFORD, AND ASSISTANT
MATHEMATICAL LECTURER AT THE UNIVERSITY COLLEGE, BANGOR
OXFORD
AT THE CLARENDON PRESS
1920
&c.PEEFACE
THOUGH
reader has not
the theory of plane algebraic curves
still attracts
mathematical students.
part.
all
many
Salmon's classic treatise supplied
desired at the time of
edition
its
that could be
appearance.
it
would
be almost impossible to trace the steps by which particular results have reached their present form.
a
field
in
which has attracted
so
many
workers. I have not
In
fact. the English
suitable books at his disposal. XX. Teixeira.
attempted to give systematic references.
But most of the contents and examples are extracted from a very large
number
of mathematical periodicals.
list at
ception of the
the end of
With the exCh. if only to bring some more recent developments within reach
of the student.
. It seemed therefore as if a new book on the subject might be useful.
The
solutions
mine
most
even in the case of examples derived from other
authors. XVII.
was published some
and has
been long out of print. Basset. XI.
of
my own
are
contributions to
for the
the subject. but the last
forty years ago. In some cases
I
cannot eveli remember whether a
.
In the preparation of this volume
Wieleitner. Loria.
result is my own or not and XVIII contain most
but Chapters IX.
I
have made
frequent use of the books written by Salmon.
for
example. into which any quartic with two
unreal biflecnOdes can be projected.
No
one can really master a branch of mathematics
it "himself.
work has been published on
The reader will doubtless detect other important omissions.vi
PREFACE
In a book dealing with so wide a subject I can hardly hope to escape the criticism that I have included just that material which happens to interest
myself.
I
have not
thorough
in-
cluded any discussion of curves of degree n for special
values of n other than
2.
except by working at
I
therefore.
and have not attempted to subdivide the
all their
type by considering
possible positions relative
to the line at infinity. But on the whole I have tried
to cover
the limited ground I have selected with
reasonable completeness. such as
touch
How many
conies
five given conies?' I have treated all curves with the same degree and singularities as forming a
single type.
I
have not seriously dealt with
'
problems of enumeration.
A
dis-
cussion of quintic curves would be very welcome.
I
have not given the properties
type. setting
him on the
right track
if
he
in difficulties.
Cassinian curves.
or 4. as he pleases. The reader
I
can select from them few or many.
At any
their
rate very little
properties. for the long lists of examples.
hope that
these will be of real assistance to the student. enabling
. 8.
of
'
special plane curves % unless they are representa-
tive
of
some general
such
as.
I give hints for solution in
most
is
cases. and have excluded other matter of equal or
greater importance.
make no apology. but
at present the difficulties
seem
insuperable.
VII.
H..
June 1919.
. XVIII.
PREFACE
him him
to check the accuracy of his results.
#10 and 11.
8 to 10 Ch.
are due to friends
and pupils who
I
have made suggestions and pointed out inaccuracies
while the book was being written. XII.
7
and 8
3 to 7
XI
XX.
owe a
special
debt of gratitude to Miss G.
vii
and giving
a guarantee that the examples are not of undifficulty.
Ch. X.
reasonable
The beginner should not attempt
parts which
to read the
book
straight through. IX. Ch.
3 to 12
7 to 10
§§
. Sadd. H. VIII. D.
Ch.
. Ch.
.
$
.
As
a rough guide I
recommend the omisa
first
sion of the following portions on
reading
VI
.
9 to 15
Ch.
$$
Ch.
Ch.
.-
must
also express
my gratitude
to the Delegates
of the University Press for so kindly undertaking the
publication of the book.
§§
.
§§
§§
.
4 and 5
Ch.
most. XVII.
§§
6 to*8
Ch. My best thanks
XXI.
Ch. but should select for himself those
he thinks
easiest
and
likely to interest
him
Ch. XIX. : .
$
XVI
Ch.
who
has given
me
I
very valuable
lists
of corrections required in the
MS.
2 '.
references.
Ex.
for '» — 3'.
Ex. Add Gettingen Nachrichten. 12.2n+1 yln+l 22jl +i' f° r ^sn+u Vsn+l) s»+l Page 24. For the solution read Use Ex. § 6. and 'a. The quartic consecutive to Q is supposed non-cuspidal.
this
book was in the press a
treatise
by
S. X.
'
'
'
N
ment on page 146. Page 369. 11 or Ch. P may be made any point of the n-ic might be incorrect. Ex. In Ch. Read '2m — 2'
•
. Page 371. 308 to the list of
' ' '
'. p.
While
2. xi (1909). Similarly in Ch.
10. Otherwise the statePage 146. must be less than n. line 33.
. Ex. By choosing t properly.
. X. In Fig. § 4. 13 it is essential that the nest should lie inside the circle and outside 12. In the theorem at the foot of the page a linear branch is counted as being superlinear of order 1.ERRATA.
Ganguli
called Lectures on the Theory of Plane Curves has appeared.
Omit(-l)". § 7. ETC. line 24.
Page Page
9. Read ' three real tangents for three tangents '.
Preface. I. Page 116. 8. Page 359.
and OM.
.'
Coordinates. 1) we draw parallels PJ¥.
.
We
If
two
ON
to (Fig. x and y are the Cartesian
through a point
P
PM
coordinates of the point
P (x. involution. contain x and y units of length. projection. y).
shall assume a knowledge of the more elementary portions of the calculus and of pure and analytical geometry including the theory of cross-ratio. reciprocation.
INTRODUCTORY
§ 1. and inversion but in this introductory chapter we shall remind the reader of some elementary results of which we shall make frequent use. fixed reference lines meeting at an origin 0.1
CHAPTER.
but considered fixed when once chosen.
. They are evidently are called the areal coordinates of P. the ratios to the triangle the triangles PBG. For instance.
ABC
x + y + z=
Instead of defining the position of by its areal coordinates.. 2). 1.
y
=
k2
. The new equation of a given curve when fresh homogeneous
. z in this We have the relation ca'se general homogeneous coordinates. y.
I 1
x = r cos 6.2
COORDINATES
If the reference-axes are perpendicular. y.
A
Fig. though these are the actual coordinates only if areal coordinates are
used.
In this k lt k2 k 3 are any constants chosen arbitrarily. PGA. connected by the relation
If
^4£C
x. y. z -we may take any quantities proportional to them when dealing with such homogeneous equationsi. (0. we may use instead any constant multiples of them and write
P
x
= k PBG/ABC. the vertices of the triangle of reference will usually be taken as (1. (0.
z
=
k.
1
. y. 6) with respect to the pole and prime vector y = 0.
2. We call x. We call r and 6 the polar coordinates of the point P(r. z of
PAB
1.
.
We shall mean by the symbols x.
PAB/ABC. 1) .
®/h+y/h+ z/ ]h=
!>
by means of which the equation of any locus may be made homogeneous. y = r sin 6 where r is the distance OP and 6 is the angle between OP and y = 0.
PC A/ABC. z the general homogeneous (not necessarily the areal) coordinates.
is a fixed triangle of reference (Fig. 0).
. unless the contrary is stated. 0. 0. 0). Instead of x. or quantities proportional to them.
. IT' and a fixed point V. Similarly for involution pencils. its projection is
.
Projection.
BD)
AB CD/ AD GB = — 1.
of a straight line is evidently a straight line.
Suppose we have fixed planes n. Then P' is called the projection of P on n'.
AD
.
3. so that harmonic. where I.
'
'
'
b2
. sections of two transversals with the rays of a pencil whose
The projection The projection
AB
. z in the original equation by expressions of the form
1-yX
+ m-^ + n^. which is called the projection of c.
l
3
iv
+ m3 y + nz z
respectively. P' traces out some locus c'.
is harmonic.
'
'
'
Fig.
. Similarly the projection of a pencil of four concurrent lines In fact. involution range It follows that the projection of the traced out by points P. In particular.
l
2
x + m 2 y + n 2 z. PP') is harmonic.
if
the range (AG. of a range (ABGD) of four collinear points CD/ GB is a range of the same whose cross-ratio is For the range and its projection are the intercross-ratio. P' such that (IJ. y. the pencil and is a pencil of the same cross-ratio.
§ 2. Similarly c is the projection of c' on IT.
vertex is V. its projection have the intersection a of the planes IT and IT' as a common transversal. V being the vertex of projection'.2
'
1
PROJECTION
3
coordinates (and triangle of reference) are taken is obtained by replacing x. If P traces out a locus c. Suppose also that P is any point in IT and that VP meets IT' in P'. J are the fixed double points of the involution. is an involution range.
PP' and RR' continue to intersect. It is possible with a given vanishing line to project two angles into angles of given magnitude or to project a given conic into a circle. R with it. In general.
If the
its
approaches
projection
P
lies
on
v. Q'.
Plane Perspective. This is only possible if PP'. and that Q. ' all points at infinity in the plane IT' lie on the straight line which is the projection of the vanishing line v '.
P
. But P is any point whatever of IT.
P
projection. since they are not coplanar. QQ'. Q and Q\ R and R' are projections of each other. 2.
v'.
P
Ex. and Q. that case in which the planes are parallel. R are taken as fixed points of IT. Since Q'R' is the projection of QR.
are concurrent. even in the say. while IT' is kept fixed. moving in the plane IT. in which the plane of the diagram is perpendicular to the
lines v. If two lines through P in the plane n meet v in II and A". therefore. Similarly PP' and QQ'. 1.
. Ex. Q. R' continue to be coplanar. Hence QQ' and RR' continue to intersect.
at infinity
.
*
Suppose that P and P'.4
PROJECTION
I
2
Suppose that the planes through V parallel to II' and II meet IT and IT' in the lines v and v' respectively (see Fig. and P as any other point of IT. 3.
[See Ex.]
§ 3. P' recedes indefinitely while if has no does not meet IT'. two figures! in IT and II' which were originally projections of each other remain projections of each other. QR and Q'R' continue to meet on a. QR and Q'R' meet on a. respect to a conic and between conjugate points or lines is unaltered by projection.
v. 2. If the plane IT is turned about a carrying P. Similarly such a statement as a parabola touches the line
It is
meet
We
'
projection touches the vanishing line and so on. Similar remarks apply to the line v'.
'
means that
'
its
'. The relation between pole and polar with. if finite points. 1. so that
. The vertex V
RR'
. or (what is the same thing) that the projection of a straight line is a straight line. R. 3.
VP
point P. when we describe any property we mean a of a curve at an infinitely distant point P' property possessed by its projection at the corresponding point of v it being understood that the property is one and P' were which would be unaltered by projection. Hence when IT is rotated about a.
a). the angle HPK projects into an angle equal to HVK. Project any two conjugate pairs of lines through the pole of the vanishing line into perpendicular pairs.
convenient to observe the convention that two planes in a straight line. Ex.
since the distance between a and v is constant and is equal to the distance between and v'. 3.13
PLANE PERSPECTIVE
turns about the line v'. as is evident from Fig.
V
Two figures which originally were projections of one another are now two figures in the same plane with the property that the line joining corresponding points of the two figures passes
ysy'
. Suppose that n turns about a till it coincides with IT.
f(x.
is
bc/y)
=
We can always choose the projection so that b — c. then. y. Then putting n cm in (ii) we see that the perpendicular from P' on GB bears a constant ratio to the quotient of the peron GB and pendiculars fro. Now (changing the notation) we may take homogeneous with ABG as triangle of reference. curve.
Let Ix
the equation of
its projection.
P
P
AB
so for y/z. If. y.
PLANE PERSPECTIVE
b(y + c)/y)
of
I 3
line
the
is
line
VP
and the
joining
H'
But
to the intersection of
by definition
y)
P'
with the axis of perspective.
+ my + n —
it
distance d from
of the point
be any line in the first figure.
Hence
d'
(c l
2 2
+ n 2 )% =
bed (P + mrf/y
(ii)
of which the Suppose that we take a fixed triangle vertex G coincides with V. coordinates (x. y..
. the axis of x being the vanishing line both for the curve and its projection.y' + b). if f (x. is the homogeneous equation of Conversely. y) is given by
The
d (I2 +
m 5 = Ix + my + n. (x'. y by ax/y. 1) obtained by a projection in which z is vanishing line. P (x. y/z)
and
GA
—
=
is the vanishing line.
=
is
the equation of the locus of P. is a homogeneous equation of a projection in f(x/z.m B and similarly for the perpendicular from P' on GA. z) of such that x/z is equal to the quotient of the perpendiculars multiplied by any constant we please on GB and from
ABG
=
P
A
. we get the equation of its projection .
2
)
The distance d' from the corresponding line in the second of the point P' (bx/y. We thus get the useful rule 2 If in the equation of a curve we replace x. and y are the 0. bc/y) is given figure clx + ny + bcm = 2 by d'y (cH + n 2 f = be (I'x + my + n).
which z
=
=
=
=
=
=
followed by an orthogonal projection.
6
(bx/y. Also the perpendiculars from P' on GB and GA are constant multiples of the Cartesian coordinates of P' referred to and GB as axes of reference. from which (i) at once
PR
follows. while A and B lie on v.
f$x/y.z) is the Cartesian equation of the curve a curve. a /y. and the axis of y being unaltered. f(x.
If/ (x. y) is the Cartesian equation of a. x projections of the Cartesian axes of reference.
V. J be the points in one plane corresponding to the circular points of the other plane (§ 5).
recedes indefinitely.
Asymptotes. and the figures have Fas a centre of similitude.
at
Fig.] § 4. The axes of reference are IJ and its perpendicular bisector.
[Project a to infinity. d/y approaches the limit zero as P approaches Q along the curve.
P corresponding to a given point
on
Given
V. the perspective is called harmonic]
Ex.
[If
Q
is
Given on v. So for the other plane.
Hence the point P' of the corresponding curve in the second which is at a distance d' from the projection t' of t.
P corresponding to a given point P. the cross-ratio of (VPRP'J
constant. 4. Such a line as t which is the projection of a tangent. v.]
Q' is at infinity
on VQ. a.
struct
2. 1.4
1
ASYMPTOTES
7
in
is
Ex. 5).
[Let I. a. whose point of contact is on the vanishing line (but the tangent and
.
and a pair of corresponding points Q and
P. but t and v do not coincide referred to in Taking t as the line Ix + my + n = (Fig. approaches t' indefinitely as P'
figure (the projection of the first).
2. If P and J" are any two corresponding points of two figures plane perspective and VPP meets a in i?.
Ex.
5.
Q'. (ii) of § 3 and P as any point of the curve.
con-
[P' lies
VP and
on the line joining
construct
Q' to the intersection of
PQ
and
a.
Now
use Ex.
Ex.]
3. If the cross-ratio is —1. The relations connecting the coordinates of any two corresponding points of two planes which have a one-to-one correspondence can be put in the form (i) of § 3 by a suitable choice of axes of reference.
Suppose that in § 3 a curve in the first figure crosses v Q and t is the tangent at Q.
We shall denote them usually by a> and u> Any line through a circular point. any line parallel to y — + •/(— \)x.
4.
infinitely distant points
go through the unreal and \/(—l)x.
as
we may put
it.
§ 5.y) 0. Applying this to the general equation of a circle
It is well
x*
+ y 2 + 2gx + 2fy + c —
x2 + y*
0.
8
ASYMPTOTES
is
I
4
vanishing line do not coincide). though most of our work will be concerned with the case n > 2.
called
an asymptote of the
Circular Points and Lines.
all circles
on y
= ±
'
Ex. 2. z of degree n (and whose Cartesian equation is therefore obtained by equating to zero a polynomial in x. e. is called a circular line.
Ex.
known that the lines through the origin parallel to the asymptotes of a conic are given by equating to zero the terms of the second degree in the Cartesian equation of the conic. the properties of conies being supposed too well known to require further investigation. The word higher implies that the degree is greater than the second. y. Hence A curve of degree n meets any straight line in n real or unreal points in general. z) = is the equation of such a curve. y. These points are called the circular points at infinity. If f(x.*
'
'
=
* note that the number of real points a positive integer see also § 7. another.
[With the line joining them as vanishing line project any conic through them into a circle. See § 2.
point not on a circular line
is
at
an
infinite distance
from
that line. i.
Higher Plane Curves.:
. Nevertheless we shall not consider conies necessarily excluded from the definition. where r
is
zero or
.]
§ 6. is called an algebraic plane curve or higher plane curve of degree n.
see that the asymptotes of all circles are parallel to
= 0.
1. 1) will be f(vx. 0. the lines joining its intersections with Xx + fiy + vz = to the point (0.
Two
Any
points on a circular line are at zero distance from one
Ex.
we
or.-
A curve whose homogeneous equation is obtained by equating to zero a polynomial in x.
.
Project a given pair of points into the circular points. —\x — y. vy.
Ex.
projected curve.
3. 3.
A
circular line
is
perpendicular to
itself. Ex. and are therefore n in number in general.
We
is
n — 2r. y of degree n).
3.00.
BQn
•
:
Oft Oft independent of the position of 0.
Ex. . 5-ic.BP... ' curve of degree n ' will often be called for short ' an n-ic '..
The
lines joining
«-ic in
meet a given
ratio
P P
1(
2
. 2 to the triangle ABO. 4.y')+
{coefficient of
a.
2
y
.
Ex. Qn'' -Bi> Ri
Sn
P1. when taken together. and that the intersections of an infinite number of n tangents lie on the (n— l)-ic. The cases n = 1
.. quintic..
..
.
2
. and close to meets the curves in ft and the tangent at Pin T. [Apply Ex.BR.
any point to two fixed points A and B Show that the . .. .
Two
P
R
P
(OQTR). A line curves touch at P.
p2
two lines are drawn in fixed directions P" and ft.. Given all but one of the 3w intersections of three lines'with an construct the remaining intersection.
.
Show
{-IrAR.]
'
ARn .
.
2..e.
. take ABC as triangle of reference. 1 or and 2 are known as Menelaus's and Carnot's theorems. is the crossthat the ratio of the curvatures of the two curves at 6.
5.
1
. ..
ratio of
at
Ex. 5..
m-ic
with the sides of
w-ic. the (finite)
If the curve meets the line at a point on the vanishing line projections of the curve and line meet in less than points. Then
1
.BQ2
.
."
in/(». AR
2
[Use Ex.
.. as forming an n-ic.
sides of a triangle
ABC
meet an
n-ic in
ft> ft) ••.
.. y) = 0. 2 to the intersections of an any polygon. Show that 0Pj OQ. 3. BP BPn CQ CQ Cft. ft. quartic.
OP . show that the n(n — 1) points thus obtained lie on an (n — l)-ic. P
2
. [Take Cartesian axes of reference in the fixed directions {oc y ).. Another notation for an n-ic in common use is G '.
1..']
Ex.
OP2
. For instance a cubic may degenerate into a conic and a straight line or into three straight lines.
let
be
1
.6
1
HIGHER PLANE CURVES
9
n
(§ 2).
The
that
.
OPn BQ. Q n
. Such an n-ic will be called degenerate. a cubic. n Two or more curves whose degrees have the sum h may be considered...
'
Thus a curve of degree
Ex. ft. P" and Q u Q2 .
is
.
A
..
Pn
. Deduce the fact that the ratio of the curvatures of two curves a point of contact is unaltered by projection. Given a set of n tangents to a conic meeting at \n\n — 1) points and a second set of tangents also meeting at \n(n — 1) points.OP
1
2
OPn =
±f{x'..]
Ex. i. curve. = AQ AQn CP BRn CPn ..
Oft AP. Extend Ex.. 7. Oft
.
AP
2
APn
through
Show
is any point not near P.
OP.
Through any point
n-ic in P.
. will be called a 3-ic..
meeting a given
the ratio
is
P
2
P.. . and the curve be f(x.
1
.
Ex.y)}. Such exceptions will be considered later. 4-ic. . 4.
independent of the position of 0.
]
x=
tTz.
This eliminant gives the lines joining the point (0.]
§ 7..
y
= (t+T)z.i+1.y sn+1
+ zl
/1
^2
.
variable line is drawn in a given direction meeting a given where the radius of curvature is p and the tangent makes an angle with the given direction.
4i.
(
{f(t).
.<p(T)-f(T).
=
xi>
2fr.
then
("2'»+l_
xx xt
Ex. = 0.
"n+lf-l g
..
bn
\
homogeneous in x and
h h
y. zN-l and by 1.
Intersections of
Two
Curves.
.. the summation extending over the n intersections P of the line and «-ic. z 2 .
••X Sn+1 / y1 yi . 0. Let their equations be
~2
a zn + a^z".y = ax + 6.
a
a
a.
a..
x.. the equation of the n-ic being y n — (ax + b)y"..
z 3n+1
=
0. Then. *s«+i)» « being homogeneous in z of degree n — 3. Multiplying these equations respectively by 1... then
8. 1) to the intersections of the two curves.
(i). the intersections of the tangents whose points of contact are given by f(t)+k<p(t) = 0.
0)
where ar and b T denote homogeneous polynomials of the rth degree "in x and y.**) (*sn+ii ys.+an
.1 + . z^ 1 we obtain n + fi linear equations in
.10
HIGHER PLANE CURVES
16
[Take the conic as the locus of (f.
n-ic at P. = 3 is well known.
Two curves of degree n and iV intersect in nN points.
in+1 xyzu in If any line meets the curve x + y 1 "+ l + z in+1 2 i)> (*2> jfe.<p(t)} + (t-l').
2 2 2.
from which these quantities may be eliminated.1
+I).
1 X.. we have
For instance.
the line
is
^# + py + »»
= 0. y.1 + a 2 z n
+.
a.
x Su+1 + y y2
[
. 1). z. where /and <f> are polynomials of degree n in t. lie on the (n — l)-ic obtained by eliminating t and T from
The case n
Ex.
.y Sn+1
5"+l)/(X2'1 +l_ v 2
..
(ii). Suppose the equations of the curves are given in homogeneous coordinates. bi
\ h
=
0. It.... Show that 2cot<£ is constant and 2 (p sin 8 (p) _1 = 0.. if to = 4 and If = 3. whatever h may be. &
.. Differentiate this relation twice with
respect to x...
. for it is readily seen to be
The
lines are
nN~
in
number
for
. z. z2 .
9.
xx x2
[If
.
<t>
A
[For a given value of x we have 'S..
For other methods see Ch.]
and v
=
are
polygon of h
If a non-degenerate »-ic 2 or else n sides. 1) to those intersections of the curves which do not coincide with (0. . 0.s A r^ and by 1. 0. Higher Algebra.
It is
impossible for every line to meet a given n-ic in
.
=
^
* For a detailed discussion by this method of the intersections of two curves see Segre.
2.
= 0. 1) is rbN—kK — Y in general. 1). =
where u = and v = 0. n
has the symmetry of a regular
k. 1). bK are iden.]
Ex.*
Ex.e. 1-50.
[The only real points on m + (-1)* v real w-ics. .
and
parallel to
any tangent
will
meet
it
in
n—2
An
«-ic
with unreal equation cannot pass through more than
w 2 real
points..
Multiplying equations
2
(i)
n~k. = 3. Similarly.. z.. For example.
N
fc
It is readily seen that the number of intersections of the curves other than (0.<v (n — k)K+ (N— K) n =
nJST— kK in x and
y. § 3. if a & and b% have r factors in common.
n
real
points.
z2
. z. 1)..
b2
A
typical term of the determinant just written v -*> fc
is
which is of degree Hence the curves meet in
case that
± v. If in this a and bK have a factor in common. we have
roots of (ii) occur in conjugate pairs. we say hK of the intersections of the curves coincide with (0. kK+l intersections of the curves coincide at (0.z and eliminating as before.
±b
na
N
n
.
tically zero. z . a h _ x and b \..
1. 0. .. we get an equation giving the lines joining (0. xxxvi (1898).
In order to observe the convenient convention that in eveiy two curves of degrees n and meet in nN points. the factor is a factor of the first column of the determinant just obtained..x
by
1. i.
nN—kK points other than (0. a and a x = 0. 1). Suppose that a a v . pp.. 0.b = 0. 0. VI.
N
"2
'3
"'i
a"
6j
&i
a3
b2 b3
aA
b3
=
0. are the real intersections of u
Ex.
3. 0. both curves pass through (0. 0. 1).. For complex
'" 1 .
^
.
We note
is
that the
number
tiN— 2r. .. and Ch.
. if n = 4. § 69. kK + r intersections coincide at (0. If a and b are zero. IX. For the general theory of the eliminant (ii) see BDcher. 1). r being
of real intersections of the curves zero or a positive integer. Giomale di Matematiche di Battaglini. § 2. 0.
"
[A line adjacent to real points at most.I 7
INTERSECTIONS OF TWO CURVES
is
11
a typical term of the determinant just written The theorem is therefore proved.
[(i) Use polar coordinates. P. assumed that the n-ic does not pass through the circular points. by analogy with the well-known 1 case in which n The n2 fixed points through which all curves of the pencil pass are called the base-points of the pencil..]
.
'
'
'
'
'
§ 8.. and v n passing through the «2 intersections of u The family of curves obtained by taking different values of k is called a pencil' of n-ics.
which touch a given
[Take z
Ex. (iii) The same is true.
Ex. in Ex.] Ex.
The sum of the angles which OPlt 0Pa
. if we replace Cartesian tangent by Cartesian normal or radius of curvature ' (iv) The centroid of all points on a curve at which the circle of and radius o at an angle a is curvature cuts a circle with centre independent of a and a... The coefficient of se** only involves the coefficients of the — 1 in F. u +
=
=
=
=
.
with the tangents at
(iii)
Plt P
2
.
The number of curves of a pencil of «ics in general 2 (« — 1).. y) =
N
N
Deduce that (i) The centroid of PJf P.. [Consider the intersections of the w-ic with circles concentric with the polygon.
If u degree. the 2 n — 3 in general. (ii) The centroid of all points on an algebraic curve at which the Cartesian tangent is of length a is independent of a.
is zero... terms of degrees «.. y) of degrees n and respectively...]
2.
v
=
and k
is
any constant.
5.
If
=
F (x. we obtain in general an nlf in this equation equation for x of degree nN.
.:
12
If
INTERSECTIONS OF TWO CURVES
I 7
n is not a multiple of h. If
is
in Ex.
= 0. . 4 is independent of a.
fixed line is
(ii)
independent of both a and the position of 0.
number
[One touching at the base-point and 2 (n — 2) others.
OP2n makes 0PU
at
with any
.
Ex.
of curves
=
as the line. They are not necessarily all distinct... 4. 1 the line passes through a base-point. The coefficient of x degree n in / and degree only involves the coefficients of the terms of _I in F..".
(i)
P..
Pencil of Curves. A circle with centre and radius a meets a given M-ic at
P" P2 . n — 1 in / and degrees N.
(ii) and (iii) follow at once from (i). the re-ie passes through the circular points.
line
is
1.
are the equations of curves of the nth is a curve of degree lev 0.
Show that
.. The sum of the tangents of the angles which 0Plt
make
2
.
The sum of the polar subtange^its of the curve
Pu P
•••
is
zero.]
It
is
N
and we eliminate y between two equations f(x.
13
Tangents. touches the curve at P. cubes.
PQ
and Q
(Xx'
(x. y)
=
(»-*& + <*-rt$ =
If the coordinates of
°-
any point
of a curve are given in the
form
x=f(t). Xz' + fiz)
=
where j—7 means the
result of putting
a:'
for
a. &c.
3'for. <f>y)x + (+f. of dt. y.
The
line
PQ
. z') is
of degree n. this line goes through the given point and the consecutive point..in^^divides
. z)
:
in the ratio
any point on a curve f(x. + fdt).
y' for
2/. where A + fi = 1.z) = is any other point. Xy' + py.
if
+ /iy.y =
t is
4>(t).z
=
+{t). The
P
Ix
oz
which
is
therefore the equation of the tangent to the curve
is
at P.<f>(t + dt). the tangent at
where
called the
parameter of the
the point
is
0.
&c. whose coordinates are
(W-
=
+ </>'dt. is fi
Xy'
+ fix.
second root is zero.
This
or
lies
on the curve
f(Xx' + fix.
Xz' + jjlz). we get the tangent at any point of the curve whose Cartesian coordinates are
+
Suppose now
P
(x'. If the equation of the curve in Cartesian coordinates 0. The point X.I 9
TANGENTS
§ 9. (/ (* + dt).
This equation in fi/X gives the ratio in which the curve
PQ.Vf)y+(f#-ff) * In fact. z') is
not on the curve.
point. the tangent at (x\ y') is f(x. condition for this is
A
PQ
° <>y
for lies on the curve.
One
root
if
is
zero. y.
Suppose
dividing
P (x\
y'. y'. y\r(t + dt)) or (f+fdt. neglecting the squares. If we put z — 1.
each conic corresponding to itself in the perspective. such that OP . 1) is
in/is zero.
n of z
If the «-ic f(x.
circle
+ y2
=
a2
.] Ex. Find the lengths of the tangents and normals from any rJbint to a curve. 2 J = 0. 1).
*
Or a sphere. u \* + 3u 1 \ 2 li + 3u 3 \n 2 + u 3fi 3 = 0.1 J
§ 10.
Show
x
that two perpendicular tangents to
=
a (2cos£ + cos2i).z)
=
passes through
(0. the constant term in zero and the tangent at the origin is obtained by equating to zero the terms of the first degree in /. Ex.3 «.
4.
if
the loci are not coplanar. z.
constant. The tangent at any point P of yx* = a? meets the axes in A. and put the equation of the curve into the form f(r. where u r is of degree r in x. y = 2tr/ (1 + t''). and writing down the condition for equal roots. and they meet the cubic again in six points on a conic.OP' — W.y. 3. y. dr
or
/= dt
0. the resulting equation that of the tangents from (x'.
(See §
with O as vertex.BQ are
constant.
=
/
Ex. 0.
Ex. y'. The tangents meet the two conies again in twelve points on another cubic also touching the tangents.14
TANGENTS
we write down
1.
in general. we have
The tangents from any point
+ 4m! 3 ) + m 2 2 (4i( z . The two cubics of Ex. 8. Ex. [Putting (i) in the form (n = 3).t) = 0.
a. Prove that the ratios AP-. 1. if the equation
If
is
(i) in /jl/K has equal roots. z') to the curve.
Inversion.
'
.
Eliminate
t
from/ =
0. 2.
. the locus of P and the locus of P' are said to be inverses of each other with respect to a circle* with centre and radius k or more simply inverses with respect
If 0.
Ex. the condition for this. being fixed. The axes of reference intercept a constant length on any tangent to x%+y§ = ai. 0.PB-.
2
y
= a (2 sin? — sin2<)
meet on the
Ex. and the tangent at
(0.
The sum of the intercepts made by any tangent
is
to
xi + yi
—
ai
on the axes of reference
5. 3.
I 9
will touch the curve.] Ex. 7. 1) and u2 z* + 2xy. 6 are in plane harmonic perspective u 3 (h 2 w 3 '
6 j^MjKj
i/
. P.
[Choose the triangle of reference so that
=
[Take the point as origin. The two conies are u 2 = and 4m m 2 = 3%2 and the other cubic is w 2 m 3 — 6j( m 1 m 2 + 4u 1 3 = 0.)
O is (0. the coefficient obtained by equating to
zero the terms involving s" _I is Iif(x. y) passes through the origin.
P' are collinear points.
Ex.
and
/ = 0. B and the curve again at Q.
6. 0.
to a cubic are six in number Their points of contact lie on a conic.
'
to
'. where x = (l-t 2 )r/ (l + f).
y). the fact that 0. P. its inverse. Hence two curves cut at the same angle as their inverses. OQ'P' are equal. Let the tangents at and P' to one pair of inverse curves meet in and the tangents to the other pair in K. If now the two inverse figures are coplanar while is taken as the origin of rectangular Cartesian axes and P is the point 2 2 2 2 2 This follows at once (x. inverse is a line through the other circular point. P' are collinear and OP OP' = k 2 from
P
H PH =
P
PK
. it is evidently its own Also. and the angles HPK. HP' make supplementary angles with = P'K. triangles HPK. or a straight line if The inverse of a straight line is in general a circle through 0.
PQ =
k 2 P'Q'/
. since the tangents HP.I 10 If
INVERSION
15
Q' are points on two inverse curves. from the similar triangles OPQ. Then. Again.0Q'. y)
=
is
the equation of a plane curve.
Hence the inverse
the inverse of
is
x + iy
k
2
=c
and may show
is
=
c(x—iy).
Hence. OP': OP'.OQ'. For instance. OQ'P' (Q not necesP'.
OF
. HP'K are equal. we see that the tangents at P and P' to the inverse curves are coplanar and make supplementary angles with OPP'. P' is (k x/(x +y ). k'y/(x + y )). a plane if
We
have a similar result
a sphere.
for three dimensions. that the inverse of a sphere with respect to lies on the given sphere. if the line passes through 0. Hence the angles OPQ. or
. k 2 y/(x2 + y 2 ))
the equation of the inverse curve. HP'K are congruent.
f{¥x/(x 2 + y2). Therefore the P'H and similarly OPP'.
P
and
Q and
0P.
0Q\
which enables us to express lengths in one figure in terms of
lengths in the inverse figure. Suppose now that two curves meet in and the two inverse curves in P'. if the line passes through a circular point.
.0P'=0Q.
is
if
f (x. the inverse of the circle
=
is
Ic*
x2 + y 2 + 2gx + 2fy + c = + 2k 2 (gx +fy) + c(x 2 + y 2 )
=
0.
in its of a circle with respect to a point lies on the given circle. Making Q consecutive to P and therefore Q' consecutive to P'.
sarily consecutive to
PQ
Hence
P) we have :P'Q'=0P: 0Q' = OP. In fact. plane is a circle. But.
3.
According as G 2 + 4iH 3 = a 2 (a 2 a32
is
— Qa^a^ + 4a
a23 + 4a1 3 a 3
— Sa^a 22
)
negative. [Let Q be a point of the curve near P and Q' its inverse.
1.
n=
2. .
-5ai + a 5 (a5 -5a1 )}
)
{a
+ 25
in the cases
(a a.
The typical equation of degree n is ~ a xn + »(?! a^.a 4 ) 2 + 16 (a x + a 3 ) (a a2 + a x a^
% + a2
{a (a
=
2
)
0.\.
Theory of Equations. This projection is called stereographic]
of Ex.
I
10
The inverse of a circle with respect to a point not in its plane is a circle..
a3
a.1 + n C2 a2 xn 2 +
.
[The projection is the inverse of the original figure with respect to V. one real root.
For convenience of reference we insert here some well-known
results in the theory of equations. the cubic has three real roots. 3.
«2 a. Now let Q approach P. for the intersection of two spheres inverts into the intersection of two spheres. If a curve cuts Ota at P. If a.. 5. 2.
.] § 11.a
16
}
INVERSION
.
+ 2a a3 -
+10a 2 +5a4) (a — 5a4) + (a6 + 10a3 + 5 %) (<x6 — 5 a
5
a4 — 2a5 a2
=
x)
0.. the inverse with respect to a point P on Oa> is at oi. that circles project into circles and angles are unaltered by the
projection. the corresponding tangent at <o of the inverse curve is the inverse of a P. positive. A sphere is projected from any point V on its surface on to Show the diametral plane perpendicular to the radius through V.
[Use the fact that all circles through to a circle cut this circle orthogonally. The quartic equation (n 4) has two invariants
=
I
=
a ai —4a1 a3 + Sa2 2
1
J=
"0
""1
""2
«1 a. 4..
•
The cubic equation (n = 3) has two invariants = a 2 a3 — 3a a1 a 2 + 2a1s *H = a a2 — a^.. Then «Q' and a'Q are inverse lines.
The product of two
roots is
—1
if
— °> a o (ao + 3a2 ) + a s fi ai + az) = 0> {a + 6a 2 + a t ) (a .]
two points inverse with respect
Ex.
Ex.
+ an =
0. or zero.
.
Prove that a circle and two inverse points invert into a circle
and two inverse points. or two equal roots. a>' are the circular points..
3 and 4). The origin is
>. * 2 b x + bx y is a factor of c^j? + 2c t xy + c.2 y
.ni
INFLEXIONS. when 6
+ bx fn.
called a point of inflexion (or simply an inflexion) in this The tangent at the origin is called an inflexional case.
the tangent meets the curve in four points coinciding with the
. we have
also
co + ~ c i
m + ct m* = 0.
It is evident that 6 + 61 to is a factor of c + 2c 1 what is the same thing. 0.
or. 3.
Then the tangent at the origin meets the curve at three points coinciding with the origin (Figs.
m+c
i
ri\-
:
If we have also
d + 3dl m+3d2 m2 + cl3 m?
C
2
-
= 0.
=
=
we
obtain
Suppose that.
Fig. namely. ETC. = 0.
19
Eliminating ni between y mx and b" + bx m. tangent or tangent of three-point contact. the equation of the tangent at the origin.
The curve crosses the tangent at the point of contact if r is odd. +1 + ... ETC.
The
follows
last
statement in the theorem
may
be proved
*
as
The curve may be written
b x + b lV = {n.
+
tangent at the origin
(b
is
x + b iy)
+ {b* + b*f
(u r + u r+1 +
.. y) of the curve on the
1._ ).
.
contact. point of undulation The tangent is called a tangent of four-point
is
In this case b x + b x y
a factor of both
. the tangent meets the curve in r points coinciding with the origin.. and is called a tangent of r-point contact '. + i'. where b x + 6j y = 0..
The
II
1
in this case. when x and y are small and approximately in the ratio b x —b But ur evidently does or does not change sign with x and y.
+ Vl + . but does not cross it if r is even.
4. + v r_ 2 (b 2 + &.
)
:
. (r — l)*11 degrees.
.
INFLEXIONS.«)*... If the terms of the first degree are a factor of the terms of the 2 rd 3 rd ..
'
Fig. + U. and has therefore the same sign as ur + (6 2 + b^)i. . The perpendicular from any point (x. 1 2 where u and v% are homogeneous of degree k in x and y.+v n ) (l + v + v 2 +..
origin is sometimes called a.
20
origin. according as r is odd or even. .:
. The truth of the statement may also be seen geometrically
)
=
+ un
-f (1
... the terms of the first
e^ + Z^xy
+
c2
2
:
degree equated to zero give the tangent at the origin.
3 2 2 s y and of o] x + 3d1 x y + Sdi xy + d3 y These results can be immediately generalized and we have Ifa curve passes through the origin.
3 for the case r 3).
'
'
On the same branch of each curve . if they may be considered as the limiting case of two curves meeting one another at r points * very close to are said to oscuP. .m3 2 + a.
and y = infix meets the curve where = x l (c + %cx -m + c 2 2 ) + x 3 (d + 3d1
m
m+3d m
2 + <£.
)
roots of this equation are zero for every value of m.
between this equation and y Eliminating the equation of the tangents at the origin
m
=
mx. + 2c^xy + c 2 y 2 + d x 3 + 3d 1 x2 y + 3d2 xy 2 + d3 y a + 6q x +
.. . if the radius of curvature of the curve is a maximum or minimum at P. In this case the curve has two real branches through the origin. osculating circle called the circle of curvature '. at each point of a curve there is an late at P. 5). we have
c^-b'Z^xy + ^y 2
=
0.
Such a tangent may be considered as the limiting position of a chord joining the double point to another point of the curve.
L
Suppose that in
=
c aj
z
The curve is now § 1 b and b are zero. line through the origin meets the curve in two points coinciding with the origin. The line y = mx is such a tangent if
c
+ 2c 1 m + e 2 m z =
0. The tangents may be real. whose centre and radius are the centre of curvature and radius of curvaThe circle of curvature has ture of the given curve at Pfour-point contact with the curve. The definition of ' r-point contact of a line and a curve may be generalized.4 (e +.II 2
DOUBLE POINTS
21
by considering a tangent of r-point contact as the limiting
position of a line meeting the curve in r points close together (see Fig. Any two curves will have r-point contact' at P. Curves having three-point contact at For instance..
*
'
. The origin is called a
Two
Hence every
double point..) + . t See treatises on Differential Calculus '. when the origin is called a crunode (Fig.. There are three cases to be discussed. The tangents at the origin are defined as the lines meeting the curve at three points coinciding with the origin.f
=
'
'
P
'
'
'
'
'
§ 2. when this point approaches the double point. see next section.
..
Double Points.
The origin is a real point of the curve. but there is no real point of the curve adjacent to the origin. e. ' isolated point ' or ' for cusp is obsolete. 7). a factor of
As
tangent
is
we show that.
'
Fig.. The tangents may be coincident. to a double point which is not a cusp ". when the origin is called a cusp (Fig. if one factor of c x + 2c x xy + c2 y2 is d x s + Sd l x2y + 3dz xy 2 + d3 y 3 the corresponding an inflexional tangent.
7.
fiecnode in this case (Fig. The origin is called a
2
. 6.
The term spinode
for crunode. If both tangents are inflexional tangents.*
' '. when the origin is called an acnode (Fig.22
DOUBLE POINTS
112
The tangents may be unreal. 8). The word node is equivalent to crunode or acnode i.
in § 1.
'
con-
. so that the terms of the second degree in the equation of the curve are a factor
*
Other nomenclatures are
'
node
'
'
jugate point' for acnode.
Fig. 6).
9. 0.
If the highest
power of z which occurs in ~
the
homogeneous
. shows that intersections of the curve with the line y
= mx
equation of a curve of degree n is z n k the point (0. 1) is a multiple point of order lc. 1).
if
Similarly
degrees. 0.
.
a curve are of degree
If the equation of the curve is given in homogeneous coordinates.:
:
. 0. points (at least) coinciding with the origin. and the tangents at (0.
9).
113
MULTIPLE POINTS
is
23
called a biflec-
of the terms of the third degree. which meet the curve in k+\ points (at least) coinciding with the origin. the origin
node (Fig. c 2 are all zero.
§ 3.
. 1) are
. In general we have the result
If the terms of lowest degree in the Cartesian equation of k. Any line through the origin meets of the curve in k points coinciding with the origin. except the three tangents at the origin obtained by equating to zero the terms These tangents meet the curve at four of lowest degree. Let us suppose the curve is of degree n.
Fig. except the k tangents at the origin obtained by equating to zero the terms of lowest degree. fourth.
second degree
is
one of the factors (or both) of the terms of the a factor of the terms of the third.
. and passes through the point Exactly as in §§ 1 and 2 the consideration of the (0.
If c c 1.
. so that the terms of the lowest degree are the terms of the third degree. the origin is called a multiple point order k (a k-ple point).
Multiple Points. a very slight modification is necessary. the origin is called a Any line through the origin meets the curve in triple point three points coinciding with the origin.
[n
=
4. crunode at (1. Ex.
r
=
3.
24
MULTIPLE POINTS
II 3
n~ h obtained by equating to zero the terms multiplying z ~ in n ' k z n . i.
Ex.
7.]
Trace roughly the curve y* + x (a:* .
Ex. [Acnode at (-1. x y\
a'y
i
i
and trace roughly the curves y> (a-x) = x\ 1 xi + yi = axyi
. cusp at (0.
r
=
4. Show that a conic can be and passing through R.
Ex.
3
2
Ex. Three tangents of n-point contact of an n-ic are taken as the Show' that the equation of the n-ic is of the form
the
+
+ (% + cz)" + (cz + ax)" *»«««-» + ( ax + sign in the ambiguity being taken if n
W
=
is
oV + bnyn + *V
odd. = sfun _ r geneous of degree t in x. E. If the terms multiplying z have a factor in common. 7 [The equation has a*" as a factor when we put x —
.
4. and PQ meets the curve again at R. = -3.]
Ex. 12.z n c r . If n is odd.2. 6
n
=
3. triple point. 9. 1.
2
Investigate the nature of the origin
ay
=
x\
+ y*)*
=
ay 1 = x\ a*(x l -9>).]
Ex. 4. 8. contact with the curve at P and Q. Trace y* = x t {x + k) when k = -1.] Ex. 3. Trace y + x{x-k){x-2f
Trace
y*
l
i
3. 0).
to 10 are special cases. cusp.
Ex. triple point (solve for y to trace)..k ~ x . Q.
The
line joining
two real inflexions of a cubic passes through a
third real inflexion. 11. drawn osculating the quartic at D.
Ex.
. S. An «-ic has the sides CA. The line through three real collinear undulations of a quartic passes through a fourth real undulation. B of a quartic meets the curve again in D. Show where u t is homothat its equation is of the form xyu^^. The line joining two undulations of a 4-ic meets the eurve Show that a conic can be drawn having four-point again in P.
Ex.2) = 0. biflecnode (turn into polars to trace).
r
=
3. 5.
2.. either the points are collinear or the three lines joining each to the intersection
. Transfer the origin to each point in turn. y. . S.. quadruple point (solve for y to trace). The line joining two inflexions A. 6.]
Ex. the three points of contact are collinear. 10. B meet the cuive again in P.
when A. The tangents at A. the equation.
5.. 0).
[In Ex. An »-ic has three tangents having w-polnt contact. Q. undulation. A and B being the points of contact.1 ) {x . 0. If n is even.]
Ex.
E
[n
= 4. 1. 0. triple point (turn into polars to trace).
(a*
1. z. and either sign if
n
is
even. + xi (x' -2x + k) = (a— 4) s =0 when fc = -1. cusp. 1. or y = 0. CB of the triangle of reference as tangents of r-point contact.
sides of the triangle of reference. 0).
!s
3
s
[Inflexion. 0.
= x\ ay* = x\ (x' + y y = ay{3x -y% a"y + x = 2a. the corresponding tangent has (r + 2) -point contact with the branch it touches.
2 in Ex. 0.
An
n-ic has r-point contact
C.
Conditions for a Double Point.
Ex.
Ex. 20. 13 w = 4. Y) be any point on the curve whose Cartesian equaTransfer the origin to this point. and the is f(x. 15.
17. There are considered as an equation in z. or
tion
0=f(X.]
Ex. XIX. equation becomes f(x + X.
13.
Ex. 6-ic.
Show
that
its
with each of x = and y equation is of the form
=
at
[See Ch. see Ch.]
Ex. y) = 0.
An
n-ic has r-point contact
two
distinct points. or 7-ic lie on a conic. IV. Show that a cubic touches the quartic at the six points in which these two lines meet the curve again. 4.
of reference
ABC at B
Show
with the side BC of the triangle that its equation is of the form
Ex.
and
25
of the tangents at the other two are concurrent.Y) + x¥ +y
V
\
x
DX 2+
y l>XlY
+y 37V
. The tangents to an n-ic from the &-ple point (0. 18. ?
. § 7).
.
I. [In Ex.II 4
CONDITIONS FOR A DOUBLE POINT
= 2. 19. A line is drawn through each of the points of contact of a bitangent of a quartic (a line touching the quartic at two points. &c.]
contact of three bitangents of a 5-ic.
P
.
Let (X. only the former alternative is possible.
What
are the corresponding theorems for a 54c and 6-ic ?
Ex. 19. Enunciate a similar theorem for a sextic having three-point contact with a line at two points. If the six points of
§ 6. 1) are found by writing down the condition that the equation of the curve.
Ex. 16. §
2.
[Use Ch.
[Use Ex. should have equal roots. The tangents at three collinear inflexions of a quartic meet the curve again in collinear points. Bt and P2 Show that the conic PiQ-JjJliItz passes either through P2 or through the harmonic conjugate of P2 with respect to B and C. 11. y + Y) = U. n(n— 1) — k(k+l) such tangents in general.
17.
§ 4. r=2. the other intersections of the bitangents with the curve are respectively collinear.]
Ex. 14. If the tangents are concurrent. quartic at Pt and 2 Q t and Q2 .
Discuss the cases n
3. 18. lying on a conic.
What are the corresponding theorems for a 2n-ic having n-point contact with each of three lines at two points for a 6-ic with three triple tangents. Three bitangents of a quartic form a triangle ABC whose sides touch the. the base points of a pencil of 3-ics.
e. y) of/ with respect to must vanish. according as >
'
\oxoy/
Similarly. Xz + /iZ)
i. Xy + fiY. Y. let (X.z) be any other point. z) = 0. If (x.
= 0. acnode.. Let P (x.
oX~~
oY~ oZ~
'
. Z) be any point on the curve of degree n whose homogeneous equation is f(x. y. two roots of this equation in X/fi are If zero for all values of x y z or
:
. if (x.
^
means the
If the
result of putting
X
for
x and
Y for
y
^.&c.
and the tangents at the new origin are
x
)x 2+ ^oXo7 +y
oY*
Hence. y)
J
=
0.
x and y up
the partial derivatives to the order k— 1 inclusive
Again. y.
This point
lies
on the curve
if
f(Xx + ixX.
The double point
is
oy ox a crunode.
:
P0
in the ratio
X.
ox'
new
origin
is
a double point on the curve. or cusp.26
CONDITIONS FOR A DOUBLE POINT
II 4
where
in
.Z) +
^x(x^ + y^ + z^)
^zix
:
+ 2zx
+2xy
+Xnf{x y iTJY) + '
z)
'
=
°"
is a double point. and let Q be the point
/ \x + /j.Y.
^f(X. y) is a double poind on the curve f(x. with a slight change of notation.
is
<
'
°
ox*
all
oy*
a &-pie point.X
V
dividing
X + /i
/x
Xy + fiY Xz + pZ \ X+ X+p )
/J.
These are all satisthe points (0.
~-
2a)(x + 3a)a?
= 0.
/ = yH/Jta-yY-lx* {x + Zaf =
we obtain
(x +
0.
y. Transferring the origin to each of these points in turn.
rj. (-2a.
1.4 (a. we see that the first two are triple points.
s 4 3 i
.
.
£ being current coordi-
= = 0.
Find the double points of
x*
(i)
(ii)
(Hi)
-2af. z) is
a double point on the curve f(x.
Hence
If (x. acnode. 4ta). We may deduce the result for Cartesian coordinates by putting 2=1.]
(y-2«)(4»-y)V
fied at
Ex. or coincident
nates. Therefore the equation of the tangents at is
*X* +y
3P + "
*Z*
+
ys
*7*Z
+
*Z*X +2x
ywjT=°=
0. or the result for homogeneous coordinates may
r.3«y .
since
by Euler's theorem on homo-
Three roots of the equation in \/fi are zero if P lies on a tangent at 0.
y.
2.
The reality.
be deduced from that for Cartesian coordinates by replacing x and y by x/z and y/z. £) geneous functions
=
0. unreality.
[Equating to zero J-
J-. or ( = 0.0). y (4a-y) 2 = 2a.:
II 4
It will
CONDITIONS FOR A DOUBLE POINT
be noticed that
27
implies /(Z.
£. and the last is a crunode. (-3a. unreal. F.
Find the multiple points of
.2aV + a* = 0.
= 0. 2a). ^-4a* -2ays + 4aV + 3«y-a =0. 0).4a).
Ex.
are real. (-3a. or cusp. z)
The double point
the lines
is
ix ~hz <>y a crunode. or coincidence of the lines is at once determined by considering their intersections with | 0. (0. the next two are cusps. xy + 12a (3x + 2y) + 108a = 0. + 3a). according as
* lx*
+ r>
y + ^^+^i^^^J^/^teJr
.
where the
to x. 2 = 0. r = node at the intersection
[See Ex.. if p approaches
. = .
the coefficients are made to vary.
=
are
=
=
and
js
=
an
»-ic with a
are 0.
3 denote partial differentiation with respect
[For the cusp.
7. if p p^. 1. 8.
0)
. z) should be a node of are fa = 0. =
t
.
=
0.
We now apply
this result to the theory of plane curves.. 0)
.
Suppose that in the equation ~ p^xn +p1 xn 1 + p 2 xn 2 +
.. If u » 0..If
Ex. y.. the tangents at O form an involution.
the
becomes indefinitely
For the equation
where
xg =
1. (1.8. ..
.. If every curve of a pencil has a double point at O. jp«2 + 2g'M« + n.
(0.. . Similarly. = 0. are zero. . 0.
=
27 ab.
Of a pencil of
.1.
is
Po+Pit+P2i*+-"+2>n£
n
=
0.]
Ex
[a
6. one of the roots of the equation
large.
.. 2. r+1 roots of the equa.]
have a node in general.. 0.
is satisfied
[/== Ex.
-Sab)..
—
tion in
x become very
large.=
have a
.
fc-ple
point at
P
and
.
«-ics 3 (n
Ex.
Now if p > 0. C = 0. 1.
S. 10.
are any other curves.
up into three lines...
b.1 )/..
^ = e' =
=
=
at P.
1). A node in addition if k
has z s
<?
=
1
or
2..
. one value of £ given by the latter equation tends to zero.]
when some
of the
§ 5. In symmetrical form the conditions that (x. S = 0.
&c.
2
/= CjS.
which gives
We
leave to the reader the modifications required quantities fu /23 .
we get
/22 /S3 =/25 2
. y. + C S
has a
fc-ple
2
+
+
C.
)
=
0.
Points at Infinity._p r all— »0.
6
= 0. ca. . yi=/2 =/3 = o.
z.
2
Si = 0.
(v)
Always a cusp
at (0. and that it should be a cusp are 2 3
/=
=/ =/
f = JuJs> —/12/13 = /2z/3i ~~fafn = J33J12 ~/31/32 —
suffixes
1.
1).Sr
=
point at P. . C.
c
=
point?
(ax + by + cz)xy a double a
(be. limit zero. (m-2)-ics. If
0.
For what values of
*
a.11 5
(jii)
POINTS AT INFINITY
k
8ft 8
29
(iv)
= -f an acnode at (1.
(»-l)'(/i a /i-/u/2 )
=/i 2
(^/ii+y/i 2
Then + s/is)-/ii(^/2i + 2//2 2 + s/23
.]
lines
is
9. Hence one value of x given by the former equation becomes very large. The cubic splits
. 2 of u and v 0.]
Ex.
Hence
fx //u = /2 //M = /3 //" = »// (» .
+p n = Then.
.y for then nx
One
= mx
=m
.x x..+kmn l x71 1
m
=
mj)
2
mn
)
)
+ . i. e.
m"...
Hence any line parallel to y = ra^x meets the curve in one and only one infinite point. degree n in n
*
m^
The reader
if
a.. are the factors of the terms of highest degree.. for which p + (% 2 ) (m1 It follows at once that any projection of the curve touches the projection of I at the point where it crosses the vanishing line.
is
'
.
In general y = + c meets the curve once and only once at infinity whatever finite value c may have. but t More accurately.
...
(y
—m
and y we obtain n ~ 1 + bxn . . ..
=
0...
3
.
= 0. = 0... .
Next suppose mx = 2 ^ m3 mi Now y = m^+c meets the curve where ~ pxn 1 + {g + rc + (m1 — 3 ...™" + (q + re +
1
1
. convention every straight line meets a curve of points and so throughout.* The line y <mx meets the curve where ~ ' (m— (m— ).
y
root of this equation becomes very large... ... ' meets the curve in exactly n — 1 (finite) points it is convenient to keep the..
. sian equation in descending powers of x
{y—m-^x)
Writing
Carte-
(y—m2 x)
. while the line at infinity meets the curve twice on y = m^x.. . . § 4).
'
'
. if the line as it turns about the origin approaches one of the lines y ="m 1 x.. I is an asymptote (Ch..
(m — m
x)
(m . + (n—l)Jemj n 2
. This evidently
will easily make the necessary modification in the argument a factor of these terms.
. + kyn ~ 1 n x) + ax + Axn -* + Bxn 3 y + . {my^-m^) c 2 }xn ~ 2 +
i .
m
)
.. .
{m-L
— m^c} a.
.. First suppose not equal to any one of 2 2 The line y = 1 x + c meets the curve where
m m
m m
.
we
write
p = a + bm +
l
r=
+ kni-^q = A + Bm 1 + b + .2 y+. y — 2 x.
m
m
.f with the exception of the line I — 3 ) (uij — m") c = 0.
.
.
...m
30
m
POINTS AT INFINITY
..
{p + (m1 —
if
m
2 ) .
where y — m.=0.
II 5
its
Suppose we have a curve of degree n. .
(m—mn —> 0. I.m2 )
..
Let us now discuss a few cases in more detail.)
xn
~2
+
. if rather
The terms of highest degree in the Cartesian equation of a curve equated to zero give the lines joining the origin to the points at infinity on the curve. form
up
this result in the convenient..
)
We may sum
inaccurate... (m— xn + (a + b<m+ .
ay2 = x s has an inflexion.m5 x + c meets the curve at only one infinite point for x every finite value of c.
. If y — m-^x + c meets the curve at two and only two in-
=m
m
m
m ^m
. {m m") c 2 meet the 3) 1 curve in three points at infinity.
y
Suppose now = 2= 3 If the line 1 4 .
[Use § 5. these two
. however.
6
.
to obtain the equation of a projection at/y as in Ch. Then every line parallel to y m-^x meets the curve in two points at infinity..
a?
3
Trace these curves. y + ix is a factor of the terms of highest degree in the equaSuch a curve is called a circular curve.
. Another method of the curve by replacing x. p 0.
1.
infinity at which lines are a pair of
parallel asymptotes. so is y — ix. the curve has an inflexion at infinity along y = m-^x at which the line at infinity is the tangent.
n—
=
=
—m
—
—
= m^
these
Hence the curve has a double point at two lines are tangents i. I. If. if point at infinity. § 3. y — m. the curve has a double point at infinity along y x at which the line at infinity 1 is one tangent. bicircwlar and so on. x x
where
it
is a factor of the terms of degree 1 as well as of the terms of degree n in the equation of the curve..
. a3 y = x* has a triple 4 a.
If a curve with a real equation passes through one circular For.Y x.
and so
on. while the two lines y +c for which q + rc + (m l . the curve has a cusp x at which the line at infinity is the tangent. tion. has an undulation. + c meets the curve at two and only two infinite points for all finite vahaes of c but one. y
* is
by ax/y. If y m. x x + c meets the curve at three and only three infinite points for all but three finite values of c.
l
—x
)
= =a ay =
a?y
3
y
has a cusp at infinity. 1 If y to-j a. Similarly we may discuss the cases in which
finite
along y
=m
=
=m
=
m =m =m =m ^m m
1
2
3
t
5
.. and %axy — x 3 + a 3 have crunodes.
Ex.
. it passes also through the other.
(a 2 point. e.]
More accurately
:
the vanishing line touches any projection cf the
eurve..
points for every finite value of c. Similarly a curve with a real equation having a node at one Such a curve is called circular point has a node at the other.II 5
POINTS AT INFINITY
infinity touches the curve *
31
means that the line at meets y = m. the curve has a triple point at infinity.
7.
. the sum is a multiple
of
2tt.
=
and bicircular n-ic are of the forms + 2/2 )»".
[Use Ch. § 4. if the product is constant.
is
POINTS AT INFINITY
2.. 9. the terms of Ex.2 + un .
3.. where u 0.
Note the cases n
=2
or 4. A quartic meets its four asymptotes in lying on a conic. Show that O is the centroid of the intersections of the m-ic with any line
The equations of a
circular
8
(a. = 0.32
Ex.
[Theconicisg.
Ex. (n — r+ 1)* degrees in its equation have respectively ~' the factors (y — mx) r (y — mx) r i (y — mx). See Ch. Conversely..]
.
I.
Ex. [Taking O as origin.. tangents at three collinear points of a cubic meet the curve again in three collinear points '.. Generalize by projection.. .
through O.
1.]
Ex.]
that a cubic meets its three asymptotes in three finite Generalize by projection. are the [The cubic is f uvw+p 0.
§ 6.2 + «B_i + "«-2 + .
The sum of the eccentric angles of the intersections
ellipse is
of a circle
and
a multiple of 2 v. v 0.= 0. Also y = mx meets this curve in n points whose abscissae and ordinates have
zero sum. 4. 6. An w-ic has n distinct asymptotes all passing through O. and (x* + Z/7X-4 + (*' + y*) V. 5. A varying ra-ic passing through n given points at infinity meets a given ellipse in 2m points whose eccentric angles have a constant sum (to within a multiple of 2n)..|=(^J
in Ex.]
Ex.
3. +u = 0. XII. A variable cubic has a cusp and three fixed asymptotes..s + . where u k is homogeneous of degree h in x and y. If an n-ic has r asymptotes parallel to y the »th (» — l) th .
8.
Two
whose centroid
asymptotes.
. the curve meets the line at infinity only at the circular points.
eight finite points
mx..]
Ex. [Any one of the given lines meets the curve in only n — 2 finite points. the curve is u n + it"_ 2 + m"_ 3 + .
II 5
The equation of an
a l x + bl y
n-ic
with n distinct asymptotes
.
curves with distinct finite asymptotes meet in points is the same as the centroid of the intersections wf the
Ex.
only... Ex.
Show
collinear points
=
=
=
=
—
Ex.3 + «W-» + . (an x + bn y + cn ) + u n .
+ c1
=
0. If a jeal u-ic meets the line at infinity at the circular points any line through a fixed point O meets the curve in n points whose distances from O have a constant product.. = 0...
n must be even.
a n x + bn y + cn
=
of the form
i
(a x + b 1 y + ci )(a i x + b i y + e l ) . If the n points can be paired to form an involution with the points at infinity on the axes of the ellipse as double points.. 10. Show that the cusp lies on the conic touching at their middle points the sides of the triangle formed by the asymptotes._ tv = On projection we get ' the asymptotes and p is the required line.
[The case n
=2
is
well inown. v = is circular.]
§ 6. i. The information that the curve has a node (not at a given point) is equivalent to one relation between the coefficients namely the result of eliminating x. y)
is
stated to be a cusp.+l)
whose equation is written down at has no double point. see that a curve
'
—
'. The information that the curve has a cusp is equivalent. n + 1 of the nth degree.. Hence the. one of whose sides makes an angle OC/n with the
fixed direction.e.
we have
a fourth relation
between the
coefficients
/jv_y =
\}>xly)
^...equation has %(n+l) (n + 2) coefficients.. (i) and to two relations found by eliminating x and y from that it should have a A. w
If (x. It has no point-singularities i.
'
. y from (i). multiple
We
2 relations.
namely
(i).. multiple points. Show that the locus of P is an «-ic whose asymptotes are parallel to the sides of a regular polygon.II 6
RELATIONS BETWEEN COEFFICIENTS
33
u
Ex. y).
Anautotomic' (not * The definition is therefore not free from objection.
'
Similarly the information that a curve has a /c-ple point at a given point is equivalent to %Jc (k + l) linear relations. The lines joining a point P to n fixed points make angles with a fixed direction whose sum is a (plus a multiple of it).
Relations between Coefficients.
i/c(A. three of the second degree... The sum of the angles which the n asymptotes of the n-ics make with a given line is the same. 11. Show that one 0..
. . y) is equivalent to assigning three
linear relations
between the
J
coefficients of f(x. e. two terms of the first degree. not exclude aenodes cutting itself) which has been suggested would.-ple point to and similarly (ii)
. curve of the pencil u + kv =
=
Ex.
/=^ = ^ = *y
7>x
.. however.
has The information that a curve with equation f(x.* The Cartesian equation of a curve of degree n contains one constant term. 12. or unreal multiple points..?v . Zy lx
2
2
V (
ii). y) = a node at a given point (x. though it has in general 'line-singularities'. Such a curve is called iwnrandom singular..
tangents.
and that no n-\c can be found in general satisfying given conditions. the case n = 2.
Hence
curve of degree conditions. if their number
exceeds \n (n + 3).
relations
in general for it may happen that the § n (n + 3) between the coefficients are inconsistent or not independent. This is legitimate.
'
'
. if a curve has 8 nodes. Expressing the conditions for this. y).
true in general. But. an infinite number of conies pass through the points for the line through these four points and any line whatever through the fifth point form
is
. being a line-pair. have then
We
^(n
+ 1)
(n +
2)-l
= ±n(n + 3)
by
arbitrary coefficients. As an easy deduction from the results of this section have
we
A finite number of curves of degree n can in drawn having 8 nodes and k cusps and satisfying
%n(n + 3)-8-2 K
other relations. Hence
'
One and only one curve of degree n can be found in general passing through %n(n + 3) given points. we have ^n(n + 3) linear relations between the coefficients of the curve's equation. + 3) — 3 8 — 4 k
. If three of the points are collinear. if four of the points are collinear. for we did not exclude in the theorem the possibility of the n-ic splitting up into simpler curves.
such a conic.. which determine the ratios of these coefficients uniquely. there are 8 sets of values of x and y satisfying (i).
tlte
number
of
For instance. Consider.34
RELATIONS BETWEEN COEFFICIENTS
II 6
One of these coefficients may be taken as unity without loss of generality. But such results are not universally true. the conic degenerate. we have 8 relations between the coefficients of / (x.
A
n is determined in general
\n
(n + 3)
We mean by this that only a finite number of w-ics can be found to satisfy the given conditions. for instance. and then the theorem is not true.
general be
other relations is \ n (n
If the nodes and cusps are at assigned points. If the conditions are that the curve is to pass through \n (n + 3) assigned points. and must be
. We have One and only one conic can be drawn through five points which is
We
say
'
'
.
]
Ex.
quartic with three given nodes
a/u* + b/v* + cluP..x1 ) + cz (z* .1). .
where S =
0.
[ax {a? -f) (x* .
5.z2) {y2 .
= =
=
are given lines through the node.. the a's are arbitrary constants.
is
Ex. 6 is is the conic aC1 Ci L li + bCs Ci L3i + cC5 Ce LBS = 0. Ex.
. 5.
35
applied
with due care as is shown by consideration of Ex.1. 1.
+6M S".z 1 )
1 1
(1. 3.
E. 5 to 7. 5 with ft = \ n (n + 3) — 5 r.II 6
RELATIONS BETWEEN COEFFICIENTS
. The equation of any n-io through r given points can be put in the form S+a 1 S1 + a 2 S2 + .
S+a S + a S +
l
+ a^S^ =
0. -1). 1).1). 4. (1. +1). 10. 12. v their intersection. Slt . x = ±y = oo ..
.y1 ) + (l1 y + m1 z) {y l -z*y + {li z + m 2 x) (z'-x'f + il. S^ are given polynomials such that S — 0. The result given multiple points.
...x + m. with nodes at ( + 2. S^ are w-ics through the r given points. 5. Campbell in Messenger
D 2
. where C = through 2.
where u
Ex.
where u
v
= 0.. and fi = \n(n + %) — r... It is assumed that there is no identical relation of the form
=0
S+-b. In general one w-ic can be drawn with a given node and passing through \ (w 2 + 3 n . 3.
7. 4. S1 = 0. 4.
Ex.
+ by (y .8) other points. Ex. 9. &c. but that through the 9 intersections of two given cubics a singly infinite number of cubics can be drawn.
Ex.]
:
* The reader xxi (1892).]
Ex. -1. §7. 0).6) other given points while two n-ics can be drawn with a given cusp and passing through | (m 2 + 8m .1. or of Chap.. 6 and Lu — is the line 12. IV.
... Obtain the equation of any «-ic with r given double points and given tangents at those double points. [As in Ex. 17 below.
=
0. 6. The' reader may find the general equation of quintics [Use Ex. To be given a &-ple point and the tangents at that fc-ple point equivalent to being given Jfc(fc + 3) linear relations between the coefficients of the curve's equation.. and
in
= |«(» + 3)-3n
may
be extended to cover the case of any
[Use § 4.y) {x^-tff = 0.
may
consult a paper by
J.
S^
=
are given M-ics with the r given nodes. is
The equation of any
1
»-ic
2 2
with r given nodes
.
=
—
u + lev
=
is
any cubic through
Ex. 5.
The equation of any
0.]
9 given points . +0^^ = 0. 1. 2. Find the general equation of quintics with nodes at (-1..
8. Show that in general exactly one cubic can be drawn through
[If « are the two cubics..* Ex.S1 + b 2 S1 +
Ex. 2. 3. (0. 8. (1. w =
are the sides of the triangle whose vertices
are the nodes.x ) (z2 .] Ex. 0.+ 2f/vw + 2 g/tm + 2 h/uv
=
0.
(a l
The equation of any cubic with a given node is x + b 1 y + z) u? + (a2 x + 62 y + z)y/o + (a s x + 6 3 y + z)
and v
u
2
=
is
0. where S. The equation of any quintic with six given nodes 1.
. 16. Annalen. 18.'
36
RELATIONS BETWEEN COEFFICIENTS
is
II 6
Ex. 1) as node and the cubic is xy(y — mx) +kz arbitrary constant. 0. n—5. Bull.
Ex. the cubics through the seven nodes. de la Soc.
y
= y'.y. xxv. 13. v. 17. and these five' nodes having double points at the other three nodes. i(x.
(0.
aU i + 2hUV+hV =
2
V=
Q=
12. 14.v. and
Ex. 15.
n-S. For instance.
= 0.
1. if P lies on the sextic.
Ex. If (x'. p. VII.
(1.s — y*)
(1. w when the coordinates of P are substituted for x.. II. y) is a centre of the n-ic f(x. If u = with cusps at u — v = 0. § 10).w) jacobian (Ch. w 3 = v2 is a sextic Ex. Hodgkinson. in x and y
Ex. x (1882). 0). If a sextic has nine nodes of which eight are fixed.— w^v) = 0. Soc. 343.]
P
:
:
:
:
.
all
the partial
. and P are {%im^ — jcjit) + &(»». C = is the quartic through a conic through five of the nodes. the cuspidal tangents touching v = 0. Proc.
An M-ic passes through r 2 J(w + 2) or £(»i+l) (« + 3)
as n is even or odd. Math. Note that nine nodes of a sextic cannot be chosen arbitrarily even though such a choice is equivalent to J 6 (6 + 3) = 27 conditions.z) [J = has a node at each of the seven points. 0. w = are cubics through the given nodes. [If «!. The statement 'a curve has an inflexion at a given point' equivalent to two relations between the coefficients in general.
when x
= x'. xv (1916). &c] cusps ?
Ex. If in Ex. z. y.]
.. the terms of degrees n — 1.
[Taking (0.
is a conic and v = a cubic.
[The curve comes to self-coincidence when rotated through 180° about a centre.
given points and has a centre Find the
. London Math.
.]
'
(mxy — m 2 a. 598. p.. however. sfcow that a singly infinite family of cubics can be drawn with three given collinear inflexions and a given node. that to be given three collinear inflexions of a cubic is equivalent to five (not six) conditions. v = 0. The. [See Halphen. «>! are the values of «. For instance.
m. The statement a curve has a tangent touching at h points equivalent to k — 2 relations between the coefficients.equation of any sextic with eight given nodes
CQ. is where ?7=0. the ninth node on a certain 9-ic with a triple point at each of the eight fixed nodes.y) derivatives of/ with respect to x. 13 d = 0.
i(u. p. all cubics through the seven nodes and of the sextic. [The preceding examples may suggest interesting investigations to the reader. 0). is u s = »2 the most general sextic with six What is the most general septimic with ten given nodes ?. 0) as inflexions. n — 3. de France. 11. The general equation of a sextic with seven given nodes is au? + bt? + cw* + 2fvw + 2gtou+2huv + dJ = where u = 0.«!. a point P on the sextic pass through another point Each such cubic meets the sextic again in points Q and Q' such that all cubics through the nodes and Q pass through Q'. 20. Math..
where k
is
an
is
Ex.
lies
_
.
vanish. 19. See Eohn. 1.
is
Ex. are two cubics through the eight nodes. If the centre is the origin. y of orders n— vanish
=
0.
n — 5. 162. Hence at their ninth intersection Therefore.] Ex. Show.]
r being degree of the locus of the centre. so does the ninth u v iv = «<! % w1 intersection.
the number of its branches. g.
The following hints are
Find where the curve meets the axes of reference. the curve is symmetrical about the origin.
for
requires.
useful in curve-tracing. Or we may wish to obtain only a rough idea of the main features of the curve (e.
approximation to the truth. If the equation
(i)
(ii)
contains only terms of odd. finding for this purpose a large number of points on the curve. Or it may imply that the curve is to be drawn with the utmost possible degree of accuracy. where pos. Similarly if only even powers of y occur. and we can spare ourselves the rather considerable expenditure of time which a more detailed tracing usually
sible. (iii) Notice if any values of x make y unreal (or vice versa).
'
we
it
shall here mean the third of these will be useful to indicate.'
CHAPTER
The Object
III
CURVE-TRACING
§ 1.
of Curve-tracing. If so. &c). or only terms of even degree. and equate to zero
. a circle can be traced by means of compasses.
how a more accurate diagram may be obtained even if most practical purposes a rough sketch of the curve will suffice. (iv) Equate to zero the terms of lowest degree in the equation to obtain the tangents at the origin.
Trace a curve -with given Cartesian equation capable of more than one interpretation. the curve is its own reflexion in x = 0. the rectangular Cartesian equation of the curve being supposed given.
The Method
of Curve-tracing.
§2. these values of x do not correspond to any real part of the curve. and to draw a sketch of the curve which gives some
'
The problem
is
. which is called a centre of the curve. If only even powers of x occur in its equation. It may mean that a mechanical construction is required for instance. the position of its asymptotes and singular points.
By curve-tracing alternatives though
' . See if the curve has symmetry.
1. (vii) Find if it is possible to get any number of points on the curve by solving equations of degree 1 or 2 at most. there may be one or more. whose index is equal to the intercept made on the axis of y in Newton's diagram by a line through (a. if necessary. origin to the points at infinity on the curve. §§ 3-. A being a numerical coefficient. if
the curve is of degree n.
Suppose that any term in the equation of a curve is Ax a y&. diagram for the curve. II. Suppose that a line * joining two or more points of Newton's diagram (not parallel to an axis of reference) is such that all other points of the diagram lie to the right and above the line. proceed as in § 4.
The above
n—r
§3. and a. if the points on this line are considered as representing terms of the same order of magnitude. It may sometimes be useful to obtain the tangent at one or more points of the curve. and. where is negative or odd. If
* If the curve does not pass through the origin. j8) referred to rectangular Cartesian axes placed as in The points thus obtained are said to form Newton's Fig. x and yl are of the same order of magnitude. Find the asymptotes of the curve. all the other points of the diagram represent terms of higher order of magnitude. the order of the term Ax a y& is the same as that of y0 +a 1/P.
all
hints are usually more than sufficient to obtain the information necessary for tracing the curve. when x and y are both small or both p Then large. if it is met by a line in r real points.
Newton's Diagram.
give any fresh information.) Use Newton's diagram to get approximations to the shape of If this the curve near and very far from the origin (§ 3). 5. Suppose now that.
is
no such
line.
the terms of highest degree to obtain the lines 'joining the. this requires the solu-
tion of
an equation of degree n — 2 at most. /3 zero or positive integers.
If
. though occasionally special devices are needed.38
THE METHOD OF CURVE-TRACING
III
2. j8) parallel to the line making intercepts p and q on the axes of x and y. (vi) Find the finite intersections of the asymptotes with the
fails to
(v)
curve
:
for. there the curve does so pass. Then. (Ch. It is also well to remember that a tracing of an n-ic cannot be correct. Consider a geometrical representation of the terms
of the equation such that Axa y B is represented by the point (a. find on which side of an asymptote the curve approaches it at the two ends.
the terms represented by points on the line give an approximation at infinity.
.
orio-in
=
the branch through the near the approximates to x2 + 5y =
origin. we get an approximation to the curve
near the origin. The lines gents are y = the origin to the points at infinity joining jr-jr are x = 0.
Similarly. 1) gives as approxi. l. and H '* ) 5x + 2y 0.
of the
Ex.
Ex. x + y = 0. Newton's diagram (Fig.
Examples of Curve-tracing.
1.]
§4.Ill 4
EXAMPLES OF CURVE-TKACING
all these
39
we
suppress
higher terms in the equation of the curve
and then divide out by any power of x or y. which is a factor of the remaining terms. Illustrations of the use of Newton's diagram are given in §§4-8. Enunciate and prove a method similar to that of Newton's diagram for approximating in three dimensions to the shape of an algebraic surface near and far from the origin. The origin is a crunode at which the tanand 5x + 2y = 0. 3. As another example take the curve in § 7. There is no symmetry. we get an approximation to the curve very far from the origin. and far from the origin
=
x(x-2y)(x + y) = 0. in the curve x*y + if — 2xy — x = Newton's diagram gives the approximation y 1 = x near the origin and a? + y = 0. The only new information is that
touching y
Fig. 2.
+ bx 3 near the
origin.
I.
Ex. 3
— x y— 2xy 2 + 5xy + 2y* =
2
0. if all points not on the line lie to the left and below the line.
The curve meets the axes of reference in no finite point except the origin.
In each of the following examples we shall only give details working when such details illustrate methods which have not been used in previous examples. Apply the method to find the shape of a surface in the neighbourhood
of a parabolic point. Ex. xy = 1 far from the origin. If a line through two or more points of Newton's diagram is such that all other points lie to the left and above or to the right and below the line.
Newton's diagram applies even
if
the original axes of reference
are not rectangular. x = 2y." t _ <*rH 2 mations near the origin x + 5 y = 0. For example.
[It
approximates to the form of the surface z
= ay
1.
Trace
a.
If the equation of the curve had contained terms of lower degree than the second. we should have retained only the terms of highest degree in y in the numerator of the right-hand side after putting 2 y for a. The curve evidently meets the asymptotes at their intersections with
3x + 4y
=
2
giving three points on the curve.
On the right-hand side of this equation put in the first approximation —fa.) We have now enough data to trace the curve (Fig. Similarly the other asymptotes are
On
=
x—
=—
Write
x = 1 and x + y =1. We will verify the diagram by finding the side of each asymptote on which the curve approaches it at either end. and we have the second approximation 2y 2. If the equation of the curve had contained terms of degree higher than the third. (See Ch. 2).
=
.. The curve becomes
2 2
2
. Ex. 3. Thus write the equation of the curve in the
there
x
We know that = 2y. § 5 or we may employ the device just used to find the approximation at the origin. for y and we get the closer approximation 2y + 5x = —--^x* as before. II. Write" the curve
. Another method is to write the equation of the curve in the form 2y + 5x = (-x3 + x 2 y + 2xy2)/y. II. now the equation of the curve in the form (x-l){x-2y + 2) (x + y— 1) = -Zx-ty + 2.40
EXAMPLES OF CURVE-TRACING
To
get an approximation to the branch touching
III 4
Tpxit2y
+ 5x=
Y.
These
form
_ -5xy-2y 2
x(x + y) the right-hand side put in the first approximation x 2y.
0= Y -5xy-l8x^ + 9x 7-x7
and Newton's diagram now gives the approximation 57+18a:2 = or 5(2y + 5x) + 18x* = 0. may be found as in Ch. we should have retained only the terms of lowest degree in x in the numerator of the right-hand side after putting — f a: for y.
are asymptotes parallel to x 0. which is the asymptote.
where the left-hand side equated to zero is the equation of the asymptotes. § 5.
5x + 2y = 0. x + y = 0.
2 35!/2 + 53M/+2J/ 2 =
lies close
0. As stated However.
for the
2y — 2 —
—
5
for
the curve.
slightly
smaller for the curve than the asymptote. if to the asymptote and to the left of it. the accurate in § 1. we do not usually require this.
tote
An
We
get
x-2y + 2=--.
y
is
Similarly.e.
.
xS-xty. The reader
may discuss
The
line
x
= ty
the other asymptotes similarly. tracing of most of the curves given in the examples of §§4 to 9 does not present much difficulty. and
2y — 2
i.Ill
4
EXAMPLES OF CURVE-TRACING
41
approximation to the curve as it approaches the asympx — 2y + 2 = is % = 2y — 2. large and negative.
Hence the curve
Pig. 2. the curve lies to the right of the
asymptote. meets the curve where
(5t + 2)
x
=
(l+t)(2-t)'
y= "
t(l
bt + 2 + t)(2-t)
Putting in any value for t we may obtain any number of points on the curve. and retain only the highest powers of y in the numerator and denominator. Put this value of x into the right-hand side to get a closer approximation. and trace the curve accurately.
5
is
For a given large positive value of y x
asymptote.
which
is
the pair of parabolas
y
2
= ±
F a tacnode in such a case. The curve is symmetrical
(x*
about the
= 5x + v -2. The intersections with the asymptotes and with 2x = 3.-8y-6. The origin is
Hence there
are
— -V
. 6. oo ) and (oo Bitangents x = y.
Ex. The curve has a crunode at (oo 0). Fig.
.
1
.
(xvi)
Asymptote's
(xvii)
. See Ch.
called
' '
two branches of the curve touching each other and touching the axis of y at the origin.
Trace
Newton's diagram (Fig.
0).
*
A
tacnode
is
the point of contact of two ordinary branches of a curve. Fig. (x-l)(x+l)(2x-y + l) + 3x+l=Q.6. and an oval lying between the asymptote and the axes of reference. 0).
x = ±2. XVIII. (xviii) Asymptotes x = ±l. 3x + y = 4.
(v)
(vi)
(vii) (viii)
2)(*-y-l)« + 8*-8y-2-. (oo 0) and a triple point at It has a branch asymptotic to x + y + 6 = (0.y = ±B. and a triple point at (0. Consider its intersections with y = tx or txy + x + y = 0. Solve for x and y. Solve for y.
+ x+l)(x-y+l)(x + y)=y. oo ). Solve for x and ' «.
(x*-l)(x-2)(y*-l)=2x* + x-4:y-2.* As in § 5 we find the parallel asymptotes y+1 = and y— 2 = 0.
§6. y = ±\. The tangents are parallel to «. XVIII.
. Compare (ii)
T
(* + the origin. 4) gives us as the approximation at the origin
4
2/
=
2x\
</2x.
(xii)
. (xiii) The curve has acnodes at (0. It is sometimes called also a double cusp but this nomenclature is open to objection. See Ch. to (x-l)(9 + l)(x+y-l)(x+y-2) (xi) The asymptotes are x^y* = 0. &c. (5. y{{x-yf-2} = x-y. (xix) Asymptotes x*= 0.
.
There
is
an
inflexion
at
(ix)
origin
and has an inflexion at the origin. (xiv) (x + l)(x + 3)(y+l)(y-S) + lSxy + 12x + 6y + 9 = 0. 10) ^) respectively. y = 2 should be found. as we shall see later. Asymptotes xhf = 0. (xv) Asymptotes x2 (25y 2 -36) = 0." J f i the axes of x and y at ! c Jf and (8.Ill 6
EXAMPLES OF CURVE-TRACING
45
iv) T he C e h a n0 al ^ymptote. Vx = y. Biflecnodes at (0. oo ). (x -1) (x + 2) (x -y -l)(x-y -3) = lla. III. 2.
and lying above the axis of a.) To get an approximation to the curve far from the origin along x + y = 0.
2
. write its equation in the form
since (x
(x + y) 2
=x
2
(y
+ 2)/y2
.
parallel to x + y. + »)'
=
*»(» +
2).:
.. § 5. OF CURVE-TRACING
III 6
There are two infinite points on the curve along x + y = 0. (Ch.
x on the right-hand side and retain the highest power of x in the numerator. This is a parabola with axis
for
Now
put
—y
Pig. and readily find the side on which the curve approaches either asymptote. 5. See Fig. 5.
46
EXAMPLES.
(x
(y + 1)
it
(y. We get as a closer approximation (x + yf =? y. II. + y) 2 is a factor of the terms of the highest degree but x + y = c meets the curve at only one infinite point for all finite values of c.2)
meets
+ yf + (y* + 2y)
we
see that
its
Ex. HeDce the curve has an infinitely distant tangent with its point of contact on x + y = 0.— 0. asymptotes where 2x + y = 0.
(ii)
x*y
= x + y". Solving the equation as a quadratic in x we see that for real points on the curve y^ — 2. Writing the eqsation of the curve in the form
(2x + y)'= 0. touching y = at the origin. Trace the curves
(i)
{x + y)(x~yf
= %xy-y
i
.
/(a.
)*
From
Kepanov
=
a
little
horn. The origin is sometimes called in such cases a rhamphoid (ramphoid) cusp * or cusp of the second species '.9).
x*-2x 3 y + 2x*tj + ij* =
possible type and the latter siderable complexity. the ordinary *->cusp being a ceratoid cusp f or a cusp
see that for real points
'
We
'
'
V^
'
'
'
of the first species
'. 8. 9.
The nomenclature
is
objectionable.
*
0. Hence the approximation at the origin is given only by those parts of the two coincident parabolas for which x is negative (Fig. whereas the former is the cusp of the simplest
--
Fig.
on the curve x must not He between and 2. write its equation in the form
y
= — x + x ± V(—2x + x
2 3
5
e
). it
implies that cusps of the first and second species are comparable. for
Pig.
is
really a singularity of con-
From
fia/upos
=
beak.Ill 8
EXAMPLES OF CURVE-TRACING
51
curve. .
E 2
. To ascertain more closely the nature of the curve at the origin.
XVII.
. we see that there is an asymptote parallel to y/x = 1 corresponding to t = 0.
(1
+ *)
(1—2*)/*. To give a tacnode or rhamphoid cusp and the tangent at that point is equivalent to assigning 6 or 7 relations respectively between the coefficients of the equation of a curve. u = v 1 2 becomes a conic of closest contact with the curve at the origin.e.
6. whence t =
x
=
oo
or
2+
</2. i. The curve meets the asymptote or £( + -/2-1).
t
= —1
. and the method of the following example will often suffice.
§9If the coordinates of a point
on a curve are given in terms y
of a parameter
t
by means
of the equations
»=/(*).
x /y--=(l+t)(l-2t)/t2
. Making the equations of the curves in Ex. [The origin is called an oscnode Contrast the case of the rhamphoid cusp in Ex. VI. § 8 (iv). As an alternative we may obtain any number of points on the curve by taking various values for t and so trace the curve. x and y are infinite when t = Since y/x = (1 + 1) (1 — 2t).
obtain the equation of the curve by eliminating t between these two equations and then trace it by the methods of §§ 4—8. = at the origin and having the same curvature there. 2 and see Ch.. putting
s
=
oo
in
. If the tangent is not given.
The curve only meets the axes at the origin.
y
= (l +
t)
2
(l-2t) 2/t.
. on putting t = 0.
EXAMPLES OF CURVE-TRACING
The curve
(M 1
53
0=
+»1 2 ) a +(Wl«'i + Ms)(% + V)+V«2+«lM4 + «5+
.
we may
Ex.]
'
'. 7. we obtain x—y = 1 as the asymptote. and then We have y = tx2 and therefore the approximaor \ 2 tions at the origin are y = — x 2 and 2y = x or oo Again.Ill 9
Ex.
.. But this is apt to be laborious.
=
m. 2 and 6 homogeneous by means of u^ = v2 we get the lines joining the origin to the intersections of the curve and u 1 =v1 If we choose the coefficients of » so as to 2 make as many of these lines as possible coincide with u t = 0. Trace
x
=
.
+«"
has two ordinary branches touching «.
i
Moreover.
Ex. the number of relations is one les's.
.
Because x—y = (1+*) (1 — 4* 2). A doubly infinite number of conies meet a curve five times at a rhamphoid cusp.
. where (l + *)(l-4* 2) = 1. and no conic meets it six times there.
The
if
line
Xx + fiy + vz
0. As a particular case.
Class of a Curve. The proof is exactly similar to that used
of
The common tangents
m
in
CO. the line Xx + fty + vz = becomes
(^A + l2 /i + l 3 v)
This
+ (m{X + m2 /i+ m 3 v) y' + (n1 X + n2 p + n3 v) z' = 0. . are found by solving these two equations for X: fi:v. This number is called the class of the curve. is identical with X'x' + fi'y' + v'z' = 0. if X'=l1 X + l2 /i + l 3 v.
v)
= 0. y'.
=
passes
through the point
{x'.
/it.
m
:
m
*
two curves of class and M. straight line has no tangential equation. v) = 0. fi. v'= n-^K + n^fi+n^
x'
. v) of a curve is homogeneous of degree m.58
TANGENTIAL EQUATION
IV
1
stitution given
For on changing the homogeneous coordinates by the subby equations (i). v... fi.
v
..
<j>(X. If the tangential equation <j>(X. v) = O' and $(A. y'.
But equations (iii) give equations which proves the result. /i'=m1 X + m 2 /i + m3 v.~z7) to the curve are given by solving for A fi v from
A
=
:
:
Ax'
+ fiy' + vz' =
0. (iii). 'Hence^m tangents can be drawn from any point to the curve.
The point
equation
of contact
of
any tangent having the point-
X'x + fi'y + y'z
=
is
<f>(X. fi. in A. the tangents from (x'. /jl. tangents can be drawn to the curve in any given direction. /x.. z')
Xx' + fiy' + vz'
=
This is therefore the tangential equation of the point.
Eliminating v from these two equations. §7.
§ 2.
'
A line
is
considered as
touching
'
a point in the sense of
§ 1 if it passes
through
it. They are piM in number. we have an equation of degree in A fi.
fi. It is to be noted that the tangential equation of a point is of the first degree.0(X.
with the curve whose tangential equation has the tangential equation
^
v)
—
¥
¥
¥
.
(ii)
on solving
for A.
i.
joining
its
-(\x + py)/v)
intersections with the line
Aa3
+ fty + vz =
to the vertex (0.
3
TANGENTIAL EQUATION OF ANY CURVE
59
where
•
^
If
. z)
= 0. z')
by
eliminating (x\
between
Xx> + fi y' + "z>=0 and
g/A = */? = g/". =
(i)
we
find the lines
f{x. to obtain the point-equation when the tangential is given.
. e. z')..
means the
result of putting A'. by eliminating x/y between (i) and the result of differentiating (i) with respect to x/y.
obtain the required tangential equation
y'. I. we may call u + kv a (tangential) pencil of curves of the m-th class. z')
and
If
7>x
if
tty
3/
oz
"
to
/ = at (x\ We therefore
y'. An alternative method of finding the tangential equation of is as follows.(\x + fty)/z) = <f>{\.
/m. Similarly.IV
. 0. z) Suppose (a/. namely the common tangents of u and v 0. we find the condition that equation 0(A.
The proof
similar to that of Ch. \i.
v
in
~
<>f
&c. y'. (i) are coincident if \x + /iy + vz touches the curve. u and v being homogeneous of in A.
To obtain
the tangential equation of
f(x. we get the required tangential equation. y. if we find the condition that (i) considered as an equation in x/y has equal roots. Hence.
I. //. that this last equation should have equal roots considered as an equation in X/fi. y. p. degree The curves obtained by taking different values of k have a common tangents. This line must pass through be identical with the taDgent
Xx + jiy + vz
—
(x''.
As in Oh.
Tangential Equation of any Curve. v.
=
=
should coincide. z') is the point of contact of the tangent
=
0. y'. v) two of the intersections of the curve with a side of the triangle of reference . v
is
for X.
=
m
m
=
=
§ 3. § 8. fjt. f(x.y. Two of the lines represented by.
§ 9. 1) of the triangle of reference.
the asymptotes of the pencil of w-ics
S+ kS' =
sets ot
Any
one of these asymptotes meets
n points with the same centroid. in rectangular Cartesian
coordinates the tangential equation of the circular points
is
A 2 + /* 2
=
0. p) = / and (/> being homogeneous of degrees m and m — 3 respectively in X and ft. + 2{GH'+G'H-AF-A'F)yz +
.
'
Take the second conic as the circular points. VIII.
2.. &.
The
circular points are
X 2 + /i 2
+ » a -2^i/cos^-2eXcosB-2Vcos
C=
for trilinear coordinates. Obtain the equation of the director-circle of a conic whose equation is given in any coordinates. Write down the tangential equation in trilinear coordinates of a conic with foci (a. e..]
from Ch. p) + (\* + n") <£(X. 3. A'\ 2 + 2J> 2 + CV + 2 + 2 (?VX + 2 22> = form an harmonic pencil is
F>
(BC' + B'C-2FF')x>+
.
Ex.g.] of the tangents from (x.. yj.y + l perpendicular. y) to y =
.
for this is the condition that \x+/j.
The tangential equation of the circular points is obtained by writing down the condition that any line should be perpendicular to itself.
§
1. A curve of class Show that the sum of the angles which points and 3 other points.. _(a 2 j3 y 2 ) and minor axis 2k. + vyj (Xa a + M /3a + vyt )
.
for Cartesian axes inclined at
Ex. [(Xa t + p/3.
1.
. the tangents from any point make with a fixed line is constant to within a multiple of ir. the circle inscribed in the triangle of reference.
[The locus of a point such that the tangents from it to the conies ^IX 2 + By? + Gi? + 2Ffiv + 2 Gv\ + 222V = 0.
. 4.. The inclination 2 2 is given by /= (X +/* ) (\x + iiy)<j>.
that the degree of the envelope
§4... The envelope touches the line' at infinity where a curve of the pencil touches it...
=
0.IV 4
E
is
EQUATION OF CIRCULAR POINTS
7"
61
ot class
/
-
T!j e envelo Pe of
2n—l.
[It follows
S=
and
S'
=
in
two
at each of the
2(n-l)
points
is
4(n-l)
m general. + .
=
should be
self-
The
circular points are
X
l!
-2X /icosa) + /i2 =
an angle
a.
m—
m
[The curve is /(X.
Tangential Equation of the Circular Points.
+
. Ex. For instance.
=i
s
(X 2 +
2 2 ^ + » -2^cos^-2vXcos£-2X M cosC).]
Ex. and becomes the director-circle of the first conic]
this
'
harmonic locus
touches the line at infinity at the circular Ex. 5. ? Deduce the equation of a circle with given centre and radius.
P
B
PQ
R
P
PQ
. * But 1 (x\ y z") lies on the polar reciprocal in this case.
the polar reciprocal with respect
[The polar of {x'.
*x+y + v=0
2y
If the condition that a curve (in Cartesian coordinates) should
is
cj>
(X.
and this touches <#>(A. the sum of the angles which the common tangents to this curve and any curve of given class make with a fixed line is constant to within a multiple of w.
1
.
The envelope of Suppose we have a certain base-conic the polar of any point on a given curve 2 is called the polar reciprocal 2' of 2 with respect to the base-conic. /i.] Ex. ji. Hence 2 is the envelope of the polars of points on 2' or the relation between the curves is a reciprocal one.
(ii)
The three-cusped hypocycloid. a point on 2 and the tangent at this point being the pole and polar with respect to the base-conic of a corresponding tangent to 2' and its point of contact. y') with respect to x"1 + 2y 1= is xx' + y + y' = 0. v) = with respect to the base-conic x 2 + y* + z2 = is
<p(x.
is
equal to the class of
its
polar recip-
§ 6. 2. if 0(0!'. But these two polars are consecutive tangents to 2' in the limit.
point-equation is given. y. y\ z') = 0.
[(i)
As
in Ex. which is always the case
. and is the point of contact of either. and Q on 2.
Equation of Polar Reciprocal. Hence the polar reciprocal of a curve can be obtained when its tangential equation is known.
<j>
The polar reciprocal of the curve having tangential equation (X. its intersections with 2 are the poles of the tangents to 2' from the pole of I with respect to the base-conic.]
§ 5.
its
when
The reader
'
Ex. v)
to x* +
=
is
<l>(x. z')
For the polar of
with respect to the base-conic
is
xx' + yy'
+ zz'
=
. Discuss the case m = 3. y'.
Hence
The degree of a curve
rocal.
5.
(x'. More generally. Also is in the limit the tangent at to 2. 6.
y)
= 0. v) = 0. Their Consider two consecutive points polars meet at the pole of with respect to the baseconic.
=
0.
* '. will notice the close connexion which exists between the algebraic notion of 'tangential equation ' and the geometrical conception of polar reciprocal '. z)
=
0.
Polar Reciprocation.:
62
EQUATION OF CIRCULAR POINTS
IV 4
Ex. The polar reciprocal of a 'Lame Curve' (x/a) n + («/b)n = with respect to {x/Af + (y/Bf = 1 is a Lame Curve. If I is any line. touch
1.
intuitive from a figure. If the base-conic has a real equation. Hence to a node corresponds in the reciprocal curve a tangent with two points of contact. For pictures of bitangents see Ch. 1 Ch. except c. point with
'
We shall now
'
A
.
which may
its
with three. four.f Hence from any point on c two tangents can be drawn to the reciprocal curve coinciding with c unless the point is at I. . 7. VI.* In the same way. Fig. which meets it thrice while of the tangents from any point on i one at C coincides with i. A more rigorous proof is given in Ch. unless P is at C. v) = ax 2 + by 2 + cz 2 + 2fyz + 2gzx + 2hxy = (ax + hy + gz. Fig. Ex. 6. and general a common self-conjugate triangle.
its tangents correspond a bitangent of the reciprocal curve and its points of contact. 9.
distinct tangents corresponds a tangent
distinct points of contact. 2.
. The base-conio.
inflexional tangent.
. be called a triple. quadruple. when three of the tangents coincide with i. X. IV. e. hx + by +fz.
and
its
Reciprocal. i.
4. quad.
<f>
polar reciprocal have in
respect to
is
The polar reciprocal of with (\.
c of the reciprocal
To a cusp C of a curve and
an inflexional tangent
flexion J. &c. its nodes. gx +fy + cz)
Singularities of Curve
=
0.
.
C
P
. f These properties of the cusp are evident from the fact that. 3.
.
. to a triple.
. to a crunode corresponds a real bitangent with two real points of contact. any conic.
.
. XVIII. for which the properties are at once established. §§ 3.3 . which meets it thrice at I. node is a point of a curve at which there are two tangents. to a cusp and its cuspidal tangent correspond an inflexional tangent and
* Sometimes called an ' ideal bitangent. except i. 5. Figs.
ruple.
consider what corresponds in the polar reciprocal to the singularities of a curve. and to an acnode corresponds a real bitangent with unreal points of contact.
§ 7. /*. They are almost Similarly for the inflexion.
tangent.:
IV
7
SINGULARITIES OF CURVE
its
63
Ex. by Newton's diagram. when three of the tangents coincide with c while any line through I meets the reciprocal curve once at 7. Such a tangent is called a bitangent. any curve approximates near a cusp to a semi-cubical parabola ay 2 = a. cusps. Therefore I is an inflexion and c the
For the cusp has the properties
. 10.
'
Hence To a node of a curve and
.
cuspidal tangent i correspond curve and its in-
that every line through meets the curve twice at C. 1 Ch.
and two real inflexions. . two of the tangents from to the curve coincide with each tangent at O. a bitangent with real points of contact.. a crunode. The singularities of the limacon are an acnode. If is a multiple point of a curve with distinct tangents. two unreal inflexional tangents..
Ex. The relations between the singularities of a curve and its reciprocal are illustrated by Fig. two unreal cusps (at the circular points). This diagram shows the limacon r = 6 + 5 cos and its reciprocal with respect to the circle 4(r 2 — 3rcos0) 55. bi tangents are shown by the dotted lines. and two real cuspidal tangents. 1.64
SINGULARITIES OF CURVE
More
generally.
.
its inflexion. [Compare Ch.
If
<£ (X. 1.
a
k-ple point
its
gents correspond a k-ple tangent
and
=
Fig-
I-
Ex.
Ex. The degree and The two class of the limacon and its reciprocal are both four.
v)
=
is
<H[>
the tangential equation of a curve. To a triple point with three coincident tangents corresponds in general the tangent at a point of undulation and so for quadruple ' quintuple.]
. § 4. the
bitangents are given by
_
d$
S/L4
_
d<£
u\
5f
Find the inflexions. points. These reciprocate respectively into an ideal bitangent. 2. II.
ii. to
IV
7
and its k tank points of contact.
3.
5.
0)
are
[Use §
Ex. and i inflexions. If the point-equation of a curve is written down at random.
Suppose a curve is subjected to r conditions then its polar reciprocal with respect to any given conic is also subjected to r conditions. and satisfies r other conditions. i cusps.
Suppose we are told that the curve
±n(n + 3)
= S + 2k + t. the reciprocal has r assigned tangents. has k that only a finite number of curves can be found with these Then (Ch. but we shall see that it has bitangents and inflexions. it has no node or cusp (Ch. t cusps. II. k cusps. 0. 8 bitangents. that the polar reciprocal has degree m.
five cusps. The singularities of y"zq reciprocals of each other. we have shown.
If a curve has degree n. class m. &c. If the tangential equation of a curve is written down at random. 8 nodes. § 6) properties. class n.
triple tangent.IV 8
SINGULARITIES OF CURVE
= xv+i
at (0. tangential equation of a curve is equivalent to writing down the point-equation of its polar reciprocal with respect to the
.
7-ic
and
six cusps. four nodes.
'
general.]
Ex.
Ex.
Ex.
A
sextic cannot have a triple point.
65
and
(0. The point-equation of its reciprocal must not be considered as written down at random '. has 8 cusps.
2 base-conic x z + y
+ z2
=
0. and satisfies r other conditions and also nodes. § 6). has r nodes. in § 7. 5
(viii).
A
cannot have a quadruple point. For instance.
The polar reciprocal is of degree m. and
§8. but will In fact the writing down of the have nodes and cusps. II. and k inflexions.
1.
is of degree n. the curve will have no bitangent or inflexion.]
A
quintic cannot
have a
triple point
and three
cusps. Note that to be given that a curve has a triple point and three cusps is equivalent to only 4 + 6 conditions. whereas a quintic can satisfy
[Its reciprocal
would be a quintic with a
20 conditions in general. r bitangents.
7.
3. and only a finite number of
. t nodes.
. if the given curve is made to pass through r assigned points.
. 4. for this reciprocal curve is specialized by the fact that it has been derived by reciprocation from a curve with random pointIn fact the reciprocal has both nodes and cusps in equation. a node. 1)
Ex.
5.
(iii)
(2».
m). and length of major axis. say of degree I. 4).
(1. point.
given point. Show that the general p curves of the family pass through any given point. 2).
= 0. v. 4).
2. 2m). 2).
(viii)
(1.
Find the characteristics of
pencil of w-ics.
(vi)
(vii)*(l. 2).
p). Show that the tangential equation of the family <£(X.
(i)
4.
(ii)
(1. z.
deduce
in(n + 3)-8-2K
1.
2n-2).
through
(x)
[(i)
(iii)
The
circles of curvature
(2. =
in the parameter a.
§ 2.
(v)
(n + m. (iii) Conies through two given points touching a given line at a
3.
Ex.
(1.
(iv) Conies through a given point. 1). 2).
(ix) Conies with given centre and eccentricity. m).
I) is
called the characteristic of the family.
(2.
4. 4).
= §m(m + 3) — t-2l. 3.]
characteristic of the polar reciprocal of the family
:
Ex.
(v) Circles
(vi)
(vii)
(viii)
touching a given circle with their centres on a given line.
Hence
|m(m+3) = r + 2i + r.
of a given parabola. (10.
which
is
an algebraic equation of degree p
[(p. touching a given line.
§ 10. 4). 1)
(1.
(4. (iv) The family obtained by translating a curve through any distance in a given direction.
(v)
[(i)
Curves parallel to a given curve.
See Ch.]
Ex. Conies with a given focus.
and
.
(ii)
(n. and touching a given line at a given point. 2).
a a
Conies with axes along given lines and passing through given point.
Ex.
as r
=
4. Conies with given vertices.
y.
(x) (4. 2.
m).66
SINGULARITIES OF CURVE
. 6).
A
(ii)
Curves of degree n and class
m
with
a
common
centre
of
similitude. XI.
Find the characteristics of the following families (i) Conies through r points and touching s lines. 1.
(iv)
(v)
(2. 6). passing given point.
a)
A singly
infinite family of curves has the equation
f{x.
IV
8
polar reciprocals exist
the base-conic being supposed given
throughout. and I touch any given line. where r + s = 4.
(4.
The
is
(I. /i.
(ix)
(2. 0. 1).
We
Ex.
(iii) The family obtained by rotating a curve through any angle about a fixed point.
(2.]
(iv)
(n. (ii) Conies touching two given lines at given points. a) is also an algebraic equation in a.
and Ch.] [(6. (1. 5
(vi). 5 to 13 should be omitted on a
first
reading.
is the class of the in general.
(iii) Quartics with three given biflecnodes and given tangents at one of the biflecnodes.]
Ex.
(ii)
(iii)
in-
Cubics with a given cusp. intersections with the locus of any line through 0.]
1.
'
7.
.
(1. Verify by considering conies through four points from or touching four lines. inflexion.
W =p+
m'
= 2p + M. (1.* Find the characteristics of (i) Cubics with a given cusp. 3).
is
a family with characteristic
v). Find contact of tangents from a given point
the Pliicker's
I).
The
families are
(a
2
+ yi )
=
2 1 y (3a . 6).
a (y + ax) 2
=
axy 1 + 2a?x i y.2).
See §
3. Find the characteristic of the conies having 5-point contact with a given cuspidal cubic. Find the characteristics of Cubics with a given node. where k = Consider the is a p-ple point of the locus.
Ex. 5.
Cubics with a given cusp.
(2.
=
is
an equation of degree
in x. cuspidal tangent.]
Ex. and given tangents
8. XIV.
a 3 a3
=
xy (x + y + S az). Ex. and the tangents to the locus. and inflexions. 4). 1). § 3. §§ 5 and
8.]
Ex.
mx. 3). 3). 4)
and ax
(3.
Ex.
while
<j>
(x. three given collinear inflexions and tangents at two of them.
[(1. (ii) Quartics with a given node.
at these cusps. (ii) Nodal cubics with. 10.]
6.
(2.
numbers of the
locus of the point of to a family of curves with
characteristic (p. two given cusps. (iv) Quartics with three given nodes and four other given points. Ex.
f2
. See Ch. 7.
The
families are
1
zy*
=
ax 3
z (y
+ ax) 2
= 2 axy
+ a?x*y. and inflexional tangent.
M
*
Ex. The complete primitive of the differential equation
which
is
algebraic and of degree
v
ft
in
—
. and flexional tangent. 3).
See Ch. inflexion. envelope of the family.
(i)
given
[(1.
§§
3.
Find the characteristics of (i) Lemniscates of Bernouilli with a given node and axis. inflexion. 11. 6.
Ex. XIII. 4)
and
(1. and point of contact. 9. tangent.
(i)
[(1. XVII. (ii) Cubics with three given collinear inflexions and corresponding inflexional tangents. point. m)
(/*. 3). nodal tangents. 6).
Find the characteristics of
Cubics with nine given inflexions.y ).IV 8
SINGULARITIES OF CURVE
67
Ex.
[(1.
The foci of the envelope are w-ple foci of the locus. I). 13. tangents from
m'.
m
[In Ex.
the envelope. 12.]
Ex. I). where M is the class of A and B are l(m — l)-ple points of the locus and the A and B to the envelope are »»-ple tangents of the locus. Find the degree and class of the locus of the intersection of tangents from fixed points A and B to a family of curves of degree n and class with characteristic (p. Find the degree and class of the locus of the foci of a family of curves of degree n and class with characteristic (p.
m
[n'
—
l(2m — l). — m(M+p)+2l{m — l). 12 take
A
and
B
at the circular points.]
.68
SINGULARITIES OF CURVE
IV
8
Ex.
+(\.
")•
* (x.
m has in
.
If co and a> are the circular points.p. b).
(i). any
is
t
curve with the same foci
f(\. and the lines Sco./m
is
f\k
-.
/*. *") = o.ii
»)
=
.. drawn to the curve from co.
and
^r(\.
general
m real
and
m —m un2
real foci.v)
+ 4>(X.
the tangential equation of the circular — 2.v)=0
is
points. and has and 1) arbitrary tangents of coefficients. These tangents meet in the real focus (a. since the line joining two real points is real.
m
m
Hence
A
'
curve of class
/i.
can in general be If such a tangent also a tangent.:
+
CHAPTER V
FOCI
§
1.ii. and <p has confocal with /
where
(j>
(X. curve whose
/=0
0=0
%m(m—
/2 =
m
=
=
0./«
<P
(a. and is x + iy a+ib. and cannot pass through co or co'. v)
=
m
2m
=
=
= %m(m + 3) — 2m = Jm(m— 1)
But (i) evidently touches the common arbitrary coefficients.v).. p.. for no tangent from a can contain more than one real focus..
If /(A..
fi. since this is the number of coefficients in \j/. If the curve is of class m.
. real foci.
Definition of Foci. the general tangential equation of the real foci have tangential equations /. Sa>' touch a given curve (not at a> and co').
v)
=
is
the tangential equation of the curve.
. S is called a focus of the
curve. As a special case. then x—iy
=
= a—ib
m tangents m from
is
to'. so that there There are no more than are real foci. is any curve of class is equivalent For to say that a curve is confocal with / = common tangents of / to saying that it touches the Hence the general tangential equation of a curve 0. the
curve being supposed real. 0.
.!
~V.]
The sum
common
.
=
PM
P
{x-x1 f + (y-y.
Pis the middle point of the
of the line joining two foci of a real line joining two real foci.. If P(x. [Taking rectangular Cartesian axes through and the fixed line parallel to the axis of x.)
q ^ -q 1 \ as a factor.•
1
and
2
Since the difference of these has X
and
are equal.
If the
middle point
P
is real. of the conic]
Ex. e the eccentricity. curve
1. p.
is
the circular points and
i/r
=
properties of the foci of the general curve are The foci of special (but see the examples below). curves may have interesting geometrical properties.
m
m ' 1 li +
q2 \
m 'i
H-
i
~
Ex. ^{x^.]
=
+ /j? (-p 1 +p i ~p i + '»)+(Pi>-Pt+Pi. That the foci as defined in this section are the 'foci of the conic. if c and c' are replaced by curves respectively confocal with them. yj a focus.
Ex..
Show that a —
is 3.
[A particular case of Ex. and that the directrix is the chord of contact.
to since SP is the perpendicular from e PM. and p x cos a. namely
.f
=
Q
touch the conic. The tangents from to a curve make with a fixed line angles whose sum is (X. + y sin a the directrix. 161.y) is any point of a conic. 3. Ex. x. suppose the terms of highest degree in the tangential equations of the curves to be t
Show
p
\ TO
-p1 \ m -V+i'2V
tana tan0
==
. the equation of the conic is
Not many
known
'
'
'
m=
=
(x— x^ + (y — i/^ 2
=
e
2
(p— x cos oc—y sin.
a multiple of n. The tangents to one curve make with any fixed line angles whose sum is a. It is seen at once that the circular lines through S. p. where the directrix.
Ex. 2. when 2.
The theorem
is
well
known
in the case
of the angles (to within a multiple of w) which the tangents to two curves c and c' make with any fixed line is not altered.
a)*. Tangents are drawn from any point to two confocal curves.•) (-g + gs -gB + )^{q -q2 + qi . 58. [See American Journal Math.
5. and xii. according to the usual definition of foci of a conic may be shown as follows.1
70
DEFINITION OF FOCI
V
any curve of
where
class
m—
<f>
=
2. By analogy we may call the chord of contact of the two circular lines through a focus S of any curve the directrix corresponding to S..
that Of — is a multiple of n. and the tangents to the other curve make angles whose sum is 0. and the lines joining to the foci make angles whose
sum
is 0. 4. The coaxial family of circles through two foci of a curve have two other foci as limiting points.
Singular Foci. The intersection of a tangent at a or vice versa. and that the projection of another focus is near the point and so for a of approach of the projection of a>u' and the curve. the curve has three other sets of four concyclic foci. bb'. the of tangents.
Ex. If four foci are concyclic. If the curve passes through a and a/. or touching <b a at a. [A curve of class m touching a>u>' at o>.] Ex. Therefore so do the pencils (abed). the centre of a circle is a singular focus. [A segment through « must have zero length. [The envelope of the family must evidently consist of circular lines.* of foci is less than For instance.
(b'a'd'c'). If the two curves of the family Pa?+Qa + R = which pass through any point are always orthogonal. Ex. The projection of a curve nearly passes through the projections and a.]
Ex. The envelope of a line of given length with its ends one on each of two given curves has no (ordinary) foci. Find a curve with no. where P. the tangents at m and co'f meet in points which are not usually included among the ordinary foci '. the pencils {abed). §7.
&c]
§ 2. other than moo'. cc'.
a>
of
Ex. The four circles cut orthogonally.]
*
A point is not called
coai'. which can be drawn to
m
the curve from m and a>. Show that the projections of four of the foci are close together and in the limit coalesce at the projection of a singular focus. y and a is a parameter.
focus unless it is finite.
t Excluding an ordinary
if this is
a'. is not counted as either a singular or
a
'
'
. The projection of a curve nearly touches the projection of a>a. IV. focus. of which 1 are real. and (m— 2k) 2 ordinary foci 2k are real (see Ch. <a' and m — 3 other points.
m
. if the curve has ordinary contact at one point with mm'. singular or ordinary. it will have (m— l) 2 foci. 4. 7. of which For example. R are polynomials in x. well-known example is the parabola.
Ex. a and m — i other points. that the projection of (m-1) foci are close to the projection of a.
number
If a curve of class touches the line at infinity com'.V2
•
SINGULAR FOCI
71
Ex.
Show
. and having cusps at a and to'. c'. and the circle has no ordinary foci.
with a tangent from
focus. is less than in the general case so that the number 2 .
2. They will be called singular foci. [If a. If the curve has /c-ple points at m and m.
A
m—
'
m—
1. The case in which the family consists of conies is familiar. b. it will have in general k 2 singular foci of which k are real. a tangent. 2). b'. then the curves of the family are confocal. d are tangents from &> and a'. c. 6. 3. (a'b'c'd') have the same cross-ratio. d' tangents from a such that the intersections of aa'. dd' are concyclic. Q. and by their intersections determine the foci.
.
parallel to the sides of a
and
..]
Ex.. (a' 2 /3' 2 ) Show that there exists a curve of class n — 1 with real foci at the former points which touches at its middle point the line joining any two of the latter points. the real singular foci of the locus are the middle points of the segments...
Ex..
is
4..
... 2.... § 2.
= 0.
6.... m). . t prove that the centroid of the points (pt lt ft).
PB2
x
remarquable de coiirbes
[In connexion with the above examples see Darboux.—x-iy)/(x r — iy. + iy. If <r = kir.
For another proof see Ch. + iy.. .
is
with y
Ex..-x + iy)
= n (x. ..
9.
s
2'. A m Bm
I
. and passing through
its
own
singular foci. 5. . 5. 2 = and 2'= n (\x.. n as singular foci. (0C it j3 2).. the locus of the foci of the pencil 2+&2' = (2m — l)-ic having an (m— l)-ple point at a and w'..
and Ch..]
2
=
[If
The lines joining any point of the locus of Ex.
13.)
foci of
is
the
same
for
a
7.
8.-x.
§ 1.]
makes
=
The tangents from the
tangential pencil 2
+ fc2'
=
real foci of 2' = to any curve of the touch a curve confocal with 2 = 0.
Ex.. y.
PA =
h
. . coincides with the centroid of the points (a'D/3'j)... y'. Discuss the case in which a is an even multiple of \ir. .
Ex. 6. Show that the locus of P is a m-circular Zw-ic.
Ex.. XXI. y) to d and h.
6. 11.
§ 8..
an algebraic equation of degree n with roots
are OL\ + Oi\ + iji\ . y) and eiia (x. 2' = is a the m-th class. and so for
. IV.
iy...
points. .-iy'.
If
f(x)
=
.
Ex. 10.
(x.
[See Ex. If o.
=
0..
Ex.
2
PBn ^
PBn-
has
A lt
A lt
The
has
. 5 to the and 2' = make angles with any fixed line whose sum
both curves to within a multiple of
n. and the roots of /' (x) »! + j|3j (X 2 + i/32 ..is constant.
.
the angle which the line joining (x.. + iy\ -x-iy). we have the locus of Ex.
..) are the foci of (r= 1.. Bn _..
4... If 2 = 0. A n PA 2 PAn = h PB* P£
.is an odd multiple of \n..
. .
m
fixed line by the tangents from any point of an m-ic through all its foci
latter curve are
The asymptotes of the
regular polygon. + pyr + 1) + (V + fi2) yjf..]
Sur une
classe
.. PAt
.+ (\a 2 + /i/32 + l)-1 + AiBu A^B^.. PB m make with a fixed line is <r..
[By §
2
=
y r ) and (a/..
PB.
1 if (x.
r
Writing down the conditions that the
should touch 2
+ &2'
=
0.-x + iy).
The sum of the angles made with a
to a curve of class is constant. locus of such that
A
P
PA. If A lt such that pa
i
. n as ordinary foci. the locus of
P
..]
PA m
and
Ex.
An
. 5.
-x + iy) (a/.
72
SINGULAR FOCI
V
2
are the tangential equations of two curves of Ex.iy) (x'.]
PBlt
[The difference of the sums of the angles which PA lf . et de surfaces algibriques.
=
[The curve
is
(Xa 1 +
^
1
subtend angles at P whose sum a. Fixed segments
+ l).. are fixed .. .
and eliminating
lines joining (x. we have for the locus
w
n(x.
Blt .. Now use Ex. § 30..
0). 0) and
and S of the limacon . 6.
1. 6 the
foci of
V
3
x=af + 2bt +
x x
.
a focus of a given curve.]
Ex. the lines Oco and 0a>' becomfc the
which
is
points
co
and
<o'.
c. if any. If we reciprocate with respect to (i.
atp+q . and S'co touch the inverse curve. Sa> and Sea' touch the curve. Ill. the reciprocal of a conic with respect to a focus is a circle. and Sea' respectively
§ 10). a>' and a node at 0.
a quartic having cusps at a>.
If
I. with respect to a circle whose centre is 0).
q being positive integers. e. the reciprocals of the foci are the lines joining the intersections of the reciprocal curve with Oco and Oco'. IX. and the reciprocal of the corresponding directrix is
the centre of the
If
circle. As an example. Ex. Find the singular foci.
ij
= AP + 1Bt+C. if of the curve.
reciprocate with respect to any point 0.
Hence The inverse of a curve with
a>
respect to
a focus has cusps at
of a conic with
and <o. Therefore The inverses of the foci of a curve are the foci of the inverse
S'a>'
:
S
The inverses of the lines joining m' to the intersections of the curve with Oco are the tangents at a> to the inverse curve is a focus (see Ch. of the curve in Ex. § 10.
7.
(i)
METHOD OF OBTAINING THE FOCI
Find by means of Ex. § 1). 3.
y
=
+ t%
8.
p and
3 at /(l
(ii)
(iii)
= atp y = = aty(\ +f). For instance.
Hence
:
The reciprocal of a curve with respect to a focus is a curve through a> and a/. § 9 shows the method of obtaining the tangents at the circular points. The real focus of the limacon is the inverse of S.
§ 4. [Ch. Hence. and the reciprocal of the corresponding directrix is a singular focus.a2 )/2 6.
.
S'a>
are the inverses of Sa>
With respect to a point 0. I.
Ex.
Inverse and Eeeiproeal of Foci.
If S' is the point inverse to
S
>SV and
(Ch. Apply it to Ex.
we
The singular and ordinary
with node
foci
r
= a + hcos6
H
((6
2
are (|6.:
74
Ex. 7 (iii). the
inverse with respect to
real foci
S and
is
a limacon
r
= a + b cos 6.
is
Hence
curve. or Ch. two tangents at a> to the inverse curve coincide.
]
3.
curve.
4 HP2
= 4bSP+2a -b\ P
i
being any point of the
[The equation with S as pole is 4V r* .
for the points of contact of
Ex.+ &cos<9 = 0. If in § 4 S and O are reflexions of each other in the directrix corresponding to S. 2r = <*. the inverse curve has inflexional tangents at a and w'
which meet at
S'. 1 in the cases &>a>0. There are real inflexions if alb lies between 1 and 2 see Ch. 1. IV.]
—
=
0.
[The equation of Ex. XVII. (ii) a is kept constant and a or 6 respectively vary.
4.
.
Then we get : ' The tangent to a conic focal distances of the point of contact. parabola.
Ex.{a + b cos B)/b sin 8. If a = b.
a parameter. 2r2 + ar(3cos 2 0-2)+2a 2 sin 2 i9 = 0.4br (a? cos 8 +
_
h*)
+ (a? -V)*-0. [0 is a crunode. r(l-fc s ) = a(cos6-k). the curve being the inverse with respect to a focus of an hyperbola.
For the inflexions 3 ab cos 8 + a 1 + 2 b 2 the bitangent 26 cos 8 + a 0. 6.' similarly other properties of the limacon or cardioid.
9.]
1
In Ex.
The
inverse of any curve with respect to a singular focus
has
also a singular focus at 0. Fig.V4
INVERSE AND RECIPROCAL OF FOCI
is
75
Ex.]
Ex. The angle <j) between the tangent and radius vector of a limacon is a maximum when $ = 8.
If
P
is
between
OP and
any point on the limacon.
[tan
= . 10.]
Ex. and Ch. cusp. ».
5.
. 5
is
not altered on replacing r by (a 2 -&2 ) 2 /46V. a = 6. 1. 8.
OSP
is
twice the angle
[Invert with respect to 0.
is
a family of
Trace the limacon of Ex. 7.
Ex. the tangent at P. 2.]
Ex. a>6>0. acnode. Find the locus of the inflexions and of the points of contact of the bitangent of the family r = a + b cos 8 when (i) b. where fc limacons with a common node and focus. ellipse. Fig.
makes equal angles with the
Obtain
Ex.
[(i)
(ii)
r2 +fercos5 + 2& 2 sin 2 5 = 0.
Ex. A limacon is its own inverse with respect to the circle through the node with its centre at the ordinary focus. the curve is called a cardioid.
Take. and a.. (3. m) add m to the right-hand side of (i). at any ordinary point of the curve. a. .. It is only at multiple points where two or more tangente coincide that we can have superlinear branches. b. the curve
y-2x+2x2 -3xy-y 2 + 4. If we take a single a-th root of unity for <o we get a ' partial superlinear branch '. In the neighbourhood of the origin it is possible to express the value of y on any branch of the curve through the origin (not touching x at the origin) in the form
=
=
y
. are constants.. . We can also have multiple points with coincident tangents formed by the contact of distinct linear branches see. y) be the equation of an algebraic curve passing through the origin. . Ill. Ch... the branch is called linear. for instance.
= ax + b<o&x a + C(a(x a +
(i).
where 1. Through an ordinary multiple point of order Jc with distinct tangents. y..x* =
the method being general. just one linear branch passes through the origin. First suppose the curve has a linear branch through the
'
'.Fig.
f
.
.
+ bx 2 + cx3 +
in the
For points on a curve near (0. Substituting for y the expansion ax
*
0. for example. k linear branches pass.* entire portion of the curve near the origin obtained by putting for co every a-th root of unity in turn is called a superlinear
is y ax. c. 5. are The positive integers in ascending order of magnitude.
origin. f We assume the above results as proved in books on the
a a>
=
branch of order a whose tangent
=
=
Theory of Functions and only give here the practical method of obtaining the expansions such as (i).CHAPTER VI
SUPERLINEAR BRANCHES
»
§
1. S.
. If a If we take the origin 1.
Expansion of y in terms of x near the Origin.
Let f(x.
.
Putting x and y zero..
x 2 x 3 . method given below we can expand y in terms of * and thence get y in terms of x. get
.*
As another example..
..
=
We
and equate a = 0. which has an inflexion at the. Suppose now that the curve has any singularity at the origin and that y = ax is a tangent at the origin. origin with y = as inflexional
tangent. p and q being positive integers prime to one another... . b =
to
G.
Put y ax 2 + bx3 + cx* + . We may therefore confine ourselves to the case in which the curve touches the axis of x at the origin. which
sufficiently small.. .
Similarly for a curve having r-point contact with y the origin we obtain an expansion of the form
=
at
axr + bxr+1 +cxr+2 + ..) — (ax + bx 2 + cx3 + . Suppose that Newton's diagram (Ch.
we assume legitimate when x we have (ax + bx 2 + ex 3 + ..
we obtain a
expansion
=
2. Ill.
.
Hence
y
-
Cx3 + (AC+G)x* +
.. Y
(0.. Suppose that
*
We
= 0.. in this equation.
we
obtain the values of
—
.... y) = of the curve put y = YXP.. § 3) gives us as an at the origin terms approximation to the curve touching y = represented by points which lie on a straight line making an angle tan" 1 p/q with x = 0.
VI
1
EXPANSION OF
y
IN TERMS OF x
77
is
equation of the curve..
.
at the origin.
.) 2 + 4x 3 =
Equating to zero the
coefficients of
x.. if we consider X.
. x = XI..
=
thus get a new curve f(Xi.
c=AG+G.. y
.
0. and have as the required
..)-2x+2x 2 -Sx(ax + bx 2 + cx 3 + .
—-j.. in turn.
m)
is
an intersection
Another method
is
to differentiate the equation of the curve repeatedly.
b
=
8....
.. YXP) as current coordinates.. Putting ax + y for y in the equation of the curve. zero the coefficients of x 2 x 3 x*.
Then the expansion
may
be written
down by
Maclaurin's theorem.
-~-j.. In the equation /(*...
y
=
2x + 8x 2 + 52x 3 +
. consider the curve y= Axy + By 2 + Cx3 + Dx2y + Exy 2 + Fy 3 +Gxi + .. c
=
52.. we obtain a curve touching the axis of x at the origin x and y being considered By the as the current coordinates of any point on the curve.
. apply Newton's diagram to this curve as before..
Newton's diagram gives the approximation y 2 origin. we put in the original curve
X=
y=Y + mXP... ..
in particular.
'
By an
we mean
b.
.
Y=
and thence
±a* + ibX±ia-?(e + ca + %b2 )X 2 + .. of the original curve..
+
ai).
. ..
into the equation of the curve and equating to zero the co2 3 efficients of X.
y= YX\
2
Y = a + eX + bXY+cX Y 2 +
This has linear branches through Putting
(0.
X=
Y=
Whence
p
1
2
y
l
= xV(m + bxl + cx% + ...
y
k
y
=
Exactly similarly for the curve
=
axk+1 + bxk y + cxk ~ 1 y 2 +
. f{Xl.
that there
is
d. m) with 0.... m) we get an expansion rn + bX + cX 2 + .).
e. in and thence y in terms of x.. and that through this point there is a linear branch of the curve.
so that our
0. and therefore put
x The curve becomes
2
—
ax s
at the
=X
2
2
.
ordinary cusp between the coefficients a..
78
EXPANSION OF
y IN
TERMS OF
x
VI
1
of the new curve with 0. where w is a primiget a complete superlinear branch
But if no linear branch of the new curve goes through its intersection (0..
origin.
Y=
X X
. suppose the curve has an ordinary cusp at the y = 0.
ill
a> 2
Replacing xl by
coxl.
.
±ai + BX+GX 2 +
..
. the tangent at the cusp being
For example... order to expand Y in terms of Substitutions similar to the above must be repeated as often is
new curve
x=Xl.
tive g-th root of unity.... we have .
±a^x^ + ^bx 2 ±^ a-i(e + ca + ^b 2 )xi + . Y + mXP) =
We now
X
as»necessary..
. Then transferring the origin to (0.
we
xl.. .. .
no relation none of
them are
zero.. ccPxi. The curve is
y
2
=
ax 3 + bx 2 y + cxy 2 + dy z + exi +
'
.. c.
.
expanding by the binomial theorem.
=
x'
1
y
1
-yt
and Y= l+aX + bX* + cX + we have a = ±1. i if + 3x*y-xy'> + 2x = 0. we may use the preceding method to expand x in terms of y.VI
1
EXPANSION OF
'
y IN
TERMS OF
x
79
origin touching
which has an ordinary superlinear branch y = 0* we get an expansion
'
of order k at the
fc
+
k
l
fc
+2
k
fc
+3
k
y
a>
=
A<ox
+ B<o 2 x
+Ca> 3 x
+.
coefficients
..).
Ordinary '. y (y-x3 Y = x 'y. because there is no special relation between the none of them..Y*X\
!
Putting
of
to
zero
coefficients
c
expansion
Ex.
[Putting y
3
. s -Zx2 yi + xi y-xS = Q.. If it does.
'
This process is known as reversion of series '.. .. we may expand x in terms of y near the origin by putting
tt
a
=
&
^
a a
12
(l
+ 4^ + -B^ +
..
2
= YX\
x
= X. b. 6 = X X
3 .
. §§ 3 to 12). The arithmetic of the methods described in this and the Devices for saving some of following section is often tedious. and equating coefficients of powers of y on both sides of the resulting identity. 2 2 = + 2/ z)(!/ + a..
Ex. where a<)3<y<. IX.c. In expanding y in terms of x we assumed that the curve did not touch x = at the origin.
a
of b
P gives
an expression
for x.. have found the expansion
Jc-th
y
y = bx + cx a + .]
2. )
!
V(^
*
'
a.
being any
If
we
root of unity.
= Y X*.
.
(i)
is
y
=x
= +|
Hence the required
2
(I
±x + x2 ± \x*+
.
of the following curves near the origin
:
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
{y~3?f = a.
Expand the branches (y-x2) 3 =ary..). ..Vy-x"i = xi y — 2y*..
we get (F-l) 2
equating
1. for a curve touching y = at the origin. for instance.. being zero. and then obtain the expansion of y in terms of x by reversion of series.
1. the labour will be described under the head of 'quadratic transformation' (Ch.
Find the expansion of the branch of the curve
(y-x2 f
near the origin.. Each possible value
in the given expression for
y
in terms of x..
We
showed in Ch. i. I. § 7)..+%
=
F(x. y)
fo-«i) (#-««)
1
(y-u n) =
meet In
while
= " in nN
.(u n -v Jf )
x
.2.f
(i) One curve has a node or cusp at the origin.(u -vN
1
)
x (u 2 -Vj) (u 2 -v2)..... and the other passes through the origin but does not touch it there.:
VI
2
INTERSECTIONS OF TWO CURVES
§ 2.
.
G«
. (ii)..+bN
=(y-v )(y-v
. The number is e.
* It is the
f Cases
221fi
(i)..
this
ult u2 . e..
81
Intersections of
Two
Curves. also that no intersection of the curves other than the origin lies on x = 0. vN are functions of x. I..
1
that
two curves of degrees
•••
N
o
0)
+ a2 y n ~* + .
ciently small. (v) have been already dealt with in Ch.
later on.
—
y
being expansions of y in terms of x for branches of the two curves passing through the origin.. i.
We assume that x = meets the curves in finite points only and does not touch either curve. when x is suffi... which will be useful
= U{ and
y
=
V.
determinant just mentioned multiplied by a constant... Let us consider the following examples.(y-v N)
=
points. § 7.
ar b r denote polynomials of the r-th degree in x v x v 2 .
y»+b 1 y»-* + b2 y»-* + .. e..2.n..
In this equation ^>(x) is a polynomial of degree nN in x* Suppose we wish to find the number of zero roots...
It is evident that the lowest power of x in Uf—Vj is x° unless y uit and y = Vj are (partial) branches through the
=
origin.
x
(<!*"-*>!)
=
0.
. of the expansion (§ 1).(u 2 -vN)
(u n -v a ). § 7. where xe is the lowest power of x in (f>(x)..
... un and which we have obtained
. the number of intersections of the given curves which coincide with the origin..
n.
Hence
The number of intersections of two given curves which' coincide with the origin is the index of the product of the lowest powers of x in all possible expressions of the form u^ Vj.N.j
=
l.. y)
=a
b
y +
n
a^
I. the product of the lowest powers of x in each of the expressions
(uf-Vj)
i
=
l.
f(x. The result of eliminating y between the equations of the two curves may be expressed by equating to zero a certain determinant (Ch. .
2
).. or in the form
(f>(x)=
K-fi) (u1 -v2 )..
Hence
. and the curves meet thrice at the origin.... u 2 = a'x + b'x + 2 v 1 = ax + Bx + ....
2
82
INTERSECTIONS OF
In this case
TWO CURVES
VI
w1 = ax + bx2 + ......} {(a'-4)a: + ... no two of the Ic + taDgents coinciding...% = a'x + b'x2 + .
(iv) One curve has a cusp at the origin.
.
.
Here
Uj
v2
=
axi +
..... (iii) The curves have a node at the origin. the tangents at the node to the two curves being the same.
and the curves meet JcK times at the origin.}
{(a'
—a)x + .
....}... .
. . . v3 = Cx 2 + v 3 = — Ax% + or v2 = Bx 2 +
..
. and k are of the type
U^
The curves
Bx2 + Cx s + .
In either case the curves meet eight times at the origin.
at the origin. . meet k (k + 1) times at the K
origin.
Ul
Here = axi + bx2 +. linear branches through the (v) The curves have k and origin.u2 = -axi + bx2 .
n(Ui—vA
=
{(a— a')x+.... — vA are of the type Here k(k— 1) of the factors in ax + bx2 + cx 3 + . * Here II (^ — Vj) has IcK factors each of the type
K
K
ax + bx 2 + cx3 +
..* (vi) The curves have k linear branches through the origin with the same tangents for each curve.
..
..... .
. . two tangents at which coincide with the cuspidal tangent.}
..... Hence n(wf -«_.}.
= Axi +
«2 = — axi+. .. Here 2 %! = ax + bx 2 + ... as might have been
anticipated. v% — a'x + B'x 2 + . and
the curves meet twice at the origin.i\ = Ax 2 + Bx 3 + ..
* This result imght have been expected ..) = {(a-A)x+ .. while the other has a triple point..
(ii) One curve has a cusp at the origin.v1 = Ax + Bx 2 + ....
. and the other passes through the cusp and touches the first curve there.
.... for each of the k branches of one curve at the origin meets each of the branches of the other curve once at the origin..
..}
{{b-B)x2 +
and the curves meet six times
{(b'-B')x2 +..." vx = bx + cx2 + ..
Show that they meet q times at 0. 2 and 3 to obtain the conic of origin with x — y + xy + x 3 = 0. Comparing the and the expansion of Ex.(>—
1)}
= k'-L]
§ 3. if the expansions of y in terms of as far as the terms in x r . 7.
x* in this
m = 1.2x s 2 [x + y=y 2 and y — x = (2x-y) . Show x for the two curves are identical have (r-f-l)-point contact at the
of y in terms of
2
3 2
Ex.
Ex. h.
meetsit in
at the origin to the superlinear branch points coinciding with the origin. 2 and §
1. § 3. IX....
is
3.
See also Ch. x2 = Sat/. But a less laborious process •will usually be the following.
TANGENTIAL EQUATION NEAR A POINT
The tangent
/3
83
of §
1
Ex.
[See Ex. We may of course find the tangential equation of the curve as in Ch. How many of their intersections coincide at ?
Ex. the curves origin. (ii) to find the lines joining the intersections of curve and conic to the origin and to choose m.
Ex.
8.
[b
3
Find the conies of closest contact at the origin with a 2 x + by 2 + y 3 = and x 3 + y s = 3 axy 2 = (ai x + a2 b2 y)x and if = Sax.] {a?x + by )
5. 6 = -2.
A
[Sr {(f. b so that five of them are y x.1)
(r+l)/r+(k-l) . n having distinct tangents (h a {k — l)-ple point such that the tangent to the branch of order rt of the first curve is a tangent to a branch of order rt — 1 of the second. and then employ a method similar to that used in § 1 for point-coordinates.. .
The expansion
2
x
i
for the conic
2
y= mx + ax + 2hx(y-mx) +b (y-mxf
y
Ex. a. ]
=
Ex.
[For the curve y
efficients of x.
Tangential Equation near a Point..
4..
= mx + ax + 2ahx + a(4:h + ab)x + 2ah(4:}i =
. Another curve has at *'ii ^z. Other methods are (i) to differentiate the equations of curve and conic four times and to identify the values of
dy
J
'
x + x + 2x + 2x + 2x + .
. h = 1.
6.
2 3 i 5
closest contact at the
co-
x 2 x3
. IV.]
Ex.
Find the parabolas of closest contact at the origin with = x + y + xy + x 3 and y = x + x2 .
(i)
Ex
-. a—1.
VI
3
1.
Use Ex..]
curve has a &-ple point with superlinear branches of order = 2j-). Two curves have linear branches touching the same tangent at 0. where p^q.
Suppose that it is required to obtain an approximation to the tangential equation of a branch of a curve near a point.
curves have linear branches through the origin.
+ Sab)x* +
.
11.. one having p-pomt and the other g-point contact with the tangent.
2
-
Two
that..
G 2
. 3 we have
<Py
cPy
'
d*y
'
dx'
dx 2
dx 3
dx*
for curve and conic at the origin ..
§ 12.
Transfer the origin to the point
The line becomes (h.
The above process gives now
X= +|aZ-26Z
"
= ±|aX
3
+ |cZ 3 -. .. + cxi + dx ±
2 3 2
. and obtain the condition by expressing v in terms of X. B. which
X
is
\^". we have on putting
.
substituting
'
e
ba>0i
a
i
it is
s
+ ca>y£" + da>s £ a + .... 0).C. in the neighbourhood of a cusp at the origin at which y = is the tangent. Then substitute the value of given by (iii) in the second of equations (ii) and we have v expressed in terms of A.)
a(r + fcA0...
for
i\
and writing £
= Xa =
(bPm0Xt3. For example.a
/*
Comparing with = 1 and X = A0.l + 6Z* + |cZ 5 +. T
where £
=X
2-
.. § 6.a) <»7j7 + Xx + fiy + v + hX = 0... we have
=
y
= ± ax% + 6a. and v by y..
1
"
&(/3-a)ffl0X0 + c(y-a)a>7X'Y+.. Ex. just as in § 1 we expressed y in
terms of x.a )
+ d8co s X s .. and we have the point-equation of the branch by § 1 in the form
y=
The tangent
$ 7 ba>Px tt + C(6~'x a +
(£.
84
TANGENTIAL EQUATION NEAR A POINT VI
Suppose the branch touches y = at the point (h. .a + cy<»7X7-«+.
s
dws xa +
is
(i)
at the point
y) of the curve
or. .
.
. in turn... We shall take ft = 1. We should touch the want the condition that \x + fiy + v = branch. IV. Xx + fiy + v + h\ — 0.Q aA0-°= -(&/3w0X0. .
If
we now replace X by a. 0).a +.. we get the pointequation of the polar reciprocal of the branch (i) with respect to the base-conic x* + 2y (Ch. J' Substituting in the first of these equations
=
W
*
X=AA + BA
2
+ CA 3 +
(iii)
and equating to zero the coefficients of powers of A. 1).a
+
cya>yXy.
3
.a) a>PXP + c (y . we get A.)x-Ocy b (j8 .
. 1
= a x + a3 x + ai !c +
.
find the
number
of
common
taDgents by recip-
Ex.
=
=
&.. coincide with I ?
[One or q according as the point-singularities do not or do coincide. § 7. Show that.. . IV. represent the same linear branch touching y at the origin.
we prove that The number of those common
in § 2.
1.
Ex. involves o 2 a" . 2.]
Ex.
.
2.
§4.
Ex. 7..
Show
1
that. in the first of these equations.. if
3
2 3 and . a..
Ex.
i/
Expand
v in
terms of X for the curves of §
i
1.
.
=
=
We may also
rocation.
2..
2. +1 a r+! . but not a..
V
of
this
branch with
respect
to
~
4
27a*
X3
166
~81tf
X4 +
--
This equation represents a linear branch with an inflexion at the origin.
Ex.. r of their common tangents coincide with the tangent at P. (i)
to
(iii). we get
.. Ex.
Common Tangents
of
Two
Curves.
As
tangents to two curves.
The polar reciprocal x2 + 2y — is .. Reciprocate and use § 2... Two curves have superlinear branches of orders^ and q(p^q) How many of their common tangents with a common tangent I. is the index of the product of the lowest powers of A in expressions of the form ui — Vj ) v u$ and v Vs being expansions of v in terms of A for branches of the two curves touching the axis of x.
[See §2..
_
166
..."
whence
4 .
X=A\
v
_
2
.. Ex. v b 2 X + b 3 \ + b i X*+ .. and equating to zero the coefficients of A. that to a cusp of a curve corresponds an inflexional tangent of the polar
reciprocal. verifying the result of Ch.]
. if two linear branches have r-point contact at P. 1. 2. which coincide with the axis of x. A 2 \ 3 .
and
§ 3.:
VI 4
COMMON TANGENTS OF TWO CURVES
85
Putting + B\* + C\ 3 +..X 27 a a
3-i^ + 81a
4
..
of a point of undulation is the tangent to a superlinear branch of the third order. From equations (ii) and (iii) of § 3 we have v + hX expressed in ascending powers of A.
deduce at once that The polar reciprocal of a superlinear branch of order a. Ch. and therefore a + a' = /3 = /3'. unless the index of every power of A in the expansion of v + h\ is divisible by a factor of /3 — a. Hence a' = 1 j8' = a + 1 as was to be proved. superlinear branch of order <x at is a whose tangent meets it in /3 points coinciding with superlinear branch of order a. we should obtain 03'-O/73' = a//3 and /3*-a'>a. j9'>/S.. § 7)... v in the tangential equation of the branch (§ 3).
We
For taking the tangent to the superlinear branch at the origin as axis of x. + l)-point contact with its
tangent. the lowest power being A# where A A^ -0( Hence a' /S — a and /S' /8.
.8' and
. is in general a linear branch having (a.
86
RECIPROCAL OF SUPERLINEAR BRANCH VI
§5. This is inconsistent with the preceding.
of § 3. y = But the origin is a multiple point of order k. (cf.
=
-
=
=
=
/3-a>a'... Therefore j8 of the theorem at the beginning of this section is a + 1. the Cartesian equation of the curve takes the form
y'X
uk .
But considering the original curve to be obtained by a second reciprocation from the reciprocal. so that the tangent to the superlinear branch meets the other branches in k — oc points at the origin.
. y for A..
For example. IV.
5
Polar Reciprocal of Superlinear Branch.
(3
of § 3.
1
. a
= uh + u k+2 +
.
If the polar reciprocal of a. then
a+
«'
=
/?
(i)
=
/3'./3>/3. It is seen at once by points coinciding with the
The polar reciprocal is obtained by putting* a. in which cases (j8 — a)//3 a'/.' at 0' whose tangent meets it in fi' points coinciding with 0'.:
. The theorem may also be established directly by the process
.
Take the superlinear branch
§ 2 that its tangent meets it in origin.
where ur is homogeneous in x and y of degree r. &c. the polar reciprocal of an inflexion is a cuspidal tangent. Putting we get an equation of degree Jc + 1 for x in general.
'.
.. Also the number of tangents from a point on the tangent at which coincide with this tangent is equal to the number of intersections of a line through 0' with the reciprocal branch which coincide with 0'.
Ex. 10.
Ex.
= 4. = 4. de France. so that a' = 2. /8' gives us Hence the relation a + a! . p. /3 tangents from coincide with the tangent at 0.
'
'
'
'
vi (1877). [For the nomenclature see Halphen. We see that the reciprocal singularity is
2.
/3
/3'
a rhamphoid of the same
Ex.
origin with y
Suppose a curve has a superlinear branch of order 2 at the = as tangent and expansion
y
=
ax i ±bx* + cx ±dxTz +
!>
. 2. Bull. If a is the order and — OC the class of a superlinear branch. The origin is Ex. 6).
Then a
=
cusp (§ 1.VI
5
RECIPROCAL OF SUPERLINEAR BRANCH
87
The number of tangents from which coincide with the tangent
to a superlinear branch at is equal to the number at
of intersections of the reciprocal branch with its tangent which coincide with the singularity 0'.. and j8— a tangents to the branch from any point on the tangent at coincide with the tangent at 0.
Write down the order and class of the branches in
§
1. Math.
. de la Soc.8
=
=
If a superlinear branch of order ocatO meets its tangent in to the branch j8 points coinciding with 0. the order and class of a branch are respectively the class and order of the reciprocal branch. type as the original.
3.
1. 2 to
5.]
Ex.
the curve from a vertex V.
meeting a given curve of the The locus of P such that
degree in
Q
x
. the h-th polar curve {for each value of k) projects into the k-th polar curve of the projection of with respect to the projection of the given curve.
and
so for
second polar '.. (n l)-th polar . 2..2..
Q2
pQ
is
OQ..
curve
called the third polar curve of with respect to the given and so on. . .. .n.. n 2.
If
we project
of
*
Or
'
'
first
polar of
for the given curve
'. If I is the polar line of 0..i* j)
with respect to the given
called the second polar curve of curve .n..OQz + . ...
is
ZOQi.curves are of degrees n 1.j.
. is called a ' pole ' of I with
We
.2.... (n— 3)-ic.O Qj OQ k/PQi PQj PQ k = k=l.
Let
be any fixed point.
Through
7i-th
draw any
line
OP
Qn
.*
Similarly the locus of
20&
is
•
OQj/PQt ..
—
—
—
. . conic. (n— 2)-ic.
if
there
is
no ambiguity
.
n—
respect to the curve. third.
CHAPTER
VII
POLAR CURVES
§1.
Polar Curves.j=l.+ OQn_r +
.
l
FQ2
PQ n
)
called the first polar curve of
with respect to the given
curve. the locus of such that
P
(i.
It.j=£ k&i. They are therefore also called 3... second. &c.i ^j)
. {n-— 2)-th. line of with respect to the given curve.PQj
P such that = (i. the polar (m-l)-ic. 1 respectively. shall see in § 2 that the first. .
)'/-
0. If
a curve has a
curves of
any point
centre 0. polar of P.
If
Pand § being (xlt y lt z and {x An alternative method in Ex. If P lies on a given line. so are the asymptotes of the polar curves of any infinitely distant point. If the asymptotes of a curve are parallel to the sides of a regular polygon. The r-th polar curve of to the (r-s)-th polar curve. 1). must be concurrent at 0.ra-ic. if
P lies
on a given
the class
is
N(n — 1).)*(*£ + ".
is
the s-th polar curve of
Ex.
6.
Any
polar curve of the
(x/a)«
triangular-symmetric
curve
+ (y/b)« + (z/c) n
'
=
and the polar
with respect
triangular-symmetric.
is
'
^-0 3y*~
'
^-0
'
3z*
'
2.
yt
«.
(0..
Q
lies
on the
(n— &)-th polar curve
of P.* the polar curves of at infinity have as a centre.
[The polar curves are
(*s +
Ex.
(-^ + -)^-°.
is
brought into self-coincidence by rotation through 180°
about
. the locus of O is an (n — 2)-ic.
ABC
A
B
B
A
[Take
Ex. 12.
Ex. for
which z
=
the polar line of
(0. 10. 5.
x)
-P.
+un =
0. there is in general one position of and 0.
§
is
node of the (kEx.
an
.
If
P lies
on the
k-th.
'
Ex. 0. 8. &c.
ABC as triangle of reference.
More
generally. 0.]
Ex.
EQUATION OF POLAR CURVES
The k-th polar curves of the
91
vertices of the triangle of reference
^"o 3a*
Ex.
polar curve of
Q
for
an
n-ie.
1.]
~'
.
3.
and
find the
number of tangents from
JV-ic.
a node of the (n — k)-th nolar of l)-th polar of Q. On the polar line of O there are three points whose polar curves have an inflexion at O. 1).VII 2
Ex.°
and
2
..
first
[Take the curve as z n +s
is
n -'i
u^
+ zn 6 us +
5.
5.
See Ex..]
Ex..
P
is
a
also the r-th polar of
[It is
The &-th polar of P with respect to the r-th polar of Q for/ = Q with respect to the k-th. (ii) If a triangle can be inscribed in the polar conic of O which is
*
The curve
O.
(a. If the polar conic of with respect to a given cubic has as a self-conjugate triangle.
Ex. the polars of with respect to the polar conies of and C...] 9.).. If and C? are given. 7 is
to take
P and § P for
as vertices of
the triangle of reference]
6. 11. the envelope
w-ic is of class
of the polar line of
P
with respect to a given [Take the line as z =
n — 1. 4. (i) If the polar conic of O with respect to a given n-ic has a self-conjugate triangle which is inscribed in a given conic.
| +. Ex.
is 7.
=
If the given curve is ' azn + (b x \y) n x
+
+ (c x2 + 2c 1 xy + c 2 y 2 )zn ' 2 +. we show in the same manner that
x + bx y
Any
an
If
r-ple point at
is
polar cwrve of an r-ple point of a given curve has with the same tangents as the given curve.+ ^. Math.
Folar Curves of a Point on the Curve. 49. 375. p.
the k-th polar curve of
(n— k)l
(n — k — 1)
v
°
ia/
+
If b
lies
is
(J-I-2)l
(^ + 2«i^ + «!») «*-*. and that they are independent of the position of 0.
is
. ix. 9). and the case of an jt-ic only meeting the line at infinity at the circular points.
a factor of
in general
an inflexion of the given curve.
on the given curve. and are the same for all w-ics with the same infinite points.
[See Salmon's Conic Sections.
Suppose the triangle of reference chosen so that
Then the
&-th polar curve for
/ ='
7>
h
is
:—^
f
=0
by
§ 2. Show that there are n such axial directions in general. a = and the tangent at = 0. de France.
Qn •••> OQ1 OQ2
Show that an
line is drawn meeting a given n-ic in 'axial' direction of the line making
=
P
§ 3..
is (0. the locus of
(of.
Through a point
Qn-
any
OQn a maximum or a minimum is perpendicular to the polar line of the point at infinity in this direction. Ex. or a triangle can be inscribed to one and circumscribed to the other. See Bull. &c]
Ex. 1).. The &-th polar curve of P with respect to an «-ic having an (n— l)-ple point at is an (n—k)-ic having an (n— h — l)-ple point at 0. de la Soc.
.
[Use polar coordinates.
(i). 0.:
z
!
2
92
EQUATION OF POLAR CURVES
is
*
VII
self-conjugate with respect to a given conic. more generally.
If two of the tangents at to the polar curve coincide (or are perpendicular). the locus of is straight lines through O. The reader may consider the cases in which the conies touch. and similarly
The polar curves of a point at which the tangent has" r-point contact have r-point contact at with the same tangent. And.]
Ex. Discuss the case in which the curve passes through the circular points any number of times. 14. §§ 373. a = and b x + b x y is c x 2 + 2c x xy + c 2 y 2 Hence is an inflexion of all the polar curves.
a 2 (n— 2)-ic
§ 9.
Qi>
.
.
13. Note the case n 2. Hence the polar curves of all touch the given curve at 0.
as is seen by taking the origin at the cusp and writing down the equation of the locus.] Ex. (n — /• + 2)-th.
.
7. which we do not count as a parabola..]
is
Ex.*
+ f2)
[If
a rectangular hyperbola is a (n — 2)-ic.y)
= un + un -
1
+
.x-^z)zx+{a-b)(^y.
.
Ex.] whose polar conic with respect to a given
(|
=
2
+
r.n x)xy has the triangle of reference as a self-conjugate triangle. 2. 1. where « t is homogeneous of degree k in x and y.. 8. There are conic is a circle.
a
The
n-ic is a
parabola
locus of a point whose polar conic with respect to a given is a 2 (n 2)-ic. . 1. The locus of a point whose polar conic with respect to a given and the n-ic has given eccentricity (=£0. x factor. The [n — r)-Va. 6 passes through each cusp of the «-ic.]
Ex.
. The number of points on an n-ic whose polar conic rectangular hyperbola is n (n — 2).
The
locus of a point
(az 8 + ty 2 + cs ).
ij.
10.
—
.
Ex. 1.
general
2n(n — 2)— 2k.
6.
+ Mi + m
=
are rectangular hyperbolas..
3.. The n asymptotes meet in a point and are parallel to the sides of a regular polygon and so are the tangents at any multiple point. polar curve of Oat is (n-l)/(n-k-l).. the equation of the curve in rectangular Cartesian coordinates
is
f(x. where
k
is
the
[The locus of Ex. the polar of the origin is
Wh
=
Ex.
[It is (a|»
=
11
+ brf + c£>)
{x 1
+ y* + z*)
Ex.. asymptote passes through the origin.
The number of points on an
is in
»-ic
whose polar conic with respect
to the curve is a parabola number of cusps.—
VII 3
POLAR CURVES OF A POINT ON CURVE
93
Ex.
^Therefore
If one
common
u k = r*> (a cos k 6 + b sin k 6). up and w"_i have a 0.
»-ic is
4. The polar conic of a cusp is the cuspidal tangent twice over by Ex.
(»-l) y+(n-k-l) ax 2 +
(|.. or 2i) is a 2 (n — 2)-ic number of points on an n-ic whose polar conic with respect to the curve has a given eccentricity is 2n (n — 2). The (n — r+l)-th. The ratio of the curvature at of an »-ic to the curvature of the k-th.. If all
the polar conies for
f(x..]
The polar
conic of
f)
with respect to
(6-c) (r)Z-Cy)yz + (c-a) {(. polar curves are non-existent. polar curve of an r-ple point of an «-ie consists of the r tangents at the point. 9.
Ex. u n ..
the locus
is
+ |!{ =
0. [If the curve is = y+axt + . 5. then u k = are lines parallel to the sides of a regular polygon.
(n— 2) s points
=j-£-
in the plane of an
ra-ic
whose polar
\<A = 2
<Sx
<rZ> t)y*
and
dxdy
=
are equations giving the points. Ex. y)
=
0.] If two asymptotes meet at the origin.
if ra>2.
being the equation of the curve. 11. + a n x parabolas and conversely every curve all of whose conies are parabolas is of this type.]
O with
respect to a pencil of
»-ics
Ex.. the curve is a (2» 2 — &)-ic with a (r— l-&)-ple point at and B. o.
Ex.. then the line cr 2 " goes through 2n+1
.
VII 3
n are Ex. .
[Take the line as z
= 0..
is
circular..
P
P
P
[If
u + kv
=
is
the pencil and z
=
<)x
the given line. 0).
is
3«
<>x
2>v
2iy
_
iv in
<>y
Ex. such that o-j is the first polar of Pj for the given curve...
.
2^=0
/=
Taking r = 2 or
Ex.
[Take k
= n-l
in Ex. If
?/ S 0.]
every polar conic
is
a
circle. . we have the two following examples. The curve is of degree and class n.
P
. oo ). 16. The tangent at O to that curve of a given pencil which passes through O goes through the intersection of the polar lines of 0#rith
respect to any two curves of the pencil.2 ". If every polar cubic
or bicircular quartic. The centroid of the intersections of this curve with any other lies on a fixed line.15. 0. the curve is
a circular cubic
and we Ex. if <»"_!
= 0.Yf of a given pencil of m-ics have a node in general. 17 P%. 16.]
Ex. 1. All the polar conies for y — a + a l x + a 2 a? + . 18.
P
P
. 3 (n .
the curve
is
a circle. cr 2 is the first polar of 2 for cr 1 cr 3 is the first polar of s for o-3 . The k-th polar curves of a point form a pencil of (n — &)-ics. As (» — l)-ics trace out the same locus.
P
P
.
0)
and
and
(0. If every polar curve of degree r goes through two fixed points and B. 14. 3. 10 such that
factor of
a. 18.
13. the line at infinity being the tangent. is any point of the line. the first polar curves of If form a pencil travels along the line the base-points of this pencil of of (n — l)-ics. If a line meets a (2» + l)-ic in 2 "+1 .. derive a series of curves <r lt <r 2 . 19.. The polar line of such a node is the same for each «-ic of the pencil and the node lies on the 2 (»-l)-ic of Ex. 17.
Ex..
is
a
un
.. the locus
.
Then
substitute in
ay ay _
lies
/
ay y
The centroid
on x
—
0.1 at (0. The locus of the poles of a given line with respect to all curves of a pencil of n-ics is a 2 (» — l)-ic passing through the points of contact of those curves of the pencil which touch the line..
—
[Taking
A
and
B as
(1. . 12.
[Choose axes of reference for the curve of Ex.. It has a superlinear branch of order » .]
A A
Ex.
94
POLAR CURVES OF A POINT ON CURVE
.
ill
iv
.
(iii) x = y.
HARMONIC POLAR OF AN INFLEXION
Show
that the harmonic polar of
3?
(0. is the tangent at the point. Prove PF. and y— z = the harmonic
2. 0) is 60a". if any.
Ex.
Then
chords through a point O of a cubic meet the curve again and Q' while they meet the polar conic of O in and fl'.1 + 2 (c x + c x y) 2»" 2 + 3 (d x* + 2d lXy + dtf) n ~ s + Since this meets the given curve at (0. Then the tangent at (0.
Ex. Q. 0. (ii) x+y = 1.
1. R.z
. Take the curve as (i) in § 3. 6. Any line through an (n — 2)-ple point O of an »t-ic cuts the curve again in Q 1 and Q%.
make
in
[Use the harmonic property of the diagonals of a quadrilateral.
96
Ex. OPQ' are drawn through the inflexion O Show that PF and QQ' meet on the harmonic polar of O.
VII 4
— 1. The first polar curve of (1. Hence G is the point of contact of a tangent from 0. 0. QQ'. are concurrent.
the
first
We
=
. 1) to the given curve is 2/ — and passes through O.
Two
P and
P
. Show also that the tangents at P.]
Ex. if (OP. OR.
« (y-*) •=& + <*) (*b* + V)= x y + y' + x + y.
Any two chords OPQ.
RE
R
R
[As in Ex.
= 0.
polar
is
at G.0. 5.
-1) (-1. O as (1.
consider now the intersections with a given curve of polar curve of O with respect to this curve. OS meet the cubic again at four points on a conic also touching the cubic at O.
Ex. Take (7(0. 1. Q.
. 0)* and any point as B(0. 0.
3. Hence a single intersection of the curve and its first
Firstly.
First Polar Curve. Q 1 Q2 ) is harmonic. we must have one of various alternatives. 0.
[The
first
polar of
O
less
the inflexional tangents at O. the two chords adjacent.]
[The polar conic of (0.
2mx—y—z =
polar.
. 1) as such a point of intersection. S.
*
3.1)
(0.
we may have a = b = 0. Q. concurrent.]
4.]
for
for
%x*
l
i
of a cubic. . — 1. OQ. A conic touches a cubic at O and cuts it at P. 0) for (x + y + zf + ^hxyz = 0. 0). Show also that the tangents at Pand Q meet on the harmonic polar of O. 0. Find the locus of P.
Ex.]
§ 5. The curve and the first polar curve have' C as an ordinary point and the tangents to the curves at G are distinct. 1) is (2mx—y—z) (y — z) <= 0. Show that OP. 1. 1).
1)
for
+ y3 +s? + Gmxyz
=
is
the same for all values of m.
(i) (ii) (iii) [(i)
Find the harmonic polar of
(0.
1.
on a curve.]
Find the tangents Ex.
1
+ 2~4). if is the class of the given n-ic
We
We
sum up our
first
The
m
(Ch. the reciprocal curve is of degree m. 1) to a?+y = 2z s 3 = hxyz. 1. 9) to
x3 + y s
2
3
=
. the
meets
Ex.
y^-b^ + iz (x + y) = i The (y + 2xy-x ) = 0.:
:
VII 5
FIRST POLAR CURVE
97
Secondly. the given curve may have a cusp at G. It follows at once that.
Find the tangents from
(1. §
1) to
ax(y*-s*) + by (z'-x^ + cz
7 or Ch. 16. 3a) to ay
(a.
first polar touches the curve at 0. and thrice at each cusp of the given curve. and has r nodes and i cusps (Ch.
e.
%xyz.
(v)
(vi)
to
= a?.r
2
-/) =
0. on a curve must be considered as coinciding with a point the tangent at 0. 1. a)
(0. 1) 1 meets the curve where (y^-Vxy + x ) contact of the tangents are therefore
2. and the first polar curve has G as an ordinary point. Hence two of the intersections of the curve and its first polar coincide at G.
(Jo. 1) is
to
{a?
+ y 2 ) z = 2x s
i
.
*
2'ilfi
H
. -1) to x'+y' +
.
(i).
8 the
number
of nodes. the tangent at G to the first polar curve not coinciding with either tangent to the given curve at G. 0.
0. 1. If the curve has t bitangents and i inflexions. e. s 3 (i) "From (1. The given curve has a node at C.
i.
[The first polar curve of (1.
I. we may have a = b = b t = 0. § 7). Taking « = 0as the tangent to the cusp.
intersections of the first polar of
Hence
If
lies
i. IV. we have
a
—b
=
b1
=c =
x
c2
= 0.
For
m = n(n-l)-2S-3K m is the number of tangents from 0.
n=
it
m(m— 1) — 2r — 3t
(ii).
This
points of
—1
+2i.
(. 1. IV.
see that the first polar touches the given curve at the cusp. x 3 + ys = a
See Ch.
the
number
of
with the given curve at points which are not multiple points of the given curve.
and k the number of cusps.
(iii)
(iv)
From From
From From
(11.
(1. (ii) From (0.
§ 2. twice at each node.* Thirdly.twice at Hence two of the tangents from (§ 3). VI. and three of the intersections of the curves coincide at G*
results thus polar curve of meets the given curve once at the point of contact of each tangent from to the given curve. is of class n. §
2).
I)
to
touch at points on the
first
polar
and meet the curve again on
§
6.
This
is
a line-pair
if
(X.
ay
a^ax
which
degree degree
is
ay
a^ay
n in n — 2.
(ii) (iii)
(iv)
(v)
(vi)
. —3. z)
=
X
IX*
"'" y ay*
2£
+2
.«-. c*x= {a±(a2 -c2 )i} {&±(&2 -c2 )4} </. 0. 6 to the case in which tangents are drawn to an »-ic from a («-3)-ple point. Z)
lies
on
ay
alF
a&a^
2>x 2>z
H=
I
ay
aj/aa. x = at2 y = a? where t = 1. xh/(Bxi -2xy + Sf) = 0.
Equation of Hessian. is with respect tof(x.]
2 or
. the equation of the Hessian of / x. Heesian we put z — 1. § 9.
ay
ay 2
<>yiiz
=
ay
as 2
o. 2. Y. (x-y)(2x-y)(8x2 + xy + Uy') = 0. Ex. y. I. Extend the result of Ch.
In § 1 we denned the Hessian of a curve as the locus of points whose polar conic with respect to the given curve The polar conic of (X.
. Z) degenerates into a line-pair. 3.98
FIRST POLAR CURVE
(i)
VII 5
[The points of contact are given by
xi + 2x*y + 2xf + y l
x/y
= 0.
.1 + a/2.
If / 0. each element of the determinant
=
is
of
is
of
Hence
The Hessian of an n-ic is a 3(n — 2)-ic.->£.
=
Ex. z. Y. y. To find the Hessian of a curve whose equation is given in Cartesian coordinates we make the equation homogeneous by Then after finding the writing x/z and y/z for x and y.
[The tangents from
(0. "JV + 2zx }>Z7>X aFaZ
+ 2xy
ZXIY
=
0.
= and K = meet.
[Putting
i
=
1
in § 6 (1) the expression in the first brackets {} is
a/a/
(
a 2/
/W
^f
dr2
?
1
dr
7>8
3rd0
^7>6J
(*f'y v/\ 2 I dr/ d<0
j
and the expression in the second brackets
^d»Vdfl
is { }
i Vf{#£ j. but also twice at each intersection /=0and = 0./
7>y
2
except that / sections of / 2 of
coincide
(0. z and add the first shall obtain the identity *
two rows
to the third.
.
2
Vda. not only at the inter= and H — 0.VII 7
INTERSECTIONS OE CURVE WITH HESSIAN
9!)
add the rows by
Multiply the columns of the determinant by x. JJ\ .dj/
3 2/
_. 0.
By
its
relation
Hessian with
H=
of § 6 the intersections of an %-ic / are the same as the intersections of
=
with
/=
-
Ix ly Ix ~by
\ly/
da.
Ex.
3/
d 2/ da?
d 2/
da. and then multiply the first two columns to the third
.
For by Euler's theorem on homogeneous functions
3/
da. 1)
.1).
Intersections of a Curve with
(i)
its
Hessian." n 3/
da.
8)
=
given in polar co-
ordinates.
the intersections of
To determine
Hessian
*
H=
3/
as on / = 0. s.
H
x.
Find the Hessian of a curve f{r.
*"
di/
3s
dzds
H2
. we shall take with any point and for this purpose we may evidently replace the
/=
and
H =0
which
by
K = 0. or when an approximation to the Hessian is required in the neighbourhood of (0.1 fill 2 2
l
dW
*/V
'
1
J
r*\l>rl>6
> 18 J
§ 7.
We
i
)
a/a/ *h = in-iY \h Zx
v
'
yf
'
2>y
%x<sy
r^V 7>xf . y. y.r^V \hyJ \ix^
-
l
^
dj/
2
This gives an alternative method of calculating the equation of the Hessian which is convenient when the equation of the curve is given in Cartesian coordinates.0.
Summing up the above results we have :
The intersections of a curve with its Hessian lie one at each inflexion.(d x + d y)y
1
2
z in
-'J
+. Hence
x2 + c.
.
=
3n(n-2)-6S-8ic.
conic of
c
is
~
The polar
If this is n 2 i.
K
H
The curve and the Hessian intersect eight times at the cusp (Ch.
keeping only the highest power of z. k of cusps..
. then.
«
l ..l)byz + 2c^xy = 0. y
K Hence K =
.
100
INTERSECTIONS OF CURVE WITH HESSIAN
VII 7
Suppose. 0.
H=
=
~ ~ / = y2 zn 2 + (d x s + 3dx x2 y + 3d2 xy 2 + d z f) zn 3 + .. Suppose now/= has a node at 0.
= (and = 0) has a triple point at at which Hence two tangents coincide with the cuspidal tangent of the given
curve. 1) and the tangent at on /
=
=
We
have
byzn
~x
/=
+ (c x2 + 2c t xy + c 2 y 2 )zn 2 + (d x* + 3 d Y x 2y + 3 d2 xy 2 + d 3 y s)+. is a cusp of the given curve at which Suppose now that Then is the tangent.
and
K = -24.6 +.. is y point is (0.. and the coefficient of z™ -1 a line-pair. Ex. VI. In this case = 8( Cl 2 ^c c2) (c Q x2 + 2c 1 xy + c 2 y2 )z Sn . VI.. either c = ~
=
=
The Hessian meets a curve only at the inflexions ple points of the curve..
K
with the same (and 0) has a node at tangents as the given curve. so that 6 = 0.
If 8 is the
number
i
of nodes. the curve .
. the triangle of reference chosen so that any 0. The curve and Hessian there-fore meet six times at the node (Ch. and
i
of inflexions of
the
7i-ic... or else 6 of the curve.e. (iii)).
In the case 6 =£
0. Ex.. c
2
and
multi-
=
-6b2 d1 )y-6b 2 d
x}z 3n
-6
K=
{(Sbc 1
+
. § 2.2 y 2 + (n. in /is a factor of the coefficient of z / and is a multiple point has an inflexion at 0. § 2. and eight at each cusp of the given curve. = (and therefore the curve and Hence the curve and the Hessian) meet only once at 0. six at each node. (iv)).
7. 4.
The Hessian
of the
'
triangular-symmetric
n
'
curve
(x/a) n +(y/b) n + (z/c)
is
=
the sides of the triangle of reference n — 2 times. 8 is the tangent at an inflexion O. A curve of odd degree has at least one real inflexion.102
INTERSECTIONS OF CURVE WITH HESSIAN VII 7
2z{x + y + (x + y) (x*-4xy + if) = 0.
Ex. If a curve has r-point contact with its tangent at 0.
[By Ex.}.
Ex.
—
[The Hessian of yu + v = 0... The number of inflexions of an »-ic is even or odd according as the n is even or odd. 8 is the envelope of the (»-2)-th polar curve of any point on the given line. the 2(n— 2)-ic is called the 'polo-conic' of the given line..=!v *!/_/_£/ y = a
i>x
!
2 (n — l)-ic
is
5j/
2
\iix<)y/
the degenerate polar conic of a point on the Hessian is unreal. + . 12 xy* = z (12 g^x 1 + 36 g.i
Jix)
(x)
(xi)
(xii)
l
s
i
1
(xiii)
(xiv)
Ex. In the case n = 3.]
Ex.
If the
locus has a node at
given line of Ex. the Hessian has (r— 2)-point contact with the same tangent at and the curve and . 6.]
l i ) i
.
=
0.
[If
/=
is
the M-io and z
ft
= the given line. The locus of the points whose polar conic with respect to a given ra-ic touches a given line is a 2 (n — 2)-ic.
10.. a x 3 — a 1 x'y — a s xy + a s ys = 2xyz. 8 (yz + a. 2 2 (6 -c-)(fy+c2)a. while in the other the «-ic has no acnode. 8.. Similarly for x n y + y'"z + z n x = 0.]
Ex.
. 3. the with the given line as one tangent at O.
5. 9. + . z\{(l+. Each side passes through n points of the curve at which the tangent has m-point contact and passes through the opposite vertex. + .
(vii)
(viii)
.2 ) 3 + 1 6 xy s (yz + a.]
Ex. The locus of the inflexions of zx^y varying values of a is 2y* + x > z = 0. S(x s + y") = xy(z — x — y). where the terms of lowest degree in Fare of the (r-2)-th degree.
Ex.]
If
F>0..xy + g^).k' )x-By}+x{k 4 x*-k (l+k >)ry + (l-k + k t )y*\ =0. X +y s + z? -9 p*(xz2 + yx' + zif) -3 (18 p s + l)xyz = 0.
8 l y (2x + y)+ax
for
[The Hessian
is
x%
(2
ax*
+ 4 xy s + xi yz + 8y
i
)
=
Now
is
eliminate
a. Hessian meet i 2 times at 0. 2 ) + 3 y e = 0. x^z + ifx + z'y = bx^y'z*. The 2 (»-2)-ic of Ex. 4 each such point counts as (n — 2)
inflexions. This (n — 2)-ic divides the plane into parts in one of which the n-ic has no real inflexion or crunode. is of the form yU+ V = 0. = {x^ + y' + z } {a{W-e^)x + ...
(vi).
a 6(n —
l)-ic. where the terms of lowest degree in v are of the r-th degree. the p. Ex. The locus of the inflexions of a pencil of n-ics Explain the case n = 2.1
.
1) to the inflexions of zx^yi + u homogeneous of degree p + q + 1 in a.
If
p=
they become
p
(p
+ \)u
= xy
oxoy
.
suffixes 1
and 2 denote partial differentiation with respect
. Suppose that X.
[Eliminate z between the equations of curve and Hessian Ex.
^=
0. 14. With the notation
2 a Ali 2
or use
x and y of
= Au p
\l
22
+ X 22 /i. The tangents to a curve at the multiple point (0.
where u
is
The lines joining (0.
/'l2
^M. q. and y.i . Show that the tangents to the Hessian at the same point are in general u and . A fc-ple point of a curve is in general a (3& — 4)-ple point of the Hessian.).VII 7
INTERSECTIONS OF CURVE WITH HESSIAN
=
103
Ex.=
'
v(. 0.]
Ex.!
c AM -
= »( A V + . v are homogeneous functions of degrees p.a^ +
v) + ( w_ l)( 6wA + ^A. 1) are u 0.
9(2+1)* ^S +i>(i' + l)r^
If
= 2 i'2^ oxoy
q.
they become
o u v-" = ox
2
0. 12. 13. ix. 0..
Ex. r not less than 2.»
=
y
y
ox*
oy 1
\oxoy)
Ex. u u #u . 14 (v). and the tangents to the curve at are tangents to the Hessian at 0.^ + V.2 X 12 ^ 12
''1
.
where the to x and y. are
=
0. 11.
rj. i/.
Hence
of
:
If the first polar curve of P has a node at Q.
g from equations
we have
iV
aZ
2
*/
ay
"
ay ay
azaF aza^
aF 2 ay
ay aFaZ ay a^az
Hence
The locus of
given curve
of
aTal
ay
zz 2
=
o
(ii). 112. It may be shown that. 9.
a^aF
the double points of the first polar curves of
a
is its
Hessian. 183.
«
we have by Ch.
3 2y
'
II. Another proof is given in § 9. §§ 8.
yy ^p^=0.
(i).
^ ^
(i)
°
£> V>
3(%-2) 2 . namely.+
VII 8
:
THE STEINERIAN
down
106
first
of the given curve.ay / = + *ZIX^ '«iiF TS Z. 2 az +y aF + aX2
s
Q *f
32
ay
2
ay a 2/ rK hY iz' ""'hZiX
1
ay
n
~^azaF
=
are the equations
(i).
d2
§
/
aZ 2
^aTaF + ^aZa* =0
ay
i
a2
/
/
^vi^ry + Tr2 + *dF2>Z' aFaZ 'aF 2 T
. the polar conic Q is a line-pair meeting at P. see Henrici. Suppose now that corresponding points P' and Q' on the
For * Salmon's Higher Algebra. 9. Z. § 4.
an extension of the theorems in
.
i
the conditions that the
l+
a2
^ ^
+
yy
'
=
°>
should have a double point at Q. London Math.". ii (1868).. Writing polar curve r of P. Y. Ex.*
Eliminating
|. § 76. since each of the three equations (i) is of degree n — 2 in X. Soc. ^ from the equations
+
»2
7.
The conditions that
P should be a
node of the polar conic
/ Z' . pp. F.
(i) and replacing £ by a. z we have the equation of the Steinerian. the Steinerian is in general of degree
Eliminating
. Proc.
^ dX+ & d¥+ ^z
e*f +
r)
s<>x + v ?)Y +
hz
'
dZ
h---
+
----
-'
using
(i). curve*
3(w-1)(to-2)-2<5-4k.
. i (1873). y\ z').¥_
These equations show that the polar line of
Q
is
is
the tangent at
identical with the line through the points and P'. k are the number of nodes and cusps of
the given
* For the effect of multiple points of a curve on the degree and claas of the Steinerian. Bull. see Koehler. (+d£) and (X + dX. 124-9. n-l.
P
P
If the
first
ofQ
is the
polar curve of P has a node at Q.
polar curves of
P and
P'
go through
Q and
Q' respectively. This Hessian and first polar curve are of degrees S(n — 2). Z) on the Hessian. Z + dZ)
Then. Y. Y+dY. de la Soc. and may be shown to meet twice at each node and four times at each cusp of the original curve (Ch. Hence:
The
class of the
Steinerian
is
where S. Math. §2). the polar line tangent at P to the locus of P.
tangent from (x\ y'. such that (ii) holds and
also
A
X
7>X
+V
+Z
7>Y
*Z~°-
Therefore the tangents from (x\ y'. z') to the Steinerian are the polar lines of the intersections of the Hessian with the first polar curve of (x'.:
8
106
THE STEINERIAN
VII
Steinerian and Hessian consecutive to dinates
(£
P
and Q-have
coor-
+ d£..
¥ . VI. de France. other than the double points of the original curve. since the
first
respectively.
^ +d Kli +
or. z') to the Steinerian is the polar line of a point (X.
rj
+ dr). which Hence to the Steinerian in the limit. pp.
v ^)
v
/
/
\
/
S2
/
a2
/\
Identifying them and eliminating X. tj.^Y)
v
<^izix~ wix) + y^^ziY.
.
¥
Therefore the given line (n — I) 2 intersections of
. The two curves (i) touch at this point. we have
tions
:
If two poles of a line coincide. y.
. v.
(i). Moreover. 0. The tangents at the point to the curves (i) are
=
<
/
x
idxa2
v
d^) + y{
a2
x
^Y..
Suppose that the line Ace + py + vz (X.
107
The Cayley an.
'
which are the
first
polar curves of
(to
(v. 1).
— p.
f=azn + hzn
Ex. Y.).
2.y*+zni
a
{d a?
+ %d1 x y + ^ di xf + ds y*) +
Use Ex. Summing up. Z). a node at Q (0. Z) with respect to f(x.
.
— l) 2
poles with respect to
Suppose now that two poles of Xx + fiy + vz qoincide at the point (X. /j. z) = 0. 1.
=
is
the polar line of
Then
/
¥
a
¥
is
.
/x
-J-
¥ = v +¥
Zz
7>y
. The Gayleyan is the envelope of the tangent at the point of contact of two first polar curves of the given curve.
and that the
. If the first polar curve of P(l.
a
Hence
:
In general a given line has given n-ie. the Gayleyan (or Pippian ') of the
'
given curve.
polar conic of
Q
is
to show that Q lies on the Hessian a line-pair meeting at P.
-1
y + cz
1
n
. the corresponding point (£. v between the relaHence thus obtained we get equation (ii) of § 8.
0. Y. Y. Z) lies on the Hessian.
0)
l
with respect to
/=
has
.
0..— X) and
(0. A ^dz
the polar line of each of the
¥ = v ^¥
dee
and
. the coincident poles lie on the Hessian and the line touches the Steinerian at the corresponding point. (X.
Ex.VII 9
THE CAYLEYAN
§ 9.
If the first polar curve of envelope of is called the
PQ
P
has a double point at Q. ^) of the Steinerian lies on either of the (identical) tangents by equations (i) of § 8...
a (U. W =
dec
the curve
7>y
iv
ix
<)x
yv yv =
iy
Zz
<>y
0.
[The
common tangent
d^x + d^y
=
0. 6.]
PQ
4. P' and Q' are consecutive pairs of points on Steinerian
at Q'.]
is
PQ
the Steinerian for the degenerate polar conic of Q. Show that O is a node of the Steinerian and that the line is a 4-ple tangent of the Cayleyan.
[If
7. curve
The Hessian and Steinerian of the 'triangular-symmetric'
Ex. The Steinerian and Cayleyan touch the inflexional tangents of the given curve.
Ex.]
and Hessian.
V = 0.
Ex. 9.
Jacobiau of Three Curves. 10.
P =
0. 12 (ii) meets the Hessian at the points corresponding to the intersections of the Steinerian with the given
conic. The first polar of any point on the tangent at Steinerian touches PQ at Q.
are the equations of three curves.
W)
=
3 (x. y.
as
it is
usually written.
P
and
touches the Hessian at Q.
.
yw yw
Zz
or.]
Ex. and the Cayleyan
into the vertices of this triangle.
at
The tangent
P to
the harmonic conjugate
is
[The tangent at
polar conic of
P is
the polar line of Q.
The
Q is
'
n(n-l)asi + 2(n-l)bzy + 2cy'
i
=
0. z)
is
called the
Jaeobian of the three curves.]
Ex.
[Take O as
(0.
Ex.
[These tangents are the polar lines of the inflexions. Ex.
If
U=
0.
3.]
§ 10. v. The locus of § 2. which
•
naz + by
=
0. 0.
P
to the
If
PQ
Q. 1)
and the quartic as s z* + z u 1 + z^xy + ti t
=
0. The tangent at Q to the Hessian is the for the tangents to the first polar of at Q. (*/«)» + (y/by + (z/c)« = degenerate into the sides of the triangle of reference.
harmonic conjugate of
[The tangents are dqX 1 + 2d 1 xy + d2 y l
Ex. The polar cubic of with respect to a quartic degenerates into a line and a conic. 8.
of
5. 11.108
Ex.
THE CAYLEYAN
The second polar of P touches the Hessian
is
VII 9
at Q.
PQ and PQ' meet
Ex. it touches the Cayleyan at Q.
Hence a first polar curve is only a particular case of a Jacobian.
Take the tangent
to
W=
W=
1)
and y
at
(0.
1)
is
an intersection of
J=
0. 0.
as
=
0. 1) Hence the Jacobian cuts once at in this case.
dently of degree
is the locus of a point whose polar lines with respect to the three given curves are concurrent. 1)
(i)
in the following cases.
.
. and the n-\a 0.
Hence the Jacobian
. if 17 and are straight lines.
VII 10
JACOBIAN OF THREE CURVES
lines of (x. Y.
^iU
"bU
Z
being current coordinates. the Jacobian is the first polar curve of their intersection with respect to 0.
Arranging the determinant just given for Jz in descending powers of z.r ic 0.
That curve of the pencil U + kV = there. . Multiplying the columns of the determinant J by x. may generalize this investigation by finding the intersections with a given w-ic of the Jacobian of an 7i. y.. In § 5 we discussed the intersections of a curve with any first^ polar. 0. It follows that the Jacobian of three curves projects into the Jacobian of their projections.
at once verify that. 1) touches
W
=
which passes
The Jacobian does not in general touch at (0. If the curves are of degrees n n2 ns the Jacobian is evi1
. = with a curve of the pencil each point of contact of
W
W= W—
. we readily find that the Jacobian passes through
(0. an n 2 -ic V 0. and taking for U and V the most general expressions of degrees in x and n 2 also arranged in descending powers of z. 0.
=
V=
W=
We
U=
=
W=
W=
<>x
Jz
=
Suppose that
(0. z)
109
The polar
with respect
to the three curves are
V IU
X.
0. we can express Jz in the form 2>U
We
%+n
2
+ n3 — 3. y. through (0.
Then
W=b
yz 71
'1
+ (c x 2 + 2 Cj xy + c 2 y 2 ) z n~ 2 +
.
0. z and adding the first two columns to the third.
8. « s -ic respectively.
V=
W
Ex.
(ii)
b
-
0.
Ex.
The Jacobian
of
ofthefamily
4.
TT=
is
also of
aVW+bWU* cUV =
the locus of the nodes
0. If O is a cusp of a curve. with y = as cuspidal tangent.
W ="0 has a cusp at 0. less the nodes of counted twice and the cusps counted thrice. If the curves are of the same degree.
4 to
7 take
O
as (0. = w2 = n s and ftx = ft 2 = ft3 O is in general a (Sftj— l)-ple point of the Jacobian.
Q
is
on the Jacobian.
1.
Ex. The number of curves of a pencil of N-ics TJ+kV = which = of degree n and class m is in general touch a given curve 2n(N-l) + m.
Some important
below. the Jacobian of the curves is an (ra.
0. 0.
Ex.
1)
(0.
a Xyple. every node of
W=
(0.
Now
(iii)
W=
b
has a node at
1).]
0. of the Jacobian does not vanish identically.
Ex. If O is a cusp of = at O. If the two other curves have the same degree. Any three curves of the family all+bV+cW = have the same = 0. the Jacobian touches the third curve at 0. — 3)-ic with as a multiple point of order not less than i 1 + & 2 + s — 2. O is a (2 ft t + fts — 2)-ple point of the Jacobian at which ft 8 tangents coincide with those of the « 3 -ic. This tangents to If
=
may
and
=
also be proved at
U+kV=0
by identifying the any point of inter-
section. V=
and
0. Jyple point of an n^ic. 0. If «. [The points of contact are the (2N + n—S)n intersections of and the Jacobian of £7=0.
the Jacobian has a (k— l)-ple point at O.
[In Ex. Now use equation (i) of § 5.
three curves pass thrdugh 0. the Jacobian has a cusp at O with the same tangent as the given curve.
0. V=Q.
Ex. touches If O is a &-ple point of
W
W= W=
.
These results are true only
if
U and V are
perfectly general. the Jacobian of this curve and any other two curves through O has a node at O one of whose branches touches the given curve at O. 1) and the Jacobian passes = twice at Hence the Jacobian meets
W
0.
through (0. 0. 5. « 2 -ic.
special cases are discussed in the
examples
If the first polar curves of
P with
respect to three given curves
are concurrent at Q.
U= 0. +«. 1).2. 7.
=
c
= «! = 0. 0.
the Jacobian of
V=
0. provided the equation Jacobian as the three curves U = 0. If
aU+bV+cW = 0. The Jacobian touches y = at Hence the Jacobian meets W = thrice at every cusp of W =
Now
(0.
Ex.
W=
0.
Ex.
If
O
is
ifc
"
.'
110
JACOBIAN OF THREE CURVES
VII 10
JJ+kV 0. If n 1 = n l and fcj = ft.
V
= 0.
3. ftyple. 0.
6. If two of them are of the same degree. their Jacobian passes through is a node of the Jacobian. + ». 1).]
W
W=0
W—0 W=0
.
if the n-ic has ordinary multiple points with distinct tangents at some of which the N-ics have ordinary multiple points of assigned order.z) = 0.y.
The number
Ex.1). The number of curves of a singly infinite family with characteristic (p. less each node basepoint six times. IV> §8. 7 the Jacobian of the ra-ic and two N-ics of the pencil meets the w-ic at the points of contact required. A pencil of N-ics has as base-points points of an n-ic.
of points of contact
is
therefore
(2N+n-3)n~6(8 + K )-2r.]
Ex.]
Z)
=
|(m . 1). a) = equation. the . take the
node as
•
Ex.]
that the result of Ex. say yjr (X.
while
r+8 + x = £N(N+3)-l. and twice at the other r base-points of the pencil. and </> (A.S £ fc (& .
[By Ex. the points of contact of a curve of the family with the line \x + fiy + vz are the intersections of this line with the curve obtained and = 0. y. n.
(IF
7>F
iF
\
To
discuss the intersections of
(0. and q curves of the pencil touch the «-ie at a point which is not a base-point. 3. The reader may find
.
. 12. Show that the of N-ics touching the n-ic at a point other than a base-point is
>
(n-N)(N-»> + 3) + 42)-2.
Ex. 8. where k is the number of cusps of the m-ic which are not base-points of the pencil.VII 10
Ex.
The two simplest
other cases. 4.
. 0. each other node twice. v.
10. class
m
F=0
[This example is a generalization of Ex. y. 6.
[By Ex.]
F=
and
|=
at » node. 13. Then the points of contact of a curve of the family with are the = with the {l+pn— p)-ic intersections of
—
/=
F
F=
. a. n. v. Any curve of the pencil meets the n-ic at p points other than the base-points. x. p. cases are n = 2 and n = 5.Ex. If 24« 2 + 25 is a square number. 11 still holds.
[Use Ex. is the family.]
[The family whose nodal locus we require is the second family of Ex. 14. I) which touch a given curve of degree n and is in general nl + mp. 4 to Jacobian of the
7 the q points of contact are the intersections of the n-ic and two N-ics with the n-ic.2)
. Show that q = 2p + 2(D.N-ics having a (k — l)-ple point at each fc-ple point of the
n-ic
Show
where
6. the locus of the nodes of 2 those | {-y/ (24w2 + 25) — 3}-ics which go through the 3» intersections of three n-ics taken in pairs is the Jacobian of the x-ics. each other base-point
twice. a) = its tangential If f(x. each cusp base-point six times. six times at each node and cusp of the w-ic.
9.1)— k'.1)
(« . v and degree I in x. Show that the result of Ex. 9 is true if the w-ic has multiple points. z.
JACOBIAN OF THREE CURVES
A
an
111
k cusps of
number
where
pencil of N-ics has as its base-points each of the S nodes and w-ic (n N) and other points of the n-ic. D = |(»-1) («-2)-8-k. by eliminating a between where yjr is homogeneous of degree p in X. 11.
and each other cusp
(k'
in
number)
thrice. For the definition see Ch.
(ii). (v).CHAPTER
VIII
PLUCKERS NUMBERS
§1. cusps. 8.
Pliicker's
Numbers. or '(i). Then. m. (ii). t. k number of n degree. k.
n
i
(ii)
(iii)
= m(m — 1) — 2r — 3t. k. Thus.
=3m(m-2)-6T-8i. — k = 3 (m — n).
t. i are called the Pliicker's numbers of the curve. (ii). we get (vi) and
. of which the most useful are perhaps
:
We
=
m=
=
=
=
=
(i)
m = n(n-l)-28-3K. for instance. class. (iii). assuming (i). the other relations may be deduced. t number of bitangents.
(ix)
%{n-l)(n-2)-8-K = f (m-1) (m-2)-T-i.
i.
shall use the following notation for any curve throughout the book unless otherwise stated number of nodes. IV.
. n2 -28-3 K = m 2 -2r-3i. 8 number of inflexions. »If take three. if we prove (i). but are connected by various relations. The equations (v) to (ix) do not give any fresh equations by consideration of the reciprocal curve.
8. 2(t — 8) = (m — n) (m + n — 9). = 3«(%-2)-6<S-8/c.
(iv) k
(v)
(vi)
(vii)
(viii)
in(n + 3)-8-2ic=
i
Jm (m + 3)-T-2t. i The six quantities n.
Hence.
n. They are not independent.
Of these
so on. we get (ix) by subtracting (ii) from (i). (i).
we
Pliicker's equations only three are independent.
The
Pliicker's
numbers of the reciprocal curve are by
Ch. Similarly. § 7
m. (ii) follows at once on consideration of^the reciprocal of the given curve. subtracting (v) from (ix). (iv) follows from (iii). (v).
These intersections lie one at each point of contact of a tangent from 0. proved (i) in Ch. All curves with the same Pliicker's numbers may be said to belong to the same type. a quartic cannot have four double points. These intersections lie one at each inflexion.
preferred. by considering the intersections of the first polar curve of any point with the given curve. But we have given an independent proof of (v) in Ch. six at 6ach node. if
D
and
D
is zero. X. which
the curve. § 8 and we may deduce (iii) from (i).
We
shall denote
%(n-l)(n-2)-8-K
by the symbol D. § 7.. We proved (iii) in Ch.
is
called the deficiency (or genus) of
cannot in Ch. which we have proved.
.
For instance. b) is a double point of a cubic. For the line joining two double points meets the curve twice at each. § 3. Again.
so expressed. We may deduce (v) from (i). For the conic through four double points and any other point of a curve meets it in nine points at least. and three at each cusp of the given curve.
y—b
. (ii). Then (ii) follows from the reciprocal curve. and eight at each cusp of the given curve. whereas a straight line meets a cubic in only three points.
D = 0. (ii). if
We
.
VIII 2
It
DEFICIENCY
113
remains now to establish three of Pliicker's equations. § 5. VII. the line
it
= t{x—a) meets the cubic again in a point whose coordinates are rational
functions of t. by considering the intersections of the Hessian with the given curve. Also if (a. if the coordinates can be
We shall prove
that.
§ 2. a cubic cannot have two double points unless degenerates. Also the pencil of conies through three double points of a quartic and any other given point of the curve meets the
. two at each node.
be rfegative point on the a parameter
.
Deficiency. (v). IV. that the coordinates of any curve can be expressed rationally in terms of while conversely. It is readily seen that a curve has the same Pliicker's numbers as any projection of the curve but that curves with the same Pliicker's numbers are not necessarily projections of each other. (iii). VII.
&c. = 0. « =
x>t and = 4(Z>-1). a quintic with two nodes and four cusps.
Ex.
= £(re + l)(re-3)(re
-12). 5. D = 0.
Enumerate the types of cubic and
and Ch. m=8. XXI. If 8 t. and 8 = t. If 2«*=i. Ex.
The exceptions are given by
8
=
t.
The number of cusps cannot exceed the smaller of and l(n-2 + 2D) \ {4re-3-(4« + l)i + 4Z>j. while m > \(n + 2) if n is even.
Prove the rest of equations
Prove that
r
(i)
to (ix) in §
1. If
and
[If
m > \ + 3) if is odd.
i). § 7. For example. and > 0. the quintic with
five cusps. whose coordinates are rational functions of the parameter of the pencil.
!!
Ex.
n+m
Ex. or a sextic with nine cusps. For a more general definition of deficiency when the curve has multiple points other than ordinary nodes and cusps see Ch.
m(m — l) > «.
(ii)
re
=
7. XVII.
1. 6Z)= (re-l)(re-6).
= 9.
Ex. then of the same type as its reciprocal.114
DEFICIENCY
VIII 2
quartic again in an eighth point. =.
L
then
k
1. =.] < 9. K = n-2 + 2D. the number of such types is the integral part of f (re 2 — 5 n + 12).
[See Ch.
<t
2
as
n + m>. For a more general theorem of which this is a special case see Ch.7. the cubic with a cusp.]
(i)
quartic. the quintic with three nodes and three cusps. m=2(n-l)-K = l(n+2 +
(re
re
D = and n >4. k = t.
»
Ex. either n = m.
m>
${l+(4n + l)*}.]
11. If
Ex. 2D = n + 2.
Ex.
§ 1. k or else the curve is a nonsingular cubic. Equation (vi) of § 1 states that the deficiency of a curve and its reciprocal are the same. the quartic with a node and two cusps.
i i.
. not all the double points are cusps. (viii). 13. or if k = i. XIII. a quartic with two nodes and a cusp.
2. [Use Ex.
= 8 + £(n
1. IX. 10.]
. m^ 2(re-l).
=
=
[Use §
f (vii). (ii) Find the type of curve for which n = m = 8 = K = r = i.
T
8
>.
4. § 3.
1!
-2re-2S-3/<)(re -9-28-3K). If 8
= n + m. (i) m = 2(n-l)+2D.
. If 28 = 8Z> = (re-2)(«-4). we have re = m.10. If re is given. £ = 0.]
i Ex.
' '
Ex. If
n>m.K
8
(ii)
If
= 3 (re-2) + 6D-2k.
If
(viii).
3. = \(n-\)(n-2)-D-K.] Ex.
i
Ex.
8 = £(w-2)(re-3)-3Z).
If
The curve
is
[(i)
Use
§ 1 (vii). §
n = m.
12.8.
Ex.6. 9.
[For
9.
if the branches are all made to approach 0. The result is plausible for a curve which has k veal branches all nearly passing through has evidently ^k(k — l) crunodes in the neighbourhood of and these coalesce at 0. shall show that
We
Plucker's equations
still
multiple point with ^k(k — 1) nodes and each ordinary Jc'-ple tangent with k' distinct points of contact as equivalent to J k' (/<. 15.
Multiple Points with Distinct Tangents.]
. the tangents to the curve and the first polar at being all distinct.:
VIII 3
MULTIPLE POINTS
R
115
Ex.
The
first
0) is
^-
= 0. equations (i).
+ zn
. and from these three equations the other equations of § 1 can be deduced.]c .
we replace the
fixed line
by
[Project the conic into a circle.
ur
is
homogeneous of degree r
polar curve of
(0. 1. The result of Ex. If the range (OPQE) has a fixed crossratio.
which has
evi-
dently a (k— l)-ple point at 0. .
Let us now suppose that the curve has one or more multiple points with distinct tangents. § 2. (ii). Suppose the curve has a /c-ple point Its at (0. or Ch. Hence by Ch. if ive reckon each ordinary distinct tangents as equivalent to
/= zn . the curve and first polar meet in k(k — 1) points coinciding with 0. 14. as required.+u n
in x
=
0. equation is
. Through a fixed point any line is drawn meeting a fixed curve in P and a fixed line in Q. § 7..
[Project the fixed line to infinity plane perspective. VI. We require to show that any first polar curve and the Hessian meet the given curve k(k— 1) times and Bk(k — 1) times respectively at a /c-ple point of the curve.) The
..'— 1) biiangents in the evaluation of S.
to possess only ordinary
So far we have supposed the curve cusps and nodes. below. 0. the locus of is a curve of the same type as the fixed curve. One of the statements in this theorem is the reciprocal of the other. and invert with respect to 0. 14 holds a fixed conic through 0. To find the number of intersections of curve and Hessian at we adopt an indirect method. (See also Ex. 2.1
u
rc
_1
+. It will be sufficient to prove one of them. if that is so.
and
y.
or note that the two curves arc in
if
Ex. (iii) of § 1 hold good when each fc-ple point is considered equivalent to %k(k— 1) nodes.
hold.]
§ 3. For.k u
where
l
. I. and D. 1). r.
§ 2. k cusps.
.
31c
and therefore the curve and Hessian meet
at 0. .:
.
[If
2.
(0.
times at
=
and
(iii).
since
k
= 0. and i inflexional tangents. 1).
3
116
MULTIPLE POINTS
VIII
is evidently dependent solely on the nature of the and there Is therefore no curve in the neighbourhood of .
We
show now more generally that
equations hold if (i) each multiple point of order k having I ordinary superlinear branches with distinct tangents is counted as equivalent to %k(k— 3) + 1 nodes and k — l cusps .
PliicJcer's
* It is supposed that the tangent touches only one branch of the curve at each of these I' points.
Ex. The deficiency is taken as %(n — 1) (n — 2) — ^2fc(&— 1).
Ex. VII. The deficiency of a curve with no singularity other than ordinary multiple points with distinct tangents is £ (m — 2n + 2).
= 3n(n-2)-68a -8ic-Sk(k-l). 0.
[£(m-2» + 2) = £(w-l)(»-2)-8. VI. I. 1).
u = are the tangents to the curve at the multiple point the tangents to the Hessian at the point are
« by Ch.
(k— 1) times
with The first polar of has evidently a /c-ple point at the same tangents as the given curve. The\i from the reciprocal curve as in § 1 n= 1) — 2r-3t.]
Establish directly the fact that the Hessian meets the curve
3Jc(k — 1) times at
a
fc-ple
point. and (ii) each multiple tangent meeting the curve in*k' + V points at its V points of contact* is counted as equivalent to %k'(k'—3) + l' bitangents and k'—l' inflexions.]
§ 4. 0. 1.
cla:
-2
•
—=
dy"
'
I
\
ixiyj
§ 7. (Ch. namely 82 nodes.
Multiple Points with Superlinear Branches. Ex.
number
m(m—
3m(m-
and we have just proved that
m = n(n-l)-28
These give
i
2
-3ic-k(k-l). loss of generality in supposing that the curve has besides only ordinary point or line singularities. 14
fc(fc+l)
(0.
. § 7 first polar meet k(k + 1) times at (vi)) so that 2/c of the tangents from a fc-ple point must be considered as coinciding with tangents at that point.
A
Hence curve and Hessian meet + 2fc(fc-2) = 3fc(Jfc-l) similar method will apply to § 4. Hence the curve and Ch. r bitangents. k = 2) — 6r — Si.
We deduce
since
c
2
= 3n{n-2)-682 -$K
result.
2
— I)
. = 7»-12. (ii) is the reciprocal of (i).]
[(i)m
(ii)
<e
t
. t = £(»j-3) (9»-16).. < = n-S. Find Plucker's numbers for a curve of degree superlinear branch of order (i) n — 1.
As
n=
while
m (m — 1) — 2r
kz
2
—3
1
2
and
m = n(n — 1)— 2S — 3k — (k
2 2 2 2
I). 8 == £(ra-3)(»-4). r 2 bitangents. 8) the curve and the first polar meet in k 2 -l = 2{%k{k-3) + l)+3(k-l)
points at 0.
§ 1). m = S(n-l). D = n-2. Ex.
of the theorem is similar to that of § 3. 4 the Hessian of the reciprocal curve meets it k — L times at its points of contact with the multiple tangent which is the reciprocal of 0.
where there is that a may be y = (Ch. b. Hence (Ch. VI. § 2.
d.4
VIII
'
MULTIPLE POINTS
'
117
By an ordinary superlinear branch of order q we mean one whose Cartesian equation near the origin (taken at the multiple point) is of the form
y no
=
12
3
x (a + bxl + cxi + dxl +
between
. S = r = \(n-V)(n-%). § 7.
n with a
single
i
= ».
c. VI. I) = 0. = = »-2.. VII. Ex. only note the modifications which are necessary. Since a factor repeated r times in u k is repeated r— 1 times in -r-^ . and
no singularities
t
2
inflexions.). As before.
which establishes the
Ex.
<?
2
that the curve has besides nodes.
= 3m (m — 2) — 6r — 8t — (k —
by Ch..
special relation
a. if the
tangent to the branch
is
except taken as
.
zero.
Again.. to prove that the number of intersections of curve and Hessian at is
6{±k(k-3) + l} + 8(k-l).
. k 2 cusps. (ii) n— 2.
we may suppose
except
before.
2
-6{%k(k-3) + l}-8(k-l). every tangent to a superlinear branch of the curve
is
The proof
We
a tangent to a superlinear branch of any first polar of order lower by unity.
6^ + 8/q + =
t!
/3
.f Since the curve and the first polar of meet altogether in n(n—l) points. could make the solution definite if we had a definition of the deficiency D.
e. and how inflexional tangents have their points of contact at ? may consider these as ordinary nodes.
(i).
to be noted that
and
(ii)
give identically
3(a + p)
=
. if for the reciprocal curve the line corresponding to passes through p intersections of curve and first polar and a.
. t for a. bitangents.
. and inflexions not coinciding with O.
many
We
.
— m(m — 1) — n — 2r — 3i = 3m(m — 2)— 6r — 8t — equations give 3(a + p) = jS +
p
2
2
.
P
m
a
= n(n— 1) — m — 28 — Sk
2
2
.
The general problem.
o-. and a times at 0. without loss of generality.
The last four The equations
dependent.
2r 1 + 3i 1
First of all
it is
=
o.
quantities 8lt k x t 1 tj could not be found if this relation were not satisfied.
are an infinity of solutions.
Similarly.118
HIGHER SINGULARITIES
§5. in order that Pliicker's equations may formally hold ? Suppose the numbers required are 81} Klt tv t r Let any first polar curve and the Hessian meet the given curve in a respectively. . twice at each of the 82 nodes. 0. cusps.
Hence the
. &c.
(ii).
and (ii) are then consistent but not inOur problem is soluble but not definite there
(i)
.
. p.
. Then we must and /S points coinciding with
:
have
28l + 3K 1
= a. thrice at each of the k 2 cusps.
j
6^ + 81! + *! =
(i)
a
.
£ = 3n(n-2)-682 -8ic2 -i 2 Likewise from the reciprocal curve
a-
. take 82 k 2 r 2 i 2 aa the number of nodes.
. and inflexions * must be considered as coinciding with 0.
as required.
2
2
ac
2
.
Similarly from the Hessian. which was
We
How many bitangents have both points of contact at 0.
VIII
5
Higher Singularities. special cases of which have been discussed in §§ 3 and 4. a only depend on the nature of the curve in the immediate neighbourhood of 0.
. bitangents. cusps. To show that it is satisfied.intersections of curve and Hessian.
/3
+
(r.
.
*
i. namely at the points of contact of the tangents from P. is as follows Given a curve with any multiple point whatever 0. how many nodes.
Let any variable transversal through G cut 2 in Q. we may take homoChoosing xy. as is easily
The dotted conic in Fig. Z). P' will trace a locus c'. Y.
'
ABC
2. P' lies
. as triangle of reference. It and let (PPf QM) be an harmonic range on the line CQR. geneous coordinates so that 2 is z
P
'
.
=
If P' is the point (X.
-^
. 1). In Fig.
P
is
(
y
. if P lies in any portion of the plane.
Suppose we have a fixed conic 2 with fixed tangents GA.
for these
. 2 is the locus of the middle point of the chord QR. traces out a locus c.
~\
.
in the other portion labelled with the
verified. GB (Fig.CHAPTER IX
QUADRATIC TRANSFORMATION
§1. 2.
same
letter. c' are said to be derived from one another by 2 being the base-conic and C quadratic transformation the pole of the transformation.
Definition of Quadratic Transformation. Suppose that The loci c.
z) = is transformed into f(l/x.
c' is
any other line. y. c' is
* It is more usual for reasons of symmetry to take the transformation in the form Y : Z = 1/x 1/y 1/s so that the curve f(x.
. then
is
the locus
c'
of P' *
\y x z/
.:
IX
z1
1
QUADRATIC TRANSFORMATION
C and
is
121
points are collinear with
=
xy. P' is
at B. or from the original definition. C. 1/y. a conic through A.
if
conjugate with respect to r
Hence.
y. a line through A. P' is at C. z)
=
the locus c of P. which is a conic
If c is
a line through B.
=
0.
c' is
c' is
If-Pison^C. B. 1/a) = 0.
P is on BG.
we
at once
derive the following
P is on AB. G.
f(x
.
the same line. if c is the line \x + fiy + vz X/y + n/x + u/z = 0. This is equivalent to quadratic transformation followed by interchange of the vertices A and B of the triangle of
X
:
:
:
reference.
If c is a line through C. B. through A.
From
If If
this result. P'isat A. c' is For instance. If c is a line through B. If c is a line through A.
p q.
If c meets
Pig.. the conic becomes the osculating conic to c' at G. 3). It follows that. c' touches at A the line through A corresponding to BP. and similarly if c touches GA.
If c touches AB.
and
.
—
n— — n—p — n—p n—p —
c
* Provided P is not at A.. B. If i? approaches P. CB respectively at k. G'j4. i. two tangents to c' at G coincide . touches CP at C* at P to c corresponds the conic
c'
osculating
c'
G and
passing through
A
and B.
For if Q is close to on c. two tangents to c' at coincide.
P
P
R
Again
CA at P. Then c meets J. the line joining A to the second Similarly if c meets CB at Pintersection of BP with 2. Also. Suppose c is an n-ia with a ft-ple point at G.
. R' touching c' at 0.
.
to
GA. — q. and a ^-ple point at J5. a p-ple point at J. B. and the limiting position of CQQ' (which is CP) is the tangent to c' at G.: :
122
QUADRATIC TRANSFORMATION
IX
1
Again
If c meets
To the tangent
at
AB at P.
For. Q' is on c' close to C. Hence c' has k.
Hence
(or
branch of
to every intersection of c with corresponds a c' through G. to the line joining of c corresponds to any point a conic through A. P is an ordinary fc-ple point of c. to every intersection of c with CA
c'
AB
CB) corresponds a branch of
through
A
A
(or
B)
. P'
is
close to
A
(Fig.
if P is close BP meet on 2. or C and so throughout. if has k linear branches touching CP at C. 3.
AP'.B.
and
vice versa.
and
conversely. n — q k other points. e.
A
and
B
a conic through
Ex.
[Take three double points as A. P' are now inverses of each other with respect to the circle.
inversion.
IX
2
INVERSION
123
n—q—k
branches respectively through C.
k
=
is
0. B.
'.
well-known theorems connected with
One
of the
main
uses of quadratic transformation
is
to
deduce properties of a curve c from those of its transform c'.
If
c is
an
—
— —
For instance. then c' is a (2n—p — q — k)-ic with an (n p — q)-ple point at G. and an (n q k)-ple point at B. Our knowledge of the theory of quintic curves (and still more of sextic) is very limited. If two curves (or branches of the same curve) have r-point contact at a point on CA. Since c' meets at { (n —p — k) + (n — q — k) + k } points.
a
circle
with centre
From
'
the results of the last section
'. c' of P.B. Also it meets of AB with the tangents to c at G and similarly it meets GA and GB at p and q points respectively other than A. The properties so derived are not usually of much elegance or importance. 1. C. it is a (2n—p — q — k)-ic. 1 are to
be projected into the circular points. g. taking n
=
2. and a q-ple point at B.
C. B in Fig. the transformed curves (or branches) have (r+ l)-point contact at A. the inverse of a circle is a circle the inverse of a curve with respect to a focus is a curve with a cusp at each circular point and so on. B. Hence
AB at the k intersections
AB
:
n-ic with a k-ple point at G.A..
p =
q
=
1. The loci c. be both nodes.
If in § 1
we
project A.
we
see that
A
the transform of a conic through and B. Hence inversion is a particular case of quadratic transformation.
'
we may
derive
many
For instance. and quadratic transformation is the generalization by projection of inversion. an (n-p-k)-ple point at A. if A.]
Ex. 2. a p-ple point at A. P' are inverse points with respect to this circle.
2 becomes
and P.
.
§ 2. it is convenient that they should bear similar relations to the curve c (e. The main advantage of quadratic transformation over projection followed by inversion is that. Properties of any quintic with three or more double points can be deduced from those of a quartic with the same deficiency.
B into
the circular points. This is exactly equivalent to the process of projecting two points into the circular points and inverting with respect to
some other point. C of § 1.
Inversion.
P
nN
ABC
nN—r —
—
N—K N
2N—K. Take the point as C. if we are to observe the convention that an n-ic and an iV'-ic meet at points.
The Number of Intersections of two Curves
a given Point. VI. and therefore times at G. 1 does not lie on a side of the triangle the quadratic transformation evidently transforms two curves meeting at into two curves meeting at P'. Quadratic transformation (or the equivalent process of inversion) will also enable us to simplify the solution of problems which have been discussed in Ch. At A.
INVERSION
IX 2
or both cusps. The transformed curves therefore meet (n k) (N—K) times at each of and B. &c.
times at C.fif-ple point on the two curves respectively. and G no two branches of the transformed curves touch one another. 1) so as to have no other special relation to either curve. and have multiple points of orders n and at G (§ 1). Suppose that the curves meet in r points other than G.:
124
. Therefore nN—r = s + kK. B. &c. Suppose then we have two curves of degrees n and N. so that
If two curves have a k-ple and a K-ple paint at G respecand the transformed curves meet at s points on the line AB other than A and B. the two given curves meet s + kK
tively. and times at G (Ch. quadratic transformation. namely.
at
in Fig.
A
A
B
—
A
nN
.
s
AB A = (2n-k)(2N-K)-2{n-k)(N-K)-nN-r = nN—kK—r. VI and elsewhere. § 2). I. which is a /c-ple and a . such as the determination of the number of intersections of two curves at a given point.) and this restriction is unnecessary if we employ the generalized form of inversion. They meet also at the transforms of the r intersections other than G of the original curves. and take the sides of the triangle (Fig. since we took the points and in a general position. the expansion of y in terms of x near a given point.
§3. The transformed curves are of degrees 2n k and have multiple points of orders n k and at each of and B. and we want to know how many intersections of these curves must be considered as lying at a certain point. § 7 Ch.
. Hence they must meet at s points on other than and B. the number of tangents from a point whose points of contact coincide with the point. where
If the point
P
ABC.
and an %-ple point at. If the transformed curves intersect at a point on AB. We may find the equation of the polar An alternative method is the curve and then use § 3. though this restriction may at times be removed if due precautions are taken. It meets the first polar of B (n-k)(n — k-l) times at A.
following. meets the Ti-ic a times at G. and (say) a. take the case of a curve passing through the cusp G of another. all the branches having as a common tangent.
so
. and at the polar of points of contact of the tangents from P. the two curves having a common tangent at G. that the lines GA and GB have no special relation to the curve.IX 4
GLASS OF A CURVE
125
We suppose. one touching AB and the other not. The transformed curves have /c-ple and IT-ple points at and therefore meet in IcK points coinciding with H. points on
m
m
AB
*
a
this assumption involves
depends only on the shape of the n-ic in the neighbourhood of no loss of generality. It has ordinary (n — /c)-ple points at A and B.
H
Hence the number of intersections
is
of the original curves at
G
1+2. take the case of curves with k and linear branches at C respectively. (n-k)(n-k+l) times at B.
C.
Class of a Curve.
K
GH
H
. taking G at H. points of contact n(n-l) times at G (Oh. The ti'ansformed curves pass through H. VIII.
tangents from A to the n-ic corresponds To each of the tangent from B to the transformed curve whose point of a contact does not lie on AB.1 =
3.
As another example. and it is not immediately evident how many times they meet at H. which meets A B at H. We shall assume the n-ic has no multiple poiDt other than C* Then
Suppose
first
A
The
P
m = n(n— 1) — a. as stated before.
§4. G. § 3). we may apply the above theorem to the transformed curves. As a simple example of the above theorem.
m
and B have no special relation to the curve. so that they have a simple intersection at H. at the of tangents from B not lying on AB.
Suppose we wish to find the number of intersections a of a curve with any first polar curve at a multiple point G of order k on an n-ic.' times at other than A and B. The original curves therefore meet at 2kK points coinciding with C. Now the transformed curve is of degree 2n — k.
or (ti — fc) (ti + Zc— 1) — a' = m = ii(?i— ])— a.
Tangents from a Singular Point to a Curve. The other tangents from C to the transformed curve are the n tangents at G each counted twice.
Now
H
H
H
Hence
The number of intersections of a curve with any first polar curve at a k-ple point G is equal to the number of intersections k{k—\) at an ordinary k-ple point together with the number of those intersections of the transformed curve with any first polar which lie on but not at A or B.:
126
Therefore
CLASS OF A CURVE
IX
4
(2n-k)(2n-k-l) = (n-k) (n-rk-l) + (n-k) (n-4 + 1) + w(to — l) + m + <x'. Hence the class of the transformed curve is
m
m — \ + p+ 2n.
AB
the curve has k linear branches with a tangent at C meeting at H.
Suppose that in the curve of § 4 X of the tangents from C have their point of contact at G. The effect of this on the class of the transformed curve is to lower it by k(k — \) and therefore the effect of the singularity at G is to lower the class of the original curve by 2k (k — 1).
of a curve with the the number of intersections at first polar of is dependent on the shape of the curve in the neighbourhood of in general and not on the position of (unless is near or on the tangent at H). as in § 4.
This gives
a
=
<x'
+ k(k — 1). the transformed curve has a /c-ple point at H.
common
AB
. and the TX tangents from C to the given n-ic which do not touch at C.
is
by consideration of the tangents from
B
that this class
m+2(n — k) + e.
where e is the number of tangents from B to the transformed curve whose points of contact are on but not at B (or A).
AB
Therefore X
= 2k+u — e.
But we
see.
or:
The number of tangents from the k-ple point C to a curve whose points of contact coincide with G is 2 k.
§5.
if
For example. and that u of the tangents from G to the transformed curve have their points of contact on AB. plus the number of tangents from G to the transformed curve whose points of
.
djB. If a tangent at a /c-ple point C of an n-ic meets the curve in k + r points coinciding with C and meets AB at H. §§ 2 and 3 as i(n— l)(n — 2)— S— k. the J-ple point at H' moves up to G and coalesces with the &-ple point to form a higher singularity at G.
AB but
For example. the transformed curve has a /c-ple point at at which in general neither AB nor
CH is a tangent. so that the original curve has an ordinary /c-ple point at G and a ^'-ple point corresponding to at the point H' very As AB is moved hack into its origiDal position close to G.] at C or H. with the proviso that an ordinary /c-ple point is to count as ^k (k — 1) nodes. Suppose now that the transformed curve and the lines GA. but an attempt is here made.
Suppose that the transform of a curve with a /c-ple point at a J-ple point at on . Ex.
'
We defined the deficiency D of an n-ic in Ch.
from
H
Hence the points
with
C.
§ 6. The transformed curve now cuts AB in k shape slightly.) First we word the definition of deficiency when no higher We say that the singularities exist a little differently. We say that G has been analysed by quadratic transforma'
tion
'. [The transformed curve meets CH n times at C and n-k-r times not But it is of degree 2 n .
of contact of
C coincide
2k tangents
Ex.IX
the
7
DEFICIENCY
on AB. the transformed curve meets CH in r points coinciding with H. distinct points not close to A or B (j of them very close to H).
'
Deficiency.
§ 7. VIII. (See also § 12. 10.
Latent Singularities. minus
the. CB are kept fixed while the line AB is slightly displaced so as not Then the original curve will alter its to pass through H. The J-ple
G has
H
H
'
'
point of the transformed curve is said to be latent in the &-ple point G of the original curve. not coinciding with A or 5.k.
127
contact are
number of tangents from
B
to
transformed curve whose points of contact are on not at B. It is desirable to give a definition which shall be valid when
the curve has higher singularities. It is difficult to do this satisfactorily without a knowledge of function-theory. where S was the number of nodes and k the number of cusps.
. if the curve has k linear branches at C with a common tangent meeting AB at H.
By § 1 the transforms of" the n-ic and (to— 3)-ic are therefore a (2n—p — q — k)-ic with (to — p — /fc)-ple. That this definition of deficiency is equivalent to the earlier
'
'
one
is
readily proved.
We
—
'
§ 8. and so for J. Now suppose k = 0. B. have assumed p. § 6). (to— p — g)-ple points at A. q.C respectively.
. as required for the
deficiencies of the TO-ic
We
*
or
1
To complete the definition add and B = 0. multiple points.has a (k of an n-ic. (k — l)-ple points at A. The transform of the (to — 3)-ic is now only of degree In— p — q— 4 (not 2to— p — q — 3.128
deficiency of
coefficients
DEFICIENCY
an
n-ic is one
I
X
more than the number of arbitrary
of
the
in
the equation
most
general
adjoined
(n — 3)-ic. If the A -ic is • adjoined' to the n-ic. G and a curve of degree
tic
We now
2(n-3)-(p-l)-(q-l)-(k-l) =
with a multiple point of order
(to
2n-p-q-k-3
%
— 3) — (p — 1) — (q — 1) = n—p — q — 1
at G.
Deficiency unaltered by Quadratic Transformation. the summation being taken over every multiple point of the n-ic. and B. l)-ple point shall state the fact that an iV-ic. But this is the number we have defined in Ch. from which the equality of the
and its transform follows immediately. (to — q — /c)-ple. (q — l)-ple. B.
show that the deficiency of a curve and its quadratransform are the same. Any adjoined g-ple. VIII. § 3 as D-l.*
By an adjoined J\T-ic we mean one which has a (k — l)-ple point at every /c-ple point of the n-ic. when the curve has only ordinary Suppose as in § 1 that an n-ic has p-Tpie.
II.B. by saying that the JV-ic is at a given /c-ple point r adjoined to the n-ic at 0'.
l)-ple point at a given state that a curve has a (k point is equivalent to assigning •§&(/« 1) linear relations
To
—
—
between the
the
coefficients of its
equation (Ch. /c-ple points at A. G. k all greater than zero. it is adjoined at every multiple point. (n — 3)-ic has (p — l)-ple. Thus the transforms of the curve and an adjoined curve of degree lower by three are also a curve and an adjoined curve of degree lower by three. according as the cubic has or has not a double point
' . if » = 1 or 2 while when n = 3.
Hence
(n— 3)-ic
has
J(TO-3)»-2ifc(ft-l)
arbitrary coefficients.
D=
'.
IX
9
DEFICIENCY FOR HIGHER SINGULARITIES
;
129
and has a multiple point of order n — p — 2 (not n — p — 1) and so for B. But if, as is lawful, we take as the transform of the (n — 3)-ic the (2n — p — q — 4)-ic together with
proof) at
A
the line AB, we obtain a curve of degree adjoined to the transform of the 7i-ic. In fact, since the transform of the n-ic is of degree 2n— p — q with (n—p)-p\e and in — ?)-ple points at A and B, an adjoined (2n— p — q — 3)-ic has (n— l)-ple and (w — j — l)-ple points at and 5. The line therefore meets the adjoined curve in points whose number is greater than the degree of the curve, showing that the curve degenerates into and a (2n — p — q — 4)-ic. Similarly, if p 0. CA is part of the transformed adjoined curve ; and, if q 0, CB is part of the curve. In practice, however, the case p, q, Jc all greater than zero is the only one we need consider. In the following we shall require to transform any curve by successive quadratic transformations into one having only ordinary multiple points with distinct tangents. To do this we can take and as ordinary points on the curve and G as any multiple point which is not ordinary '. The process of transformation is repeated till only oidinary multiple points are left.*
2n—p—q—3
A
p— AB
AB
= =
A
B
'
§ 9.
Deficiency for Higher Singularities.
The definition of deficiency can be applied to curves with higher singularities when we have defined what we mean by a curve adjoined to an n-ic with such singularities. Suppose that in § 6 a curve has a (j — l)-ple point at H, then its transform would be considered as adjoined to the given curve at the /c-ple point C. If the ./-pie point is not an ordinary one, it could be still further analysed by taking C at In this way we can find what is intended by a in §6. curve adjoined to the transformed curve at H, and then the transform of this adjoined curve will be adjoined at G to the given curve. __ According to this definition we shall still have deficiency unaltered by quadratic transformation. For instance, if the given curve has k linear branches touching CH at C, the transformed curve has an ordinary &-ple point at H. A curve adjoined to the transformed curve has Hence a curve adjoined to therefore a (k — l)-ple point at H.
H
H
* It is fairly
evident that this
is
possible.
We
do not give here the
formal proof.
:
130
DEFICIENCY FOR HIGHER SINGULARITIES IX
9
at C. the given curve has k — 1 linear branches touching Also the fc-ple point C lowers the deficiency by k(lc — 1), since lowers the deficiency of the transformedthe /c-ple point at curve by %k(k — l).
GH
H
§10.
Intersections of a Curve with Adjoined Curves.
To state that an i^-ic is adjoined to an n-ic at an ordinary k-ple point G is equivalent to assigning \k (k— 1) linear relations between the coefficients of the equation of the i^-ic. The number of intersections of the n-ic and iV-ic at is k(k— 1), which is twice the number of linear relations just referred to. suppose, as in § 6, that C has a latent ^'-ple point. The
Now
transform of the i^-ic has an equation whose coefficients are subjected to %j (j — 1) linear relations, since it has a (j — l)-ple Hence the coefficients of the equation of the JV-ic point at H. are subjected to \j (j — 1) linear relations in addition to the £k(k— 1) relations which state that C is a (k — l)-ple point of the iV-ic. Also, since the transforms of n-ic and JV-ic meet j (j — 1) times at H, the n-ic and N-ic meet k (k— 1) +j (j — 1) times at C by § 8, which is again twice the number of relations between the coefficients of the i\T-ic due to the fact that it is adjoined to the n-ic at G. is not an ordinary /-pie point, it may be analysed in If its turn till the n-ic is finally resolved into a curve with ordinary multiple points only.* Hence
H
The number of intersections of an n-ic and adjoined N-ic
at the multiple points of the n-ic is twice the number of relations to which the coefficients of the equation of the N-ic are subjected owing to the fact that the N-ic is adjoined to the n-ic.
The number of
for
relations in question is
i(n-l)(n-2)-D;
an adjoined (n— 3)-ic has
deduce that
is
D—1
arbitrary coefficients in its
equation.
We
If a pencil of N-ics
adjoined to an n-ic
and
the base-
points of the pencil other than the multiple points of the n-ic are also on the n-ic, any N-ic of the pencil meets the n-ic in %(n—N)(N—n + 3) + D variable points (n > N).
* This statement assumes that a curve and an adjoined curve are transformed into a curve and an adjoined curve i. e. N = n — 3. But the number of intersections and of relations between the coefficients due to the adjoining at C is evidently independent of the value of N.
;
IX
11
ANOTHER TRANSFORMATION
—
131
Of the fixed intersections of the n-ic with a variable A~-ic of the pencil (to — 1) (to 2) — 2D lie at the multiple points and * -JA(iV+3)-l-i(«-l) (to-2)
Ex. 1. A pencil of JV-ics has as its base-points points of an n-ic, being adjoined to it at all its singularities with the exception of k cusps and certain of its ordinary multiple points. If any JV-ic of the pencil meets the n-ic at p points other than base-points, while q N-ics of the pencil touch the n-ic at a point other than the base-points, show that
q
= 2p + 2(D-l)-K'.
;
[The n-ic may be transformed by successive quadratic transformations into a curve with ordinary multiple points and p, q, k, D are unaltered by this process. Now use Ch. VII, § 10, Ex. 12.]
Ex. 2. A pencil of iV-ics has as adjoined to it. Show that
its
base-points points of an »-ic and
is
(n-N)(N-n + 3)+4:D-2
of the JV-ics touch the n-ic at points other than a base-point.
§11.
Another Transformation.
form of transformation alternative to that in § 1 is obtained by taking as our base-conic a pair of lines through B harmonically conjugate to BA and BC. As before, we take and any transversal through G cutting these lines in Q and and P' on the transversal so that (PP\ QR) is take points
A
R
P
harmonic In Fig.
(Fig. 4).
lies in any pprtion of the plane, P' lies in the other portion labelled with the same letter, the hyperbola in the figure being the locus of the middle point of the
5, if
P
chord QR. ' Choosing homogeneous coordinates so that
s 2 we triangle of reference and the two fixed lines are a? find that, if P' is (X, Y, Z)"P is (X, Y, X*/Z), since CPP' is a straight line and the pencil B(PP', QR) is harmonic. is f(x, y, z) 0, the locus of P' is Hence if the locus of
,
ABC is =
the
P
=
f(x,
*
y,x 2/z)
=
0.
have
N must be such that this quantity is positive or zero. We may always N = n— 1 orm — 2, when «>2. For N to be n-S, we must have D>1.
K2
132
ANOTHER TRANSFORMATION
IX
11
It is at once
by taking
§ 1,
'
shown that this transformation may be obtained in succession three transformations of the type of the base-conics being respectively
x2 — xy + z 2
= 0,
x2 — y 2 + z2
=
0,
x 2 + xy — s 2
=
and the pole of the transformations being (0, 0, 1). The properties of the transformation are somewhat similar
to those of the transformation of § 1. The reader will have no difficulty in verifying the following
To each intersection of c with BC corresponds a linear branch of c' touching AB at B. To each linear branch of c through B (not touching AB) corresponds a linear branch of c' through B, the tangents to the two branches being harmonic conjugates with respect to the two given lines. If c is an m-ic with a &-ple point at G and a ^-ple point at B, c' is a (2n — k-q)-ic with an (n — g)-ple point at G and
IX
12
APPLICATIONS OF TRANSFORMATION
linear branches through B,
133
n-k
n,-k-q
of which touch
AB
^The theorems in
tangents from a singular point of intersections of two curves, transformation of this.sect on the
to italics in §§ 3, 4, 5 relating
the
number
and class still hold for modifications The proofs are very similar, and the necessary apphcatmn^o w_ill require the may le left to the /eader. He which linear WcheB theorems to the case of a point at
the
touch.
each of these sections This has been given at the end of
of illustration.
by way
Fig. 5.
ThP transformation of « For !, nf a curve
the dethis section does not alter
SH
(§8).
•
the type of §
1,
equivalent to three transformadeficiency none of which alters the
it is
iturn of the
Applications of this Transformation. method of § 11 i. that the posiThe main advantage of the *or instance, disposal. "t +>,p^lin line GA is still at our
§ 12.
j»
1.
PSci Is a tangent
wS
not the
,
A
'
case In §
Much simplification in arithmetic may
other than
We may h Cthro without loss of generality which
A
or
be thus secured.
o, B, at ,' fo, .** not at A Je Ke rfTf-d^LTn ot e
«.
rt "f 5
S read d
B
'
for
'
other
tlia
11
B
'.
In
134
If
APPLICATIONS OF TRANSFORMATION IX GA
is
12
a tangent at -C, the transformed curve passes It is convenient to take GBA instead of ABC as This is triangle of reference for the transformed curve. equivalent to interchanging x and z in the equation of this transformed curve, which now becomes / (z, y, z 2/x) = 0, the 0. original curve being / (x, y, z) Writing it in the form f(x/z, xy/z 2 1) = 0, we see that, if the Cartesian equation of the original curve is f(x, y) = 0, the transformed curve is f(x, xy) = 0. Points oif{x, y) — near the origin correspond to points of (x, xy) = near the axis of y, and points of one curve near / the axis of x correspond to points of the other curve also near the axis of x. If the nature of any singularity of the transformed as origin curve is not immediately obvious, we may take
through A.
=
,
H
H
and repeat the transformation.
Ex. 1. If CA is the tangent at the cusp C of a curve, AB is an ordinary tangent to the transformed curve at A. More generally, if CA is a tangent at C to a superlinear branch of order r, AB is a tangent of r-point contact to a linear branch at A.
[y
icy
r
=axr+ + bx 'y + cx''-' y*+
'
,
i
...
+kyr+1 +
...
...
becomes,
....]
on
putting
for y
and dividing by xr r y = ax + bxy+
,
+ kxy'' +1 +
Ex. 2. If CA is a. tangent of r-point contact at C to a curve, a tangent of (»•— l)-point contact at A to the transformed curve.
Ex.
3.
CA
is
A
common
tangent.
quadruple point of a curve consists of two cusps with a Find the effect of this point on class, deficiency, &c.
[Taking C as multiple point and CA as tangent, the transformed (Ex. 1). have curve has two linear branches touching at shown that the effect of such a ' tacnode as is to lower the class by 4, and also the two tangents from to the transformed curve touch &^A. Therefore the effect of the quadruple point on the class is to lower it by Again, the tacnode lowers the deficiency by 2, so 18. 4 + 2 + 4.3 A that the quadruple point lowers the deficiency by 2 + 4-4.3 8. curve adjoined to the transformed curve at has a linear branch touching at A, so that a curve adjoined at C to the given curve has a, cusp at C with CA as tangent. The tacnode is latent in the quadruple point C, which may be considered (Ch. VIII, § 5) as equivalent to 8 nodes and k cusps where 8 + <c 2. 8, 2 8 + 3 k 18, or 8 6, k The tangents from C which touch at C are 2.4 + — 2 6 in number.]
'
Hence C is a double point with a latent cusp. The existence of a rhamphoid cusp lowers the deficiency by 2 and the class by 2 + 3 = 5. An adjoined curve has a linear branch touching CA at C. The rhamphoid cusp may be considered as formed by uniting an ordinary node and cusp. Since the reciprocal of a rhamphoid cusp is also a rhamphoid cusp (Ch. VI, § 5, Ex. 1), it may be considered as formed by the union of a bitangent and inflexion. The number of tangents from C whose
points of contact
lie at
C is
4.]
-
Ex. 5. Discuss the nature of the origin for y 1 = x 2n+1 [Put yx for y and use induction. There is a latent cusp and n - 1 latent nodes. The origin lowers the class by 2.n + 1 and the deficiency by n. An adjoined curve has a linear branch with y = as a tangent of n -point contact.]
Ex.
6.
Discuss the origin for the curves
(i)
y^
+ ixVy + x
1
^=
.
0.
(ii)
yv
=
x"ti l
.
(iii)
(y-xv)v
= xPf1
[Put yx for y and use induction.]
Ex. 7. Show that the effect on Plflcker's numbers of a double point at which two linear branches have r-pomt contact is the same as that of
r nodes and r bitangents in general. [Use induction.] Ex.
8.
Find any latent double points of the following curves
origin; where
u^x+y
u*
it?
at the
1
.
{i)
(ii)
= y*(px* + qy*). = xy
3
.
(iii)
(u-OCy*)
{n-Py ) (u-yy2 )
1
= k(xu-y
3
)
2
-
[(i)
Two
latent nodes.
(ii) A latent node and cusp. As another example find the double points latent in its Hessian at the origin.
(iii)
Ex. 9. JTbe order of a superlinear branch at C (§ 11) is equal to the order of the transformed branch plus the number of tangents from B to this transformed branch which coincide with BA.
If 2 a >/3 > a., [Put yx for x in the expansion (i) of Ch. VI, § 3. apply reversion of series (Ch. VI, § 1) and the theorem at the end of Ch. VI, § 5 to the transformed expansion.]
' '
Ex. 10. If
I is
the order of any superlinear branch of a curve,
2(D-1) = m-2n + 2(l-l),
the summation being taken over
curve.
all
the superlinear branches of the
[Repetitions of the transformation of § 11 will transform the curve eventually into a curve with no superlinear branch, for which the result
136
is
APPLICATIONS OF TRANSFORMATION IX
But Ex. 9 proves
12
readily seen to be true. by such a transformation.
m-ln +
'S.
(I
— 1)
unaltered
We may
if
define the deficiency as
we
Ex.
choose.
11.
This
is
£{m-2w + 2(/-l)} + l, an alternative definition to that given in
2,
§ 7.]
Prove Ch. VI, §
2,
Ex. 7 by quadratic transformation.
[Use Ex.
Ex. 12.
§ 3,
and induction.]
Two
1
curves having at
11.]
K(K >
at C. case in
C
k)
with a
common tangent meet
superlinear branches of orders k and k (K+ 1) times at C.
[Use Ex.
Ex. 13.
and
Two curves have How many of their
which two or more
a common tacnode C with a common tangent intersections coincide with C? Discuss the of the branches at C osculate.
Ex. 14. Discuss similarly the case in which both curves.
C
is
a rhamphoid cusp of
)
should
*(<)
<t>(tj
^(0
f(h)
f(t2 )
=
o
.
.*
* We assume throughout that in general to each point on the curve corresponds a single value of "t...
fc-fg) (*. and that the parameter of the point considered as lying on the other branch is ty The equation (iii) must be satisfied for all values of t 2 since the points with parameters t and t x
find the real nodes.
(iii).
Making ^ approach
£C
we have
2/
the equation
/(<)
4>{t)
<t>'{t)
+(t)
*'(*)
=
o
..
.
to
This will give us equations in t and t x which are equivalent two equations to solve for t and £.
t. which we will call the 'parameter of the point. See § i. y.
. t lt t 2
(ii)
/(*)
for the tangent at the point
with parameter
The condition
be collinear
is
that^the points with parameters
f(t)
/(t.
Suppose the coordinates of any point (x.-<)(*-<!)
f(t 2 )
4>(t 2 )
that the parameter of a point considered as lying on one branch of the curve at a node is t.CHAPTEK X
THE PARAMETER
§1.
To
we suppose
coincide. by the equations
The equation of the
t.
t. 13.
_
Point-coordinates in terms of a Parameter.
line joining the points with parameters
£j is
evidently
y
z
f(t)
t
=
(i)-
t
-t
£.. Ex. z) of a curve given as functions of a quantity" t.
'
*
\r
point with parameter T. If the values of
a crunode
. But
. if 2
is real. .
/(*)
+"
r
V'
=
.f(t) For suppose the tangent (ii) meets the curve again in the
Then
*(<)
«/>'(*)
*(*)
*'(*)
0.
for
f'Wf(t)
=
4>W<l>®
= M)/W)
lc
•
•
(
iv )-
For if the values of the parameter at a double point are iandij.
. i.
. y.f{t)}/(T-t)\ &c.f(t)}/(T-t)> =f\t)
which proves the
* It is
result. t the node is a real acnode. one approaches t. *'(*)/*(*) become equal to the limiting value of (k— l)/(*i — <)• The inflexions are given by the equation
F(t)
= \f
1/
*
</.
.* the node is x on giving a small increment to the value of t or 1 1 we obtain a real point on the curve close to the node. suppose the tangents at the double point approach coincidence. If the values of t and t are conjugate complex quantities. e.
(v).
i//
are real.
Now
value of
if
T
the point of contact approaches an inflexion. without
altering its value.value theorem.
L^.138
USE OF A PARAMETER
similar
(i)
.
fihVM =
Therefore
0(*i)/*(«)
= f(Wf(t) =
(say).
.
Now make tx approach t..
1
-t])/f(t).
(/>. The cusps are given by the equations
determinate. z in
is to equate to zero the coefficients of for the line joining two coincident points is in-
t and t thus obtained are real.
f(t)
f(T)
0(D
+(T)
row
of the deter-
We may
minant by
replace the elements in the third
2{f{T)~f{t)-{T-t).
supposed that/.. denote f(t). by the mean. 2 {f(T)-f(t)-(T-t).
where /./'. Then
all
(k-iy^-t) = {/&)-/(*)}/&-«)/<*) =f(t+o[t where 1 > 6 > 0.
f(t)/f{t) and similarly *'(*)/*(*).
X
1
A
method
x.
•
2. If /> 0> are polynomials in t. are also given by The cusps are given by the values of t satisfying * both F(t) = and / * + F'(t)= =0.
+ ^(t) + "ylr(t) =
n
If this equation gives values of t corresponding to exactly distinct points on the curve.
If
a tangent to the a = by equation (ii) of
is
Xx + fiy + vz = curve in § 1.
of §
is
1. Then
An
f(T)
4>(T)
*(<)
tf{t)
+(!)
*(*)
+'(t)
{T-tf
f(t)
=0
. for
0. the curve is of degree n.X2
The
cusps.
= ff-ff. we may
^
§
1. and the inflexions by the single
= ^
roots.'"
while the parameters of the inflexions satisfy F(t) = 0. fi. but not F'(t) in general.
Line-coordinates in terms of a Parameter. For instance. Just as we obtained the parameters of the nodes and cusps and the degree of the curve.
(i). v=w-f'<i>
A.
The degree of the curve noting that the line
meets the curve where
\f(t)
(if it is
algebraic) is obtained
by
\x + jiy + vz
=
0.
alternative method of finding the nodes and bitangents Suppose that the tangent (ii) of § 1 meets the as follows.
USE OF A PARAMETER
139
which f'/f = 0'/0 = yjr'/yfr.
f{t)
Write down the condition that
this equation in
T has
equal
. so by using the line-coordinates we may obtain the parafneters of the bitangents and inflexional tangents and the class of the curve. v of any tangent to the curve are expressed in terms of the parameter t.
take
W-0>. the inflexional tangents will be given by A '/A = (J-'/fi which correspond to equation (iv) v'/v
Hence the line-coordinates
=
. the cusps are given by the repeated roots of F(t) — 0.
F(t)
=
f
&
y
/'"
$'"
xj. curve again at a point whose parameter is T.
l.140
roots. we get an equation giving the on the curve from which two parameter T (^ t) of a point coincident tangents can be drawn not coinciding with the
P
tangent at F.
If we write down the condition that (i) considered as an equation in t has equal roots.
USE OF A PARAMETER
X2
the obtain an equation in t whose solutions are contact of the bitangents and the parameters of the points of points of contact of the tangents from the cusps.
.
We
Fig.
v If the tangents coincide in a bitangent.
. As an alternative method notice that the tangent at the point with parameter t meets the curve again at a point with parameter T where
2{t-l)T i + 2(t-l)(2t + l)T-t[2t+l) =
.1 and ^ give the parameters of the points of contact of the
bitangent. bitangent.
Ex. degree.
inflexions. v = -\ as the values of u and v required.
= t+T. Hence t = .X
an
2
It is clear that
USE OF A PAEAMETER
141
is either an intersection of the curve with inflexional tangent.
t
The equation of the
!
line joining the points with parameters
and
T is
6PT (t+T+l)x+{t+T-2tT)y
The
t
coefficients
(
all
+ 6a{(l-2tT)(fi + tT+T i) + (t+T)(t-T)i } = 0.
=
m(2m + 1)(m 2 -2m-2)
=
0. we see that T = 0.
1 y = -6(t + t)a. for instance. vanish if t+T=-l. the denominators of these Eliminating v we get fractions vanish.
= \ —1 +
The
2
>/%) as the parameters of the node. i.
T meet
at the point
y
•»(2w 2 -4d2 +
-2u + 2uv + 2v + l
where u
m-2»)
9a
-2u + 2u'v-u + 5uv-2v + v
s 1
l
tT. or is a node.
is
This equation is of the fourth degree in t.
Find the double points. so that the curve fourth class.
and obtain « = -|. In the latter case the coincident tangents from are the tangent to the other branch of the curve through the node.
3a.
. the tangent at any point is obtained by putting 1 for z in equation (ii) of § 1.
of
1. If in the original values of x and y we replace t by \/T and repeat the process.
. If the curve is referred to Cartesian axes. e. The inflexions are given by
2t 2 -2i-l
=
or
*
= l(l±v/ 3).
(ii). t = a cusp.
gives occurs twice on the left-hand side.
t. the coordinates of
P
P
any point being
the results of §§ 1.
. This gives 2tT = -l. The equation (v) of § 1 giving cusps and inflexions is
of the
f(2P-2t-l)
Since the factor
t
= 0.
+ l)x-{t-l)y-9t(2t2 -l)a=0. where
This equation is of the fourth degree in Any tangent to the curve is
s St (2t
so that the curve is a quartic.
and
class
x
[The curve meets
=
(2t-l)a/t 1
~Kx
+ py + a = {2i-l)\-6(t i + f)^ + t 2 =0. 2 evidently still hold with slight modifications .
. t = <x> also gives a cusp. tangents at the points with parameters *.
)
[Each intersection of two constituents of the «-ic counts as a node of the n-ic]
* If the n-ic this statement
was degenerate.
2. Hence in this case
D>0
. &<?.
we have subjected the (n — 2)-ic to 2^k(k-l) + (n-2) + D = |(m-2)(%+l)
II. An (n — 2)-ic can be drawn with a (k — l)-ple point at each /c-ple point of the w-ic.]
Ex. IX into a curve with ordinary multiple points without altering the deficiency.]
An n-ic cannot have k + k 1 + k s + ki + k5 >2n. §
6
But the
7i-ic
and (n — 2)-ic meet in
n(n-2)-2k(k-l)-(n-2 + D) = D
other points.
2216
L
.
3.
if
l
points.
cannot have two multiple points of orders k l and & 2
.
1
D...
in(n-l)-(n-r)-(D + Dl +
+£>. and also passing through n — 2 + D other fixed points of the n-w. t The summation is extended over all the multiple points of the n-ic.
multiple points of orders k 1.
1.
if &!
D>0.
An
n-ic
+ k2
> n.. Ex.
[Consider the intersections of the w-ic with a p-ic through the triple
points. 1. 1 . 3 and Ch.
also
Ex.
[Consider the intersections of the n-ic with the conic through the
points.. § 3
D = i(n-lH»-8)-2**(*-l). It should be noticed. If the curve has higher singularities.
.
.
If
an
«-ic
degenerates into r curves of deficiencies
D1} D2
ithas double points.*
It follows that
n(n — 2)
The deficiency of a non-degenerate curve cannot be negative.
[For otherwise the line joining the points would meet the curve in
more than n
Ex. since the n-ic and (n — 2)-ic meet
points. VIII.
4.k
!i
. that there may be other restrictions on the nature of the multiple points of a nondegenerate curve in addition to (see Ex. 5 to 7 Ch.. k 3 .
..
If a non-degenerate n-ic has
\p (p + 3)
triple points. 2. IV. k i. We have so far supposed that the w-ic has only ordinary nodes and cusps. Ex.k i . For by Ch.t
and by Ch. XVII. and would not be correct. however. Now let it have ordinary multiple points. it may be transformed as in Ch.. § 6. so that the previous argument still applies. the (n— 2)-ic might be part of the m-ic..:
•X 3
DEFICIENCY NOT NEGATIVE
in'
145
other points.
conditions.]
Ex.). § 7.
it consists of n straight If it has \n(n-l)-l nodes. obtaining an equation of degree Of the roots of this in x with coefficients rational in t. In a similar manner the ordinate of is expressed rationally But. curve of zero deficiency is often called a rational or The former name comes from the fact unicursal curve.
DEFICIENCY NOT NEGATIVE
«-4
X
3
If a degenerate ra-ic has \n (n . where u
. it consists of n-2 straight lines straight and a conic. e. Suppose Consider now an n-ic f that a variable iV-ic is subjected to certain conditions. If it has \n{n-\)-2 nodes. it consists of lines and two conies.
Unicursal Curves.
the curve of § 1 is of zero deficiency. The latter name comes from the fact that the curve can be drawn by a pen which never leaves the plane of the paper
This implies that. the coefficients can be expressed linearly in terms of a single quantity t. giving the abscissa of P.1) nodes.
Dividing the equation by the factors corresponding to the abscissae of the fixed points. If Cartesian coordinates are used (a similar process holds good for homogeneous coordinates).
the jV-ics form a pencil.
lines. and the remainequation of the w-ic ing root is the abscissa of the other intersection
=
=
nN
nN—
P
and
i\T-ic.
. that the coordinates are expressed rationally in terms of a parameter. Thus the equation of the iV-ic is of the form are two such iV-ics i. 4.]
§4. Hence
P
P
The coordinates of any point on a curve of zero deficiency can be expressed as rational algebraic functions of a parameter.:
146
Ex.
5. Suppose also 1 times at these fixed points. Since the %N(N+ 3) ratios of the coefficients in the equation of the i\T-ic are subjected to %N~(N+3) — l linear relations.
[See Ex.
f(t). and v u + tv 0. may be made any point of the n-ic. by choosing t properly. that the i^-ic meets the n-ic We assume for the present that such a variable iV-ic exists. we have an equation of the first degree in x with coefficients rational in t. we may eliminate y and u + tv between/ 0. such as passing through fixed points of the n-ic or having assigned which conditions are singularities at fixed points of the n-ic in all equivalent to |iV"(iV"+3) — 1 independent linear relations between the coefficients of the equation of the iV-ic.
=
nN—
=
=
=
. be taken as polynomials in t. in terms of t. 1 are abscissae of fixed points.
with zero deficiency.
if
and
\jr(t)
may
A
. <p(t).
i. i. §3
.
* See Ch. is an acnode. the point moves continuously as t varies (provided is finite). the n-ic.
times at the fixed points as required. §i\T(i\T+3)-l and the if-ic meets the n-ic
2k(k-\)+n-3. and subject the JT-ic to the conditions that it shall have a (/c — l)-ple point at each /c-ple point of the n-ic. P is also given by t a — /3i. and also pass through n — 3 other fixed points of. II. Or as an alternative we may transform the n-ic by § 10. or to insert the acnodes in other words.JV-1
. e. is a double point. First suppose the n-ic has no multiple points other than n — 2.
We now show that there always exists an i^-ic with the properties stated at the beginning of this section. where a and /3 are real. and subject ordinary nodes and cusps. This would not hold if t were complex but.e. «.
Lastly suppose that the n-ic has higher singularities. and use the result of Ch. e. IX. 9.g.
147
except to pass from one end of an asymptote to the other. curves of deficiency other than zero may consist of a single circuit . i.
nF-l
as required.
l2
.
2JJb(A-l) + »-3.+ (Hi gives evidently an unreal point of the curve. We take the iV-ic as an adjoined (n — 2)-ic passing through n — 3 fixed ordinary points of the n-ic. Since any value of t close to tx . XX. Of course. §§ 1. f It is often convenient to take them as points on the n-ic consecutive to double points.
2)=$(n-l)(i»-2). coefficients in the equation of the iV-ic are connected by
N=
i(»-l)(m-2) + (n-8). y i x+ (x — 1) (x2 + 2x + 2) = 0. + pi. § 6 are connected by
linear relations. the curve consists of a single circuit.-2£&(*-l)
= 0.
P
P
. i. if a real point on the curve is given by t — a. Then the coefficients in the equation of the JV-io by Ch. the coordinates of a point P on the curve being continuous functions of t.* In fact.X
4
UNICURSAL CURVES
. Next suppose the «-ic has ordinary multiple points as well as ordinary nodes and cusps.
i. e. We take the i^-ic to the conditions that it shall pass through each of the %(n — l)(n — 2) nodes and cusps of the n-ic and also pass Then the through to — 3 other fixed points f of the n-ic. $N~(tf+8)-l and the N~-ic meets the n-ic 2 x %(n-l) {n-2) + (n-3).e. We take ZV = n — 2.e. VIII.e.
times at the fixed points For by Ch. there is no real part of the curve in the neighbourhood of i.
P
=
P
P
P
.
linear relations
.
[The tangent at
D = 0. VIII.
+& =
giving
a
2V
= iN(2n-N-S). the highest common factor is (t . § 5.
by
l/(t
the curve has a cusp given — OC). while
Ex.tt> + i. t — (X.
7.
Ex.
2(i
and N-ia are « double
raiV-l. changed in sign by interchange of t and t t
. if T is the ratio of the coefficients of any two powers of tx and is not a constant.
b
= N(N+3-n)-l. <t>. i? are polynomials of degree n with the coefficients of t n " x zero.] y»& = xv+z. Suppose that to each point on the curve corresponds r values of t..]
Ex.l) Hence is (p + q) Wx =py + qtP+iz. 1..
.
(f>. But the expressions whose highest common factor was taken are only See Liiroth. (t .
See Clebsch. 8i Three points on a curve have parameters t lt t z ts Transform the parameter so that the points have new parameters 0.] to make zero the coefficient of I "
. K = i=p + q — 2. p.
Conversely. where /. § 9.
UNICURSAL CURVES
or
151
u—1
n — 2. The coefficients of any two powers of t in this are symmetric functions of tlt t 2 .
11. . and that the highest common factor of
:
=
. 13. 4>. so that to each point on the curve corresponds a single
is
value of T.1
... Then.1. Ex.)..
p. where /3 is chosen
Ex. = m=p + q.
m
are given
by
§§ l'and 2. The curve of Ex! 9 has a cusp.]
.
If in §
4 the
»-ic
has only ordinary cusps and nodes.
[The new parameter
is
(t-t-y) (t2
: :
-t3 )/(t — t 3 )
: :
(£ 2
— tj). 1.
See also Ch. take as a new parameter T=fi + 7 -1 in a.
Hence
> »-3. Then
a + 6=iJV(JV+3)-l.
:
y
:
z
=/(<) :<f>(t):^
(t)
[n
and
12.]
:
:
Ex.
X
4
Ex. For other examples find the Plucker's numbers of the examples in
Ch.
yjr
Find the Plucker's numbers of x being polynomials in t.
/(*)*(«i) -/&)*(') and *W*(*i)-*(*i)*(0 arranged in powers of t lt the coefficients being polynomials in t.
«-ic
N must be
points
[Suppose the fixed intersections of the and 6 other points of the w-ic. Annalen.. lxiv (1865).
f. b = r =>\{p + q-2)(p + q-2. Math.tj (t . 44.2 ) .). t. In the «-ic x y z =f(t) <b(t) \|? (*).
Ill. curve with a cusp can be put in this form.
ix... 9. of t. . and therefore their ratio has a single value for each point on the curve.. to each point on the curve corresponds in general a single value of t but this is not universally the case. the sum of the parameters of the intersections of the n-ic with any algebraic curve
is zero. Oe«e. yjf are polynomials. 163. any unicursal
if
[The cusp is given by t = oo Conversely. In the curve x y z f(t) <j> (t) : yjf (t). 10.
[If to a point on the curve correspond the values t lt t 2 .]
.
n
(P. oo
. t r .
.
. x:y:z can be expressed rationally in terms of the parameter T.
Ex.
Find the Plucker's numbers of
D = 0. where /.
and sin so that the parameters lt <£".
Putting
t
=
tan -
. rational' expression in terms of cos <£ and sin </> The following theorem is of interest
The coordinates of any point on a unicursal n-ic (n > 2) an acnode can be expressed rationally in terms of cos $ .]
§ 5. 0.
<j>
<f>
.
H and K being fixed points. Here the most convenient parameter is the eccentric angle of any point.
HB
KC
Ex.
it
by
Suppose that the triangle of reference
(1.]
Ex. of the inter2 sections of the n-ic with any r-ic satisfy the relation
with
<j).
. 24.
Ex. 25. B.
We may
is taken so that suppose the parameters of
. The case of the ellipse is well known. it may be more convenient to express them rationally in terms of cos <£ and sin </> or of cosh $ and sinh (f> where <f> is a new parameter. 22. The locus of the poles of any normal to a given conic is in general a unicursal quartic. q.
we can immediately change from
.
A
Any
[Its
node
is
at the vertex of the pencil. r are linear in x.. y.
<t>i
+ $2 +
and
•
•
•
+ §nr
cos
iri.
the
rational expression of the coordinates in terms of t to the or conversely. 21.:
X
5
COORDINATES OF A POINT
153
Ex. 23.
sin
"")•
(p
For an n-ic with a crunode
cosh
<j)
and
are replaced by
and sinh
0. The locus of the intersection of the tangents at corresponding points of homographic ranges on two given conies is a unicursal quartic.
AP
[A unicursal cubic and quartic]
Ex. find the loci of the intersection of the line through A perpendicular to with the tangent and normal at P. Show that the^ locus of the intersection of and is a unicursal quartic. pt? + 2qt + i-=Q. Q.
—°
</>
(wwd. If A is an end of an axis and P any point on an ellipse. Consider the case in which the conic is a parabola.
[Th* tangents can be put in the form Pl 2 + 2Qt + B=0. s.
. where P. 0) is
the acnode..
range of points on a conic is homographic with a pencil of line of the pencil meets the tangent at the corresponding point of the conic on a fixed unicursal cubic..
lines.
Instead of expressing the coordinates of any point on a unicursal curve rationally in terms of the parameter t. p. A triangle ABC of fixed size and shape turns about C.
Coordinates of a Point in terms of Trigonometric or Hyperbolic Functions.
=
1
and — 1. we have at once . + (t* + 1) ([6oA i + c If s h is the sum of the products of the roots of this equation at a time. For <p t 2 in this case also.) (Sl -s 3 + s 5
'i/]
li
-r(a
-a a + a4 -.
If a variable N-ic (N<n) passes through |iV(iV+3) — fixed points of a unicursal n-ic (as in § 4) and meets the n-ic 2 times at these fixed points....+<i) n
)
= = (-a + a -a
1 3
..
nN—
*
coinciding with one of the fixed points..
Hence... = 1...-ic. any n collinear points Similarly the sum of the parameters of the intersections of the n-ic with any r-ic is a constant (mod..
(mod.
y
=
(t
=
(t*
+
l)(c tn -*+
+ 1) (b t n ~ 2 + .
if (x.
.
or
^+
where
tan a
(
(^2 ... Let <f> 1 (f> 2 be their parameters and let /3 be the
. <j) 2 . if any straight line meets o^.. One (or both) of the N-ics here obtained may meet the w-ic at a cusp instead of touching it.) = 0..
..1
154
COORDINATES OF A POINT
X
5
the acnode to be i and-— i (see end of §4). That this constant is zero is evident by taking the r-ic as r straight
lines..
x=(a
t«
+ a 1 P.1 +
z
.)-i-(a
-a2 + ai -.. and put t — tanh <£.. equations of the form
Then we have
. two of the family of i^-ics touch the n-ic in general* For the n-ic and iV-ic meet at two more points. y.)-=-(«" — a 2 + a i — 6 + a/n. n)..
2
+6"_ 2 ). For the corresponding result in the case of a cusp see
If the point (1. + c n _ 2 ).
= — c^ +
by
Replacing
of
ftg — a + .... 0) is a crunode. <f> n the n-ic in the points for which <j>
t&n(<l> 1
+
<l> i
+. so that..) = (-al + a3 -a + .). +an) 2 3 i»" + [6 l/t + c^] i»" +. The n-ic meets any straight
line
Xx + fiy + vz
where X(a
<"
J
=
1c
+ a.
we
take
its
parameters as
§4.... 0. it).. x:y:z are expressed rationally in terms of cos <f> and sin <j>. z) is
First take r
any point of the -n.i' .
+a"). we have the sum of the <£
.
At a point not
—
</>
.).. 10..
parameters
(mod. Ex.1 + .
tt).
6
+. if we put tan </> for t.)..)-r(l-s2 + s4 ...
§6... +<f> n
=
a.
. VII.
This process gives a method of expressing the coordinates of a point on a unicursal curve rationally in terms of a parameter alternative to that given in § 4 which.
2{J(»-l)(»-2)-2} =n(w-3)-2. IX.X
6
COORDINATES OF A POINT
155
of the parameters of the nN— 2 fixed intersections.
if
a and ac — b are
2
positive
and similarly in other
cases.
N
.
If the JV-ic touches the n-ic.
<f> 2
it). whose coefficients are rational in t. 11 and 12. is gives the two N-ics which touch the n-ic. 1 and 2.
. if homogeneous coordinates are used). If the multiple points of the n-io are all ordinary nodes or cusps. Ex. may be much easier in practice.
. Since of the second degree in t. the N-ic may be taken as an (n— 3)-ic through all the double points of the n-ic but two since
. § 10.
x where L.
by the sub-
=
2((Xc-6 2 )lr-r(T 2 -l). out by the factors corresponding to the nN — 2 given intersections. we have two equations for y which have a
M are rational in
X
common solution of the form M' + L'X~z.
i(n-3)n-l =i(n-l)(n-2)-2.
:
-f/3 follows from Ch.
^ = and or = -f/3 + 4tt. t. and Substituting either of these values of x in the equations of n-ic and N-ic. if less simple in theory. giving the two variable intersections of n-ic and JV-ic.
in terms of a
stitution
new parameter T +b
for instance. Ex. The coefficients in the equation of the i\T-ic may be expressed linearly in terms of a parameter t On eliminating y between the equations of the n-ic and iV-ic we get an equation in x Dividing (or in x z. Solving it we have
. Hence the coordinates of any point on the n-ic may be expressed rationally in terms of t and an expression of the
X=
X
form
(at 2
t
But
and
+ 2bt + c)k (at i + 2bt + c)^
at
can always be expressed rationally
. since the value of may be lower than in the method of § 4. Ch. Then
sum
</>i
+
<t>2
+P = °
(mod. we have a quadratic equation in x. where M' and L' are rational in t.
*!=
This result also § 10.
= M+LX?
is a polynomial in t. the parameter of any point on the n-ic being taken as in § 5.
The second N-ic of § 6 is the line at infinity for this line passes through the infinite cusp. when we eliminate y between the equations of the w-ic and iV-ic. In the argument of § 6 must be n — 1.
S
. Then. Suppose that just as in § 4 a variable .
[A variable line x + y + \
curve where
=
through a node at infinity meets the
2(X-2)(X + l)y
' '
= -X + 3X + X (3-X)*. . in the equations of
.y^f .]
Ex. 5. Dividing the equation by the factors corresponding to the abscissae of the fixed points we have an equation of the second degree in x with coefficients rational in t whose roots and Q.
N
Ex. [As in
§ 4. the N-ic may be taken as a straight line through the (n— 2)-ple point and so on.%a? + 6 xy + y* =
rationally in terms of a parameter.
3. we get an equation in x 2 of whose roots are the abscissae of fixed points.
2.y ) (x + Sy) .
Apply the method of
§ 6 to the
examples in
§ 4.]
Put
\=t — 1.the two variable intersections P.
Express the coordinates of any point on
(x
+ yYz2
=x
3
(y + z)*
rationally in terms of a parameter.
Ex. Ex.
Ex.
1. 4.156
COORDINATES OF A POINT
.
X
6
If the %-ic has a (n — 2)-ple point. and' Q of n-ic and iV-ic. that the iV-ic meets the n-ic But suppose times at the fixed points (not 1 times as in § 4). while the remaining two roots are the abscissae of . or n—3 in the case in which all the multiple points of the w-ic are ordinary nodes or cusps./V-ic u + tv is subjected to certain conditions which are in all equivalent to \ N(N~+ 3) 1 linear r relations between the coefficients of the equation of the A -|c.{x* . 3.
Curves with Unit Deficiency.
[y = \x meets the curve where \x/z = — 1 + (X+ 1)^.
7. and we have (t 1 — 1) x = y = (t s -l)z as in §4.
Consider now an n-ic with unit deficiency.
Express the coordinates of any point on the curve 2 1 (a.
t
M and L are rational in
and
X is a
x
we
substitute either of these values of
polynomial in t.
4. Ex.] Ex.]
§7.
2
Ex. Suppose its solution is are the abscissae of
=
—
nN—
nN—2
nN—
P
x
where
If
= M±LXi.
Putting X = 3 — (t+ l) a we get the coordinates given in § 4.
[As another example the reader may take the general conic or the quartic with nodes at the vertices of the triangle of reference. n — 2.
XV. Ex. cnu.
a & g3
1
«"
we
have. 1). Ex. Ch.. or Goursat. p. § 10.
the values of t given by are the parameters of the points of contact of those of the variable iV-ics which touch the n. If then we assume the existence of the iV-ic. positive. But by Ch. IV.
L
=
w
^
«2
^=
1.
we take a
as a constant defined by
3
.
where M' and L' are rational
in
t. defined by
. VII. dnu or if preferred. putting p 2 and 1. IX.ft B + 4osft + a 1 is See Halphen's Fonciions elliptiques'. well known that t and can be simultaneously expressed as rational functions of the elliptic functions snu. Hence is a polynomial of degree four in t.
For instance. § 10. Ch. §§ 2 and 4. §§340 and 362. on putting
ai a
{a
t i
3 2
. Traite A' Analyse. II.
i
P'u-p'a
p-u — pa
a<?
+ 4.
* If o is negative. put t r _1 + fc where' ro fc* + 4<i 1 fc 5 + 6o.aI t + 6a 2 t + 4'a st + ai }i =
{pu-p(u + a)}*
Hence: The coordinates of any point on a curve of uniUdeficiency can be expressed rationally in terms of elliptic functions. 120. we get is also of the form
whose common root
M' + L'Xh. Ch. Ch. and proceed as above.-ic at a point other than one of the fixed points. §§ 12 and 16. XVII.
a
2
. a
p'a
a 2 a3 — 3a a1 a 2 + 2a 1 3
*0
. we have
Now
X=
=
D=
X
may
The coordinates of any point on a curve of unit deficiency be expressed rationally in terms of a parameter t and an
expression
X
of the form
{
t*
+ 4a 1 £ 3 + 6a 2 i 2 + 4>a s t + a4 }i
It is. Corns d'Analyse. we see that the number of such points of contact and cusps is four. II. in terms of Weierstrass's elliptic function pu and its derivative p'u.'1.
=
'
.
where
a"
2
3
a o2 9*
=
a ai -4a1 a3 + 3 a
2
. and the parameters of the cusps of the n-ic which are not included among the fixed points.
pa = — (a a2 — a-f). 11 and 12 (cf.
X
(p'u) 2
=
4(pu) s -g2 pu-g3
if
. The reader may also consult Picard.a
:
X
7
CURVES WITH UNIT DEFICIENCY
two equations
in y
157
n-ic and iVic.
158
CURVES WITH UNIT DEFICIENCY
—
X
7
To prove the existence of the N-ia.
x
= ty.]
.
terms of
[Consider the intersections of the cubic with a line through a fixed point of the curve.
rc-ic
[Consider its intersections with a line through a node with an (» — 2)-ple point. § 7 (i).
Express rationally in terms of elliptic functions the coordinates § 5 (i). we show that when the w-ic has only ordinary multiple points and cusps. if iV
—
=
D=l.
x
= tz. § 14 (ii). the coordinates of a point on any curve with unit deficiency can be expressed rationally in terms of elliptic
*
also argue as follows
:
functions.
3
2
1
)
)
0. IX.
Ex.
[Consider the intersections with
x=ty. 2.
y
=
tx. XVIII. Express rationally in terms of elliptic functions the coordinates of any point on
(i)
(ii)
(iii)
(iv)
(v)
(vi)
= y(y-x){y-k' x). {y-x) = (y + x) (a'x^ + Vf).
of
4.
any point on the quartics of Ch. Express the coordinates of any point on a quartic of unit deficiency rationally in terms of elliptic functions. however. if the %-ic has only ordinary double points.
z
=
t(x + y).
i(»-l)(»-2)-42*(fc-l)=
1.]
Ex. an (» 2)-ic with a (fc-l)-ple point at each /c-ple point of the 7i-ic and passing through m 2 other fixed points of the n-ic satisfies n — 2 and the conditions imposed on the i\T-ic. we may often take a lower value of JSf. y z = i» -g xz -g z x + y + z + 6mxyz = 0.
Express the coordinates of any point on a cubic rationally in
elliptic functions.x + cz (x* .
In general we may take for the iV-ic an adjoined (n — 2)-ic passing through n — 2 other fixed points of the n-ic* In practice.
z
=
t(x + y).
This follows from the second theorem of Ch.
n-2 + 42*(*-l) =! ^iV(JV+8) — 1. But we may Since the coordinates of any point of a curve with unit deficiency having only ordinary multiple points can be expressed rationally in terms of elliptic functions.
.if) =
xz I
i 2
(at + b i )
. we may take for the i\T-ic an (n — 3)-ic through all the double points but one.
s
s
t
%
i
i
. In fact. § 10. 1. ax (y2 . 3. Ex. For instance.
§ 15.
1.z2 + by (z .]
Ex. (x + y + zf + ^kxyz = 0. and any curve with unit deficiency can be transformed into such a curve by rational change of variables (successive quadratic transformations).
and
so for
any
Ex. tc-2 + 2A(&-1) = «iV-2.
If gf 27grs 2 the curve has a single circuit given by real values of the parameter u.p 2u. p u. 2. 6. y. y
=
=
where/.
(ii)
(iii)
(u . If # 2 3 >27p 3 2 the curve has a second circuit given by values of « for which u — &>' is real. A
.
pu.\
Ex. where 2o>' is an unreal period of Weierstrass's function. <
(iii)
y
=
tx.
[As in §
in
5. .
using Abel's theorem on the roots of an equation rational
pu
and p'u. When the coordinates of a point on a curve with unit deficiency are expressed rationally in terms of Weierstrass's function of a parameter «.. p'u.
and
B
being
Now
we
in
differentiating repeatedly the relation
(p'uY
express
2
= 4(pu) -g pu-g
3 2 3
.
3
. y\r are each of the form polynomials in pu. terms of pu.
5.
=
1. « 2 m 3 are of degrees
..
X
X
[(i)
Consider the intersection of the
w-ic
with a (« — 3)-ic through the
.
But there are exceptions
e. we have <)>(u).X
of
8
Ex. z) are homogeneous coordinates of any point on the curve.. where a polynomial in t.fy
2
)
(u .iff. The coordinates' of any point on an w-ic of deficiency 2 can be expressed rationally in terms of a parameter t and Xi. p"u. we may suppose the sum of the parameters of the intersections of the curve with any other curve to be zero.]
Ex.. p'n.
an
n-ic
with
a (n
— 2)-ple
point.
CURVES WITH UNIT DEFICIENCY
Express rationally in terms of
159
elliptic functions the coordinates
any point on a a (i) s?(a? + y )=x y>.. (ii) u
= ty*. p'u pu.
7.
nodes.
3
y. where is a polynomial in t of degree 5 or 6.
= A + Bp'ii.
If (x. z * f(u) f(u).
Ex.
linearly
&c. <j>. where u yz + x*. p'u.. where %. p u.g.. the curve is of unit
:
We now
deficiency. The coordinates of any point of a curve are expressed rationally in terms of Weierstrass's function.]
prove the converse theorem If the coordinates of any point on a curve can be expressed rationally in terms of pu and p'u. 8.
.
=k(xu.pu = -hp'"u. For instance pho = ^(2p"u-g2 ).
(ii)
As in
§ 4. .
Ex. 7.
[Consider the intersections with (i) yz = tx(y-z). .yif)
in
x and
z u s + zxyu 2 + x y 2 iii
2
=
0.Oty 1 ) (u
2
. It is in general impossible to express the coordinates of a point on is a curve of deficiency greater than 2 in terms of t and Xi.
.
.
The points of confact of any tangent through the point
(x. y. The curve meets Xx + /xy + vz = where
\f (u) + fi(j> (u) + vf(u)
=
0. 148) will show that the deficiency But /..160
CURV.
and so for ji.
<j>. y. Theory of Functions.. z are linear functions of pu and p'u. The argument of § 4 (p.]
Ex.
. y.
= 0. v.
§ 116.
= f $-/<(>'. so that A when reduced to linear form becomes of the type
A + A pu + A
l
p'u +
. and so for </>(u) and ijr(u)..ES
WITH UNIT DEFICIENCY
X
8
Hence we may take 2 f(u) = a + a pu + a 1 p'u+. 1)
:
:
*
Forsyth. ^ will not usually be rational functions of a single parameter. p'u.
and has
therefore n zeros.hi.
/*=*'/-*/*.
In Q'yjr — <j)%lr' the terms ^n p^-^u p&'Vu cancel.
+ A in _ i p^ n '^u. [We proved that with any choice of triangle of reference x. z) is
given by
\x + /iy + vz
where
(see § 2).
Show by means
as (pu.
v
A
= *>-**'. Hence three linear functions of x. Hence the class of the curve is 2n at most. is zero or unity. z can be chosen in the ratios pu p'u 1..+ an _ 2 p^.
The left-hand
side of this equation has
n
poles.* Hence the curve is of degree n.
of § 8 that any point on a cubic may be taken by a suitable choice of triangle of reference. Hence in general D = 1.
for instance its polar reciprocals.
M
.CHAPTER XI
DERIVED CURVES
§
1.
and 2 denoting partial differentiation with respect to the evolute of an algebraic curve is algebraic*
is
equation
just written.
In
determining
shall
curve curve
we
make
§ 2.
2216
and then replacing
obtained by eliminating x and y from the three equations f and rj by x and y.
Evolutes. it is sufficient to find the number of its tangents passing through a single point.=y+m 2 +f2 2)/(W2fi2-f22fn-A%2)-
•
suffixes 1
x and
* Its
y. the number of its intersections with every line is the same. &c. Since the centre of curvature at any point of the curve
f{x.
The evolute of a curve is the envelope of its normals. The case of the polar reciprocals has been already sufficiently discussed. the locus of the intersection of each normal with the consecutiye normal.
2
rj). it is sufficient to find the number of its intersections with a single line. Hence. Similarly to find its class. pedals. in order to determine its degree. if the derived is algebraic. We shall also determine the multiple points and any other peculiarities of the derived curves which may be of
interest. orthoptic locus.
Derived Curves. when the type of the original curve is given.
. y)
=
is (£. 2
where
2 2
i
=
X+fl(fl +f2
WflfJl2-f* fn-fl f22)> r.
Feom a given curve may be derived other curves by various geometrical processes. inverses. In this chapter we shall consider some of the other derived curves. evolute.
the Pliicker's numbers of any derived use of the principle that. In particular we shall concern ourselves with determining the type of the derived curve (as denned by its Plucker's numbers). and the locus of the centre of curvature at each point of the curve.
. and the line
which
. Hence R is a cusp of the evolute. Then P 1 R' and P 2 R' are the normals at Px and P2 so that through R' three consecutive tangents R'Plt R'P. Hence
n'
= Sn +
i. and meets mm' 3n times at these
. In this case let Q' be a point on mm' near P. k'.162
EVOLUTES
XI
2
t. Then
Suppose that
o>. and let Q'Pj. P'P2 can be drawn to the evolute^ all ultimately coinciding with mm'.
The only other
to the
t
infinite points of the evolute are those due inflexions of the given curve. m. QR) is harmonic.
P
P
P
P
Fig. t'.
The tangents
from R.
P
PR
i.
Let (mm'.
—
m' are the circular points at infinity and that the tangent at to the given curve meets toco' in Q.
e. while (o>o/. one for each of the n intersections of 0)0)' with the given curve.. D. S'.
1.
touches the evolute. m. and of its evolute by n'. m'. Q'P2 be the tangents from Q' to the curve which touch at 1 and 2 adjacent to (Fig.
m normals corresponding to the m tangents to the given curve
QR)
is
from a point Q on mm' are the
o>o>'
harmonic. and QR coincides with o)o)'. 1). We see then that the evolute has n cusps at infinity at which mm' is the tangent. D\ We now determine n'. k. where
to the evolute
(mm'. If is on oxb' it coincides with Q.
cusps. . and mm' meets the evolute in three points at R.
denote the Pliioker's numbers of the curve by n.
We
S. is the normal at P. m'. i'. Q'R') be harmonic. which is therefore a tangent to the evolute. in terms of n. at which mm' is a tangent.
t.
XI
2
EVOLUTES
163
counts for n tangents, since it has n points of contact, namely the cusps of the evolute on a>a>'. Hence
m'
Again,
it
= m + n.
evident that in general no two consecutive normals coincide hence the evolute has not coincident tangents at consecutive points, i. e. has no inflexional tangent. Therefore
is
;
+ p Ex. 3. Find the Pliicker's numbers of the evolute of a*y = xp i, p and q being positive integers prime to one another. [m'=p + 2q, D' — O; n' = p + 2q if p > q, »' = Sq if p < q. For other examples take the curves of Ex. 2.] Ex.4. The number of normals which can be drawn from any point to
a curve
is
m + n.
[The number is m'. Verify the result by consideration of the fact to the intersections to a curve are lines joining that the normals from of the curve with the curve obtained by rotating it through a very small angle about 0. See also Ch. XII, § 5, Ex. 12.]
Ex. 5. How contact ?
many
circles of curvature of
a given curve have four-point
ma>'
[Each such^circle
is
given by a cusp of the evolute not lying on
3 (ri — m')
Hence the number
K
'
is
-n =
+ i-n
=
hn — Sm + Si.~\
Ex.
6.
How many
lines are
2
normal to a curve at two points ?
[r'-\n(n-l)
=%(m
+ 2mn-4m-ic).] M 2
164
•
EVOLUTES
XI 2
Ex. 7. How many pairs of concentric circles exist, each of which osculates a given curve ? Ex. Ex.
8.
The
deficiency of a curve
and of its evolute are equal.
foci
9.
The evolute has the same
and
directrices as its involute.
[A tangent passing through a> or a' coincides with its normal. The curve and evolute touch at the point of contact of such a tangent.]
Ex.10. There are »»(»» + »— 4) normals to a curve which are also
tangents.
[The
o)
common
or a.
tangents of curve and evolute which do not pass through See Ex. 9.]
Ex. 11. Find Pliicker's numbers for the locus of the extremity of the polar subtangent of a given curve, being the pole.
[The locus respect to O.
is
the polar reciprocal of the evolute of the reciprocal with Consider the case of a conic with as focus.]
Ex. 12. Find the Pliicker's numbers of the locus of the harmonic conjugate of a variable point P on a given curve with respect to the intersections of the tangent at P with two fixed lines.
[Reciprocate and project the reciprocals of the fixed lines into a> and
a'.]
Ex.
13. If
(toco',
where
Ex.
at
<o
a curve touches aa' at P, the evolute has an inflexion at R, PR) is harmonic, a>a>' being the tangent at the inflexion.
a curve passes through a (or a), the evolute has the tangent
14. If
as inflexional tangent.
aa>'
Ex. 15. If a curve touches
in g points
and passes / times through
i'
each circular point,
«"= 3» + i-3(2/+£),
What
Ex.
a>a>'
m'
= m + n-{2f+g),
=2f+g.
modification must be
14.]
made
in the results of Ex. 4 to 7 in this case ?
[Use Ex. 13 and
16. If
the curve has aa>' as inflexional tangent, the evolute has as a tangent at a point of undulation, and so on.
and
Ex. 17. The number of normals common to two curves of degree n. w, class m, m 1 is mn n. 1 + J
mm
l
+m
[The normals are the finite common tangents of their evolutes. n t = 0, m t = 1 we have the result of Ex. !.]
Putting
§3.
Inverse Curve.
to the circle
The curve inverse to f(x, y) = with respect whose centre is the origin and radius k is
point of order n at each of the circular points co, co' * and has also an w-ple point at the origin 0. (See also Ch. IX, §§ 1, 2.) To find the Pluck er's numbers n', m', 8', k', t', i', 1 of the curve inverse to a curve with Pliicker's numbers n, m, 8, k, t, i, D, we proceed as follows. First of all, it is obvious from the above that n' = 2 n. Secondly, a cusp of. the given curve inverts into a cusp of the inverse curve, and vice versa, so that k' = k. or on <oa>') Again, a node of the given curve (not at inverts into a node of the inverse curve. Also the TO-ple points of the inverse curve at 0, co, and co' are each equivalent to \ n(n — 1) nodes, so far as their effect on Pliicker's numbers Hence are concerned (Ch. VIII, § 3).
n tangents each counted twice (Ch. VII, § 5) and the tangents from Hence m' = + 2n. to the original curve. The same result follows from consideration of the tangents from a) (or a/), remembering that a line through co inverts into a line through co'. The result If = D is a particular case of the theorem that two curves with a 1:1 correspondence have the same deficiency (Ch. XXI, § 3). We have proved earlier (Ch. V, § 4) that the inverses of the foci of a curve are the foci of the inverse curve ; and that, if is a focus of the curve, the inverse curve has cusps at
to the inverse curve are the
m
a>
and
co'.
Ex. 1. If the original curve has a fc-ple point at at each of a> and ta', we have
n'
and a #-ple point
= 2n-2p-k,
6'
= $(«-2p) (n-2p-l) + (n-p-k) (n-p-k-\) + h, k = k
;
and the inverse curve has a (m-2p)-ple point at points at each of <» and <»'.
Ex. 2. What modification touches (bo)' ?
*
and (n-p-k)-j>\e
3, if
is
required in the result of §
the curve
only
In fact any line x±iy = c meets the inverse curve whereas the curve i9 of degree 2«.
;
at
n
finite points
166
Ex.
3.
INVEESE CURVE
A
curve of degree
XI
<u
3
2N
with
.ZV-ple
points at
and
a' inverts
into a curve of the
same
type.
Ex. 4. Find the number of circles of curvature of a given curve which pass through a given point 0. Discuss the case in which lies on the curve.
[Invert with respect to 0.]
Ex. 5. Find the number of circles passing through a given point and having double contact with a given curve.
Ex. 6. Find the number of circles passing through two given points and touching a given curve. [Invert with respect to either point.]
7. An »-ic is self-inverse with respect to a circle j. Show that n even, and that the curve has a Jw-ple point at each circular point. Show also that the curve has \n(n — 2) foci lying onj in general, which are the intersections of j with the locus of the centres of a family of
Ex.
is
circles
having double contact with the
n-ple points at
...
,
.
ra-ic.
*
Ex. 8. A 2 w-ic has P meets it in Qlt Q
a
,
<» and <a'. Any transversal through Qtn Show that the product Pp =PQl .PQt :...PQ, u
;
is
independent of the direction of the transversal with respect to we have
p'
[(i)
and that on inversion
P ,=k
in
.p P /p
Use polar coordinates,
(ii)
OP" Take OP as the
1
.
.
transversal.]
§4.
Pedal Curve.
If OY is the perpendicular from a fixed point on the of a given curve, the locus of Y is the tangent at any point pedal (first pedal) of the given curve with respect to 0. The pedal of the pedal is called the second pedal, the pedal of the second pedal is called the third pedal, and so on.
P
The angles between the radius vector and the tangent corresponding points of a curve and its pedal are equal.
«orf
For let the tangents at consecutive points P, J" of the curve meet at T, and let OY, OY' be the perpendiculars from on
these tangents (Fig. 2). Then since 0, Y' Y, T are concyclic, the angles OTY', OYY' are equal. But in the limit these angles are the angle and the and the tangent to the pedal at Y. angle between
,
OPY
on
OY
The pedal
is the envelope of the circles described
the radii
vectores as diameters.
For by the
at Y.
*
last
theorem the
circle
OPY
touches the pedal
n',
We now
proceed to find the Pliicker's numbers
m', 6
V
,
k,
4
XI
PEDAL CURVE
167
t, t', D' of the pedal, the Pliicker's numbers n, m, S, k, t, i, of the curve being supposed given. The pedal is the inverse of the polar reciprocal of the curve with respect to a circle whose centre is 0. For if Q is the pole of in Fig. 2 with respect to this circle, OYQ is a straight line and is constant. . OQ have then only to interchange and n, 8 and r, k and i in .the expressions for n', m', obtained in § 3. We have
we note that it is evident geometrically a branch of the pedal through corresponding
Pig. 2.
to the given curve. Hence to each tangent from is an m-ple point of the pedal. meets the pedal Again, any line through times at and times where it intersects the' perpendicular tangents of the given curve, so that n' = 2m, Also the tangents from to the pedal are the tangents at each counted twice and the perpendiculars from to the n asymptotes of the given curve, as will be evident from the first theorem of this section.
m
m
m
m
Hence m' = n + 2m. The same theorem will show that
to each bitangent of the
given curve corresponds a node of the pedal and to each inflexional tangent corresponds a cusp so that k' = i. The reader will readily prove that to each of the tangents from co to the given curve corresponds a tangent at co to the
;
m
4
:
168
PEDAL CURVE
XI
pedal, this tangent at «o and <oa> being harmonic conjugates with respect to Om and the tangent from co to the given curve.* Hence a> and a/ are m-ple points of the pedal ; so that
S'
= |m(m-l) + r.
r+ccos(6-OC)=f(8). The pedal of a parabola with respect
.
Ex. 1. If the polar equation of the pedal of a curve is r=f(6), the polar equation of the pedal with respect to (c, oc) as pole is Ex.
to the
2.
vertex
is
a
cuspidal cubic.
[The pedal with respect to the focus is the tangent at the vertex use Ex. 1. Otherwise: Any tangent to y 1 = 4asc is x — ty + at' = 0, and the perpendicular from the origin is tx + y = 0. Now eliminate t.]
now
Ex. Ex.
3.
The pedal of a
1.]
circle is a limacon.
[Use Ex.
4.
The pedal of rm
—
m a cos mS
is
obtained by changing
its
m
into
m/{m + Y). Ex. 5. The pedal
quartic.
of a conic with respect to
to a focus of a curve
centre
is
a unicursal
Ex.
6.
The
line joining
is
bisected
by a singular
focus of the pedal.
foci of the pedal are the feet of the perpendiculars from on the lines joining the intersections of the curve with the circular lines through 0.
The ordinary
Ex.
7.
Any tangent
is
to the curve of Ex. 1
is
x cos a + y sin
a>
=/(<»).
The normal
— xsinu> + y cos a =/' (<u), and the corresponding normal to the evolute is X cos a + y sin a = — /" (<a). The coordinates of the point of contact of the tangent and of the centre
of curvature are
(cos
o)
./ — sin
<i>
./',
sin
o>
./
+
cos
to
./')
( - sin a ./' — cos a ./", cos to ./' — sin a ./"). and The radius of curvature is /(<») +/" (a). Ex. 8. Tangents are drawn in any direction to a given curve.
Show
that the algebraic
of the radii of curvature at the points of contact Show also that the centroid of the points of eontact coincides is zero. with the centroid of the corresponding centres of curvature, and is independent of the direction of the tangents.
[If the polar
sum
equation of the pedal
is
i
where u k
radii
... +« m = 0, homogeneous of degree k in cos 6 and sin 6, the sum of the d2 u — of curvature is by Ex. 7 —uu which is zero.]
r
m + rm-1 u + rm -<u + x i
is
-^
*
Draw
at,
close to
a diagram in which the tangent to the given curve at construct Y, and proceed to the limit.
P
passes
5
XI
ORTHOPTIC LOCUS
169
Ex. 9. The sum of the perpendiculars from the point of Ex. 8 on the tangents drawn in any direction is zero, and is the centroid of the foci.
[We have u x = 0,
Ex. 11.
to
if
we take
as pole.]
its
Ex. 10. The deficiencies of a curve and
pedal are the same.
Show
that, if a curve touches ona at
O has
in general
H, its pedal with respect an asymptote perpendicular to OH, whose distance
from
is the same for all positions of 0. Discuss the Pliicker's numbers of the curve.
Ex. 12. Discuss the Pliicker's numbers of the pedal of a curve with multiple points at O, w, and a>'. Ex. 13. Find the Pliicker's numbers of the second, third, a curve.
[See Messenger of Math., July 1904, p. 50.]
...
pedals of
Ex.
i'-th
14. If the curve a is the r-th pedal of the curve 6, 6 is called the negative pedal of a. .Find the Pliicker's numbers of the first negative pedal of a given curve. [It is
the polar reciprocal of the inverse curve with respect to
is
0.]
Ex. 15. Show that the inverse to the r-th positive pedal negative pedal of the inverse curve.
the
»"-th
Ex. 16. The locus of the intersection of any tangent to a curve with the line through making a fixed angle with the tangent is similar to the pedal.
Ex. 17. Find the Pliicker's numbers of the locus of the centre of a circle passing through a given point and touching a given curve. [Invert with respect to 0.] Ex.
18.
Find the
Pliicker's
passes through a given point
similitude.]
numbers of the envelope of a circle which and whose centre lies on a given curve.
have
as centre of
[The envelope and the pedal with respect to
§ 5.
Orthoptic Locus.
The locus of the intersection of two perpendicular tangents of a curve is called its orthoptic locus.' If the coordinates used are rectangular Cartesian, we may obtain the equation of the orthoptic locus as follows. Let be the tangential equation of the curve, and let /(A, /j.) on be the tangential equation of a point Xx + fiy + 1 to the the orthoptic locus, so that two of the tangents from homogeneous curve are perpendicular. If we make /(A, /*) in A, fi by means of Xx + y.y + 1 0, the resulting equation in to the given X//jl gives the slope of the tangents from The product of two of the roots of this equation in curve.
—
=
R
R
=
=
—
R
— X/fi must therefore be —1. If we write down (Ch. I, § 11) the condition that this should be the case, we obtain a relation
5
170
ORTHOPTIC LOCUS
a)
XI
between
locus.
and y which
is
the point-equation of the orthoptic
We see at once that the orthoptic locus of an algebraic curve is algebraic. If the equation of the curve is given in parametral form, we can express the equations of the tangents at the points with parameters t and T in the form
*+/(% = <K0
and x+f{T)y
If these are perpendicular, /(£) .f(T) + and from the three equations just written we can express the coordinates x, y of any point on the orthoptic locus in terms of a single
= ^(T). 1=0;
parameter.*
If
RP, RQ
are perpendicular tangents to a curve, the circle
PRQ
touches the orthoptic locus at R.
Fig.
3.
For suppose R, R' are consecutive points on the orthoptic 3), while Rp, Rq and R'p, R'q are the perpendicular tangents from R and Rf. Then p, q, R, R' are concyclic. If we proceed now to the limit, p, q, R' approach P, Q, R reand the theorem is proved. spectively
locus (Fig.
;
Ex.
1.
Find the orthoptic locus of the following curves
Parabola,
:
(i)
(ii)
Central conic,
Circle,
(iii)
(iv) if
=
x3
.
(v) 3 (x
(vi)
+ y) = xK
i
x 1 y1 -4a(xr + f) + l8a xy-27a*
=
0.
»* It is usually advisable to express t, T in terms of u *= t + T, three equations then give x, y conveniently in terms of u or v.
while the interperpendicular sections of an inflexional tangent with the In general there are no tangents are cusps of the locus. Now suppose any line whatever through co meets the given and Q.
Again. It is readily seen that there is no other point of the orthoptic locus on coco' in general. or vice If this is the case.
172
PLUCKER'S NUMBERS OF ORTHOPTIC LOCUS XI6
and a tangent the circular points co and co'. is equal to the class of the envelope. while the tangent at to the orthoptic locus passes through co' and PQ through co. coco') are harmonic. we obtain the number of tangents from co to the envelope.
m m
m(m—
R
P
. when becomes the point of contact of either tangent. Now there are 1) such pahs of tangents so that the orthoptic locus has a ^m(m — l)-ple point at each of co and co'. coPQ will touch the versa. Suppose now E. coV) is harmonic. Proceeding to the limit we see that to each pair of tangents to the given curve from co corresponds a branch of the orthoptic locus through co.&n& therefore ^approaches
E
F
R
co.
E
F
R
tangents in question.. q become the points of contact P.
|m(m-
n =
m(m — 1). a tangent from on from to the given curve are perpendicular and meet at the orthoptic locus. The line TV cannot pass through co unless coPQ touches the given curve at but not at Q. while the tangent to this branch and the line coco' are harmonic conjugates with respect to the pair of
Then
R
. so that
. Q of the perpendicular tangents RP. Now in general the envelope will not touch coco' and hence the number of tangents from co' to the orthoptic locus.
harmonic
also approaches co. 3 RR' passes through co'. coco') and R'ipq. It suffices to find the number of tangents which can be drawn to the locus from co'. other cusps. whose points of contact are not on coco'. while R(coco'. p.
P
. By what has just been said. Let the tangents at P and Q meet in T curve in and consider the envelope of the line TV where T(PQ. however. EF) remains become consecuand unless the tangents from tive. To find the remaining tangents from co'. suppose that in Fig. Proceeding to the limit. pq passes through co. every tangent from co' to the envelope which does not coincide with coco' will be a tangent to the orthoptic locus. so that k = mi. the theorem of § 5 shows that the intersections of perpendicular a bitangent of the given curve with the tangents are nodes of the orthoptic locus. To find this class. Of these tangents 1) coincide with the tangents at co'. We now proceed to the more difficult task of finding m'. Then since R(pq. RQ.
D' from Ch.
«'
= («» — &)i. and touches it at the feet of the
m(m + n —
4:)
normals which are also tangents. = im{(m + l)(m-2) + 2T} *' = m<. &c. neither of which coincides with the tangent at the point. If in Ex. PQ) is harmonic.]
There are \m (m + 1) (m — 2) (m — 3) points from which two pairs of mutually perpendicular tangents can be drawn to a given
Ex.
If
cu<»'
is
a
fc-ple
= (m — h)
(m — 1). If to the curve other than ww. the general cubic has an orthoptic locus of degree 30. YZ) is and Ex.
segments joining the
m
real foci of
Ex.
We now
n'
5'
obtain
r\
i. Jare cusps. m' = m(m + n — 3). = m(3m + /c-6). Hence each tangent from co to the curve is an (n — 2)-ple tangent to the envelope .
[The curve meets the orthoptic locus at the points of contact of the tangents to the curve from o> and <o'.
3.
curve of class m.
2
§ 1
and
see that
= m(m— 1). the locus has (m— 2) (m-3)-ple points. so that the class of the envelope is (n — 2). For instance. If the curve touches ma' at and as (m — 3)-ple points.
n'
5. We have then
m
m'
= m(«— 2) + m(m — 1) = m (m + n— 3).
m'=(m — 2)(m + n — l). VIII. »»') is harmonic. the orthoptic locus simplifies materially.
&>'.
and the mr nodes
Ex.
k'
=
(m-2)
— 1). 2. There are m(m — 3) (n — 2) points of a curve from which two perpendicular tangents can be drawn to the curve. 4. 4 the curve has oxa' as inflexional tangent at Y. t' = im{m(m + n) -(6m + 6mn + n )~m + 22 + 2S}. We have
n
= (m— 1) (m — 2). Discuss the cases in which Y. a> and o>' as harmonic. But as will be seen in the following examples. 6. m! = (m — h) (m + n — 3 — k). It also cuts o)a>' at one other point. Z)' = |(m-l)(m-2) + mi)
!
2
2
2
i'
In general the orthoptic locus of a curve is of relatively high degree. if tbe given curve is specialized by touching the line mm.
Y
Z
Y
Z
Z
.
The singular
foci of the orthoptic locus are the
\)
\m(m-\)
middle points of the \m.
Ex.XI 6 PLUCKER'S NUMBERS OF ORTHOPTIC LOCUS
173
envelope at a point U such that (all. the tangents (m — 1 J-ple point.]
a.
There remain
m(m — 3) (n — 2)
intersections of curve
and
locus. at to the locus are the tangents from
Y
Y
Z
Z
Z
Ex.(m —
the curve.
tangent to the curve. the as an orthoptic locus has a and to' as \ (m — \) (m — 2)-ple points and is a cusp at which <aa' is a tangent. inflexions.
tf.
(i
Ex.
where (o>o>'. 7.
[The nodes of the orthoptic locus other than derived from bitangents of the given curve. &c. the locus has (m — 2) linear branches touching wm at Z. If a curve touches <b<o' at and (YZ.
1.
m — 3.
Ex. 4 (iii) has t»a>' as quadruple tangent.6
.
Isoptie Locus. of the points of contact dividing a>a' harmonically. while n = 4.
The
locus of the intersection of
which are inclined at a fixed angle a
two tangents to a curve is called an isoptie locus. = 3. But this is not usually the case. 10.
If
bitangents of the curve pass through
and
= =
(m + 1) (m — 2). The orthoptic locus is a straight line when the curve (i) is a parabola.
dividing
toto'
harmonically.
. while n (ii) has toto' as ordinary and 3 inflexional tangent. while n 6.
is
[The locus
the orthoptic locus of the evolute.
=
m—
oxo'
harmonically.
11'
m = m (m + » — 5) + 4.
ic'
k
—
to
mi. 4 (iii) has toto' as inflexional tangent at two points
=m=
n
=
m=
.
Ex. 3 still holds. are distinct and both algebraic (being in fact concentric circles). two pairs of points of contact dividing
Ex. (m +
(m — 2). = m(m + n — 4). The orthoptic locus
is
a parabola
. while n = 4. while n 3 (ii) has o>to' as triple tangent. f For the circle the loci of the intersection of two tangents cutting at angles <X or ir-a. two 4. implied in Fig. The orthoptic locus is a circle (ii) is a central conic.
1.
Ex. while n
=
8. For the parabola the two loci are the two branches of the same hyperbola. (ii) touches wto' in two points dividing toto' harmonically. 12.
toto'
—
m=
. 15.
m=
5.
curve (i) is a circle.
Ex.
m'
= mi — 4.
The and
investigation of its properties is similar to that of §§ 5 The construction for the tangent to the isoptie locus 6.j
1
Ex.
Ex.
to 11
on the curves of
§ 5.
The
locus has no cusp.
Ex.
Ex.2) + mi. If inflexional tangents of the curve pass through
m'
1)
and
to'. 14. while
=
6. If the curve passes through w and a branch through to touching the curve at branches of order 2 at to.
to. 11. Verify the results of Ex.
m—
5.
when
the curve
(i)
has
to<o'
as inflexional tangent.f
* The orthoptic locus is a cubic in fourteen cases.
. (iii) touches toto' at
when the
to
and
to'.
9.
= m (m + n — 3) — 4. The degree and class of the locus of the intersection of two perpendicular normals are
(m — l)(m + n — 2)
and
(m — 1) (4m + K — 6). and a quartic in thirtyeight cases. The locus is an algebraic curve if we consider tangents cutting at an angle n — a as included
among
those which cut at an angle a.
174
PLUCKER'S NUMBERS OF ORTHOPTIC LOCUS XI
a>'. 8.
m
Ex. We have
n'
the orthoptic locus has and m — 2 superlinear
k
to
= m{m-\). The orthoptic locus is a central conic when the curve (i) has as bitangent. the points of contact dividing toto' harmonically.
ni'
=
2 (m
to'.*
§ 7. 13.
16. while n 5.
3. 2 On any line through there are n1 points such as n. Tangents EP..
. which is a multiple of 2ir/k.]
0. while the Pliicker's numbers of the cissoid are and n2 . Find the Pliicker's numbers of the envelope of a chord of a given curve subtending an angle of given magnitude at a given point 0.
= P
P
Ex..
+ y2 -
-b2) 2
4
(b
2
x 2 + a 2 y 2 -a 2 b 2 ). meet on a curve which is part of an isoptic locus and which is similar to tthe pedal with respect to 0.
any line is drawn cutting Suppose that from a fixed point Take fixed curves 2j and 2 2 respectively in P 1 and P2 The locus a point P on the line such that OP = 0P1 —0P i of P is called the cissoid of 2j and 2 2 for the pole 0.
Cissoid. t' = m{2m(m + — (8m + 8mw + m2)-2m + 12 + 25}. 5. and at the 2»t(m + )i-4) making an angle OC with the points P such that a line through touches the curve elsewhere. [Reciprocate with respect to 0. x . 1.
2 ta.]
Ex.. Therefore the orthoptic locus is a straght line or concentric circle.
175
of an isoptic locus
For the
find
we
n'
= 2m(m-l).
6.n OC (x*
x'/a?
+ y'/b2
2
= =
1. Show that the tangents inclined at an angle OL.
Ex. The isoptic locus touches the curve at the 2 m points of contact of tangents from a and a (cuts if a %ir). and the normals at P and Q meet at S. neither touching at Q. /=2mi. m.]
Ex. [The isoptic locus of the two curves taken together.
the instantaneous centre of rotation of the rigid body PQE.. = m(2m-3)(m — m-l) + 2mr.
m
.
.
Ex. m' = m(2tt + 2m-4). EQ to a curve are inclined at a constant angle.
m
.
[S
is
Deduce the orthoptic locus of a parabola or central
conic..a
XI
8
Pliicker's
CISSOID
numbers n\ m!.1) + 2mD An isoptic locus has m(m— l)-ple points at and
5
/ 2
TC)
2
2
t'
2
a>
«>'. less the isoptic loci of the two curves taken separately. Show that -RS is the normal at
R to the locus of R. 2. Suppose the Pliicker's numbers of 2 X and 2 2 are %. = 2m(3m + /f-3). D' = (m .]
Ex.
. one drawn to each of two given curves.
A
curve has
fc-fold
symmetry about
at the ends of the radii through
m
§ 8.
two
.
.. Find the locus of the intersection of two tangents inclined at a constant angle.
. The line RS is parallel to the axis or passes through the centre.
3
[y
Find the isoptic loci of 1 and y — 4 ax — 4 ax = tan2 OC (x + a)2 and
4. It cuts the curve at the tangent at 2m(»-3)()i-2) points Q of the curve from which two tangents can be drawn inclined at an angle OC.
.
2
to an intersection of 2j to the cissoid . and so for the
to
is
m
2
evidently tangents
22
.
Hence
k
=n
1
K 2 + n2 K 1
.
^
apply Ex.
and
OP = k OP + k OPi + k3 OPs
1 2
. to the cissoid at count aa 2TOjTO g tangents from 0.
Also the Uj^n^ tangents. Hence n l 2 + n2 2n1 n2 tangents -can 1 + be drawn from to the cissoid.
4.
been assumed that neither curve passes through the
circular points. is a line and K.
Ex.]
[First
Ex.
=k =
2
£. 2
. Extend to the case of given
. and so
for
22
. 3 to the locus of Q. where OQ = k1 OP1 + k2 OP" and then apply Ex. or
m
i
m
m=n m
1
!
+ ti2 m1 + 2)i1 n2
.
.
2.
l
P2 P3
. find the Plucker's numbers of the locus of P.
The
ra
2
linear branches touching 2 X at each of
its
Find the Plucker's numbers of the locus of P if in §8 OP = k1 OP1 + k1 OP2 where k x and k 2 are constants.
3.
Prove that the'cissoid of algebraic curves
.
Ex.
1.
[Consider the cissoid of the two curves obtained by increasing the radii vectores of 2 X and 2 2 in the ratio l/&j and — 1/&.
Ex.
where
PPj. or if the curves meet at infinity ?
. each of the »ij tangents from to 2j a w2 -ple tangent to the cissoid.PjP. that all the intersections of the curves are &c..
Note the case k x
Ex. the line joining evidently a tangent at
cissoid.
meets the cissoid at
Moreover.
It has
finite.
XI 8
line
P
is
and n2 such as P2 x n n2 points other than
1
Hence the
0. What modification is necessary in the results of § 8.]
cissoid has infinite points. are constants.
Again. 3 to the loci of Q and Ps . k 2 k. so that
and 2 2
is
a
-n^-ple point of the
meets the cissoid in
Therefore any line through
2%«
points. 5. if either of the given curves passes through Q.]
If a radius vector
meets three given curves in P.
is
algebraic.
N
curves.
[This was assumed in § 8.
Also it is evident from a diagram that a line joining to a cusp of Sj passes through n 2 cusps of the cissoid. respectively.
Hence
n=
2%-n.176
CISSOID
.
one of which passes through a bicircular quartic with a node at 0. the tangent at P to the cissoid passes through Q.
If on the radius vector OP of a curve we measure off PQ equal to +h.
s?.
Ex. 11. The cissoid of two circles. and collect terms involving an odd power of k on one side of the equation and the terms involving an even power of k on the Now square both sides.
P to 2^ S Glt G and OG=OG -OG
2 2
. P2 and in the perpendicular to OP through
.
transversal
P.
[If O is (0.
+ k"OP2
Ex. 12.]
. the curve is y 1 (a—x) = The area between the circle and cissoid is something Hence the name cissoid from the Greek kio-o-os.
Ex. 0).
cissoid
meet
Prove that
l
2
. is
A
Ex.9
XI
Ex..
[The polar subnormal
'
is
-^
uu
m polar coordinates. How many chords of a given curve through a given ratio at O ? Ex. 0) and A is (a. Turn the given equation into polars. 11 as axes of reference.]
. we may obtain the equation of any conchoid as follows. Replace r by r + k.]
Pliicker's
numbers of the locus of P. [Take OP and OG in Ex.
6. We have the required equation of into Cartesian coordinates. find the cissoid of the line through perpendicular to OA and the circle on OA as diameter. where k is constant. the locus of Q is called a conchoid of the curve. For instance. Expand the powers of r + k which occur by the binomial theorem.
1
OP=k OP
[See Ex.
8. A are two points. If the tangents at Pj P2 to' 2 X 2 2 meet at QT.
and the
G. circle with centre If we are given the Cartesian equation of a curve.
T and
Pj bisects
§ 9. It is the cissoid of the curve and the and radius k.]
'
like
an ivy
leaf.
are divided in
How many
chords of a given curve through
line ?
are divided in
a given ratio
by a given
Ex.
Conchoid. If 0. 7.
CONCHOID
Any
2
. and
P
OPP P
1 1
2
through
Find the
3. the conchoid.
0. 9. The normals at P.
'. and turn the equation back other.
177
meets a given curve in
if
. consider the case of a curve of degree 2n and and at the with multiple points of order n at class circular points co and a>\
m
. 10.
the locus of Q is said to be parallel to the given All parallel curves have the same normals and the curve.]
§ 10.
[{(x2
+ y2 )(ax2 + 2hxy + by2 + 2gx + 2fy + c) + k2 (ax2 + 2hxy + by 2 )} 2 = ik2 (x2 + y2 ) (ax2 + 2 hxy + by2 + gx +fy
Consider the case a
=
b
=
1. Find the nature of and at infinity. [Use Savary's theorem on the centre of curvature of a roulette. 4. Hence the at P and Q
PQ —
If along the
pedal of the parallel curve is the conchoid of the pedal of the given curve for any pole 0.*
Ex. An ra-ic does not pass through 0. What are the asymptotes ?]
its
Ex.
1. We proceed to find Plucker's numbers for any curve Suppose that the given parallel to a curve of given type. How many lines can be drawn through a given point on which two given curves intercept a segment of given length ?
'
. in the case f But they may each be distinct
*
2
The
2re-ic
points. or a. and ra(2n + l) The 2«(m-1) remaining intersections are the ends of the chords of with length 2ft.]
Ex.
meets the conchoid obtained by changing ft into 2ft in 8n 2 the above reasoning 2w lie at 0. same evolute.
n2
. an »-ple point at each of o> and <»'.]
and a
circle
through 0. Find the conchoid of a [A limacon r = k + b cos <?.]
Ex. for instance.]
Ex.
2
. and a tacnode at each infinite point of the n-ic.
'
3. 5. 6. a>. It has a node at tacnode at infinity with the given line as tangent.
the conchoid
(x — a) 2 (x2
is
)
+ y2
=
k 2 x2
.
It is called the
conchoid of Nicomedes '.
[If the line is
x
'
=
a.
Find the conchoid of a straight
line.
2.
h
=
0.f are parallel and at a distance k apart. at a'.
Ex. the 2»-ic through algebraic curves .
Parallel Curves.)
XI
10
PARALLEL CURVES
179
points of the n(n — 1) chords of the 2n-ic which pass through and are of length 2 k.
[Consider the intersections of one curve with a conchoid of the other. n(2n + l) at a. conchoid at [The conchoid is a 4«-ic with a 2 «-ple point at 0. The centres of curvature at corresponding points of all possible conchoids of a given curve lie on a conic.
Find the conchoid of the conic ' ax 2 + 2hxy + by2 + 2gx + 2fy + c
= 0. The loci of the two possible positions of Q are The tangents in general parts of the same algebraic curve.
By
of a circle.
normal at P to a given curve we measure +h.
m'
=
2m.
0. whose tangential equation
therefore
is
'Ki+k^+n ^'
2
i
+ k(\*+n*)y
.
k'
'
=
{0} + {2c}. k. 8' = {m(6m-5)} + {m(m-l) + 2r}. m' = 2n + 2m. k.
— is a tangent = 0. Here in the expressions for S'and k' the term in the first brackets { } refers to the multiple points at 0.
co and a/ being m-ple points of the parallel curve. i.
§9
m'
= 4m. &>'
k
= {2m} + {2i}.
m'
=
n + 2m. numbers of any other curve.
i. cY = {0} + {m(m-l)-2T}. * That the class of the parallel curve is twice that of the original curve is also evident from the fact that to each tangent to the given curve in a given direction correspond two
parallel tangents of the parallel curve at a distance k from it.1)} + {t}. co. to each inflexion of the given curve correspond two inflexions on the parallel curve. the tangents
co. »> a> being m-ple points of this pedal. m' = 2in-4m.
and that
numbers are n.
co.
XI 10
its Pliicker's
curve does not pass through 0.
i
=
2i. l.
r
= m(m— 1) — 2r. then
to a curve
with tangential
\x + fiy + 'l+k{\* +
is
^=
a tangent to the parallel curve. which we are considering. t.
For this inverse Oco and Oco' are m-ple tangents. (Y.180
PARALLEL CURVES
co. Similarly. m.
or
co'. For the conchoid of the first pedal we have by
.
For
at
'
are 2m-ple points.
is the the curve parallel to
the given curve. t. by ti'. e.
this conchoid 0.
If
Xx + fiy + l
\l)
equation /(A. on'. co' and the term in the second brackets { } refers to multiple points else-
where and so in what follows.
*'
8'
=
=
{0}
+
R
{|
m (m . + 2m. coinciding in pairs. 8. of the inverse The polar reciprocal with respect to
first
negative pedal of the conchoid.
For the parallel curve we have fiually
n'
= 2<n.
We
shall denote the Pliicker's
for the
moment
By § 4
n'
for the pedal
with respect to
=
2 m. co'
For the inverse of this conchoid we have by § 3 n' = 2m.
lie
Ex. Ex. 4.]
with centre moves in its own plane so that Ex.
parallel to the parabola y*
1!
Find the tangential and parametral equations of curves = 4 ax. and draw them. 0.]
Ex.
b>k>Wfa.
a given curve.
ABCD
D
D
.
I
k
=
b*/a. a>k>b. Find the envelope of a family of circles whose centres a given curve and which touch a given circle. a node where (k + 2 a) t* = k — la. focus being the focus of the parabola.
'
y~\-p~
_4a^
2U_
l+t*'
Changing the sign of k is equivalent to changing t into — 1/t. Distinguish the cases k>2a. 18
and
§ 10. § 8 (IV).
no inflexion. Show the lines through trace traces out an isoptic locus of the curve while A.
l! .
properties of
The parallel curve can readily be shown to be the envelope of a circle of radius k whose centre lies on the given curve.
Show
curves. k > a /b. A. the conies of closest contact being (y±k) = 4 ax. C.
Ex. This gives in practice a convenient method of drawing curves
parallel to
Ex.!). and six cusps where
at (oo. It has degree 6 and class. The curve is the reciprocal of the quartic of Ch. k = b. k = 2 a. It' passes through the circular points. We
leave
to the reader the investigation of the parallel curves by this means.
. Show that their new envelope
radii of a singly infinite family of circles are all increased is parallel to their old one. >k>a.
x
4 at 2 kjl-t 1 ) (l-« 2 ) 2+ \ + fi. k < 2 a.
It
has also
2a(l+?) s
It has
= k(l-t
2 s ) .
!
i
Ex.
a?/b [{(«
-4!)\1 +(b
3. 6.
Find the tangential equation of curves parallel
x*/a* + y*/b 2
to the ellipse
=
1. (\ '+/.
2 2 [(a M -X) 2 = i \ (\ + A ! ). if 2 tan. B. It has two linear branches osculating each other
2 0).]
Ex.
on
[Combine
§ 4.XI
10
PARALLEL CURVES
181
This equation is rationalized by a process similar to that used to obtain the polar equation of a conchoid in § 9. that out the isoptic locus of a parallel curve . -i> -lf = 4i.1 * is the angle which a tangent to the parabola makes with the tangent at the vertex. XVII.
2.
its
singular
The
by a constant. h = a*/b.
[Use the tangential equation.
that the foci of a curve are singular foci of all parallel
1.
4. Discuss similarly ay 1 — X s and a'y = x s . B.
. A rhombus parallel to the sides both touch a fixed curve. 5.
Draw
the parallel curves distinguishing the cases
h<b}/a. Also. C. and that the normals at to the two isoptic loci are concurrent.]
2
'
h=a.
4.
§ 1 will be found in the following examples.
n'
m'
= 2» — 2
i'
=
= m + 2n-4k. and has an m-ple point at a and a>'.]
We see that the
Ex.
a and
<u'.
Find the Plucker's numbers of the locus of the centre of a touching a given curve and orthogonal to a given circle with
is
centre O.
. The tangents from to the envelope are n bitangents given by taking P at infinity and the 2 m perpendiculars from to the 2 m common tangents of j and s. are the normals at Q x and Q 2 to the envelope. 1 to 4. How many circles can be drawn orthogonal to a given circle and bitangent to or osculating a given curve ? [Such a circle is given by each node or cusp of the locus of Ex.
the locus of the points Qlt Q2 on the perpendicular tangent at any point P of s.
locus has the same Plucker's numbers as the reciprocal of the given curve. and those of the derived curve under consideration by n'. &.*
may
of even degree n. t. m.
Many other cases of derived curves might be given.
* If
the curve has a *>ple point at each of
ft. To each tangent from to the given curve corresponds an intersection of the locus with am. which will be found useful later on. Hence «' = m. i. But those already discussed must suffice. consider other
>
meets the given curve corresponds a tangent to the locus perpendicular to the line. The attention of the reader is especially called to Ex. Each bitangent of s gives two nodes of the envelope and each inflexion gives two cusps. k. m'.
[We consider the case in which the given curve having a |»-ple point at w and a. The reader
cases.
2(3ot + k). The foci of s are the singular foci of the envelope. and the 2 n intersections of ^ and s are ordinary foci of the envelope. The circle with centre Y and radius equal to the tangents YL and YL' from Ytoj passes through Q1 and Qt while PQX andPQ. k. The Plucker's numbers of the given curve are denoted in the examples by n. P and Bare
is
The envelope
OY from
to the
. Further illustrations of the principle at the end of. r.
= m(m — l) + 2i-.
numbers of the envelope of a circle with its and cutting orthogonally a given circle j
8'
i'
t'
[n' = 2m. m' = 2(n + m). 8. = n> + 4:mn + 2m?-l0m + 28. i
Ex. If OR is perpendicular to LL'. 3. Hence mf = ».
k.
=
=
(m-l) + 2X>. which are equidistant from r"and inverse forj..
XI
11
Other derived Curves.
D'
k'
=
2i'. 2.
182
OTHER DERIVED CURVES
§ 11. The envelope is self-inverse with respect toj. Find the Plucker's centre on a given curve s whose centre is 0.
circle
1. 1.] Ex.
in
To each of the n points
which a given line through
Hence
i
=
k. To each cusp of the given curve corresponds an inflexion of the locus.
facts enable us to construct any number of points on the envelope. and the others by the segments in question. Prove also m' = 7 m — 3 n + 3 k. and has a node at each intersection of the given line with the curve.2) 8.]
. The cusps and points of contact of tangents from a> and <u' [3 m + k.
[»'
=
2«i.
[(i) The envelope cuts OH orthogonally at and has an asymptote perpendicular to OH twice as far from O as the corresponding asymptote of the pedal of s with respect to 0. What modification must be made in the results of Ex.
(iii)
So>
5.. /*) +/3 (X.
the envelope
is
0=0-2/! (*. Find the degree of the radial of a given curve .]
Ex. M ) +/. e.]
Ex. 4. homogeneous of degree > in X and /i. See § 4.. If j is x* + y 2 = k and s has the tangential equation
where
/.
[n'
Find the locus of the middle point of a chord of a given curve
fixed direction. Of these nodes of the locus (n — 2)8 are given by the nodes of the given M-ic. meets the give branches of the radial through 0. /*)
is
= a+A (X. Find the Pliicker's numbers of the locus of the middle point of the segment intercepted by two fixed lines on any tangent to a given curve.2) (» . 9.
y)/(a?
*)/(*•
+
2
i/
+ fc) 2
(^y)/(a 2 + 2/ 2 + ^) 3 + -a. Ex. 11. 5 8' = | n (» .
6.
and Sa' are
inflexional tangents of the envelope. i. D' = D. k' = (n — 2) k. the equation of
+ f + h) + 4/s
(or.
=
Ex. (X. Find the degree of the locus of the end of the polar subnormal of a given curve.1) (n .? are drawn. The locus meets the given line at its intersections with the tangents to the curve from a and a.]
= \n(n — 1). (X. k = n.]
Ex.3) + (n . 10. the locus of the end of the line through a fixed point O parallel and equal to the radius of curvature at each point of the curve. 8.]
!
-8/3
Ex. 7.. [2(m + n). 3 if the curve s (i) touches aa>' at (ii) goes through a> and ra'.]
•
Ex.
drawn in a
Consider the infinite m' (» — 2) m. Consider the intersections of the locus with the line at [2n + m. points on the locus and the tangents in the fixed direction.]
Ex.
infinity..) + . when either s or its polar reciprocal with respect to. (iii) touches the tangents from a focus S at two points on j ?
H
. f.
«'
=
t. Any line through radial in m more points. (ii) The tangents at <o and a to s are cuspidal tangents of the
envelope.
XI
11
OTHEK DERIVED CURVES
183
These
corresponding points on s and its polar reciprocal with respect to j.
m'
= 2 m + n.
How many
lines
a given n-ic intercepts middle point ?
\n
can be drawn in a given direction on which (n — 1) segments two of which have the same
[In Ex. Find the degree of the locus of the centre of a circle touching a given curve and a given straight line.
]
Ex.]
. If P is any variable point on the first polar curve of a fixed point with respect to a given curve. in a given ratio. R. 11. 14. Find where the polar line meets the locus again. replaced by the normal at P in Ex. Ex. Discuss the locus of a point from which equal tangents can be drawn
to two different curves.
Find
X
[It is
easy to find the intersections of the loci with the line at infinity.
the tangent at
P
is
• Ex.(»»
Their degrees are
+ 2n-6)
and
2
(«-2) ii (»-3). 6.
[The circular points are \m(m — l)-ple points of the locus.184
OTHER DERIVED CURVES
it
XI
(ii)
11
Ex.]
if
Ex. 12. 13. Find the degree of the locus of X. the locus of intersection of the polar line of P and the polar of P with respect to a fixed conic is a curve of degree + 8 + «.
m
[The polar line of P passes through which is a (» — l)-ple point of the locus. and for the other points at infinity see § 2.
X
X
PQ
again in Q. The tangent at P to a curve meets the degree of the locus of if (i) divides divides QR in a given ratio. 11.
«. Find the degree of the locus of the intersection of two equal tangents to a curve.
of a curve of degree n contains y.
(®r» Vr> z >)-
The curve
will pass through
them
f{xv yv gj
=
0.. Suppose we are given r points
(*1» Vl> Z l)>
(
XV
. In this case an infinite number of Ji-ics will pass through the ^ n (n + 3) given points.. z)
y.
point.
. Next take n = 2. namely the line through the four points and any line through the fifth
.
and only one curve of degree n passes given points (see Ch.
. that the equations (i) are not
independent.
2/2.CHAPTER
XII
INTERSECTIONS OF CURVES
§
1. y.
=
0. §6). Equations (i) have one and only one solution for one and only one straight line passes through two given points.
equation f(x.+2)-l = %n(n + 3)
coefficients. They will determine the
J(«.
=
one term of zero degree in x and
n+l
contains
%(n + 1) (n + 2) coefficients. if
independent ratios of the
r
= \ n (n + 3). the conic through the points is degenerate..
1)
(n + 2) coefficients
+ l)(«. two terms of the first terms of the n-th degree. First take n = 1.
. Hence f(x. fix.
f(xr y r z r )
.. there are a singly infinite number of conies through the five points. In this case \ n (n + 3) = 2. z). being the line through these three points and the line joining the other two points. II. z)
The
degree. however.
.
(i)
These are linear equations in the ^(n + of f(x.
Conditions determining a Curve. If four of the given points' are collinear. Let us consider two special cases.
. y.
. If no four of the five assigned points are collinear.
.
Z 2 )> ••
if
. one and only one conic goes through them.
Hence
through
It
in general one
^n(n + 3) may happen.. If three of the five points are collinear.
y2 z2 )
=
0. Here %n(n + 3) = 5..
In general one and only one cubic passes through nine given points. z) linearly in terms of %n(n + 3) — r arbitrary independent quantities.:
186
CONDITIONS DETERMINING A CURVE XII
1
Lastly.
take the case n = 3.
. Therefore the cubic passes through the nine fixed intersections of u and v eight of them being the eight given points.
.
w =
s
are s fixed n-ics through the given points.
. [These are the only values of p and q such that
\p(p + S)>pq
§ 2. there are a infinite number of conies through the five points.
k1 w1 +. . Suppose eight points on the cubic are given.
fixed point.. and k is
We now
a parameter not involving x. the ratios of the coefficients can be expressed linearly in terms of J to (« + 3)— s quantities. If only s of equations (i) are independent. if all five given points are collinear.
.. 2 ai. . p = 2 and 2=1.
+ ks ws =
. we must have p = q «= 1.
It is necessary now to consider whether there is any exception to the statement that all cubics through eight points pass through a ninth. Here \ n (n + 3) = 9.y.
1
. so that an {%n(n + 3) — r}-ply infinite number of n-ics pass through the given r points in general. Hence All cubics through eight fixed points pass through a ninth
=
=
. where u = and v = are fixed cubics. z. or p = q = 2.
kt are arbitrary constants.
. namely the line through the points and any other line
doubly
whatever. The coefficients of its equation can in general be expressed linearly in terms of a single quantity. The argument which established this result
* In this case any two of the nine points are said to be conjugate to one another with respect to the other seven.
where kv k2
.A. Hence the equation is of the form u+kv = 0.
and
iq(q+
3)2s_P2-]
Cubics through Eight Points. The equation of the n-ic is then
If of the form
r< %n(n + 3).
. p = 1 and q = 2. an infinite number of cubics pass through the nine given points.2 +
. But if the nine points are the intersections of two cubics. If p and q are positive integers such that a p-ic and a q-io can be described through any pq arbitrary points.
coefficients
equations (i) express the ratios of the otf(x.*
§3. and
w = 0.
Ex. y. w% = 0.
2 They therefore pass through a ninth point. which are such that any cubic through seven of them passes through the eighth. ABODE goes through F. the line-trio
(2)
(3)
the line-trio
C^.
GH
F
.
then
A 1 A 2 A 3 and A^^. G2 pass through the lt 3 2. The case in which three or more of the given points are collinear is dealt with in a similar manner. Hence C3 lies on cubic (2).
The equation of the cubics will involve not but two or more parameters in this case and the pre-
ceding proof breaks down in consequence. But Hence the conic does not lie on GH. 1). . It will be found that the eight points lie on one or other of two straight lines. But C3 cannot lie on the lines A 1 A 2 A 3 or B1 B2 B3 since neither of them meets the cubic in four points. But cubics (1) and (3) pass through C3 .
If two straight lines meet a cubic in
B-i.
=
not independent.
Hence all cubics through eight given points always pass through a ninth fixed point.more than six points.
G C2
X
(Fig.
.
.
.
For the three cubics
the given cubic. First suppose no three of the eight points collinear.
points.
1
2
3
very close to one another. and similarly through G and H.
.
lines
Taking the
A A A BBB
x
2
3
. r 8) are
=
.
2. B2 C2 A B3 C 2 A A 3 Blt B B eight points A
A^G^ A
X
2
3
. F. D. H. The eight given points lie on a conic. one (or both) of which is part of every cubic through the eight
. eight points are such that all cubics through any seven of them necessarily pass through the eighth which implies that the equations (i) of § 1 (n 3. C&.
.
. Therefore G3 lies on
.
3
3
.
Let us
now
3
.
.
one. A 2 B2 A 3 BS meet the cubic Clt G2 C3 are collinear. and may under special circumstances pass through an infinite number of other fixed points.
XII 4
CUBICS THROUGH EIGHT POINTS
187
will not be valid if the. Let the eight points then be A.
consider some applications of the theorem of §
.
while the lines
again in Glt
(1)
C2 C3
.
. The cubic consisting of the conic ABODE and the line passes through seven of the points and therefore through the eighth point F.
A A A B^B^. E.. G.
.
B B
2
. and any cubic through them consists of this eonic and some straight line for a nondegenerate cubic cannot meet a conic in. B. G.
3. while and and FA at N.
7
.:
:
188
CUBICS THROUGH EIGHT POINTS
XII 4
and defining the tangential of a point 1 on a cubic as the point in which the tangent at A meets the curve again. CD.
The
finite intersections of
a cubic with
its
asymptotes are
collinear.
If L.
3. § 5.
II.
its
intersections with the line at infinity. BC. we have The tangentials of three collinear points of a cubic are
collinear.
1. § 3.
C.
.
Ch. 9
. then L. of course.*
Fig. FA.]
Ex.
Ex.]
[This is a particular case of the collinearity of tangentials of collinear lie. The points of contact the inflexion.
CD
points on a conic.
Ex. This is Pascal's well-known theorem. on the harmonic polar of
Ex. (2) line-trio AB.
BC
and
EF
F are
at
M. M.
l.
meet at
collinear.
§ 4. 2. DE. The points of contact of the tangents from an inflexion of a cubic are collinear.
Ex.
D.
II. are
AB
N
DE
trio
[Consider (1) conic and line LM.
Suppose now A y and A 2 are inflexions of the cubic. E. while and B1 B2 B3 are close to one another. and we have in the limit The line joining two inflexions of a cubic passes through
A A A
l
X
another inflexion.]
*
EF.
points.
(3) line-
See also
§ 5.
Cli. Then G G2 is 2 3 close to them.
Ex. B.
[They are the tangentials of
See Ch.
A.
VH.
each doubly infinite in triple contact with a given non-singular cubic.
1
is
a
A
EF meet the cubic again at three
[The second cubic of Ex.
AB. 6 take
E and F at the circular points.]
. Q.
Ex.
Show
number.]
that the tangents at
Q.
a cubic at P. C and
D
Show
Ex. D. Discuss the number of such sextactic points the cases of a non-singular.]
Show
that the tangentials
of P. or ousgidal cubic. having
that there are three families of conies. M".]
Ex. PD. C. EFN. 6. A conic has four-point contact with a cubic at A and meets and F.
[See Ex.]
Ex. If two cubics have the same asymptotes.
is
the line at infinity twice over.
m
[A particular case of Ex. C. If is meets the cubic again in N.
on a conic. E.
Show
that
PQ
passes
through an
[See Ex.
6. B.
[The conic of Ex. 13. N. PB.
(3) line-trio
PA.
The number of points
is
27. their finite intersections are collinear.]
A circle meets a ch-cular cubic in the finite points A. The sextactic points of a cubic (points at which a conic Has six-point contact) are the points of contact of the tangents from the inflexions. that AB and CD meet at Q on the curve. The theorems in § 4 and many of these examples are particular cases of this result.
A
conic osculates a cubic at
inflexion. 3.
6. Q.
[Consider (1) the given cubic.
C. 9.
8. CD.
Ex. 4. When they are chosen.
7. 10. LM is parallel to the
real asymptote of the cubic. is
conic meets a cubic at A. Then there are three possible positions for R. If six of the intersections of two cubics three are collinear. Q. R.]
P
Ex.
A conic touches B are collinear. PB. Discuss the case of a unicursal cubic.]
E
EF
Ex. R are known. the other
[Consider the two given cubics and the cubic consisting of the conic and the line joining two of the remaining intersections. CD. 0.
[The tangentials of A.
B
are collinear. B. PC. 6. 14. Show that collinear points L. The tangents from a point P of a cubic are PA. CDM. 5. excluding the case in which P. 6.
the line-trio
Q. 12.
8. If AB and CD meet the cubic again in L and M. Enunciate the theorem obtained by respectively consecutive in Ex.] Ex.
D
consecutive in Ex.]
Ex.
[In Ex.XII 4
CUBICS THROUGH EIGHT POINTS
lie
189
Ex. B. R are
collinear.
making
A
and B. the tangentials of P.
ABL.
P
and
Q. F. 11. and Q may be chosen arbitrarily. 13 meet on the
curve.
'
Show
P
and Q in Ex. D. the curve again at the tangential of the tangential of A. nodal.
N
[Take A.
Ex. (2) polar conic of
P
and
line
AB. 4
Ex. 4 particular case.
B on the curve and PA. P. DSA. E.]
ABE. with respect to the cubic through ' Let A. C.
(3) line-trio
PQT. BBC.
(ii)
two
Ex. 17. be put into homographic correspondence with the lines through so that the lines HE. Q'. The point is called the point opposite to with respect to the cubic] A. ADF.
•
[Take C. Prove that QB is divided harmonically by the cubic and its point of contact with its envelope. F. C. correspond respectively to the conies ABCDE. 22. G. 17. BC. EF. 20. BS of a cubic meet at on the curve. Chords PQ. H. G.
[Let
ABC be any three
fixed points of the cubic. B. Prove that Q'B and QR' meet on the cubic. S and any given point of the curve as vertex. they meet again on FG. Show that. Show that the tangents at A and B meet on the cubic. B. and Consider the cubics (1) given cubic. FG. if they touch at A.
.4
190
CUBICS
THROUGH EIGHT POINTS
XII
Ex. The tangents at the four points of contact all meet on the cubic.]
19. they osculate there while. P'A. P'B meet the curve again at Q.
Ex. A conic passes through four fixed points A. (3) conic ABCDE'F' and line EF. Q. 16. C. and the circle of curvature at B passes through A. C. ACG. Two cubics go through the vertices A. 18. B. B'.
[See Ex. The circle of curvature at A to a circular cubic passes through B. A variable point Pon a cubic is joined to two fixed points of the curve. 20. ECD.
If
any line through
the conic ABCPQ meets the 3-ic again in a (Ex. (2) line-trio SET.
T
DA
A
[Consider (1) given cubic. 17) and the line OPQ and the conic ABCDPQ have fixed point a one-to-one correspondence. PB. Let the line through corresponding in the
D
HG
H
D
H
. 17.]
D
ABCDEF
F
Ex. that the line joining the other two intersections of the conic and cubic meets the cubic again at a fixed point.
[Consider (i) two cubics and line-trio cubics and line-trio EBA. D. CD. CQD. D of a quadrangle and through the diagonal points E. B. C.
meet the cubic again in E. 23. going through P. of a cubic. C. 21.
D
Ex.
meets the
D
3-ic in 0.]
Ex. Show that a quadrilateral can be inscribed in the cubic with the sides AB. Q. the conies through A. if they touch at E. S' through A. cubic the locus of the intersections of a pencil of conies with a homographic pencil of lines whose vertex is any given point
is
Any
of the cubic.
D
Show
E'. ABCDG. Through four given points of a cubic four conies can be drawn touching the cubic at some other point. [Let conies S. ABCDF.
APB. Prove the accuracy of the following ruler construction for the point opposite to A. F'. B.
[See Ex.
D in
Ex. and the joining lines meet the curve again in Q and JR. B. P' on a cubic are joined to points A. C. Two points P. I. B. (2) conic and line E'F'.]
Ex.
[Make P' consecutive
to
P in
Ex. F. B.
. 15.]
Ex.]
Ex. B. 17 at the circular points.
1
. Show that if and Q. C.
3.
The theorem that all cubics through eight given points pass through a ninth fixed point can be extended to curves of higher degree in the form In general all curves of the n-th degree through \ n (n + 3) — 1 fixed points pass through \ (n — 1) (n — 2) other
fixed points. Deduce Ch.-§3. i.
the lines
B3
Z)"
and
D
A A B B C^C D D B Cs and C D and D
1
2
. The theorem is only true 'in general' if ti>3. 7 hold good in the examples given in these sections but it must be admitted that this assumption is hardly rigorous without further inves.
t
lie
on a conic.
and
t
. We shall assume that the theorems of §§ 5. D. 2. where k is a parameter .
1. Ex. 4 are special cases of this result. 10 from Ex.1 + 1 (w . [Reciprocate and use Ex. 1.
4
. are the points opposite to A. H.
Ex. and the curve and v 0.' Ex.
Clt C2
.
D
lt
X> 2
. the conies ABCDa and EFGHO meet OS at the
ninth
point of all cubics thvough A. 24.
quartic. Ex.
.
we have
tigation. Four normals are drawn to a conic from a point.
.
§ 5.2) = TO
=
=
=
. .
common
Deduce a
Ex. (2) the lines A X A 2 B t B 2 CY C2 DjD2 (3) the given conic and the conic A & Bs Cs S A A are three quartics through the same thirteen points.
A3
and
. For in general by § 1 any such curve has an equation of the form u + kv 0.]
Ex.1) (« .
l
2
meet the quartic again
t
. and from the centres of curvature at the feet of the normals two more normals are drawn.
Then IT meets 2 again at 0. 3. Ex. E. D and to E. F. B. II.
A
t
. C.:
XII
5
INTERSECTIONS OF TWO m-ICS
191
homography
E. F.
H touching at H the line
to the conic
ABCDI
meet in I' the conic 2 through corresponding to the conic ABCDH. G. construct by ruler the intersection of the cubic through them with the line through two of the points or the conic
through
five. G.
. Show that these eight normals touch a conic. ruler construction for the ninth intersection of all cubics through eight given points.
. Therefore they have three more points in common. Given nine points.
conic meets a quartic at
1 2
. passes through all the n 2 intersections of u Noting that 2 •§*l (% + 3) . 2.
2
.
Intersections of
Two
n-ioa. 6. G. 8.
Ex. 25.e.
D
.
s
Show
that
A A s At B3 Bt Cs C
at
.
the required result.
.
JB
?
.
2
. 2. 4. 9.]
[(1)
The given
.
A A B
1
. The tangents at four collinear points of a quartic meet the curve again in eight points on a conic. F. H. only if the given points are such that not all n-ics through \n (n + 3) — 2 of them pass through the remaining point. B.
t
.
.
5. C.]
through (n —
Ex. (2) n lines. 2. D. 6. 11. (3) given and cubic through nine of the twelve points.
See Ch. (3) line of collinear points and (n — l)-ic through l(n-l)(n + 2) of the n(n — 1) points. More generally. 9. two quartettes of lines meet in sixteen points of which three lie on another line. § 6. A straight line meets a quartic at A.
n''
+ 2m — 1
finite
points pass
Show that the theorem of § 5 is not always true when n > 3. If a conic passes through the points of contact of a bitangent of a quartic.
A
n—
1
inflexions of
an
n-ic passes
through an
[Take the conic of Ex.
line
D
[Consider the quartics (1) given quartic.
/=
twice at each node and
thrice at each cusp. I. the remaining n (n —p) lie on an (n—p)-ic. meet the quartic again lie on a cubic.]
Ex.
[Consider the m-ics (1) given n-ic.
which meets
2. n-th inflexion. (2) pair of conies.
and
[On (#-£)_£ dy
=
(y-ij)
_Z
ox
.192
INTERSECTIONS OF
TWO
r-ICS
XII
5
Ex. the directions of whose asymptotes are the axial directions of / = 0. B.
8. Generalize Ex.]
Ex. 12. The feet of the normals to a curve f{x. VII. Any n-ic through the feet of \n (n + 3)-l normals from to a given n-ic passes through the foot of every normal from (n 2).
(n
7.
5. The result also follows from Ch. 1 as a special case. This includes Ex. the line twice over and the n lines as the tangents at its intersections with the n-ic.]
drawn.
conic meets an n-ic in 2 m points. (3) conic and (n — 2)-ic. These are divided into lines joining each pair meet the n-ic again in n(n — 2) points on an (« — 2)-ic. Ex. The number of such normals is m + n. 9. [Consider the n-ics (1) given n-ic. 13.]
pairs.
Ex. All ^-circular 2«-ics through 2 l)' other fixed finite points. y) = of degree n class m from a point (£. if np of the riL intersections of two n-ics lie on a p-ic.]
Ex.
[Limiting case of Ex. a cubic will touch the quartic at its six other intersections with the quartic. 13]
Ex.
general.
A
Show that the n
line through Ex. For the p-ic and an (n p)-ic through ^(n—p)(n—p + S) of the remaining points form an n-ic (degenerate) through
In
—
np + %(n-p)(n-p + 3)
= %n(n + 3)-l+l(p-l)(p — 2)
. (2) n lines. B and a conic through C. The case n = 3 is well known. §
Ex. rf) lie on an n-ic through O and the pples of the line at infinity.
>
§6. 8 as. Show that the twelve points in which a conic through A.
Through each of n collinear points of an n-ic a straight line is Show that they meet the n-ic again in n(n — 1) points on an
— l)-ic. 10.
[If
Ex. it may be shown to be untrue for the thirteen other intersections of any two 4-ics through three
collinear points. the theorem is readily seen to be not true for 4-ics through the other thirteen points.
A conic through. 2. R'. §
7. C. D. 2. Show that their eight other intersections with the quartic lie on a conic.
2216
subjected. § 2.
nN intersections of an n-ic and N-ic will pass through the
N=
(r
— iV")(r — iV+3)
also
arbitrary points on the n-ic.
Take
nN-%(n + N-r-I)(n + F-r-2)
and
JV-ic. P'.
Ex. Take %(r— n)(r—n + 3) arbitrary points on the JV-ic and
intersections. Ex. If a conic meets a quartic in eight points and a conic touches the quartic at four of the points. r>JV. Two conies are drawn through the points of contact of two bitangents of a quartic. 1. 4. Obtain other theorems as special cases. 5.
is
of the intersections of n-ic
* For a discussion of limitations to which this result Bacharach. Q'.
Ex. But none of them can lie on the p-ic. Note the cases n = 3 or 4.] Ex. A polygon of 2ra sides is inscribed in a conic. [See also Ch. Annalen xxvi (1886). Any conic through P. other than the vertices of the polygon. r^n + N— 3. C.]
:
Ex. Q. D. B. XIX. Any quartic through the intersections of a cubic and quartic meets the quartic again in four collinear points. Hence they must all lie on the (n —p)-ic. Hence by § 5 all the remaining points lie on this degenerate n-ic.
An
In
of the
general
extension of the theorem of § 5 is the following any r-ic through all but
i{n + N-r-l)(n + N-r-2)
remaining
If n =
i
r^n.
[Consider the two ra-ics formed by taking every alternate side of the polygon 2ra of their intersections lie on a conic. Ex.* we have the theorem of § 5. a conic touches the quartic at the other four.] Ex. lie on an (» — 2)-ic.the points of contact of two bitangents of a quartic meets the curve again in A. A conic meets a quartic in P. Q. S'. S' meet the quartic again in eight points on a conic. S. 275-299. B. 6. R. Q'. 7.:
XII 7
INTERSECTIONS OF ANY TWO CURVES
193
of the n 2 intersections of the two given n-ics. 3. Show that a conic can be drawn touching the quartic at A. S and any conic through P'. Ex. see
. pp. for the p-ic cannot meet the n-ic in more than np points unless the n-ic degenerates.
Intersections of
any Two Curves. R'. provided
r. R. which we suppose not to be the case. Math. Any quartic through the intersections of a conic and quartic meets the quartic again in eight points on a conic. The intersections of the sides. 3. [This and the following examples are particular cases of Ex.
4. B'. 1. Then (i) shows that the curve X passes through L.
6.
intersection of the curves and n be the points together with other points which we call collectively ' the points L '. JV-ic is the quartic and the conic through the points. points.
. 2 as the polar conic of 0.
points
which we
call collectively
'
the points
0'. The tangents at the six intersections of a conic and cubic meet the cubic again in six points on a conic.
when
there can be no
. D.
[w-ic is given cubic.
meet the curve again
See Ch.
Gn Gv to be such that CiCx + Cm % + On Cr = where l + \ = m + = n + v.
. C.
m
Let the complete
Suppose that
the the the the curves I and meet in the points and N~.
Ex. curves A and meet in the points L and M'.
The n-ic together with the
— n)-io
through the
J(r — n)(r — n + 3) points form a degenerate r-ic through the \ r (r + 3) — 1 points and so for the N-'ic together with the (r-N)-ia through the | (r — N~) (r — iV+3) points. FF' meet the cubic again in six points on a conic. E.
.]
§ 8. curves A and n meet in the points L and N'. CC. B. Therefore every ?'-ic through the fr(r + 3) — 1 points passes through every intersection of this pair of degenerate r-ics which proves the result.
I. r-ic is
A
other cubic and the line through
two of the
three. C.
denotes a homogeneous expression of The curve Gr = is a curve of degree r. the lines AA'. F'. If a quartic meets a cubic in A. y. D\ E'. z. A'. F. DD'.
number
. n all pass through
G^.
Theory of Residuals.
'
Suppose that
degree r in
Gr
x.
'.
the curve r
r. Suppose that the curves m. 2. curves I and n meet in the points and M.194
INTERSECTIONS OF ANY
total
TWO CURVES
will be
XII 7
The
number of points taken
(r
found to be
|r(r + 3)-l.]
five
of
Ex.
Ex. BB'.
[«-ic is cubic. 3.
CA Gm
(i). EE'.]
quartic through the intersections of two cubics meets either cubic again in three collinear points.
[Take the quartic in Ex.
I.
§ 9. The tangents from a point six points on a conic.
Ex. r-ic is the six lines.
in
[Take the conic of Ex.
confusion with the
.
. JV-ic is quartic.
m m
.]
to a cubic
Ex.
which we
Suppose
shall call
Cj.
fj.
1 as
a pair of adjacent conies.
v respectively. Now by (i) the curves I and m. fi. and so on. we obtain relations between the curves and their . the complete intersection of any two curves represented by faces of the cube (not opposite) is the points whose symbols lie at the ends of the edge common to those faces. x = 1. Proceeding thus. z = 0.
o2
.
&c. n are so that the curve v passes through the points N. complete intersection of the curves
m
O'L' of u and
v. y = 1. m. Suppose they meet in points on I and 0' on X. MN' of /x and n.
*-x
Each curve passes through the points whose symbols lie at the corners of the face representing that curve. meet in points each of which lies on one of the curves n or v. Thus the curve I passes through the points OL' MN. z = 1 represent the curves I. Thus OL is the and n. the curve X passes through the points O'LM'N'.
U
.XII 8
THEORY OF RESIDUALS
/j.intersections which will be clear from Fig. \.
195
By (i) the curves and v meet in points each of which lies on one of the curves I or \. But the only points common to the curves I. In this diagram is shown a cube with unit edge whose laces x — 0. y = 0. n. Moreover.m. 2.
e.
[Tf
CN =
and Cn
=
0. form taken all together the complete intersection of Cn with a given curve. Show that these three (n-N)-ica form a pencil.AT-ic three ra-ics are drawn. n Cv are given polynomials such that l + \ is not and n. yjr of 0. Then use
</>.
=
=
=
=
=
We
=
=
M
=
=
=
M
=
M
=
=
=
=
=
+ M=
Similarly
O.
yjf
be the given w-ics
become when we
substitute in
them the coordinates
/". but at no other point of the given n. 1. 2. Through the remaining intersections of each pair of n-ics is drawn an (n-N)-ic.f
i.given point three given n-ics taken in pairs form a pencil.
— —
XII 8
Ex. Q.
what /.
=
with
O and L
o.
the three «-ics are
Q.
O+L
since
= 0. The »-ics through any . meaning that. As in § 8 we may consider and as the complete intersection of Ci and Cn 0.
O and L
are residual.
.
Cn +CN T =
and the(w-JV)-ics are B
= r.
^
M<tA-^A) +0oW'o/-/o<M +^o(fA-i>of) = o-]
Ex.
[Let
oi
and the intersections of
.
A
=
B.* &c. This will be expressed by the notation
+O
=
where Cp less than
m
Gm GK
. <£". Gm For the proof of this important theorem see § 10. Suppose any curve GK through L meets Cn in the points L and N'. taken together...-ic. and GK passes through the intersections of Gn and other than 0. while the curves (7j meet at 0.
Cn +CN A =
Cn +CN B =
r
0. they are the complete intersection of Gn with some curve (Gl 0).196
THEORY OF RESIDUALS
/= 0. care is necessary if some of the points are multiple points of the M-ic. namely C^ 0. assume its truth and deduce some consequences. it is always possible to find polynomials C>.
0. if groups of points P. subject to certain limitations as to the positions of the points 0. t And in general. we write P+ Q+ R + .
= A. B. and O and L as the complete intersection of Cm and Gn 0. Then the points and N' are the complete intersection of Gn -with * some curve.]
Let us now consider Gn as a given w-ic with any points taken on it. Gv such that
=
We
CjCa + ®mPn 0.
cm =
form the complete intersection of G"
* For instance. assume that. Gm each of the points 0. =z o.. . Through the nN intersections of an n-ic and an . The points O and are said to be residual groups of points for the curve Cn 0.
are the JV-ic and
re-ic..
[P+P! = 0. + PS
-
=
Q.
L and
M are
an
n-ic. —
.'
... show that this point is the same whatever the curves used and whatever the value of r.
in number. at some stage or other the algebraic result must be interpreted geometrically. they
coincide. we
see that symbolical
relations such as + 0. If the points
P are p
n lr
.
In
we proved
M
to 0. If P2I consists of a single point.
The
If
result
above obtained
may
then be worded
any
coresidual groups of points for points residual to L are residual to M. since L and were both residual residual to L were residual to M.
XII 9
THEORY OF RESIDUALS
197
The points L and are said to be coresidual groups of points for the curve Cn meaning that they are both residual to the same points 0.
— n a .
any points N'
Symbolically. of course. Through points P on a given n-ic any curve is drawn meeting the n-ic again in points P1( through t is drawn any curve meeting the «-ic again in 2 through P2 is drawn any curve meeting the »-ic again in P3 . if OL.]
* 'The points Q and the points P taken twice from the complete intersection of the »-ic with some curve.
P=P
Hence
P^-i +
2r
.
is
points.
fact.
P
2
.
we have m (»j
+ n.-
:
. 2.
we deduce from
that
+L
= 0..
. Such a treatment of the symbolism is a great economy of labour.
M
L = M. then we have Ex * if ptr and ^as are single 2S
=P
%
P=P
P^O
give
(say). &c.
Remembering the notation
L = M.
and the curves are of degrees
m. . if it is to be of any value. We denote this by the notation
. [Any line through P other than PQ meets the n-ic in n — 1 points which cannot form with Q the complete intersection of the 4re-ic with any
curve.]
Ex..
Two
single non-coincident points P.
Q cannot be
coresidual
(»>2).. In other words. and 2P + Q to mean + + Q 0* &c. can be added or subtracted just as if they were ordinary identities.
=
=
— M=0
P P
=
. O + M = 0. OM. Multiplication by an integer will be lawful..
1. for that is only equivalent to repeated addition but division is not allowed.. so are MN'. It is only necessary to consider such a relation as L to be equivalent to L M.
P
P
.
.
We
Ex..
1=
M+N'=0. LN' are all complete intersections of the n-\c with some other curve.)— jp + 1 =
0.
2
P.
«j.
that. must be careful to confine our algebraic processes to addition and subtraction. It enables us to some extent to replace geometrical reasoning by elementary algebraical work though.. L + N'=0.
Suppose other curves used.
%.. and so on.
P! +
-
P = 0.
9. Show that the conic through Q and the cubic through R meet in three points on the quartic. so do the intersections of the 4-ic and the two
conies are collinear.
5.
'
8.]
pairs of points on an «-ic (n no three of the four points are collinear. 3.] 0.
P
[Use Ex. R +
P+Q =
P+R =
S=0
Ex. through them draw a conic meeting the 3-ic in one more point R.198
THEORY OF RESIDUALS
XII 9
Ex. Let the line joining R meet the 4-ic again in two points S.
remaining
The four other
and 4-ic lie on a conic. Three cubics go through seven points. Let the 3-ic and 4-ic meet in six other points Q. meet cubic 3 in H.
1.]
Ex. and 2P+Q = 0. through these points draw a 3-ic meeting the given 3-ic in seven more points. Then on cubic 3 we have
1
3. Show that the lines joining the remaining intersections of the cubics taken in pairs form a triangle whose vertices lie one on each of the three given cubics. B + K=Q.
Two
> 3)
cannot be coresidual.]
lying on an (» — 1) (m — 2)-ic. see § 4. K.
if
[As in Ex. 0.]
Ex. so that by Ex.
.]
Ex. Ex. through them draw a 4-ic meeting the 3-ic in five more points. 6.
meet the curve
[Denoting the n points by
P=0
Hence
Ex. Show that any conic through the four points initially taken meets the 3-ic again in two points collinear with R. H=K. Two trios of points on a quartic cannot be coresidual (unless they coincide).
8. and the n{n-l)(n-2) points by Q. P+B = 0. The tangents at n collinear points of an again in n[n — 2) points on an (w-2)-ic.
If six intersections of a 3-ic
six. 1 H and K coincide.
For the
P and the n(n — 2) points by Q. and the intersections of cubics 2 and and 3 by A.
Hence
Ex. by
2P+Q=0. Ex.
The reader may enunciate similar theorems. [Denoting the points of contact. Through seven points P of a quartic a cubic is drawn meeting the quartic in five more points Q. Show that the remaining intersections with the w-ic of a p-ic through the inflexions lie on a (p-Sn + 2 S + 6)-ic which passes through the intersections of the n-ic with its nodal tangents other than the nodes. which lie on the P. Then give Q + S=Q. An w-ic has h nodes and no cusp. 10.
B
These give
P+A = 0. Let the lines joining the points A. 2. 7.]
?j-ic
Ex.
Ex. For the case » 3 see § 7.]
From any point n (n — 1) tangents are drawn to a non-singular Show that they meet the curve again in n (n — 1)(«— 2) points
=
=
P=
first
polar of 0. 11. 5.
remaining five intersections of the 4-ic with Therefore the trios coincide.
[One
trio is residual to the
any conic through the other
trio. Through is drawn a quartic meeting the given quartic in nine more points R.
4. and Q 0. See also § 5. 3. case n = 3.
[Denote the six intersections of the 4-ic and the first conic by i*and the other two intersections by R.
[Denote the seven points by P. A + H=Q.
[See Ex. B respectively.
Q=0. Through four points of a given 3-ic draw a conic meeting the given 3-ic in two more poinds.
»-ic.
then we can put/ in the form Afy + B-f. so that 1=0. XVIII. I+R Q (since the points I and R lie on the Hessian). We may suppose the axes of reference taken perfectly generally. 6.]
0. S 1. 13. Q + R 0. Then P+I=0.
. as the X shall suppose that the intersections of $ 0. which is an equation giving the abscissae of the intersections = and \j/ = 0. Ex. Show that the remaining intersections with the w-ic of a p-ic through the points of contact of the tangents from and the intersections of the m-ic with the tangents at (other than 0) lie on a {p-n + %)-ic. Then from the Hessian 1 + R 0. it follows at once from the process of finding the highest common factor that each remainder found in the process.
reach a remainder R involving x but not y. Ex. and in particular the last remainder R.
Ex. of which R coincide with the nodes.y) 0.]
=
=
=
Ex. v/r are ordinary distinct points of 0. as a factor.
and where
§ 9 we made use of the theorem that. is of the form A</> + fx\jf.
A
We
-C
. but not (x — af. 6. The sixteen inflexions of a quartic with a biflecnode lie on another quartic. if the curve passes through the intersections of <j> (x. and taking/. and from the nodal tangents R See Ch. b) be an intersection of <p 0. <$>. [Denote by I the inflexions and by R the intersections of the 4-ic with the nodal tangents. the quotient is v and the remainder 6. so that
till
and B are polynomials in x and y. letting I be the points of contact of tangents from and replacing the Hessian by the first polar of 0. y) y) \jf(x.
=
=
=
. To and \jr we may carry out the eliminate y between (j> process of finding the highest common factor of $ and \j/. so that P+ Q 0.
=
=
§ 9. 11. where A. The remainder R must have therefore of x — a. when A/ is divided by ^. and /* are polynomials id x and y. are here using Cartesian coordinates. and give a proof of the result with this limitation. Suppose that. Moreover. and \jr Let (a. \jr 0. 0. <p 0.XII 10
THEORY OF RESIDUALS
199
[Suppose the re-ic and p-ic meet at the inflexions / and other points P.
In
/ (x. while the nodal tangents meet the w-ic at points Q and R. An »-ic has a single node 0. and the eight inflexions of a quartic with two biflecnodes lie on a conic. considered as polynomials arranged in descending powers of y.
[As in Ex. Equating remainder to zero we get the result of elimination required.
We
=
=
/=
—
=
=
=
=
=
we
this
(j>
v
and
being polynomials in x and
y.j
§10. ty CK Gm Gn of § 9. 12.
b) lie on A Then 6 A/— v\jr shows that all the n intersections of
—
A
n—
R=
=
=
=
=
R
—
=
=
=
n— =
=
= = and x = a lie on = 0.
or 6
R as =A
a factor. cannot meet = a in more than %— 1 points unless — a is a factor of 8.
and therefore
.
if
that is proved. But = 0. $ = 0.
(\cj>
+
fj. The relation \<j> + /j.\jr shows that every one of the or <j> with x a lies on A n intersections of i/r 0. b).A <f>)
=
Now. For the case in which the intersections of / = 0. 6) lies on / = and = 0.
(v
+ A ^)/A must
be a polynomial
B
. being of degree n — 1.\jr).
since
R
+ A n) = A<£ +
(v
jxt/t
R
independent of
i/f
y. But the results there given can also be established in most cases by other methods.
A
being
A/ =
or
+A
and
(A</>
+ nir)
is
A (/. we prove that 6 has x a similar argument will then apply to each factor of R. we have 6 a polynomial.
*
paper in the Mathematische Annalen from which the above proof has be'fen have applied the residuation theorem in the 9 to such cases. for (a.
may
refer to Noether's
p. has not if we have chosen the axes of reference generally. and x a other than (a. while the other % — 1 lie on = and A = 0. But.
.*
We
examples of § without a more careful investigation.
a.
140. Hence A and
\jr \jr
factor.
or
f=A<f> + B^r as required. We should then have
= AR
v\jf
\lt. adapted. i^ = are multiple points of any of these curves the reader
\|f
\j[f
i//-
a. and the result follows.
any
common factor of A and would be independent of y.
xl (1892). Suppose \jr of degree n in x and y'smd therefore 6 of degree 1.
a factor. have no common a function of x as factor.200
It only
THEORY OF RESIDUALS
remains to show that 6 has
XII 10
For. But (the axes of reference and i/r with being general) the only intersection of <£ 1 intersections of i/r abscissa a is (a. Hence the 0. To show that has R as
as a factor. which was not strictly lawful
The reader may consultthe bibliography
in the Mathematical Encyclopaedia. since has x a as a factor.
Since a line joining two double points of a curve meets it in at least four points. Hence there are only three types of cubic.CHAPTER
XIII
UNICURSAL CUBIGS
§
1.* whose Pliicker's numbers are given by the table
Type
.
Types of Cubic. a cubic cannot have more than one double point.
according as the cubic has a cusp.
l.202.
Since the sum of these is 2tt — <f>. according as
h2
=
.']
. § 4). acnode. VII.' In fact. and then inverting with respect to the acnode we get Through any point of an ellipse three leal circles of curvature pass other that the circle of curvature at 0. The real asymptote of a cuspidal circular cubic is equally inclined to the tangent at the cusp and to the line joining the cusp and the focus. or hyperbola. and the inflexional tangents of
real
The
the cubic invert into the other osculating circles of the cubic which pass through 0. [Take the case of the acnodal cubic.
and unit
we
obtain the conic through
ax2 + 2hxy +*by 2 + 2gx + 2fy
.
infinity. 2. 1. and therefore the three inflexions are collinear (cf.
. e.
[Inverting with respect to the cusp and remembering that the focus inverts into a focus (Ch. A cubic with an acnode has three real inflexions.
<
.
1(4*-*). By a real projection it may be transformed into a circular acnodal cubic.
GEOMETRICAL METHODS
XIII 2
Then.
Ex.
iV"-<t>). Ch. V.
Fig. § 4) we have : ' Any tangent to a parabola is equally inclined to the axis and to the focal distances of the point of contact. ellipse.
asymptote of the cubic inverts into the circle of curvature of the conic at 0.
This conic is a parabola.
=
0. if the origin is taken at the double point of the cubic. Similarly for the crunodal cubic]
Ex.
or
>
ab. the eccentric angles of the three points of contact of the circles of curvature are
: '
-*<#>. or crunode i. the three points of contact are concyclic with 0. the curve is now a circular cubic with an equation of the form
ax 2 + 2 hxy + by 2 + 2 (gx +fy) (x2 + y 2 ) = Inverting with respect to a circle with centre
radius. if the eccentric angle of is (f>. and a cubic with a crunode has one real and two unreal inflexions.
0.
and that the circles on such chords as diameter are coaxial.
S.
If O is the centre of a fixed circle touching two given circles externally. Obtain properties of a nodal circular cubic by inverting other properties of the conic. point of contact of a tangent from O to a family of confocal conies is a circular cubic through the foci with
a node at
O. and after reflexion passes through a fixed point B. 13.]
x [a? + y ) + ax 2 + 2 hxy + by* = are perpendicular. find the locus of the inverse of with respect to any other circle touching the given circles externally. that the middle point of the chord lies on a fixed straight line. inter-
\x + fiy +
to the origin envelope. 0).
Consider the case in which O is on an axis or at infinity. if O.
Ex. Generalize by projection. is reflected at P from any sphere with a given centre O. B
are the origin.
.
s
and any
.
-6 sin
a.]
Ex.
0). we get the tangential equation
1
=
and
1
of the
Ex.
subtending 90° at
Ex.
Ex. Ex.
line
Pi.
or
is
0Q n
zero.
The
cissoid
becomes a parabola with
If a line
vertex 0.
—
AA = AO A'O
'
. 12.
6. The locus of the.
[Part of a circular cubic with a crunode at 0. 7. S'
and the
SO ± S'O *
Ex. Find the locus of P. 10. A. A cubic has three asymptotes and a node 0. P2
.]
Ex.
circle
is the node of a circular cubic with real foci 5.
P P
2
. A ray of light proceeds from a fixed point A. 11.
Show
Ex. The locus of the centre of a circle whose circumference passes through two given points and meets a given line not coplanar with the points is a nodal circular cubic. 1 [(x' + y ) {2/ (6 + a) cos a + a: (6 -o) sin a} =2abxy.
Any
.
8. z = 0.
O envelops a
sections of
[Expressing the condition that two of the lines joining the. The lines joining the node of a nodal circular cubic to the real foci are equally inclined to the real asymptote.
0QS 0Qt
.
make with the tangent
[Invert with
at the cusp
respect to 0. meets the cubic in Q and the asymptotes in P.
through
.
show that
0P1 + 0Pi + 0P3 =
'
oq. If OSS' meets the curve again at A and A'.
a sin
Oi.]
. 4.
meets the cissoid x(x i + y'i ) = ay 1 with cusp in circle meets it in Q lt Q2 Qs Qt Show that the sum of the cotangents of the angles which
Ex.
OPlt 0P 0PS
2
.
OQi. 3. Consider the case of confocal parabolas.
. 9. A chord of a circular cubic with a node subtends 90° at 0. (a cos
Oi.XIII 2
GEOMETRICAL METHODS
203
Ex. (6
cos a. The chord of a circular cubic with a node
conic.
P
3
.
i have the Cartesian equations 2y*(x-l)-2y(xi -2x)+x*(x-2) = 0. Hence the equation of the cubic is zy 2 axs Putting az for z. since z meets the curve three times at B.
XIII 3
Cuspidal Cubics.
.
=
. and the only term involving z is zy 2 since y 2 are the tangents at G.* Then there is no term in the equation of the cubic involving z 3 or z 2 .204
CUSPIDAL CUBICS
§ 3.
»
A
real cuspidal cubic has one cusp
and one
inflexion. 2. Take the triangle of reference HG so that and are is the inflexion and G the cusp. we see that
=
=
=
=
. 64y*(x-i) + 16x'y + xi (5x-12) = and x(x 2 — xy + iy*) + iy (2x — y) — respectively. 3. the equation must reduce to X 0. when we put z 0.
A
choice of
cwpidal cubic can be put into homogeneous coordinates. while the tangents at and C (Fig.
0.
s Also. 2). 2.
Fig.
and
these must be real.
B
AB
A AG
B
.
the form zy 2
=x
3
by a real
* The curves shown in Pigs.
the
we have sum of their
parameters
Similarly
is zero. or simply the tangential of P. The tangential of the first
' '
tangential is called the second tangential.
:
and
conversely. the tangential of the second tangential is called the third tangential.
If the points
with parameters
t lt t 2
. and
conversely. I.
1. for the tangent at meets the curve in the points P.
(-2)H. If the parameter of is t.
or into
yx*
=
as
. P. Q is called the first tangential.
tLL
(
^-2)
M
t
=
oo
.
and
conversely. the
=
x 3 are
sum
the intersections of this cubic of their parameters is zero.
=
.
If the tangent to any cubic (unicursal or otherwise) at P meets the curve again at Q. If the points with parameters t x t 2 t6 lie on the line
—
.
Ex.t 5
t
e
lie
on the
conic
ax2 + by2 + cz2 + 2fyz + 2 gzx + 2 hxy
these quantities are the roots of
at 2
=
0.
t
s
are the roots of the equation in
Xt
+ /i + vt 3
— 0. the
so that
t
1
+ t2 + t s
=
0. the parameter of the first tangential Q is — 2t.y + vz
t
=
0. Q whose parameters have a zero sum.
collinear. 1.
successive tangentials of a given point approach the cusp as their limiting position. t 3 ).]
.
t i .
2 If nine points of zy with another cubic.t 3
.
+ b + ct 6 + 2ft 3 + 2gt* + 2M =
t
s
Since the coefficient of
2 If six points of zy
in this equation is zero.
.
Any
The
cuspidal cubic can be projected into
a2 y Ex. t being the parameter of the point. § 3 that any cuspidal cubic can be projected into the semicubical parabola ay 2 x3 Any point on the curve zy % x 3 may be taken as (t.
and the parameters of the
inflexion
and cusp
•00
are zero and infinite. and so on.
0.:
:
XIII 3
CUSPIDAL CUBICS
205
It follows from Ch.
=x
s
.
Hence
2 If three points of zy
=
x3 are
sum
of their
parameters
is zero.
2. The parameter of the n-th tangential of is evidently
P
P
P
.
\x + /j.
t
lt t 2
.
lie
=
xs
on a
conic.
Through a point P of a cuspidal cubic a conic is drawn having 5-point contact at P'.
in three points. Show that the line PQ passes through the inflexion and is divided harmonically by the inflexion and the tangent at the cusp. Ex. 16
(i). inflexion. a tangent is drawn touching at P2 and P" approaches the inflexion as its limiting position.
3. the inflexional tangent. 8.
=
[A parabola having double contact with the given curve. i )
s
is seen by putting 8. conic passes through two given points of a cuspidal cubic.
From a point P on a cuspidal cubic a tangent from P. 9. §
2. The tangents from P to a cuspidal cubic meet the curve again Show that the tangents at these points meet in a point Q and that.
Ex. 4. If in
Ex.
. and so on. 11 the cross-ratio of (PQES)
inflexion.
.
P
P
.
P
T
Ex. similarly P2 is derived from P1.
7. 5. cuspidal tangent. The loci of the poles of the inflexional and cuspidal tangents with respect to the conic of closest contact at any point of a cuspidal cubic are cuspidal cubics with the same cusp. E.
flexional tangent.
is
any constant other
than 9/8. The conic of closest contact at P to a cuspidal cubic meets it again in P. 6. Find the limiting position of P< n >. conic osculates a cuspidal cubic at and Q. Find the limiting position of P".
Find the locus of the intersection of tangents to the semix s at points subtending 90° at the cusp. 13. X. the cuspidal tangent in Q. 16. and cuts it again at Similarly 2 is derived from t Pj.
Ex.
XIII 3
drawn touching
so on.
A
P
P
to a
Ex.
12.]
Ex.
closest contact at any point of a cuspidal cubic (other than the cusp or inflexion) meets the inflexional tangent in unreal points and the cuspidal tangent in real points.]
Ex.206
Ex. and so on.
P
. The tangent at to a cuspidal cubic meets the curve.
See also Ch. 11.. 7 to 13 by reciprocation. osculates it at P. Find the limiting position of n 3 from 2
A
P
P
. 1. SHbw that the cross-ratio of the range {PQRT) is 9/8.]
Ex.
CUSPIDAL CUBICS
is
.
Ex. z in the general equation of a conic and comparing the resulting equation for 8 with
as
(6-tf(8-t') =0. respectively.
14. Ex.
(t.
Show
that
[The parameters of
P and
P" are in the ratio
1
:
(-£)".
Obtain theorems from Ex. 10.
15.
[The parameters of
P and P
1
are in the ratio
1
:
—5. and so on.
. cubical parabola ay 1
Ex. find the locus of S. similarly P" is derived from P.]
cuspidal tangent. 1. [A cubic with the same cusp.]
Ex. The locus of the intersection of two perpendicular normals to a semicubical parabola is a nodal cubic.
. so does Q.
at P.
and
in-
Ex.
and inflexional tangent. Find the locus of the intersection of the tangent at cuspidal cubic with the tangent at the second tangential of P.
The conic of
[The conic of closest contact at
i 2
with zy 2 — x* is 4ht*x + 5t?y*-z -40?yz + 15t*zx-24:t?xy = 0. y. 6* for x.
Ex. if P moves along a straight line.
18. the coefficients of x2z and 2 Choosing suitable homogeneous z have the same sign.
The
P and
locus of the focus of a parabola passing through fixed and having PQ as normal at P. is a straight line.
t. and CB is the harmonic conjugate of CA with respect to the tangents at C.
2at).
[Any point on the curve is x Ex.
Ex. and Ch. Ex. e.
[If the parabola is i/ 2 = 4ax and perpendicular from the focus S on PQ.
§ 4. The locus of a point P.XIII 4
NODAL CUBICS
207
Ex. The equation of any w-ic with a tangent of ra-point contact and a superlinear branch of order re — 1 may be put in the form
zyn-'
= xn
. is the cissoid of Ex.
P
is
(at
1
. 19.
PQ
.]
PZ= a {! + ?)*. 21. their parameters have zero points with parameters t. CB is the harmonic polar of the inflexion A.
Ex. 17. and the tangents at G form a harmonic 9 xz 2 . P and Q is
+
2 < ). i. Find the circle of curvature. Show that
(i)
(ii) (iii)
+y =
i
)
ay* subtends 90° at
The The The
locus of the middle point of
PQ
is
a straight line. y
=
. The sum of the abscissae of the feet of concurrent normals of a (x — yf = x d is constant.
The equation of any acnodal cubic can
z (x 2
be
put in
the
form
+y
2 )
=y
(3x 2 — y % )
by a suitable choice of homogeneous coordinates.
fi
=
4a
(1
+ i2 )i
eliminate a and
Ex.
locus . the 3 2 coefficients of x and arc/ must be zero.of the intersection of the tangents at
locus of the intersection of the normals at
Pand Qis a circle. The chord PQ of the cissoid x(x 2 the cusp. — 3< is + lt*)x = 6t*y-¥a enveloping 3 5 y 2 (a-x) = Vz 3 .
Ex
points
20. 10. Take the triangle of reference ABC so that C is the node. remembering that the tangent at A meets the curve thrice at A. the coefficients of z equation of the cubic are zero. of which one at least must be real. A real nodal cubic has a real node and three inflexions.
Nodal
Cutaies. Since A is an inflexion.
[If four points
=
a/(l
+
fi). which moves so that the Bum of the angles made with a fixed line by the tangents from P to a semicubical parabola is constant.
a straight
line. yz 2 xyz in the pencil with xy 0. 8. A is a real inflexion.
The chord joining the
(l
on the curve are concyclic. since G is an acnode.
y
=
alt (1
See
§ 2.]
Q. 18.]
Ex.
envelope of the
common chord
of a cissoid and
its
sum.
.
while
SZ
is
the
SZ=at(l + t
Now
i
)i. XI.
. § 8. Since C is a node. 22. Moreover.
z = The coordinates
is
of
any point on
sinh fa
the line of inflexions. we obtain the parameters of the
.
is
77-.
and
conversely.
namely
0. fa. we may say that the whole curve is divided into two portions by the crunode.
-| 77-. 77) . iti) for (mod.
=
0.
{mod.
The condition for three points on a line or six points on a conic is the same as that given for the acnodal cubic..
The sum of these
. fa. except that the coefficients of x 2 z and y % z have opposite signs.
Similarly
lie
An
we may show
conic.
tan fa. owid conversely. tan fa are the roots of 2 (cos <f> + sin 2 fa (A cos <f> + u sin fa
+ v sin
considered as an equation in
tan fa
tan<Jb. are collinear.
(cosh$. One contains the
A
. The inflexions have parameters 0.
on a
+ fa + fa: + fa + fa. The Hessian of
z(aP-lf)=y(aa? + y*)
is
(iii)
z(x2 -y 2 )= -3y(3x2 + y 2 )
. 2
+ y 2)
by a suitable choice of homogeneous coordinates.
ir) . -n). s as found before by means of the Hessian.
(iv). fa lie
209 on the
line
Xx + fiy + vz
=
0. \iti.
on the cubic
with parameters fa. fa.
=
inflexions.
As before. fa.
Hence
crunodal cubic has three collinear inflexions of which only one is real. fa + (£ 2 + $ 3 Putting fa.
<f>
(3 cos
2
$ — sin 2 fa
tan fa
. %iri. If we suppose that the points of (iii) at the two ends of an asymptote are the same. substituting (mod. fa
that
(i)
If six points
fa.::
:
XIII 4
If the points
NODAL CUBICS
with parameters
fa... $ 2 <£ 3 (mod.
Hence:
*
acnodal cubic has three real collinear inflexions . tan fa. fa. + fa
be
=
fa..
Therefore
+ tan fa + tan
fa
= tan fa
tan
<f> 2
Hence
i/ three points on the cubic (i) with parameters fa. fa all equal.
The equation of any crunodal cubic can
z (x 2
put in
the form
—y =
l )
y
(3a.. (iii) are
sinh3^>). The proof is almost exactly the same as that given for the acnodal cubic.
§7r.
We may
mention the form x(z 2 ±y 2 )
=y
3
(v)
obtained by taking A as the node. cosh
<\>). is obtained by giving the value 6 + ^-ni. § 1 for
the non-singular cubic.
cos
<£)
or
(sinh 3
<f>. The line of inflexions is 4x = ±Sy.210
real inflexion. The form (v) is suggested by that given in Oh.
Many convenient standard forms of the equation of a nodal cubic exist.
<\>
cot
<f>
=t
or coth
<j>
=
t. by putting
— oo
to
+ oo
.
. The coordinates of any point on a nodal cubic can be expressed rationally in terms of a parameter t. 4). other than (i) and (iii). where 6 takes all real values between — oo and + oo (Fig. XVI.
Fig. G as an inflexion. the harmonic polar of G.
sinh#.
NODAL CUBICS
<j>
XIII 4
and is obtained by giving all real values from The other portion.
AB
as
Any
sin</>. 4. and CB as the tangent at G.
The condition for three points on a line or six points on a conic is the same as that given above. point on the curve is
(sin 3
<£. which we shall call the loop.
Ex.1) a uu z = -(n. we may use the theory of covariants. Show that. § 7. the
product
collinear points a conic is + 1. 3.
or one
(x + a ) real inflexions according as it has
3
Show from the form y
(3
2 a. R. X. Ex. Q. prove that from any point on one part two real tangents can be drawn to the curve.
3
2
. Show that no real tangent can be drawn to a crunodal cubic from a point on the loop and that from any other point of the curve two real tangents can be drawn. 113.]
Ex. §
2.. Show that the equation the form x s + y s = Bxyz. 6.
2^ + ^ =
(mod. 4. 4. one point of contact being on the loop
.
— 1 and the
can be projected into the folium of
Ex.
.' 4 the product of the parameters of three collinear points is — 1.
[The tangents at the finite inflexions of y(x i + al ) = which is the harmonic polar of the inflexion (oo 0). line meets a nodal cubic in P. xxxii.
sin fa)
constant. Messenger Math. Show that.
. Considering an acnodal cubic (supposed continuous at the two ends of an asymptote) as divided into three parts by the three real inflexions. verifying the for any special form of the equation of the cubic]
= 3.2 a A . 16
(ii). 5.
1
=
a that a nodal cubic has three an acnode or crunode. VII. The conic of closest contact at P with a crunodal cubic meets the line of inflexions at unreal or real points according as P does or does not lie on the loop. A t ) y = 0.
a?
meet on x
=
0.) x + 3 (a A 3 + a % A x .
1
+ f). and the lines joining these points to the node make angles 6 1 and 2 fa and </> 2 fa and fa^ with the tangents at the node.
sin fa
.]
Ex. Or again.
Ex.
ttj).
its
ty
are the parameters of a point and
tangential.
sin
<f> 2
. Show^that the tangents at two inflexions of a nodal cubic meet on the harmonic polar of the third inflexion.(sin #2
. 14 (v) in the
(a^x*
l x
case
n
p.
inflexion.
sin fa) 4. 7.2) a m of Ch.]
.
NODAL CUBICS
A
nodal cubic can be projected into 2 a (x 2 + »/ 2 ) = x s or into y (x" ± a )
1
211
=
3
u.
[Use the fact that in Ex.
result
For another proof see Bromwich.and Ch.
not.4
i
XIII
Ex.
The
line of inflexions of
i
is
+ 2a xy + a2 y2) z + A^x3 + %A x2 y + ZA^xy 1 + A s ys = .
'
of any crunodal cubic can be put in
curve
is
taken as
is
(3
1.
[See Ex. 8.
9.
.o^ 2 ) z + 3 (a A 2 + « 2 A . 4 (n a 2 [It is the (n .
2.]
Ex.]
3
d?V [— = 2 a
_
+
«')/(*"
+ a 2 ) 3 vanishes at an
Ex.
Ex.
1. A. one point of contact lying on each of the other
two
parts.
and the other
[If
<f>.
f2
.2 a.
(sin 6X
is
.
Ex. if any point on the product of the parameters of three of the parameters of six points on Show that any crunodal cubic Descartes x 3 +y 3 = Saxy.
=
{sin 2 a
+ sin 2 .
(iii)
As
P
[Take the cubic of § 4 (i) the case of the cubic (iii) is similar. and 'QR at and the line of inflexions at F. 15.
it
[Eliminating z between the equation of the cubic and the polar conic we get the rays of the pencil. § 11. R. 14. Find the
locus of P. and the tangents at Q' and R' meet on the curve at P'.212
Ex. I.
Ex.
QR and
Q'R'
*
meet
Ex. 11. — 10.
.
at
is
NODAL CUBICS
A family
line. The tangents at Q and R to a cubic with node the curve at P.
. Prove that the points of contact with the cubic of the six other common tangents of conic and cubic lie on another conic. and the line of inflexions the axis
of perspective. 17.
If the tangents at on the line of inflexions.
The
— £71-.
(ii)
EP)
is
harmonic.
whose envelope
is
s2
=
a. S are respectively 0. If is harmonic. the line of inflexions being the chord of contact. Now use Ch. equation of the family in the form
3
/fc
Or put the
33
=
(x + y + 3kz)xy. If the pencil is equianharmonic. 13. The conic through touching the cubic at Q and B touches the
on OP.
of P.
XIII 4
two of the inflexions are
a straight
of nodal cubics has given inflexions and the tangents Show that the locus of the node also given. the locus is a cubic]
Ex. The points of contact of the four tangents from P to a nodal cubic subtend a pencil of constant cross-ratio at the node.sin
§
2 (a
+ 0)} x +
{cos
2a + cos 2 + cos 2
+ 0)}
For the cubic of
4
(iii)
replace sin and cos by sinh and cosh. Show that
meet on
E
OP
meets
(i)
OQ and OR
and
(OF.
Ex. parameters of P. Show that
(i)
(ii)
The conic OQRQ'R' meets the line of inflexions on OP and OP'. the node being the vertex of perspective. 12. The line QR meets the curve again at S. \(ir — 4>). The line joining two points of the cubic of parameters a.]
§
Ex.
a
The
line
QR
is
z + sin 2 f x + cos 2
y
= 0. Q.
varies QR envelops a conic touching the tangents at the node.
line of inflexions at a point
(iii)
P
and P' meet on the
curve. A conic touches the tangents at the node of a cubic.
3. A nodal cubic and its Hessian are in plane perspective. 10. The other intersections with the curve of any line through a fixed point of a nodal cubic subtend an involution at the node.
+ y*. 16.
[The harmonic polar of the third inflexion by Ex.
Ex.]
Ex. (3 is
z
4
(a
(i)
with
y.
at
also with respect to
are harmonically conjugate with respect to the tangents OP and OS. the locus is the line of inflexions. The tangents at Q and R to a cubic with a node meet on the curve at P.
.
. R.]
Ex.4
XIII
NODAL CUBICS
.
213
Ex. 21.
3x
Ex. the chord meets the cubic again at a fixed point. Prove a similar theorem for the six points in which the cubic of Ex. and that the centre of this circle lies on a fixed line. Show that the locus of the centre of the circle is a fixed line. 18.
A
=
Ex. If a chord of x s + y 3 == Baxy subtends 90° at the node. Show that the points of contact of tangents from any point on = 2a to a{x2 — y 2) = xs lie on a circle through the origin. circle through the node cuts a (x2 -y 2 ) x s again in four points. Q. Q.
2 If f l t it t 3 are the is (a(l + t ). line cuts a{x i + y i ) = Xs in P. parameters of three colliriear points. which shows that the lines joining the points to the origin make angles with x —
[Any point on the curve
whose sum
is Jn-. 19. A straight. 22. t^ + t^ + t^ = 1. Show that the lines joining P. 18 meets any conic. It to the node make angles with any fixed line
whose sum
is
constant. at(l + fi)). and the sum of the angles which the lines from the node to these points makes with a fixed line is constant. 20.
Ex.
Project such a line to infinity. Place the asymptote vertical (Fig. 1. remaining parallel to the asymptote. 1). we see that there are lines meeting the curve in only one real .CHAPTER XIV
NON-SINGULAR CUBICS
1. cubics without a double point and of unit deficiency. e.
9 (i + 4) y 1 + 2ixy + x (x2 + 2x .
A
Fig. XIII. and the curye will have one and only one real asymptote. § 2.
Cubics with Unit Deficiency.
In this chapter and the two following we shall consider the properties of non-singular cubics.
.16)
=
0. As in Ch. and suppose a line I starts from coincidence with the asymptote and travels to the right. i.point.
namely the nine intersections Of these an odd of the cubic with its Hessian (Ch. The curve can have no other portion for. Hence there is a continuous portion of the cubic called the odd circuit * approaching the asymptote at its ends. namely the distant point and as a poirit travels along the curve from . The cubic has nine inflexions. For this is true of one point of the circuit. two real tangents can be drawn from to the odd circuit. the term even circuit' is applied to that part of the curve (if any) which can be projected into an ovaf. since both cubic and Hessian are real. Similarly if I travels to the left. the rest of the curve being the odd
it
. if two ovals existed. For any line through must evidently meet the odd circuit and must meet the even circuit in a point other than (see fig. unless the line is the tangent at P. when I travels still further to the right (or left).
'
'
circuit
'. 2
. and contained between two lines Zj l parallel to the asymptote. The curve will then have a portion called the even circuit * or oval contained between two lines l s Z4 parallel to the asymptote. and twocircuited cubics having an odd and an even circuit. t The names unipartite and bipartite are also used instead of onecircuited and two-circuited
' ' '
'
'
'
'
'
'
'
'
'. to P. that they are collinear. it touches the cubic again. number must be real. Of the three points thus obtained no two can coincide.f From a point on the even circuit of a cubic. a straight line cutting both would meet the curve in four points. § 7). If a cubic has three real asymptotes. Moreover. VII. for the curve in general approaches the asymptote on opposite sides at the two ends.
. From a point on the odd circuit two real tangents can evidently be drawn to the even circuit (if any). a real tangent from the point remains real and an unreal tangent remains unreal. since the curve has only one asymptote.
Hence cubics can be classified into two main divisions one-circuited cubics having a single odd circuit. 1). eventually these two points coincide and I is a tangent.
It may happen that.:
XIV
1
CUBICS WITH UNIT DEFICIENCY
215
At first I meets the curve in two real finite points. As I travels yet further
must again become a tangent. As I travels. and that they lie on the odd circuit of the cubic.
.
P
P
P
P
P
A
A
* The names odd or even circuit emphasize the fact that the circuit is met by any straight line respectively in an odd or even number of points. meeting the asymptote in a finite point. We shall show that exactly three are real. no real tangent can be drawn (other than the tangent at P).
. since and its interthe odd circuit does not cross itself * between the circuit has at least one section C with the asymptote.
. § 4). Ie I3 I6 IS !$!$. 2).
fl
. &c. VII.
.
Secondly. But every straight line must evidently meet the odd circuit in at. suppose that ij. The circuit has then two real inflexions. Then 1^.
A
Fig. 1 that. Then I-^I^. while 7j/7 and IS I5 meet the curve again in real inflexions Is and J which we may suppose to be their intersections with 72 /5 and Zj/7 since these latter lines also meet the curve again in inflexions. I2 ^4' ^z^-i meet the curve again in real inflexions I& I6 J7 .
* Contrast the case of the crunodal cubic in Ch.
. .
fig. The line joining them meets the circuit again in a real point which is also an
inflexion (Ch.
.
To show that the curve has no more than three real inflexions. I2 Iz are the inflexions just found and that Ii is a fourth real inflexion (Fig. meet the curve
. 4. tangent at such an inflexion would meet the cubic at three points coinciding with the point of contact.
Similarly it has an inflexion inflexion between A and C. and therefore no straight line can meet the even circuit in
more than two
points. least one point. . 2.216
CUBICS WITH UNIT DEFICIENCY
XIV
1
For the Firstly. XIII. no inflexion can lie on the even circuit. between B and G. suppose A and B very distant points on opposite ends of the odd circuit. It is evident from fig.
Circular Cubics. Then the cubic takes the form (i). and the tangent as the axis of y.
Ex. 6. the cubic is the locus of the point of contact of a tangent from to any conic through A. We may project the unreal intersections of the line and cubic by a real projection into the circular points co. 1. Hence we can always project a cubic into a circular cubic.
.
We pointed out in § 1 that we can always find a line meeting a real cubic in only one real point.
[Use polar coordinates with
as pole
. as the direction varies. such that the lines joining P to three fixed points meet three fixed lines in collinear points. They are parallel to the real asymptote.
5. is a cubic through the fixed points and the vertices of the triangle formed by the fixed
Ex.
lines. Take tbe point of contact of one of these tangents as the origin. ca. Obtain the results of § 1 by proving that at least one line can be drawn through an inflexion meeting the curve in only one real point
and projecting that
line to infinity. 2. x(x2 + y 2) + ax 2 +2hxy + by 2 + kx
=
. Ex. i. § 8. 00.
We
. Show that the locus of the middle point of PQ is a cubic with a node at and asymptotes parallel to those of the given
cubic.
§
1
Ex. Through a point of a cubic any line is drawn meeting the cubic in Pand Q.]
The locus of a point P. e. a cubic passing through the circular points to.
The inverse
of this with respect to the circle with centre the
* may apply the above proof to show that an acnodal cubic has exactly three inflexions which are real and collinear. OD are the tangents from a point of a cubic to the curve.
217
again in real inflexions contrary to the fact that the curve has no more than nine inflexions real or unreal*
Ex.XIV
2
CIRCULAR CUBICS
.
by projecting a
real inflexional
Ex. Obtain the results of tangent to infinity. 4.
i
§
2.
Ex. OB. B. C. If OA.
. Show that the locus of the centroid of the three intersections with a given cubic of straight lines drawn in a fixed direction is a straight
line
.
or see Ch.
and
that.
3. From the real point at infinity on a circular cubic two or four real tangents can be drawn according as the 'cubic has one or two circuits (§ 1). D. a/. XI. the locus envelops a conic. 3.
is
sometimes used in the sense of
'
its
own
. The remaining
circles are all
*
The word
'
inverse
or
'
anallagmatic self-inverse '. three of the circles with respect to which it is self.inverse are real. and thence that any one centre is the orthocentre of the triangle formed by the other three. found to be the cubic itself.
mutually orthogonal. since the cubic is. If the circular cubic has two circuits.
This system of circles is obviously self-inverse with respect any one of the circles.:
218
origin
CIRCULAR CUBICS
and radius
It* is
XIV
2
ky/(x 2 + y 2 ) for y. Hence
A
to
circular cubic is self-inverse* with respect to four circles. and
is
obtained by writing Tcx/(x 2 + y 2 ) for x. It follows that the radical axis of any pair of the circles must pass through the centres of the other two. Therefore the
Pig. 3
(i).
The two circles with respect to which a circular cubic is selfinverse and which have their centres on the odd circuit meet the cubic circuits. two of the circles are and the other two have unreal centres. 3.
circuit. 1. [The points are the points of contact of tangents to the cubic from the centres of the circles.
.
has a real centre on the side of the oval
near to the odd
real. 3
(ii). This will be clear from Fig.
Pig. The tangents to a circular cubic from a point where the tangent is parallel to the real asymptote are all equal.
Ex. at two or four real points according as the cubic has one or two The other two circles do not meet the cubic in real points.XIV
2
CIRCULAR CUBICS
but
it
219
circle is unreal. 2.] Ex.
ix(x i + yi ) + ix{x + 5y) + i0y-71x =
0.
If the circular cubic has one circuit.
4. then with the notation of that section. If we allow the use of unreal projection. for the cubic is of class 6.
The cross-ratio is real. and 8 is a focus. a cubic may be (projected into co and co'. There
. any two points of. or into a bicircular quartic symmetrical about two perpendicular lines. S. since
(PQRS) have the same
The tangents to a cubic from any point of the curve (other than the tangent at the point) form a pencil of constant crossratio. Hence m'S also touches the cubic. while the cross-ratio of a pencil formed by four real lines is real.
Suppose j is one of the four circles with respect to which a circular cubic is self-inverse.]
is self-inverse
Ex.
the pencils
co.
circles. Then the inverses of the cubic and the line coS with respect to j are the cubic and co'S respectively. Now four tangents can be drawn from co to the cubic. This is evident on considering the pencil of tangents from a real point at infinity on the cubic (see Fig. /c 2 + 6/c + 6 0. (ii)
the intersection of the cubic with a circle the intersection of two such circles. the cubic being thus projected into a circular cubic.
it is self-inverse. Hence we h%ve the important theorem
cross-ratio.
Foci of Circular Cubics. other than the tangent at co.
§ 3. Q.
co
(PQRS) and
P. Let a tangent from the circular point a) to the cubic meet j at 8.
If P. if the cubic has only one circuit.:
:
220
CIRCULAR CUBICS
XIV
2
is
Ex. if the cubic has two circuits unreal or equal to —1. Q. For the cross-ratio of a pencil formed by two real and two conjugate unreal lines is unreal. Its inverse with respect to any other point is a
bi circular quartic. and the sixteen foci lie by fours
S
are the foci on
j. Any circular cubic can be inverted into a circular cubic symmetrical about a line.
a/
circular cubic is self-inverse with respect to each of four circles. if the cross-ratio is 1.
Ex. R. the cubic has the same cross-ratio as the pencil of tangents from co'. Hence
Any
mutually orthogonal on these four circles.
[The pole of inversion
for
is (i)
which
5. R.
A
nodal circular cubic
and a cuspidal circular cubic with respect
to
with respect to two one circle. unless the pencil is harmonic . 1). It will be seen in § 5 that.
—
=
.are Therefore the pencil of tangents from co to all on the circle j. The inverse of a circular cubic with respect to a point on it a circular cubic. co'. 3.
... and therefore Hence its envelope it is self-inverse with respect to (iii).
(iv).
real. If t is any given root (i) is self-inverse with respect to (iii).
If a real focus S of a two-circuited circular cubic lies on a circle j for which the cubic is self-inverse.
..
and
. (iii)
is
one of the
circles
with respect to which
(i)
is
self-inverse.. Ex.
Its envelope is found by writing down the condition that It coincides with (i) pro(iv) should have equal roots in m.
.. Consider the circle. the other real foci are obtained by inverting S repeatedly with respect to the
other circles for which
leave.
is
out.
4
of
(v).
.
*
is
As just pointed
which
. if the pencil is
.. Ex.
1. 3. Similarly we show that.
all
and therefore
Take now the origin at the singular focus of the circular cubic and the axis of y parallel to the real asymptote.XIV
exist
3
FOCI OF CIRCULAR CUBICS
both one-circuited
(k
221
(k=
is
—3—V 3)
= —3+
</3)
and two-circuited
cubics. It follows at once that the equation of the cubic is (x+p)(x2 + y 2) + ax + by + c = (i).
2tm 2 (2t-x-p)-m(4ty + b) + x 2 + yT+a + 2t(x-p) + 4t 2 =
.
and which cuts orthogonally the circle x 2 + y 2 + 2 (p-2t)x + by/2t-(St 2 -4tp + a)
Its equation is
(iii). The cross-ratio of the pencil of tangents from any point of the curve is the cross-ratio of the pencil subtended by the four real foci at a> or at any point of the circle j.
y
2
=
4t(x + t)
. if all four real foci lie on j.*
is
S
self-inverse.
=
.. Then the terms of the highest degree in the equation of the curve meet the curve in only are x (x2 + y 2). if the cubic is one-circuited. the cross-ratio of tangents only possible for a one-circuited cubic. two real foci lie on each of the two real circles with respect to which the cubic is self-inverse. and the lines x 2 + y 2 = one finite point apiece.
The circle (iv) is orthogonal to the circle (iii).
harmonic.
lies
the real asymptote.
This case requires further investigation
see § 3.
(v).
These inversions four real foci lie onj.
vided
64t i -64pt 3
+16(p2 + a)t 2 + 8(c-ap)t-b 2 =
. . 2m£)
on the
(ii).' unaltered. § i. for
which the
cross-ratio of tangents
—1. whose centre (tm 2 parabola with focus at the origin
is
where x + p
=
— t..
tbV). 0) and the real ordinary foci are
' ' '
The parabolas are
'
(0. and the distance of the centre of j from the asymptote is equal to the semi-latus-rectum of the parabola. 1-6). -2-1).
(i)
In Fig. 3
and
(ii)
the foci are
sh6wn by the small
x
. and whose centre lies on a fixed parabola with its focus at the singular focus of the cubic and with its axis perpendicular to the real asymptote of the cubic.
circles o.
--5). 3 (i) the singular focus of the cubic is (0. In Fig.by putting t x
=
They
are by definition foci of (i).
2
(a:
while the corresponding focal parabolas are
The
= 40(x + 3) and (2y + 5) = -40 -2).
(2-31.
1).
. 3 (ii) the
circles for
which the cubic
is self-inverse
are
4x2 + 4:y 2 -40x + 2Sy-99
and
2
4x 2 + 4y 2
(2y +
5)
= + 4>0x + l2y-59 = 0. and the harmonic points (ff _V<r)> (-ft. are the centre (2-305.* The focal parabola through the real foci is 4-y 2 4>x + l.
at
(75.
circular cubic may be considered in four ways as the orthogonal to one of the circles j envelope of a circle which with respect to which the cubic is self-inverse.
the cubic
is
. we have: from the asymptote x+p.
asymptote 200a?
=
=
self-inverse. The other focal parabolas are at once obtained from the fact that their latera recta are twice the distance from the asymptote of the centres of the circles for which the cubic is
. —715) of the circle through the real foci. It will be noticed that the distance of the centre of (iii) Summing up.
In Fig.1
222
FOCI OF CIRCULAR CUBICS
XIV
3
The centres of those four circles of the family (iv) which degenerate into line-pairs (circles of zero radius) lie both on t in (ii). (I.
tact of tangents parallel to the asymptote
by the dots
whon
* This follows from the fact that foci invert into foci inverted into itself.
(4-16. % "5
A
H
bolas
called focal parabolas or deferent paraof the cubic.e. (iii) and on the parabola obtained. asymptote is x = 0. The intersections of the 'parabola with the circle j are foci of the cubic. the centres of the circles with respect to which the cubic is self-inverse. i. M) of the quadrangle whose vertices are the real foci.
The points
which the tangents are parallel to the
real
361.= is 2£. the singular focus
by a
cross
the points of con. and the origin lies on the curve.
/30. with focus
[The corresponding focal parabola. No other circle passes through four foci. -4).
[The pencil
(deba). dd' are also concyclic]
Ex.
(iv)
da'. the four circles circumscribing the triangles formed by any three of them pass and their centres lie on a circle through 0. (iii).
Through the intersection
asymptote any chord
OPQ
[The singular focus of the cubic of Ex. unless the pencil of tangents from a point on the curve is harmonic. the circle (i) contains four real foci.
(a'b'c'd')
is
If (abed)
ratio.
has the same cross-ratio as (abed). can be put x(x* + y 2 ) + ax* + 2hxy + 2gx + 2fy = 0.
3
1. XI.
4. In a one-circuited circular cubic the two real focal parabolas In a two-circuited circular have concavities in opposite directions.]
at its Ex. see Ch.
£V220. ad'
are such that the.
WW. &c] through
Find the focal parabolas of 3x(x'i + f) + 2x*+6xy + 6y'> = 4x and the circles with respect to which it is self-inverse. 0)
and radii
Ex. The normals to a circular cubic with singular focus intersections with a circle for which it is self-inverse touch a parabola and axis perpendicular to the asymptote. (ii). § 11.
Ex.
cc'. so that the foci aa'. b.
Ex. The centres of three circles for which a circular cubic is selfinverse are the vertices of the common self-conjugate triangle of the fourth circle and its focal parabola. Show that P. d be tangents from o> to a circular cubic.
7.
radius of curvature of a circular cubic is a maximum or its intersections with the circles for which it is self-
Ex.
(-§. bb'. cc'.groups (i).
(i)
aa'. (a'b'c'd') and (chad) have the same crossbb'. cc'. In Ex.
cd\
(iii)
ca'. — h). 2.
1).
(1..
J. Ex. so that ca'. bb'.
Ex. 8.
.
FOCI OF CIRCULAR CUBICS
d'
223
Let a. Ex. or else the circles (i) and (ii) contain two real foci apiece. cubic the focal parabola through the real foci has its concavity in a
direction opposite to that of the other focal parabolas. cb'. and let be tangents from m. 1 the centres of the circles with respect to which the cubic is self-inverse are the centre of the circle (i) and the intersections of the pairs of lines joining aa'. ac'.
[The circles have centres (—1. bd'. Many properties of these normals can be deduced .
a'. 3. (bade). iV(~m
-J). (iv) each lie on one of the circles for which the cubic is self-inverse. 6. dd'.
in the form
[Take the intersection of the cubic with the real asymptote x
origin. Q are equidistant from the singular focus. 4
is
(— \a. bb'. db'.]
=
as
of a circular cubic with its real of the cubic is drawn. 9. Then the foci
c'.
3.
5. If aa' and bb' are real. (cdab) harmonic. ac'.
dc'.
aV. c.
The equation of any circular cubic.
V.
(ii)
ba'. lc'. for instance.
The
a minimum at
inverse.
XIV
Ex.
da". Ad' lie on the circle j of § 3.
(0.
0).224
Ex.
(0.
-y
%
identify
(y. 0).(0 + y) (y + (X) (tX + 0)} = 4 {(a2 + 0* + y2 + 0y + ya.
[See Ex. 15.
Find an equation giving the other focal parabolas.]
(iv)
with respect to m\
Ex.
[The locus of the centre of a varying parameter.
Ex.]
Ex. 14.
The
foot of the perpendicular
centroid of any four concyclic foci of a circular cubic is the from the singular focus on the real asymptote.
FOCI OF CIRCULAR CUBICS
Two
and their
11. 0).
[The corresponding focal parabola meets the circle at the real foci. and that the inverse of the cubic with respect to either of these points is a circular cubic with an axis of symmetry passing through two real foci and bisecting at right angles the line joining the other two real foci.
Show
and singular
focus. 0).]
lines joining four concyclic foci of a circular cubic in pairs are equally inclined to the real asymptote. 19.]
we get the
abscissae or
Ex. and that the inverse of the cubic with respect to any one of these points is a circular cubic with an axis of symmetry on which the four real foci lie. and the required points are the points of contact with the circle ofHhe common tangents to circle and parabola. Show that a real circle with respect to which a one-circuited circular cubic is self-inverse meets it in two real points. 12. 0).
=
B
(j3
+ y)(y + a)(OC+P) {8x(x* + y') + 4 (a* + 2 + y* + 0y + yOC + a/3) X . 0).
foci.
and choosing
[The directrices are found by differentiating m so as to make (iv) a line-pair. 10. The circle inverts into the
axis of symmetry. 13. + a0) 2 + 4 a0y (a + + y)} x\ [Find the foci of x {x 1 + y*) + ax + by + c = in the usual manner.
[Parallel to the axes of the
two parabolas through the
foci. 0). Show that its equation is (0.
We
have
a+ + +8 =
0.
that the cubic
(i)
x2 +y2 =
of § 2
self-inverse with respect to
is
and that the corresponding focal parabola (y + hf = b(2x + a).]
Ex.
(iii)
[Eliminating y or x from § 3 (ii) and ordinates of the four foci on (iii).
XIV
3
circular cubics can be found with four given concyclic real asymptotes are perpendicular. The centres of the four circles for which a circular cubic is self-inverse lie on a rectangular hyperbola whose asymptotes are the real asymptote of the cubic and the perpendicular from the singular focus.]
Ex.
Ex.
16. and
them with
(a.
is
if
t
is
taken as
Show
h. have
common. A two-circuited circular cubic has x as asymptote and three real foci A (a. 15. Show that the circle through the real foci of a two-circuited circular cubic meets it in four real points. a focal parabola in
that all circular cubics having the same real asymptote and self-inverse with respect to the same point.
&c]
.
(8. The directrices corresponding to four concyclic foci of a circular cubic form a pencil of the same cross-ratio as the pencil subtended by the foci at any point of the focal parabola through them. 17. C (y.
The
Ex.
Ex. 18.
(iii)
is
2(x+p)y + b =0.
Hence. the coefficients of x*. Since (74. if and B are any two points on the cubic. distance of P from a fixed point multiplied by a fixed line varies as the distance of P from
another fixed
line. B. C. where A.
A
. Find the locus of
(i)
P if
CPU
have a constant
ratio. Any circular cubic passing through its own singular focus may be considered as the locus of a point P such that PA PB = PC PD.
The tangents of the angles APB.
z3
+ z 2 (x + y) + xy(ax + by + 3cz) =
tz
0.226
FOCI OF CIRCULAR CUBICS
XIV
3
Ex. GB are the tangents at and B.
Pencil of Tangents from a Point on a Cubic. the cross-ratio of the pencils of tangents from and B are the same. 28. Any circular cubic passing through its own singular focus considered as the locus of the intersection of any member of a given family of coaxial circles with its diameter through 0.
.
Hence the tangents from
(ax 2
B
2 2
)
to the cubic are
+ 2 cxz + z
-4 bxz2 (x + z) =
. Let the tangents at and meet at G and take ABC as the triangle of reference. 30. B.
may be
[Take
as origin
and the
x'
circles as
+ y^+px+a
=
t(py — b). The theorem is so important that a straightforward proof is given here. if
+ 2ct + l)yz + bty* =
0.
§ 4. the equation of the cubic becomes
A
A
A
B
. choosing suitable real homogeneous coordinates.
The product of the tangents from
to the product of the tangents from
(iii) The square of the the distance of P from
P to two fixed circles is equal P to two other fixed circles. only involving real quantities.
(ii)
D being fixed
points. 29.]
Ex.
Ex.
The theorem that the four tangents to a cubic from any point of the curve (excluding the tangent at the point) form a pencil of constant cross-ratio was proved in § 3 by means of unreal projection. It is sufficient to show that.
A. y 2 z in the equation of the cubic are zero. y 3 x 2 z.
0. C. . D are fixed points.
and therefore touches
{at i
+ 2ct + \f-4>U{t + \)
= 0.
The
o
line
x
=
meets this cubic where
(t+l) z 2 +{at 2
it.
[The four tangents from a> invert into tangents from a.(2a -oab + 2b*)(a + b)-27a?b 2 + 36ab(a + b)c -12(2a 2 + ab + 2b 2)c 2 -36abc 3 + 24.
Ch.]
Ex.
2"= 12 {a 2 -a& + 6 2 + 3a&c-2(a + 6)
and 6. forming a pencil with the same cross-ratio.
(x
+ y + zf + 6 kxyz =
of these tangents
is
227
given by
. or vice versa. the cross-ratio of tangents is the same for the curve and its inverse.
B
1.
Q 2
. only if
A
and
B
lie
on the same circuit of the
cubic]
Ex. Discuss the case tacitly excluded in A goes through B. J=4.XIV
By
5
THE CUBIC
I. The equation of the inverse curve is § 2 (i) with h 2fc.
We proved in § 1 that a non-singular or acnodal cubic has Choose the tangents at these three real collinear inflexions. and The real foci are readily found. inflexions as the sides of the triangle of reference.
and
Ex. prove that the intersection of two such circles cannot be a focus. and choose real homogeneous coordinates such that the line of inflexions 0. § 11.
§
A
7.(a + b)c i -8c e Since these expressions for / and J are symmetrical in a
2
. the cross-ratios of the pencils of tangents from are the same. Another proof of the theorem is given in Ch. Can the theorem of § 4 be deduced ?
P and
P of the cubic
P
Show
P
[Strictly speaking. §
Ex. by using the fact that touches the cubic at and passes through the points the polar conic of of contact of the tangents from P.
The Cubic
(x
+ y + z) 3 + 6kxyz =
0. and k 0. is x + y + z The most general equation of a cubic is
—
3
xyz
= ax
+ by s + cz 3 + 3yz(fy +pz) + 3zx (gz + qx) +3xy(hx + ry). 1.
2.
A
and
B
with the pencil
that the pencils of tangents from two adjacent points have the same cross-ratio. 3. By taking as the intersection of the curve with a real circle for which it is self-inverse.
4 in which the tangent
at
[Compare the pencils of tangents from drawn from any other point of the curve.
the cross-ratio
I 3 (0+1) 2 (0-2)2 (0-i)»
where
=
27 J»(^-0
c2
+ l) 3
+ c 4 }. and that the four real foci of a harmonic onecircuited cubic lie on a circle for which the cubic is not self-inverse.] one-circuited if 4&
=
=
>
>
§5. If a circular cubic is inverted with respect to a point O on the curve. The curve is harmonic if ah 2 a .
. XVI.
However. For what value of h ihe cubic (x + y + z) s + 6kxyz
[(i)
=
the pencil of tangents from any point of (i) harmonic.
Symmetry
of Cubics. but this not true of a curve of the third degree.
is
A curve
Firstly.]
Ex.
§ 6.
2.]
Ex. 2.
on the lines
0.
[See Ex.]
s
Ex. The point and tangential equations of an acnodal cubic can be put in the form X-i+p-i + v.
we have
:
A
a
cubic can be projected so as to be
is
its
own
reflexion in
polar.]
Ex.
that in general two cubics with given inflexional tangents at three given collinear inflexions touch a given line. Under what circumstances is the triangle formed by the real inflexional tangents of a cubic self-conjugate with respect to one of the polar conies of the cubics ?
[The polar conic of
(1. 5.
H=(x + y + z)(2yz + 2zx + 2xy-x -y
The
inflexions lie
-zi + 2kxyz =
)
0. 4.
(ii)
k
= -i.
=
0.
line. Given the tangents at three given collinear inflexions of a cubic. 7.
3. the cubics are
m
{l-^x
+ m^y + iiyz)
(l2
x + m 2 y + ii^z)
(ls
x + m 3 y + n 3 z)
=
{(hhhfi x + [mym^m^l
y+ (n^n^i z} s
. xi + yi'+zi = 0. 1. the locus of the remaining inflexions is two straight lines.
For suppose /
any
real inflexion
and
I
its
harmonic
.
[If the line passes through an inflexion or the intersection of two of the given tangents.! = 0.
when h
= —3. 1. there is only one cubic]
Show
Ex. 3) are the inflexional tangents and + n( z = iy the three given points are the vertices of the triangle of reference.
Find the
is
inflexions of the cubic
/= (x + y + z)
[The Hessian
+6kxyz
t
=
i
0.XIV
6
SYMMETRY OF CUBICS
is
229
Ex.
of the second degree has always symmetry. 6. Show that nine cubics can be drawn in general to pass through three given points. 1). (ii) equianharrnonic ?
& 2 + 6& + 6
2. a cubic may always be projected into a symmetrical curve. Mention any exceptions.
any value of the cube roots being taken. given three inflexional tangents and the fact that the three corresponding inflexions are collinear.]
f-3H=4:(x + y + z)(x + ay + a>''z)(x + ca*y + B)z) =
Ex.
[If l t x + (i = 1.
These two theorems hold for nodal and cuspidal cubics also.
Aa/
c. p.. 412)
Any cubic (other than a crunodal or cuspidal cubic) can be projected so as to have the symmetry of an equilateral triangle. if we project / to infinity and the angle between I and IR into 90°.
Again.
In fact project the cubic so that the line of real inflexions
is
at infinity.:
230
SYMMETRY OF CUBICS
XIV
6
Let any line through I meet the cubic again in P.
A
cubic
can
be projected so as to be
symmetrical about a
centre. PQ will be bisected at right angles at R.
-
I-
Pig..
*
4. Then (IR. is the result due to Clifford (Collected Works. Q and I in R. More interesting.
and the triangle formed by the
real inflexional
.
7
will be the middle point
ry. PQ) is harmonic. Hence.
*»
=
I..\
For
or
if
we
project
I
to infinity. perhaps.
8.
.XIV
6
SYMMETRY OF CUBICS
231
:
—i 8)
k K
= —2*3 56'
Fig.
Fig.232
tangents
SYMMETRY OF CUBICS
XIV
6
is equilateral. we proved in § 5 that the equation becomes
+ y + s) 3 + 6Jcxyz from which the symmetry is obvious.
fc
= -'72
-.
m = -l. taking areal coordinates.
The case in which the inflexional tangents are concurrent wants separate discussion. Then. 10.
(x
=
0. In this case project the curve so that the line of real inflexions is at infinity and the angles between the inflexional
.
e. e.
and
m= —\
'S
when nWlicii k
—
9 2
. 4(m — l) = fc(4m — 2m + l) .
.. y curve in no finite point the equation of the curve is of the form
tangents are 120°.
y(3x2 —y 2)
or in polar coordinates
=
a3
.
. have (4m 2 + l)J 4(m-l) s
=
m
We
-2m
=
-
§7.
The cubic has one or two circuits according as y z meets the curve in one or three real points. z = X+Y-2mZ..
=
—
.
XIV
7
THE CUBIC
x s + y 3 + z 3 + 6mxyz
=
233
Take their intersection as origin and one of them as the axis of x. according as h is greater than or less than — f Diagrams of the cubic" for various values of k are given in Figures 4 to 10.
x3 + y 3 '+ z 3 + Omxyz
=
by a real choice of homogeneous coordinates.
y
3 2
(ii)
This will be lawful except in the cases
m=1
(x +
when k
=
0.. The and the dotted lines are explained in § 7..
. &c. i. 2(1— m)(2m + l)X = (2m + l)x + y + z.
r 3 sin 3 6
= a3
from which the symmetry
is
obvious.
m-
{2(2Jfe
+
9)*+(2ifc)*} -> {2(2/c
9)5-2(2/c)i}.
(ii).
We
by a x
proved in § 5 that a cubic can be put in the form
(x
+ y + z) 3 + 6kxyz =
(i)
real choice of homogeneous coordinates. and -since the asymptotes meet the 0.
= X-2mY+Z. Then the asymptotes of the curve are y + -/3 a.
Then straightforward reduction gives
3 y + z) + 6kxyz
=
Hence
if
we
choose
m as the real root of
+
^^
+1
(X 3 +Y 3 + Z 3 + 6mXYZ). In this equation put
= -2mX + Y+Z. i.
where
i.
Every real non-singular cubic can
put in
the form.
The Cubic x 3 '+y 3 + z 3 + 6mxyz
be
=
0.e.
z 0. not crunodal or cuspidal or having three real concurrent inflexional tangents. 1).
&c. that the equation of a 9p + 1 cubic can be put into this form in twenty-four different ways. 0.
and equianharmonic
We
p=
Ex.]
For the caso of any cubic.
4.y.
The
inflexions of the cubic of Ex. can be put in the
form
[Replace
x3 + y 3 +z = h (x + y + z) s x.
.curye or by putting
. noting that <f> is only real (and not — 1.
[Consider the tangents from
(1.
Ex. § 291. XVI. 0. 15. 5. 1.
tion
(ii)
The curve has two
as
may
circuits if < — \ and one circuit if be -deduced from § 6 by means of equa.
As an
Ex.
m
V
v*
=
0.
1. the
0. 0) to the. 1.
Find the cross-ratio of the tangents from any point of y*z + x'y '+ z^x + Gpxyz = 0. show in Ch.z by y + z. or 2). can be put in the form y^z + yz2 + z 2 x + zx 2 + x?y + a>y2 + 6lxyz = 0. see
Elliott's Algebra ofQuanUcs. 3 lie
on
(x + y
+ z)
(x
+ ay + a> 2 z)
(x
+ a> 2 y + a>z)
=
0.
. z + x. \.]
l
harmonic
s
=
2l6p" + B6p 3 +
=
(0. Show that the equation of any cubic.
2.
lines x 0.]
!i
.24m
2
A/«/ (A
3
+ ^ + v 3) . y inflexional tangents being y
The
=
=
=
+z
=
each pass through a real inflexion. x + y in § 7 (i). not crunodal or cuspidal. 0). § 6.
(ii)
in the expression for the cross-ratio found in that section. Find the cross-ratio of the tangents from any point of the curve. Ex.* The tangential equation of the curve is
\ e + fi 6 + v a . 0).(32m3 + 2)
(ji*
v3
+ v 3 X 3 + A 3 /* 3
)
. exercise the reader may obtain this result from the expression just obtained for (/>. &c. 3.
if
[Consider the tangents from
or 0.]
Ex. Find the cross-ratio of the pencil formed by the tangents from any point of the cubic of Ex.
.
above. if the cubic has two circuits.(48m 4 +-24m)
A
m>— \.
The pencil of tangents
if
is
0. 3.236
THE CUBIC
x3 + y 3 + z 3 + 6mxyz
=
XIV
8
This may be proved as in § 5 by considering the pencil of tangents from (—1. ifl=—l.
[Consider the tangents from
*
(
—1
.
The cubic
is
acnodal.
Ex. Show that the equation of any cubic.
k
= 4(m-l)
3
-T-
(4m 2 -2m+l)
.
any cubic through
eight inflexions
is
f+hH= 0].
Hessian meet where xyz = i. the polar conic passes through the four intersections
of
£ks + ijy' + (z*
=
and
%yz + r)zx + £xy
= 0. £). z).
[The roots of the equation
3 8ifcf
2m3 + 6Mm* + l =
in
+ l<
m
are all real.
3. e. If the pencil of tangents from any point of a cubic the cubic is the Hessian of its Hessian.]
Ex.
. as proved before.
line of a point on a cubic with respect to any syzygetie cubic passes through the tangential of P.:
XIV
9
SYZYGETIC CUBICS
'
237
§ 9. 5.
The harmonic polars of the inflexions are the same for all the syzygetie cubics of the family. The polar conic of a point P with respect to any cubic of a syzygetie family goes through four fixed points lying on that cubic of the family which passes through P.
zx3 — zy3 ). yz 3 — yx 3 .
cubics found by varying in the equation (i) have the same inflexions they are called a family of syzygetie cubics. y.
. A real cubic is the Hessian of one or three real cubics according it has one or two circuits. [If P is (£.]
Ex.
[If
The polar
(x.
[8m 6 + 20m 3 — 1
=
0. 2.
.
The curve and
its
We see that A cubic has the same inflexions as its
The family of
all
. 1. All cubics through eight given inflexions of a ctabie pass through the ninth inflexion and are syzygetie with it.
(ii). if
0. Since (ii) is of the third degree when considered as an equation in m.
Ex.
The Hessian
is
of
x*
+ y 3 + z 3 + Gmxyz
= =
.
Syzygetie Cubics.
non-singular cubic
is the
Hessian of three syzygetie
as
Ex.
Any
cubics.]
. The Hessian of any member of the family
m
belongs to the family. the inflexions of the cubic lie on the sides of the triangle of reference. 4.
Hessian.
(i)
at once found to be
x3 + y 3 + z 3 + GMxyz
where
M = -(2m
3
+ l)/6m 2
.]
is
Ex.
harmonic. is the cubic and ff = [If / = its Hessian.
P
P is
the tangential of P
is
(xy^ — xz3 .
We may call
the cubic 'harmonic'. i).
Q. Both conies touch 6 and c at their intersections with o.+y^y^y^ + zx z^zs
I.
§
%
Ex.]
is
Ex.
cubic
If
The polar I of with respect to the polar conic of Q for a the same as the polar of Q with respect to the polar conic of P. Its real inflexional tangents are concurrent at a critic centre.
Ex. If the pencil of tangents from any point of a cubic anharmonic. If it is taken as triangle of reference. and Q lie on the same cubic of the syzygetic family. and P is one of the critic centres referred to in Ex.
is
Ex.]
'
Ex. 0). We may call the cubic equianharmonic '. ysi*j). b. The tangents at two inflexions of a cubic meet on the harmonic polar of the inflexion collinear with the two given inflexions. 7. the coefficient of xyz in the equation of the 3-ic is zero. [m s (m s — 1) = 0. c. The sides of the triangle are the degenerate Hessian. 1)
and
(1. or (2) the cubic is equianharmonic.]
'
equi-
Ex. (0.
[If a. VII. 10.
Ex.
fixed and the cubic is any member of a syzygetic goes through a fixed point. One species has a Hessian consisting of three real lines. R is symmetric. show that the cubic is § 8.
y
=
0. 14. 1)
respectively.]
12.
P
family.
z
= 0.]
. 1. Under what conditions cubic a coincident line-pair ?
the polar conic of a point
P
for a
[Either (1) the cubic is cuspidal. If
any
line
meets
X s + y3 + a 3 + Gmxyz
(*i»yi.]
'
Ex. the intersection is (1. Ex. 0.238
SYZYGETIC CUBICS
XIV
is
9
Ex. Obtain the equations of the sides of the triangle for the cubic of and p = — 3~$ . If R is any point on I.
=
(^3»y3 . 7. Each real inflexional tangent meets two unreal inflexional tangents at a critic centre which is a vertex of the triangle formed by the Hessian.
with
[If
its
The intersection of the tangent at an inflexion of a cubic harmonic polar lies on the Hessian.
in
13.
If all polar conies for
triangle.
=
0. m).
9. 6. 0). m. 11. [to = and w? — 1 in the two cases. 1.
and
the conies are
x'
i
x*
+ 2myz=0
+ 8myz
= 0. The critic centres are (1.
. The tangents at the inflexions on 6 and the tangents at the inflexions on c form two triangles whose vertices lie on a conic and whose sides touch a conic.
Ex.
8. The inflexions of a cubic lie by threes on the lines a.
(0. 0. If this point is on PQ. Ex. (0. the relation between P.
Note
list
of Errata'. 8. in the cases p = equianharmonic of the second and first species respectively.2i)i
(«j.%)i
then
[See Oh. c are
x
=
0. 1) is the inflexion. 6.
x^^x. 15.
P
and Q are
I
P
[PQR is an ' apolar triangle of the 3-ic.
and Ch.]
a cubic have a common self-conjugate the cubic is equianharmonic. the Hessian degenerates. 2. There are two species of equianharmonic cubic. and Pis the cusp or the intersection of the cuspidal and inflexional tangents. The other species has a Hessian consisting of one real and two unreal lines.
§ 3. —1.
0.]
. &c]
[If the sides of the triangles are
x +y +z
=
0. 22 pass through the same two unreal points on the line of real inflexions. 0.
18.1. o>
1). 17. z = 0.1.0). The pencil of four lines joining an inflexion of a cubic with the other eight inflexions is equianharmonic.]
Ex.0). Show that these six tangents touch a conic. 20. and the remaining inflexions lie on two lines through A. and that the six tangents touch a conic.
are
(1.
Two lines each Show that the
+ y 3 + z3 + 6 mxyz + hx (x + 2 myz)
1
=
is
a
line-trio.
[All the conies of Ex. (-1.
(1.
is (0. From each intersection of two real inflexional tangents of a two-circuited cubic two real tangents can be drawn to the oval.1. with their harmonic polars are
[If the inflexions are
the inter-
sections of
x+y +z
=
(-2.0).]
passing through three inflexions of a cubic points of contact of the tangents from O lie by pairs on three lines through the remaining inflexions.»l.-2.
[Ex.
The double points of the involution
(a>.
(-o.
Ex.]
The three
inflexions with their inflexions. -1). 22. The harmonic polars of three collinear inflexions of a cubic pass through the same critic centre A. to any one of the collinear inflexions is divided The line joining harmonically by the cubic.]
Ex. y = 0.
intersections of the line through three collinear harmonic polars form an involution with those
(0. and no two sides of the same triangle pass through the same inflexion. 0).
— 1.]
Ex.
x = 0. The three lines joining two real critic centres and passing through a real inflexion meet the real inflexional tangents at those inflexions and at six points lying on a conic.
[If
the inflexion
(0. Show that from each vertex of the triangle two real tangents can be drawn. 4 to 10. 15.
(w 2 w.
2
. that the six points of contact lie on a conic. 19. and that the axes of perspective each pass through three inflexions.0. that the tangents meet the cubic again in six points on a conic.1. that their points of contact lie on a conic.
meet at
\a?
O.0). and x + wy + aPz = 0. 1. Two triangles are such that each side of either triangle passes through three inflexions of a cubic.1.0). (1.XIV
9
SYZYGETIG CUBICS
239
Ex. The three real lines each passing through one real and two unreal inflexions of a two-circuited cubic form a triangle. 21. -1. 20 to 24 are obvious from Fig. if
8m
Ex.1)..
. x + a'y + az = the axes of perspective are ax + y +s = 0.
A
[Take the collinear inflexions as those on x
=
0.
1).
Ex.1).
1).
the lines meet z
—
at
(-1.
Then
A
is
(1.
Ex. and that the intersections of each pair of tangents with the third inflexional tangent lie on a conic.
.-2). 20.1). 21. Show that the two triangles are in perspective in six ways. 16.
(-a.
s
(&
+ 3) 3 + 27(fc + l)=0.
240
SYZYGETIC CUBICS
XIV
9
Ex. 24. 28. and the chords of contact of conjugate pairs of the involution pass through I. meeting
at
(a. c).
/=
H=
/= 0.]
. Soc. and the nine inflexional tangents are concurrent at a point of the quintic. 15.]
doubly infinite family of quintics passes through the nine Ex. §
4.
' '
Ex
l{l'
=
'
l). touches the Hessian at three points.
[(i)
See Ch. (b + c)x= a(y + z). The inflexional tangents are (b+cu>)x = a(y + az). 23. 6.. [The polo-conic of \x + fjy + nz = 1 is XX' (mrx — yz) + . It is possible to project a family of syzygetic cubics so that each curve has the symmetry of an equilateral triangle. If the equation of a cubic is given.] Ex.
Show
that
I
(i)
The polo -conic of
and V
is
the locus of the pole of
I
with respect
to the polar conic of any point on V.. Each such cubic has three collinear inflexions and corresponding inflexional tangents in
common
with the given cubic.
Ex. (b + ca 2 )x= a(y +
a>
2
z). and (ii) A syzygetic family of cuspidal cubics may be projected so as to be all congruent and have a common asymptote which is the cuspidal tangent of each curve. +
and
(fiv
X'a. The six tangents drawn to a cubic from a point on the harmonic polar of an inflexion I form an involution.
A
[The quintic is ayz(y s + s3 -2x*) + bzx(z 3 + x*-2y s )+cxy(x 3 + y s -2ss)= 0..
the locus of
'
The polo-conic of the lines I and I' with respect to a cubic is a.
it
will be a line-trio
Find k such that f+kH=Q on which the
quartic through the twelve critic centres of a cubic has Ex.
See Roberts.
[Let be the Hessian of coincides with its Hessian then inflexions lie.
xiii (1882). 30.]
Ex. + p'y + v'z = + n'v) (mx' r. 27. show that a cubic and (i) A syzygetic family of nodal cubics may be projected so as to be all similar and similarly situated with the node as centre of similitude.m'yz) + .
&c.
lie
The polo-conic
of
I
I touches the polo-conic of I.. 26.+
. inflexions and the twelve critic centres of a cubic. p.. (ii) The polar line of any point on
(iii)
and vice
versa. Defining a syzygetic family of cubics by the equation
= its Hessian. [There are also twelve cubics such that the tangents from any point on one of them to the given cubic form an involution.. The configuration formed by the nine inflexions and the twelve lines joining them is dualistic to the configuration formed by the nine harmonic polars of the inflexions and the twelve critic centres. 26. + . XIII.]
Ex. prove that the inflexions can be found without solving any equation of degree higher than 4. The polo-conic of I is the locus of a point whose polar conic touches
25. Each such quiittic has an inflexion at every inflexion of the cubic. inflexion at each critic centre.]
.
A
an
[The quartic
is
ax
3
(y
— z*) + by (z* — xs ) + cz (xs — y s =
)
0. London Math.
where
/=
is
H
f+kH=Q.. 29.]
Ex.
=
0. point whose polar conic has I and V as conjugate lines.
(ii)
Project into (x-k) y*
=
1. which
on the polo-conic of I and V.. Proc.
CHAPTER XV
CUBICS AS JACOBIANS
§
1. means the
putting x' for
x. y' for y. y'.
These polars are concurrent
(at
Q
. z)
with respect
to the conies are
W
7>U
<>x
-
7>U
7>y
IU
<sz
and two similar equations. V = 0. where ^-—.
Suppose that U = 0. = are the equations in homogeneous coordinates of three conies not passing through the same four points. z' for z in
result of
—
<sx
&c. The polars of the point P (a/.
Jacobian of three Conies.
V . with the Jacobian of IT = 0. Z).V+^W.
this case the equation of their
U=
=
W=
:
The Jacobian of any three conies of
the family
XU+pV+vW =
(which do not pass through the same four points) = 0. for the polars of any point with respect to all conies through four given points are concurrent. is
= ^(X.
*U Tz +
1
W+
iz
v
3F =0 n
*z
-
Eliminating X:pi:i> between these equations.lT+^V + h.
i>'s.
\U + ^V+v^W)
z)
d (x.e. we see that (X. Y.
.W^O. V = 0. Y. if
7>U
i. y. a line-pair meeting at (X. Z) must lie on the Jacobian . and the centre of any degenerate conic must lie on the Jacobian. the lines and are har-
AP
A
AQ
. Z). the equations give one and only one value of the ratio X: fi:v.W)
S(a3. V 0.
where the
are constants.
.:
242
JACOBIAN OF THREE CONICS
of the conies
XV
1
The Jacobian
\ 1 U+p1
1
V+v W=0. If the conies do not pass through the same four points. z)
K
as
is
Hi
*(U. their Jacobian is identical with the Jacobian of the original conies 0. Hence 0.
If these conies all pass
through the same four points.V + fi. Q are conjugate points on the Jacobian.
W
is identical
is
The conic X U + fiV+vW = has a node at (X. as is geometrically obvious.
X.
X 2 U+ft 2
V+v 2 W=0. /is. and is any other point on the Jacobian.
at once verified
by the
rule for multiplying determinants. and if this is the case.
If P.
*3
H
In
Jacobian vanishes identically.
X 2 U+fi 2
V+u
2
W.
Hence
Any point on the Jacobian of a family of conies is the centre of one and only one degenerate conic of tJie family.
X's.
X
ZX
+
7>V
>
X
*X
+
3TT
V
*X =
>
n
'
7>U
IV
iW
»
x
.
A i hi
A"
fi 2
"i
v2
"s
=
0. y. T.
£z)
=
ax(y. CA.
[Into the
common
pole and polar. incident line-pair at infinity is the rectangular hyperbola through the to the conic together with the line at infinity. 2. C in common. AB. 5.
Ex.]
common
self-conjugate
Ex. C.
(i
=
1. their Jacobian degenerates into the lines BC.
3.
their Jacobian has
[Take
A
as (1. If three conies pass through the points A. IV.]
e2
. B. For AQ is the polar of P with respect to the
degenerate conic.
and a
coincident
line-pair
and a
conic. their Jacobian
common
polar and two lines through the pole.
\xi + y l +
=
0. If three conies have a eom'mon self-conjugate triangle.
Ex. If three conies have a degenerates.
Ex.]
Ex. their Jacobian has a node at A and one branch of the Jacobian touches the two conies at A. 7.]
conic. If three conies have points B.
0)*. 0. 4.
Find the
circle orthogonal to
x 2 + tf + Sx + \h = 0. 6. 1.
7.
z2
=
by (x .]
Ex. Ex.
Ex.rjz).]
Ex.
circle orthogonal to
The Jacobian of three them all. Ex.
4. If three conies pass through A and two touch at A.] feet of the normals from
The Jacobian of a
P
[The Jacobian of
ax*
is
+ bya
=
z>.]
sides of the common self-conjugate triangle of any two conies may be obtained by forming the Jacobian of the two conies and their harmonic locus.
A
in
common.
[Into the line
11. and a coEx. The Jacobian of two conies degenerates. a circle with centre P. their Jacobian degenerates into the line BC and a conic through B and C. their Jacobian degenerates into the sides of that triangle. 10.
The
triangle.
Now
put z
=
1.]
'
Ex.
[The two conies and their harmonic locus have a See Ch. If three conies have a point a node at A. 2.
[See Ex.
2x-2y-7 =
0.XV
1
JACOBIAN OF THREE CONICS
243
monic conjugates with respect to the degenerate conic with centre A. x i + y i + 2x-\2y + 2l xi + if-2x + 4:y + \ = 0. 3). 8.
x i -^y l -2^xy-2r yz + cz 2
t
=
0.
9. § 4.
x*
circles is the line at infinity
and the
[Take the circles as
+ y* + 2g x + 2f y + c =
i
i
Q.
[See Ex. and a
coincident line-pair at infinity. forming the harmonic locus of the conic and the circular points. T ne locus of the point of contact of two members of the = is their Jacobian. polar curve. 0.
rj. XI V. of the point
Then the
P
(£. § 9. The Jacobian degenerates into the axes and the line at infinity by Ex. e. the
b
=
first
the equation of a cubic.
[Take
A
as the point (1. z) polar conic. Hence:
§ 3. If the Jacobian of three conies passes through the intersection of two of them.]
Ex. 0). 15. y.
13.]
We
Ex.
non-singular cubic. i.
[Into the chord of contact and a conic dividing the chord of contact harmonically. its director circle. The Jacobian of two concentric circles and any conic rectangular hyperbola and the line at infinity.
=
Now any cubic * is 0.
9. thus obtain the tangential equation of the circular points.
The Jacobian of two conies having double contact and any
third conic degenerates.]
A
Ex. 14.
e. Then.
See
Ex. family X U + /x V + v
W
[See Ex.
[We suppose that we know the equation of one circle and of the line at infinity. z) the Hessian of three different cubics
i. by Ch.244
JACOBIAN OF THREE CONICS
XV
1
Ex.
. 16.
is
Suppose f(x. the two conies touch at A. 15. 12. 13. y. The axes of a conic (using any coordinates) may be obtained by writing down the Jacobian of the conic. 11.
ly
v=
30
0.
Any Cubic
as a Jacobian of three Conies. we have the director circle.
() is
(i)-
i>x
<>y
* <>z
The Jacobian of any three polar conies Jacobian of
Ix
is
the same as the
»/_" o.
is
a
Discuss the case in which the conic
is
a
circle.
and
is
therefore
32
/
ay
<*xhy
_ay_
(>X <>Z
<>x-
ay
f>y <>x
ly*
iylz
=
o.]
§ 2.]
Ex.
ay
<>z
ay
lz>
ix
<sz<)y
•which is the Hessian of f(x.
cubic is uniquely given when three pairs of conjugate points are given.
A
Expressing the fact that the conic
ax* + by 2 + c2 2
+ 2fyz + 2gzx + 2 hxy
=
has the three pairs of points as conjugate pairs. Z). Hence
:
Q
should be
The polar conic of a point on the Hessian of a given cubic a line-pair meeting at the conjugate point of the Hessian considered as the Jacobian of the polar conies of the given
is
cubic.
If
Consider a given cubic as the Jacobian of a family of conies. so is the and R' in Fig. e. and P'..
XV
#
3
RULER-CONSTRUCTION FOR CUBICS
245
Any cubic may be considered as the Jacobian of a family of conies.
Suppose
now
that the polars of
P with
dz
respect to
ox
oy
are concurrent at the point Q (X.
Ruler-construction for Cubics. PQ' and P'Q. so are the intersections of
Q' are conjugate pairs of points on a PQ and P'Q'.
P
and Q are
/
2
.
y/
a2/
.
*
/
. so that conjugate points of the Jacobian. 1 are conthird pair of vertices. a quadrilateral are conjugate with respect to a conic. are conjugate points on*
'
P
R
the cubic.
"
*f
"
. and one vertex into the centre.
>
a2
/
2>'¥
_n
32 f
But equations (ii) are the conditions that a double point of the conic (i). and that in three different ways. and For But when two pairs of opposite vertices of so are Q and Q'.
>
a2
/
_
n
.. we get three
* Hesse's
circle
It may be proved by projecting the conic into a Theorem.
and J" are conjugate for any conic of the family. Y. Then
a2
. i.* Hence jugate for any conic of the family.
.
§ 3.
Denote conjugate pairs Q and Q'. of points on the cubic by
P
P and
P\ Q and
cubic.
Hence
f \U+fiV+vW =
A
B
C
D
AB
D
. where A. SS' meet at P'.
a cubic (Fig.:
:
246
EULER-CONSTRUCTION FOR CUBICS
XV
3
linear relations between a. g. C and are the three conjugate pairs of points. v are constants.
at
The tangents at conjugate points Q and Q' of a cubic Trtteet a point P on the curve.
therefore. and B'.
Ex.
by
the construction
may
be
finite.
See Ch. The Jacobian of this family is the cubic required.
Q and R
are consecutive points on
the cubic.
9. Since the four points of contact can be grouped into two pairs in three different ways. From G and C". c. XVI. and D' we obtain similarly a fifth conjugate pair and so on.
of the four tangents from any point the points of contact are two conjugate pairs Q and Q'.
general construct ruler only. then the intersection of and A'B'* and the intersection D' of AB' and A'B are a fourth conjugate pair.
*
f
We have tacitly assumed that D does not coincide with C or We do not necessarily obtain the whole cubic in this way.
number
C'. and QQ' meets the curve again at the point P' conjugate to P. we can in an infinite number of conjugate pairs by
1
Suppose that in Fig. we see that
If. then the conjugate points Q' and Bf are also conWe obtain then the result secutive points on the curve. /i.
The points of a given cubic can be divided into conjugate pairs in three different ways. h and thus reduce the conic to the form 0. 8 and tf. b. Suppose that and A'.
.
For
special
positions of the three given pairs the
of conjugate pairs obtained
§ 6.f
Given three pairs of conjugate points on a cubic.
we draw
P
2). and the lines QQ'.
P'Q' and with respect to P'S.XV
3
RULER-CONSTRUCTION FOR CUBICS
247
This also follows from the fact that the cubic is the Hessian of three different cubics. XIV. P' (pp'.
[See Ch. 1. The real points of a cubic with one circuit can be divided into The real points of a cubic with two conjugate pairs in only one way. The tangent at P' is the harmonic conjugate of PP" with respect to the line-pair QQ\ SS'. P'S'.]
. In one way circuits can be divided-into conjugate pairs in three ways. This shows that the line-pair is QQ'. QS) is harmonic and rp' is the tangent at P' in the limit.
. By §2 the polar conic of with respect to a cubic for which the given cubic is the Hessian is a line-pair meeting at P'r Also by § 1 this line-pair is harmonically conjugate with
P
Pig. For if p is a point on the cubic close to P.
respect to P'Q. SS'. and is therefore the Jacobian of three different families of conies. §
1. In the other two ways one point lies on each circuit. both points of the pair lie on the same circuit.
Ex.
if we take the vertices as (1. 3.
where &
= 0L + \ir. 4 the tangents at the vertices of the quadrangle meet on the curve at (a.
Cayleyan of Cubic.
Ch. If three polar conies of a cubic are rectangular hyperbolas. Show that.y 1 ) [The quadrangle is QQ SS' in Fig.
Suppose now that P.
P
(i)
and
are
P on the Hessian of Ch. XIII.]
meet on the curve.
lie
4. Any two conjugate points 4 (ii) of the cubic of Ch. XIII.0).b. b.c)
meet on the curve at
(1/a.
[The cubic is the Jacobian of conies having the given pairs as conjugate pairs.]
See also
Ex. 1/c).0).
Ex. cos by sinh and cosh.
7.
Ex.
of contact of the tangents from O. 5. sin a. Show that the cross-ratios of the pencil of tangents from any point of the cubic of.
-3 sin 3 a)
and
—3 sin 3/3). c).1).
If the tangents at four points of a cubic
[Consider the polar conic of
(a. P and P'. and the polar conic of P degenerates into a line-pair through PFor the cubic of Ch. 2.
[See Ch. every polar conic is a rectangular hyperbola and the Hessian is a circular cubic whose singular focus lies on the Hessian.
. XIII. 4 are (c2 — a?) -r (c a — b 2 ).
(a.]
§
Ex. 3. b. 10. VII. It may be considered called the Cayleyan of the given cubic. while
§ 4.]
Ex. 1/6.
Ex.]
1
= 0. § 4 (iii) and its Hessian we replace sin and = OL + ^ni. the
cubic has the equation
ax{y*-z>) + hy (a» .]
Ex. 9. c). 2).Ex. CC) is an involution pencil is a cubic having the given pairs as conjugate pairs of points.
Ex. &c. Given three pairs of points A and A'. + 1).
(0. 8. CandC'. sin 0. Any transversal through a fixed point O of a cubic meets the curve again at P and Q. VII. Ex.
The tangents from P to the Hessian touch at two points collinear with P. the polar conic of any one has the triangle formed by the other three as a self-conjugate triangle. §
5. BB'.
XV
3
Ex.248
RULER-CONSTRUCTION FOR CUBICS
. 2
(vi). 2. It
also touches the cubic at the tangential of
O and
cuts
it
at the points
. The double rays of the involution are the line-pair of the family of conies through P.0. the locus of P such that P (AA'.
(cos/3. Show that in Ex. § 4
(cos a.0. + 1.a?2 ) + cz (x* . 6. P' are conjugate points on the Hessian The envelope of the line PP' is of a given cubic (Fig. Show that the Iocub of the intersection of the and Q is a trinodal quartic.1.
(0. §
[The circular points are conjugate points on the 'Hessian. tangents at
P
[The locus has evidently nodes at the points conjugate to O in the three ways of dividing the points of the cubic into conjugate pairs. A quadrangle is inscribed in a cubic so that its diagonal points on the curve. and that the ^angents at
(1.
1 suppose R. Q'. we have. But by the harmonic properties of the quadrilateral (P'E. §§
. Q' on the Hessian of a cubic are harmonic conjugates with respect to the point of contact of QQ' with the Cayleyan and the remaining intersection of QQ' with the Hessian. zx.
*
We
see that the Steinerian
8.
For in Fig.
2mgfiv — r)v2 — {/jiz
=
of a cubic coincide. For in Fig. of yz.
is
i(x
2
+ 2myz) +
i
(y
+ 2mzx) + ({z i + 2mxy) =
If one of the lines into which this conic degenerates
\x + py + vz
the conic
is
=
0.
and Hessian
See
Ch.
The tangential equation of the Cayleyan of x3 + y 3 + z 3 + 6 mxyz —
W
\ 3 + u 3 + v3 + -(l-4
!
m )\uv=0.
Tangential Equation of Cayleyan of Cubic. QQ') is a harmonic range in the limit. ^) on the Hessian
degenerates.
For the polar conic of
P is the line-pair QQ'. 8&. and the intersection of QQ'. RR' is the point of contact of QQ' with its envelope.
This polar conic
T]
is 0. 9.
Any two conjugate points Q.XV 5 TANGENTIAL EQUATION OF CAYLEYAN
P
249
as the envelope of the line joining any point on the Hessian to the centre of the degenerate polar conic of with respect to the given cubic*
P
the tangents from
The Cayleyan of a cubic is of the third class.
P
E
§ 5. Then f is the third intersection of QQ' with the Hessian. QQ'. 2 P' to the Cayleyan are PP'. VII. rj.
SS' by §
3. R' the pair of conjugate points on the Hessian consecutive to Q.
3
We proved in § 4 that the Cayleyan touches the lines into which the polar conic of a point (£.
(Xx + fiy + vz) (£x/\
Comparing the
coefficients
+ rjy/fi + (z/v) = 0. xy. Each of the lines of the degenerate polar conic of any point on the Hessian of a cubic touches the Cayleyan.
and two similar equations.
]
Ex. 2.
is
Or use the
The envelope of a line divided in involution by three given of the third class.
i.
conies
2.
which
is
the tangential equation of the Cayleyan.]
Ex.
{p
Ex. P consfecutive to an inflexion in
1).
8.
5
Eliminating
f we get
3
/j. and Cayleyan. tangential equation of the Cayleyan.]
Ex.
Ex. VII.
\3 +
+v 3 +.
+ v)
("
+ ty
+ /*) =
•
The Cayleyan touches the
[The polar conies of the inflexions touch the Cayleyan.
The
[The Cayleyan of a cubic having the given conies as polar conies.
inflexional tangents of a cubic. result may also be obtained by considering the tangents to the locus from any intersection of two of the conies. The Cayleyan of a non-singular cubic is a curve of the sixth degree with nine cusps. 3. The cuspidal tangents are the harmonic polars of the inflexions of the given cubic. The poles of any line with respect to a cubic form the vertices of a quadrangle whose sides touch the Cayleyan and whose diagonal points lie on the Hessian. 6.
Discuss the case
m=
in §
5.
Or verify that the two
curves touch at the points (m.t>. § 2.]
Ex. 5. m.
same way
[The cuspidal tangents and cusps of the Cayleyan are obtained in the as the inflexions and inflexional tangents of
x s + y 3 + z s +Gmscyz
=
0.
1. 9. Ex. Ex.250
TANGENTIAL EQUATION OF CAYLEYAN XV
^.
Fig.
&c]
Ex. For a particular case of the reciprocal of this theorem see § 3.
. If A and A! are conjugate points on the Hessian.
4.
Exactly similarly we show that
The tangential equation of the Cayleyan of
(x +
is
y + z)
(*
3
+ 6 kxyz =
2 (k + 4<)\(ji.
The Cayleyan and Hessian touch
at the point conjugate to
any
inflexion
[Make
P of the Hessian. Ex. 5. the polar of A with respect to the polar conic of A! for any syzygetic cubic touches the Cayleyan.(1-4to 3 )A/*i/=
0. 7.
[The polar conic of any point on the line goes through the four poles by Ch. Ex. Hessian.. Reciprocate the mutual relationships of cubic. 3. The locus of a point whose polar line with respect to the cubic touches the Cayleyan is a sextic.
12.]
. 11.
.
2 = 9(y2 ±x2). v between the six linear equations thus obtained. Ex.x2 ) + cz (x 2 -y i ) = Q. we have the tangential equation of the Cayleyan of the cubic
i/. and the intersection B of the harmonic polar of C with the line of inflexions.
and
£y.
.
. //. for the coordinates of a pair of corresponding points on the Hessian or use the method of Ex. any inflexion C. 10. z 2
.
yz. 9. 10. X'.
Ex.XV
5
TANGENTIAL EQUATION OF CAYLEYAN
2 y z
251
Ex. Find the tangential equation of the Cayleyan of
(i)
= 4:a?-g
(iii)
2 s 2 (ii) z x = y{y-x) (y-k x).
[Equating the coefficients of x2 y 2 . The Cayleyan of a cuspidal cubic degenerates into the cusp and the intersection of the cuspidal and inflexional tangents. [The Cayleyan of z(x 2 ±y2 ) = y (3x2 + f) is z Ex. i xs*-g 3 z ax (y 2 .+
+
Cy>
and eliminating f f. zx.
xy in
>?y-
(\x + fiy + vz)(\'x + n'y + v'z)
.z 1 ) + by (z2 . The Cayleyan of a nodal cubic is a conic having as the vertices of a self-conjugate triangle the node A. See §3.
0) }> (1.
cnw
clnu). the point of contact of a tangent
§ 2.
=
z 2 x^(y — ocx) (y — (3x)(y
— yx) —
0. Z by x. we have the required
now
The point
from
0.
Z2 X = Y(Y-X) (F-FZ).
<x. the equation becomes
=
=
. Suppose a > /8 > y.
Cubic with two Circuits. y. a—
z=(<x-y)*Z. ft 0). Since the polar conic of (0. Y.
. x "
£=* =
a—y
*« 4
and the equation becomes
Replacing X. yz 1 in the equation are zero. and are real since the cubic has two circuits. 1). Since the equation of the cubic must reduce to y 3 when we put x in this equation. 0.
Any may
=
2 y (y—x) (y—k x)
be taken as
(sn 3 u.
Take a triangle of reference tangent at a real inflexion 0.
ivith
2
The equation of any cubic
the
two circuits can
where r>/c 2 >0.y
CHAPTER XVI
USE OF PARAMETER FOR NON-SINGULAR CUBICS
§
1. 0. 0) are the points of contact of the tangents to the cubic from (0.
Coordinates of any Point on
point on the cubic
z2 x
a. the coefficients of z 3 y 2 z. 1) is zx 0. (1. —
tx
y
y=Y +V^L.
A
Standard Equation of the Cubic with two Circuits. Choosing homogeneous coordinates such that the coefficients of z 2 x and y 3 are equal and opposite.
by a suitable choice of homogeneous coordinates. and
ABO
AB
is
such that BO is the the harmonic polar
of 0.
be
put in
form
z2 x
= y (y — x)
(y
—k
x).
(1.
(1. y.
The points
Put now
X
= -A-. 0. 0) is
z. snw. form. the coefficients of x*z and xyz are also zero.
* Porsn(»+ K'i) = l/fcsni>. u + u + v5e =
6
e
0. C.
[The parameters of the points of contact are
-\u. where
Ex.da(v + K'i) =
ionv/snv.*
1.
v
+ K. Hence the points with parameters v 12 vu vB6 are collinear.ratios is & 2 [Consider the tangents from the inflexion C]
.ui .
1.
(c
3
. The four tangents from any point of the cubic form a pencil.
u1 + u2 + v12 =
0.
.
= v + K'i.
.
U1 + tt2 +''U.3 + tt4 + tt5 -(-tte =0.
[The method of §
7. 2. the three lines just mentioned. cn(» + K'i) see A.
d denote
cn(
sn( — \u).
Ex.
. k 2 k"sc).
If
u1 .
Ex.
side of the triangle of reference.
They
therefore all pass through the point with parameter
v5e (Ch.
&2 s s -k*scd).
where
s.
Parameters of Points on a Conic or Cubic.
ccP.
But the given cubic. + u6 + « 6
t
= 0.
. is
also available.
v l2 v3V
.u3 . §2). the conic and the line through the points with parameters v n and v3i are three cubics through the points with parameters
ult u2 u3
.
(1.
Use
§ 2 to
prove that the inflexions
lie
by threes on twelve
to the curve
straight lines.
Ex.
— \u).
and so always unless the contrary is stated.u 2 .
'
dn
(
— 1«). XII.
.254
CUBIC WITH
TWO
CIRCUITS
XVI
2
v
is real
The even circuit of the curve is given by u and lies between and 2K.
(d\ k*c*d.
.
v
+ K'i.]
+ K + K'i
Ex. § 18.
-k'*sd).
. 2 K'i
. The points of contact of the four tangents from the point on the curve with parameter u are
(s\s.
u 1 + «2 -t-w3 + w. c.cd).
. Hence
=
. one of whose cross.
.
-\n + K.
v
meet a
and verify that the cross-ratio of the pencil formed by the tangents from a point on a cubic is constant.]
§ 3. Dixon's Elliptic Functions.
u3 + ui + v3i
= 0. 4. Obtain the cross-ratio of the range in which the tangents at the points with parameters
v.
-\u + K'i. and v 12 + vu + v56 0. 3.
-\u + K + K'i.
= idnv/ksnv.
. and
k 2 +k' 2
=l.u 5 u 6
is
are the parameters of points on a
conic.
ub u e
. For suppose the three lines joining the points with parameters u t and u 2 u 3 and u v u 6 and u 6 meet the cubic algain Then in points with parameters v 12 v u vS6
The congruence
taken modulo 2K.
to4 .
xyz. Choose also the homogeneous coordinates so that the tangent at G is Then.
§ 4.
where k2 + k' 2
=
G as a real inflexion and AB as the harmonic Bnt take A.
us + u 9 + r =
0.
z 2 (y
— x) =
1.
(y
+ x)
(k x*
2
+ k' 2 y 2 )
. and that the two lines joining the points with parameters u 6 and u7 u s and u9 meet the curve again in points with parameters q.
More rigorous
is
the proof by Abel's theorem.
k. if we replace u by
The point on
u + 2m (K + K'i) + 2 n (K .XVI 4 EQUATION OF CUBIC WITH ONE CIRCUIT
255
Vu
Uj. term in z 3 .
Suppose that the conic through the points with parameters us u3 ut uB meets the cubic again in a point with parameter p.
* Ch.
have the same §ign.
of the twelve intersections of the given cubic and the quartic consisting of the conic and the two lines just mentioned.
A
Standard Equation of the Cubic with one Circuit. Ex.
. Ex. . Then
. (2) the
As
in §
take
polar of G.
u 6 + w7 + q
=
0. § 7. or y2z. § 7. Now replace z by
In
this
a and
b
generality in supposing
(a
+ b)iz.
of the cubic
is
is
in the required form.
u1 + u2 + u3 + ui + u6 +p =
0.
pair of unreal intersections of with the cubic. There is no loss of them both positive. and takes the form
AB
=
z 2 (y
— x) =
2 (y + x) (ax
+ by 2 ). for we can interchange x and y if they are negative. the equation of the cubic contains no x y. and the equation Any point on this curve
(snu en it.
(l
+ cnu)dnu).
i<_
u2> u 3> ui' u s> ue> u i> u$> u9
are the parameters of the
0. XII.
the modulus of the elliptic functions being the cubic is unaltered. Ch> X. . Therefore the remaining three lie on a straight line. and therefore
Now
=
u +u 2 + u3 + ui + u5 + u 6 + u1 + us + Ug=
1
0.
u1 + u 2 + u3 + ui + us + u 6 + u 7 + u s + w9 =
.
i. as in § 1.* Hence p + q + r 0.
snw. x2z.
be
The equation of any cubic with one circuit can
the form
put in
1.
intersections of the given cubic with another cubic. r.57. nine lie on another cubic.K'i). B as the double points of the involution determined by (1) the intersection of AB with the tangent at G and the real intersection of AB with the cubic.
The
real inflexions
is
part of the curve
have parameters 0.
when the cubic becomes
=
. l>fc*>0. we get k = 0.
elliptic. The inflexions have parameters
|e (K + K'i)
+ |e' (K-K'i).
2 y (y-x). §K. The real given by real values of u lying between
and 4K.
0.]
§ 5.
If (x. y = z = 0.
when
the cubic becomes
z2 x
= y(y—tt) a
with a crunode at
(1.
k
=
x—y
=
z
=
.4
256
EQUATION OF CUBIC WITH ONE CIRCUIT XVI
w 2 u3
.
Unicursal Cubic. 0). 1.
2*
II. 1.
0)
1.
. z) is a double point of
zi x
then (Ch. e'
=
0.
As in § 2 the parameters u 1} are connected by the relation
l
of three collinear points
u + u 2 + u3
the congruence
= 0.
[C is the inflexion. y.
2
»
+ (l + k 2 )i/-21c 2 xy
Rejecting the
either
= -3y* + 2(l + k )xy-k*x* = case x — y = z = 0.
with an acnode at
or
(1. 0. %K. noting that for three collinear points en (u v + u 2 + it 3 ) = +1. Show that the equation of a cubic with two circuits can be put in
the form
where z*{y-x) = (y + as) (y^-l^x").
being now taken modulo 2 (K + K'i) and 2(K—K'i). except that the congruences are taken modulo 2(K + K'i) and 2(K—K'i}. 2).
Ex. The proof is as in §2. § 4)
=
t
y{y-x)(y-lci x). and express the coordinates of any point rationally in terms of
functions. and A and B are the double points of the involution determined by the intersections of the harmonic polar of C (1) wifh the odd circuit and the tangent at C. (2) with the even circuit.
The connexions between the parameters of the points in which the curve meets a conic or another cubic are the same as in the case of the cubic with two circuits.
(e.
z2 x
zx
=
0.
and
M = 2 (K + K'i).. when the cubic becomes z 2 (y-x) = y 2 (y + x).
(y
0. 1. since the cubic is now unicursal. 0.
Hence many of the results established for non-singular cubics hold good with slight modification for unicursal cubics. y.
Applications of the Parameter. for the sum of the parameters of six points on a conic
A
=
=
This gives
u = I Jf +
where
for the cubic
|' iV. as could have been
z2
foreseen. 0). namely. namely. or if k = 1.
We
In Ch. N=2(K.
= 0. x = z =
when
(y—x)
the cubic becomes
= x 2 (y + x) with a crunode at (0.
with two
circuits. .
i
. the points of contact of tangents from the inflexions. We shall leave the reader in general to find out for himself what modification is necessary in any particular case. If the cubic has a double point. 2. 6)
. 1.
+ x)
x2 + k'2 y2 ).
=
1
with an acnode at (1.K'i)
for the cubic with one circuit. 0) 0.
( e .
sextactic point of a curve is one at which the conic of closest contact has six-point contact. we must have &u 0.
§ 6. N = 2K'
3. But this formula includes the inflexions (the conic» being the inflexional tangent taken twice). when e and e' are
22)6
S
.
We now apply the parametric representation of a point on a cubic to the investigation of various properties of the curve.
XVI
6
APPLICATIONS OF THE PAEAMETER
(x. XIII we expressed the coordinates of any point on a unicursal cubic in terms of a parameter in such a way that the sum of the parameters of three collinear points = 0. but not that of a cuspidal cubic. e'
M =2 K. see that the standard equation of a cubic found in either § 1 or §4 includes the case of an acnodal or crunodal cubic. z) is
257
Similarly
either if k
a double point of
(k
2
z 2 (y-x)
=
0. 4. If u is the parameter of such a point. the elliptic functions degenerate into trigonometrical functions.
A cubic has twenty-seven sextactic points.
=
where h2 + lc' 2 y = z = 0.
. 6. namely.
pairs. 4. § 3).e. inflexions. 8). we must have 9w
A
=
and therefore
u=
^M + -V.
If the tangential of such a point has the parameter
u+u+v
Hence
=
0.
=
.258
APPLICATIONS OF THE PARAMETER
There remain twenty-seven values of
XVI
u
v. It will be remembered
ways
are three different
If. is the parameter of such appoint. e'
=
0. — 8 it or 9u 0. If it.
A cubic has seventy-two coincidence points.
=
=
If the tangents at' the points with parameters u.
. 7. since v u. i. the parameters of a %(M + N).
giving
sextactic points. the points coinciding vjith their third tangential.
or v
= — 2u = — e — -iV.
Hence when the points
jugate pairs (Ch. 1.
Ex. If one polygon of 2 m sides can be inscribed in a given cubic with the sides passing alternately through P and Q. -If
e'
the tangential of a sextactic point
is
an
inflexion.
of a cubic are
grouped into conpair differ that there
by \M. we confine ourselves to the real points of the we have only one method of pairing the points of the cubic with one circuit.
\N-. where P and Q are given points of the cubic. 5.
of grouping the points into conjugate
however. or
XV.
v
£ = u + %M. coincides with its third tangential if u
As shown above.v meet on 2v the curve. and three methods of pairing the points of the cubic with two circuits. an infinite number of such polygons can be
inscribed. 2u and therefore.
number
coincidence point of a curve is one at which an infinite of cubics can be drawn having nine-point contact with the curve. the points have the same tangential.
.
cubic. the second tangential (tangential of the tangential) has parameter — 2 (-2«) = 4u. 1. 0. 2. u + ^N.
Excluding the case in which e and e' are multiples of 3.
6
(6. we have seventy-two coincidence
u
the tangential of the point with parameter has parameter — 2«. 3.
or
u + ^M+^F. and the third Therefore a point tangential has parameter —2 4u ~ — 8w.
which gives the
points.
6
both even.
then P.
. For instance.. B t A%.
Ex. The line joining two of the six points now obtained meets the cubic again in a fresh point. "^ wl) w 2n-l + ^2n w 2 ^~ W S'J Ex. A n Bn are given points on a cubic such that one polygon of 2 m sides can be inscribed in the cubic with its sides passing in order through A B lt A it P2 . q are the parameters of P. R (and Q.]
=
= =
.
«!+«! «» + «"+!.
Ex. then an infinite number of polygons can be thus inscribed.such a polygon. Ex.. s'2
u.
..
Ex.. and so on. and p. [If u lt w 2 .
Show
n
that for the polygons of Ex. Q' are two Steiner meet the curve again in a Steiner pair. If all but one of the 3» points.
(i)
2. The reader generalize the theorem.
—
—
x . consider the case
.. . Q. (iv) Any two sextactic points form a Steiner pair for inscribed
inflexions
5. 0. \£ A-y.
= 2. If P.. Q is a Steiner pair for a polygon of 2w sides. «....3. J. R) is a Steiner pair for a polygon of in sides..
.
sides
meet the curve again
9. Show that the points thus obtained have parameters given by the expression mu + nv+pw. A n B". If P.
where
P
x
and Q x are
form a Steiner pair for inscribed hexagons. B2 . on the curve where PQ X and QP1 S meet the curve again in P2 and Q 2 and P0a and QP2 meet the curve again in P3 and @3 The conies of closest contact at P and Q meet on the cubic.
If If
P and Q
(ii)
the tangents at P and § meet on the curve are conjugate points on the cubic.q) 0. Q and P'..
. the lines
PP' and QQ'
Ex. «<2 + '«S=«n+l + «n+2.
XVI
6
APPLICATIONS OF THE PARAMETER
259
[Suppose «. CWte.
1
. w. 6. 106.. >
>
Ex..
Three triangles can be inscribed in a given cubic whose
in three given collinear points. . u ln are the parameters of the vertices. 7. which are the vertices of a polygon of 2n sides and the intersections of opposite sides of the polygon. 4...
2 + « 2n + «i
= 0.
lxiii I (1864).
.. P and § are called Sterner s points. so does the remaining point. Then we must have
the congruences.
q + u2 + us 0..
pairs.
\
dodecagons.
8. and PQ meets the curve in the tangential of R.
Ex.
See Clebsch.
p..
Three non-collinear points are taken on a cubic with parameters The lines joining any two of these meet the cubic again in three more points. 2 + m4 + ms which are consistent if n (p . ••• consistent which is evidently the case. we have to prove
. The two points in which the lines joining any point of the cubic to a pair of Steiner points meet the curve again are also a Steiner pair. . lie on a cubic. Pascal's theorem is a particular case of this result. =
=
so that
n
3. .
. 5.
=
may
U 2n-S~^ w 2n-2 M 2«. v.
(iii)
Any two
If
£4
PQ and §P meet
.
PQ and
the tangentials of
P and
S
t
§Pj meet on the curve
§. u 2n are the parameters of the vertices of .
Deduce the fact that the sextactic points are the points of contact of
tangents from the inflexions. B'. All cubics having eight-point contact with a given cubic at a given point pass through the third tangential of the point of contact. XII.
triangles
Ex. Three of them on the odd circuit.
whole
.
Su=Bv=3w = u + v + w. 18. D' meet on the curve. n. w are commensurable with any periods. and six on the even circuit.
See also Ch. 141.
Six of the coincidence points are real. PC. C. p numerically less than those considered.
Now assume the are among those obtained. and that the line joining A. both inscribed and circumscribed to the cubic-
Ex. The tangents at A.
Ex. it meet* the cubic again on the line joining the point of contact to its second tangential. Ex. or D' passes through one of four points P. and use induction.
p+v+w=
give
Ex. C.
w + 4m + k>EE0
give
v
= w. If a conic has five-point contact with a cubic. PD meet the curve again at A'. Any side is divided harmonically by the cubic and its point of contact with its envelope. D'. cvii. p.]
p = 4:U.p are positive or negative integers such that
a multiple of
m+n+p—1
Under what conditions
the total
number
(u>
of points obtained finite ? *
[The points with parameters
— v—w.
Ex. v. B.260
APPLICATIONS OF THE PARAMETER
3. Q. C'..
§ 4. such that the tangents at P.
Deduce the
tangential. See Hurwitz.
Ex. and the* number of conies through three given points of a cubic which osculate the curve.
—2 m. If
[Au + v + w=0.
lie
13. S meet on the curve. The lines PA.]
. or D to A'. B'.]
a conic has four-point contact with a cubic. v.
— w—u. 16. Orelle. The coincidence points are the vertices of the 24 '. 17. D on a cubic meet on the curve. C. 10.
P is
Ex.]
Ex. 14.]
and similarly
Ex. B'. B. The number of points is finite if u. 15. 21. — 2»
(v
+ iv) +
+ u). result true for all values of m. PB. Q. the line joining the other two intersections of the conic and cubic passes through the second tangential of the point of contact. An infinite number of triangles can be inscribed in a cubic such that the tangent at each vertex meets the opposite side on the curve.
fact that a coincidence point coincides with
its
third
* If not. and any other point of the cubic. —u—v. B. S on the curve. the construction gives the whole of the odd circuit or the of both circuits. 11. Nine or three of the sextactic points are real.
They
all lie
on the
odd
circuit.
[5w + »
=
0. is
XVI
6
is
where m.
Find the number of conies through four given points of a cubic which touch the curve.
[If u.Hart '. Show that the tangents at A'. C. n.
12. u + v— (u + v + 2w)= —
2u>.
[Easy also by quadratic transformation. if it exists.
w
are the parameters of the vertices. B.
pp.
e
a curve very close to the being a small constant.] must satisfy the relation
<f>
. 26.
Ex. de France.
[The harmonic polars of the inflexions. n'. n = 3.
[As in Ex. c are constants not involving n. contact of any such conic. m. The number of coincidence points on a curve.]
Ex. The locus of the sextactic points of a pencil of syzygetic cubics nine straight lines. iv (1876). m. 22.
+u 6 = 0.
. 359. OC+OC'). b. ffx
degenerate curve ffx
of n.XVI
is
7
ANOTHER STANDARD EQUATION
261
Ex. 19.. If six points of a cubic lie on a conic.
OL'
we must have
<fi
(n. For a more general result see Halphen. In this equation g2 and g3 are any constants. xv (1879). m.
How many
Ex.
XIV. 24.
20. Hence for all values
=
/= 0. where a are the equations of two curves. OL') = (j>(n + A'. a). OL = 18. 22. <£ = 3 = n = 3. Annalen. Bull.
3
.
<j>
<j>
Ex. m = 3. We may get their equation by identifying P with its
points
. OL.
Ex. so do the six conjugate points.
§ 7. If a second cubic passes through nine points of a given cubic. Prom the number of sextactic points of a any curve has 12m — 15« + 9k sextactic points.
=
m'. de la Soc. a = 10. it meets the cubic again in the points of contact of a triply-tangent conic of the same family.]
Ex. c are at once given by noting that = 27 n = 3. is
33m-42>i + 27K. b.
cubic. 25. a) + (n'. = an + bm + cOt. Let it be <t>(n.
+(u i + \M) =0. 23. Now a. Math. Three families of conies can be drawn touching a given cubic If a conic is drawn through the three points of at three distinct points.
OL. = an + bm+cOL..
/. p. deduce that
[We assume that the number is the type.. The theory of functional equations gives readily that where a.
»
A cuspidal cubic has no A nodal cubic has three
are real ?
sextactic or coincidence points.
{u 1
+ \M) + {u i + \M)+
.
3.
Another Standard Equation of the Cubic
be
The equation of any cubic can
2
put in
2
3
the form
y z = 4x -g2 xz -g3 z by a suitable choice of homogeneous coordinates. Ex. 59-85. the notation being that which is usual in the theory of elliptic functions..
sextactic
and
six coincidence points. m = 4.
[If
Mi
+ MjH-
.
cj>
same
is
e
for all curves of the
if
same
Bm + K.
third tangential in Ch. 21. m = 6. another cubic passes through one of the points and the points conjugate to the other eight. m'. m. Math. The locus of the coincidence is eight equianharmonic cubics see Halphen.]
Ex. a = 12. m + m'.
§ 9. Now
is
= 0.
Hence the equation of every (non-degenerate) cubic can be put in the given form.
at (1. m.
s
= m (x — i-z) the curve where x = £z and
y — r\z
(m'
i
where
i 2)
+ (4|-)»2 ) xz +
£-2mn + 4:l -g
2
z*
=
0. we get the
ax? + by 2 z
required form. or acnode.]
from any point of the curve
if
is
cross-ratio of the tangents
harmonic
gs
=
0. if (x. y. Use the equation of a cubic given in § 7 to prove that the pencil of tangents from any point of a cubic has a constant cross-ratio.
the coefficient of
x2 z
zero.
The cubic has one or two circuits according as the harmonic polar y of the real inflexion meets the curve in one or Z 3 three real points that is.
roots.
=
0).
. the harmonic polar of B.
2. according as g 3 is < =. and the polar line of the intersection A of this tangent and harmonic polar. II.
%
1)
be a point
P on
the curve.
The line is therefore a tangent if this equation in x/z has equal which gives m* _ 24 £m2 + 32 ijm + 1 6 g .
the cross-ratio
§ 11)
of the pencil of
(<p+iy(<l> -2f(<i>-iYP
where
Ex.
The
line
through
P meets
4a.262
ANOTHER STANDARD EQUATION
XVI
7
Take as sides of the triangle of reference the tangent at a real inflexion B.
and equianharmonio
g2
=
0.
/=
The
if
16&.
yz
leading to
= 12x -g z = y + 2g and to g = 27 g x:y:z= .
=
=
.
We
see then that the cubic is of the form
Replacing
+ cxz 2 + dz 3 = 0. Since the line meets z the tangents from P is given by (Ch.
3
2
2
3
.
=
.. the coefficients
. cusp. Since the polar line of
is
(1.
I. § 4) shows that. The usual condition (Ch. z) is a double point of the curve.
[Let
(£.48 £ 2 = 0. according as <7 2 — 27#3 is less than or greater than zero. or >0.
3
:
:
2.0) is yz = 0.# *
2
2
2
2
2
xz + Sgs z 2
= 0.
Ex. xy and y 3 in the equation are zero. Since the polar conic of (0. the coefficients of x y.
= 2i(^-cp+iyj\ ^=64^. 1. now x by ( — 4b/aftx.
of xyz
and yz 2 are zero. 0)
is
x
=
0.. 0.
found that in this case the double point is a crunode. 1. 6 when Since the equation of the cubic must reduce to x' 2 2 we put z in this equation.
It will be readily
.
e'= 0.2). except that the congruences are taken modulo 2o> and 2a>. where u is real. given by
where
4 (pu)
!
:i
pu
2
Weierstrass's
elliptic
function
(p'uf
=
-g pu-g.
If the points
with parameters u.
—
\pu—pv)
. iu
u*.
.
are collinear..
pw
This gives
p'w
u+v+w
where
2o>.1.
we may
take any point on the curve as
is
(pu.
. -^->
The
is
real inflexions
have parameters
—
o
o
the real period.
Hence
v w + ?'lizl'u + P'U ~P'V + P -P' ' — pv — pw = — pw — pu = ~ pu — pv
'
from which
The
(i) immediately follows.. If g 2 3 >27g 3 2 the cubic has also an even circuit given by
. L'
v. § 72).•
.
-
.
parameters of the form u + co. In fact (Dixon's Elliptic Functions.
u+v+w = /p'w — p'u^
if
0.
pu+pv + piu = 4
.
Coordinates in terms of Weierstrass's Function.
)
'
XVI
8
WEIEESTRASS'S FUNCTION
263
§ 8.pu=\. 2a>'
=
(mod. 2o/).
2 y z
If the cubic
=
^x' — g i xz 1 — g% z z
'
is
non-singular. The standard form of this and the preceding section may be used as in § 6 to establish properties of the cubic instead of the standard forms of §§ 1. The connexions between the parameters of the points in which the 'curve meets a conic or another cubic are the same as in the case of the cubic with two circuits in § 3.
1)
. 4.
pu
pv
p'u
p'v
1 1 1
=
2co. p'u.
(e.
&
) i(*\pv-pw)
t
/P'v — p'w^ *
—
=ii~ *\pw-pu)
4
.
where
2a>
The odd circuit is given by real values of the parameter.
(i).* inflexions have parameters
(eeo
+ eV).
* It is at
once proved that
all
the plus or
all
the minus signs must be taken.
0.
are the periods of pu.
or 3. we may draw up a table showing all possible types of quartic as follows
Type
A
.
Types of Quartie.
quartic cannot have more than three double points (Ch. 2. § 1. VIII.:
CHAPTER XVII
UNICURSAL QUARTICS
1. so that its deficiency may be 0. § 2). 1. VIII. By the aid of Ch.
The limaijon
r
=
a+
6 eos
for a/b
=
|. § the conic
5). §. however.g. G each property of the curve can be duplicated by reciprocation
.
The method of projection and inversion will. In fact. is a cusp) . It is not readily the double points are of different kinds But for instance.
(Ch.
XVII
2
GEOMETRICAL METHODS
265
process which comes to the same thing is to project into the circular points and then invert with respect to A. . a cusp. both cusps or
if
applicable
all
In the case of a quartic with a node A and two cusps B.
f
.2fy + c
=
0. be suitable only if B and G are double points of the same kind.
since every circular line meets it in only
two
finite points
Fig-
1. and from each property of the conic a property of the quartic is deduced. an ordinary node.. if J. 1. 2.
both ordinary nodes.
An
inverse of this with respect to the origin
is
ax 2 + 2 hxy + by 2 + 2 gx +. II.
e. it is readily seen that the equation of a quartic with double points at the origin tnd circular points is of the form
A
B
and
c (x 2
+ y2) 2 + 2 (g'x+fy)
(x 2
+ y 2 + ax 2 + 2hxy + by 2 =
)
0. when the quartic becomes a conic (a parabola. quadratic transformation is still available. and a flecnode.
if we project the cusps of a quartic with a node and two cusps into the circular points. 4""j and the origin the centre
x
.
which
A
c (x
2
+ 1/2 )
=x
z
(i).
2.
is the inverse of a parabola with respect to its focus. For instance. But the parametric equations of a three-cusped hypocycloid
are
the
= %a (2 cos + cos 2<p).
(ii). if we project two of the cusps into the circular points. § 4. 2)» For the polar reciprocal of the curve with respect to a cusp A and having will be a cubic touching.
e. C into the circular points.e. we may obtain its properties by reciprocation from the known properties of the nodal cubic..
i. Hence. Or again. Three-cusped hypocycloid. If we project B. the curve becomes a
with three cusps
limacon with a
the cardioid
finite cusp. Or again.
The properties of a quartic
may be investigated geometrically in various ways. -j. 1. it becomes the inverse of a conic with respect to a focus. V.
266
GEOMETRICAL METHODS
XVII
2
for the reciprocal of such a curve is a similar quartic.
r
= <z(l+cos#). if we project the points of contact of the bitangent into the circular points. 1
= a + b cos 6. the quartic becomes a conic with as focus since the inverse of a curve with respect to a focus has cusps at the circular points. and then invert*
to A. and
with respect
A
especially Ex. §tt. and conversely (see Ch.
shows
five
limacons with the same b and
a/o
%.
A
being the origin.
of
. the circular lines at the line at infinity as inflexional tangent... 1 to 6 in that section). 2. as is evident from its Pliickev's numbers. y = Ja(2sin0— sin20) cusps being = 0. the curve becomes a three-cusped hypocycloid (Fig.
=
-g. a limacon with polar equation of the form
r
Fig. i. Its equation can therefore be put in the form Pig.
' Now invert with respect to A. 1. ABCBXJ pass through the same four points.BS) is harmonic.touch the
theorem from Pascal's theorem
similarly. If and BM' meet CA in and m'. B. C of a trinodal quartic curve. 2.]
Ex.
Ex.' It is a straight line (the polar of A). 4. C and that the envelope of the conic ABCPQ is a quartic with A as node and B. xxxiii.
A
quartic has nodes A.
drawn
BM
MN
m
CN
BA
[Project B. Then we have Find the locus of the middle point S of a chord PQ of a bicircular quartic meeting the quartic again at its finite node A. Another proof is given in § 5. through the intersection of and M'N'. ABCQT. or use the equation of the bitangents given in § 3. Q.]
Ex. A quartic has nodes A.
B.
.
. and we have Through a fixed point A a line is drawn meeting a fixed conic in P..
[Projecting the two nodes into the circular points and inverting with respect to the other node. These theorems are due to Jolliffe. Derive a [Projecting and inverting we have Brianchon's theorem. T. • the conic LMPABC. B. Through A is drawn any line meeting the. C. Ex.
BM' and CN. osculating the curve at A cut at L and touching the curve at P. we have ' The intersections of a conic with a pair of directrices are concyclic. and so on for the points P. B.
6. The four points of contact of tangents from two nodes of a trinodal quartic lie on a conic through those nodes. BS). quartic again in P. Show that LM. C. C into the circular points and invert with respect to A . Show that the locus of Sis a conic through B. C are conjugate chords of
A
quartic has three nodes A. C. we obtain
<f>
Sb2 {(x-a)* + y 2 }
+4a(x~af =
as the polar reciprocal of the hypocycloid with respect to a cusp. S. Show that the conies ABCPS. C. by differentiating with respect to in the usual manner. Hence the locus of the middle point of the chord of the bicircular quartic is a circle through A and the locus of S for the original quartic is a conic through
:
' :
'
..']
:
Ex. CN'
Show that two bitangents to the quartic pass to the quartic. V.
and . and CN' jneet in n and n'. Q and S is a point on PQ such that 1/AP+ 1/AQ = 2/'AS find the locus of &'.
A.
C
conic through A. then two bitangents pass through the intersection of mn and m'n'. Writing respect to the circle
(£c-a) 2 + i/ 2
=6
2
and finding
its envelope. C as cusps. The first conic meets the second again in P. B.
are
3. Six conies pass through the nodes A.
Ma
PA
A
The conies through A. B. Q and BC in B. But this is the curve (i). BC) are harmonic.
chord
PQ
meets
BC in B
and (PQ. B. A quartic has three nodes A.
Tangents BM. B.
B. Find the locus of S if (PQ. Messenger Math. the second meets the third again in Q.C]
Ex.2
XVII
GEOMETRICAL METHODS
down
the polar of the point
(ii)
267
with
the curve. 5. [Project B and C into the circular points. A (PQ.
A quartic has nodes A. touching the quartic.C.C. B. In general there are two quartics with three given nodes. Show that the remaining intersections of the conies are collinear.
Ex. 9. Show also that the tangents to two such quartics at an intersection divide BC harmonically. Q meet at T and cut the bitangent IJ at Prove that
Any
The tangents
is
(EF.
.
C. Ex.
Ex. the locus of the remaining intersection of the conies is a conic through A. 13. C. 11.
(iii) The chord of contact of a conic through and C touching the quartic at two points passes through one or other of two fixed points.]
Ex.
8.
A
quartic has a node
A
and two cusps B.
The locus of
T
is
a conic through
I.
Ex. C and the tangent at A is Show that the locus of the point of contact of a tangent drawn from a fixed point on BC is a pair of lines through A. C touching the curve at P and Q meet again on a fixed conic through
A. XIII.
(i)
(ii)
Q. or may obtain other theorems by reciprocating those given above.
If
AD is the
C
meets and (ii) If tangents from B and C to the curve meet at the curve at P.
D
AD
B
(iv) If PAQ is a chord of the quartic. 15. Quartics have given cusps A. B.
(iii)
The
line joining
T
to the intersection of
PF
and
QE
passes
through the point of concurrency of the cuspidal tangents. Conies are drawn through these nodes touching the quartic at P and Q. 12.
The tangents at the cusps of a tricuspidal quartic are concurrent.
Find the envelope of the line through any point P of a which forms a harmonic pencil with the lines joining
P
Ex.
fixed points cut
Quartics with three given nodes and passing through four other any line through a node in involution. the conies through A. IJ)
harmonic. The result is also evident on projecting the quartic into a cardioid or three-cusped hypocycloid. B. B.
Ex.
J
touching the quartic at
three points. F.]
[Project
B and C into the
The reader may
Ex. § 4.
tricuspidal quartic to the cusps. Through the cusps of a tricuspidal quartic three conies are drawn touching the curve elsewhere. passing through four other fixed points and touching the circle through the nodes (the point of contact not being a node). the locus of D is a conic through B and C.
Prove that
chord conjugate to BC of any conic through A.268
GEOMETRICAL METHODS
XVII
2
Ex. B.
14.' Obtain other properties of the tricuspidal quartic by reciprocating the examples in Ch.
P and
E.B. t
also given.
circular points and invert with respect to A.
'
[The reciprocal of_: The inflexions of a nodal cubic are collinear.
tangent to a tricuspidal quartic meets the curve again at at P. If PQ passes through A. PA and the tangent at P divide BC harmonically.
(i)
10. derive other properties of the quartic similarly. 7.
19.
Ex. Hence the quartic takes the form
(i). P being any point on the
curve.
. AS' are equally inclined to the tangents A to the quartic. or z.
y
Another method of obtaining properties of the trinodal
quartic is to take the three double points as vertices of the triangle of reference. In the equation of the quartic there can be no terms involving the third or fourth power of x. In Ex.
§ 3. .S'P)/AP is constant. (AS'.
H
the
cofactors
of
. If Sand S' are the foci of a bicircular quartic with a finite node A. F. A bicircular quartic with finite node can be regarded as the envelope of a circle passing through whose centre lies on a fixed conic.1
-z
b c 2h 2g 2f = —"+-" + -A + ^L + + yi i
Z
yZ
zg
. 16. 18. y.a
g.
0. 17.
from
Ex. SP±AS . 17 the lines AS. G.
We
shall
denote by
A
G.?' at 0.XVII
3
TRINODAL QUARTIC
foci
269
and common
finite
Ex.
If
this section. ay 2 z 2 + bz2 x2 + cx 2 y 2 + 2xyz (fx + gy + hz) = 2 2 2 we divide by x y z we get the equation at the head of
. It can also be regarded in two ways as the envelope of a circle orthogonal to a fixed circle j through whose centre lies on a fixed conic touching.
Ex.
s
"
" m The Quartic
. Two bicircular quartics with the same node cut orthogonally.
— X
. .
cusp. The fourth node is (F. each node is either an acnode. 3.
Ex. 7.XVII
3
TRINODAL QUARTIC
of the
line
271
For the lines joining the intersections quartic to (0. 1. or is a crunode such that the tangents at this crunode and the lines joining it to the other two nodes form two non-
overlapping pairs of lines. In this case all four bitangents are real. &c.gpv)xy 3 + 6a /i2 y*
i
2. Ex. the tangents from the nodes are all real or all unreal.]
a
and
6. If a quartic with three real nodes has a real bitangent.
show in
[The envelope is 1=0.
6. c have the same sign. Show that (2l+3J)/\/i.
modification occurs in the tangential equation of the curve.
x and y conjugate
imaginaries. The envelope of a line divided by a trinodal quartic in a equianharmonic range is a curve of the fourth class with four nodes.
Ex. 1) are
and
6bX 2 x* + 12 (h\ 2 + b\p-f\v)x*y + 6(a\ 2 + b/j.
and the
cross-ratio
I.
§ 11). if one or more nodes become cusps ?
What
Ex. 9.'i + ct> 2 ~2fiJLi>-2gi>\+4h\n)x' y 2 + 12 (h fi + aX/i. The envelope of a line divided harmonically by a trinodal quartic is of the sixth class. the two curves having the same tangents at these nodes.the inflexional tangents of the quartic.
[The axis of perspective
is
_
x/F+ y/G + z/H =
]
. 4.
8.
Find the bitangents
if
one or more of the nodes becomes a
What
modification must be
made
in Ex.
two
c. h. and
Ex.
[The envelope
is
J= 0.
=
.
Ex. 0.
(p
of this pencil is given
by the above
equation (Ch. The diagonals of the quadrilateral formed by the bitangents Gx + Fy — Cz 0.
2.
/and
g. H). B. The triangle whose sides are these are diagonals and the triangle whose vertices are the nodes are in plane
=
perspective.
1
in this case ?
Ex.v = are the nodal that the twelve common tangents of this curve and and inflexional tangents of the quartic.
If one of the nodes
becomes a cusp. We shall touches the nodal tangents of the quartic] § 4 that 2 =
Ex. The other common tangents of the curves are the inflexional tangents of the
quartic.
A
which only 'one
is
real has
[We may suppose
z real. G.]
1=0
:
is a curved of the third class and Ex. and the quartic may be put in the form
y z + z x + x y'' + 2xyz (fx + gy + hz) [For a real bitangent a. b. real quartic with three nodes of real bitangents.
C is
zero. Three of these nodes are the nodes of the quartic and the two curves have the same tangents at these nodes. It has the nodeB of the quartic as biflecnodes. The other common tangents of
the ciirves are .]
i t
1 1 2
=
0. 5. one of A.
= 0.
+ lG{q(p-s) + s(q-r)} 2 =
For the converse take
0.
r
= =
0. z. +
*/c).the conic touching QB at P'.
[For the quartic of
§
3
(i)
they are
(
+ i/a. r. B. The four conies through A. The bitangents touch the quartic at real points
positive
are
and
2 (a/2
three or one
+ hg* + ch* .
&c. 10. Show that. the first degree in x.
[The centre of perspective
2(jp
2
is p = q — r. 14. s.
is
Ex. the equation of any trinodal quartic can this form. Find the locus of the nodes of a trinodal quartic.
APa? + Brh? y + Cn*f + 2 Fmnyz + 2Gnlsx + Hlmxy m.B. BP at Q'. represents a quartic with bitangents
= 0.
2
=
ay*.
Ex.
The points of contact of the four bitangents
2 2(p 2 + q2 +r + s )
all lie
on the conic
=
{p + q + r + sy. q.
in §3(iv).] if a. p p q = s. C and the points of contact of a bitangent touch in pairs at C.
.^dbcf.
2
= 0. 17. CB pass through a pair of vertices of the quadrilateral formed by the bitangents. 12 are the vertices of a quadrangle whose harmonic triangle is ABC. Three bitangents of a quartic with nodes A. 11. 6. and the two tangents at C are harmonic conjugates for CA and CB.
16. =
r.
=
=
=
be put in
[The equation
may be
2
2
put in either of the forms
s
2 2
{2(p + q + i + s )-(p + q + r + s) {(p-s) + (q-r)} i -8(pq + rs){(p-s) + (q-r)}*
2
}
=
Gipqrs. &c. r q.272
TRINODAL QUARTIC
XVII
3
Ex. where p. y.
Ex. c
Ex.Cz 2
conies. . 13.]
Ex. 12. 19. and the + 22 + r ) =(p + q + r)i]
!
conic
is
Ex.
Ex. */ ± b.
where
t
varies
.] =
s
s
are of
p=
and with nodes
0. if AP meets QB in P. One of the common chords of any two of the four conies of Ex. 15. C form a triangle Show that the triangles ABC and FQB are in perspective. The quartic
1
the envelope of the conic
'2
where
I.
= 0. given the four bitangents and two points on the curve. + signs being taken. q r s. The equation */p + >Jq + */r+ */s 0. The four centres of perspective obtained in Ex. 18.
[The conies are
±
*/a yz + */l zx ± */c xy
Ex. 12 passes through a node. The quartic
the envelope of the conic
fix 2 + 2 1 (ayz
+ hzx + gxy) -By 2 + 2 Fyz .
is
=
0. PQ at B' passes through the points of contact of the fourth
PQB-
bitangent.
C and
[6a. n vary subject to l + m + n = 0.
The double rays of the involution formed by the tangents at the lines CA.]
Ex.
and of two similar families of
.
p — u — ^bcx + */ca y+
i
</abz. p Conversely.abc) > ( + >/af± <Jbg±Jch.
Their properties are given in the following theorems. F.
The identity
of
§ 3 (v) is useful in the
examples. Show that the locus of the other focus is a quartic with nodes at A.
. U = afyz + bgzx + chxy. and one focus lies on a fixed conic S.
A
Ex. C. By varying the relative positions of S and ABC. Soc. C. To save space we use the following contragtions
/
u=fx + gy + hz.
[The envelope of an asymptote or axis is a three-cusped hypocycloid. iv (1903).
\
2
2
2
2
By
u.
The n the
conies T theory. 22. p. pp. be at once deduced.
[Use Ex. Math. 20. B. that their envelope and the locus of their centres are quartics with nodes at A. CO with the sides of the triangle a conic of given eccentricity is drawn meeting 'the sides of the triangle again at D.4
:
XVII
TRINODAL QUARTIC
273
Ex. Various forms of the equation of each conic are given.
tri-
There are many conies of interest connected with the nodal quartic
ay 2 z 2 + bz 2 x2 + cx2 y 2 + 2xyz(fx+gy + kz)
=
. II. iii (1902). A = F\ + Gfi. 2 = a\ 2 + bfi 2 + cy 2 . + Hi>. Conies of given eccentricity pass through fixed points A. Given the nodes and five other points of a unicursal quartic. Amer. K = ay z + bz x + cx y + 2 xyz (fx + gy + hz). 103. and vice versa.]
Ex.
2 we denote
the same as in §
3. construct the tangents at and from the nodes and any number of' other points on the curve. as they are needed in the examples. BE. B. Show that AD. C. BO.
Conies connected with a Trinodal Quartic. 23. C.
2 2
fc
— 2abfg\u. Through the remaining intersections of AO. E.
If the point-equation of the conic is obtained. T = a2f 2 \ 2 + b 2 g 2 n 2 + c 2 h2 v 2 -2bcghnv-2cahfv\
T = ghyz + hfzx +fgxy. 22. M=f(gG + hH)x + g(hH+fF)y + h(fF+gG)z. B. 489. 21.
(i). of course. A.2f/iv -2gv\-2 h\/i. CF meet in a point whose locus is a quartic with nodes at A.]
§ 4. 154.. conic touches the sides of a given triangle ABC. ABC is a fixed triangle and a fixed point.
=
and U = (i. e." See Annals of Math. T = 0) are of importance They pass through the nodes of the quartic
. we may find every possible shape of a quartic with three real double points. Trans. B. the tangential equation can.
Show
Ex.
we prove as
in the case of
A t and A 2 Bx and B2 Theorem VIII that G^C^ is
.
If the tangents touch the quartic in
C1 and
2
.
3. In Figs.Gxz + Hxy =
t
.
In the following examples the conic through the points of contact of the bitangents.
=
0.hH) xy + af 2 By 2
. 2 + By 2 2 Hxy and the lines
—
bg 2 Ax 2
+ 2fg
(A .
=
(Cz 2 -Fyz-Gxz
Fig. 1) to the intersections of this line and the quartic being the tangents from the nodes -4a.Fyz . is called Conic I and so for the other theorems.
.
' '
.*?) (x-±)(Ax-y -40) = 10(x 2 + 2t/ 2 -12a.
Then
since
(By 2 -2 Fyz + Gz2) (Ax2 -2Gzx + Gz 2)
-AK
+ H^y) 2
. 4 are shown typical unicursal quartics.
3
(</
2
.
A
-j
x+
B = fF+gO — z —y
——
the lines joining (0.
lt
2
2
and also on
k (Gz 2 — Fyz
— Gxy + Hxy) /gG + hH B G
=
w
A
hH+fF
G
\
which reduces
to the given form
on taking
k
(fhF+ghG+fgC)/fgW.278
TRINODAL QUARTIC
.
. which is referred to in Theorem I..) 2
2
. 3.
XVII 4
.
the points
A A B B lie on the conic Gz . 0.
is
[Evident from their tangential equations. 2. I'roye also a similar result for the points of contact of the tangents from the nodes. and .
..]
The conies
III. the six lines AA\. Show that if in Ex.
279
intersections of A X 2 and BC.-LB are collinear.B and B C.
The chord of contact
0. The tangents a* the nodes A. 1 (either case) the line A X A 2 meets the quartic again in A\.B.
A Fx + BGy+ CHz =
and
its
pole
is (F. and their intersections with BC. &c. CA.
§ 4. 0.
G. VI touch
at the
same two
points.
Ex.
3. BB\.XVII 4
TRINODAL QUARTIC
A
. and C2 Show that the a it 1
B^an&CA.4{if
-lx*) (a-
4)
(20x + y + 160)
=
10{2x 2 -y i + 8x) 2
. A\. 1.]
. CC\ touch a conic. AA' t BB\.
Ex. respectively lie on a conic.
.0. CC\.
V. C of a trinodaU quartic meet the curve again in A 1 and A .
[The required lines are hcghx + cahfy + abfgz
=
and
ghAx + hfBy+fgCz
=
0.
AB
[The equations of the lines are given in
Ex. H).]
Pig.
Ex.
form given. The conies f7=0.
Show that the common tangents
AS + kh?
=
are the
(1
common tangents
of the quartic and the conic of the conic and the two cuf ves
(4
of the third class
.
Ex. B.
XVII
VI.
U=
Ex. 6. IX pass through the same two points on the quartic.
5. and a common chord of and VIII. [k = 0. VII. meet the conic again at its intersections with the diagonals of the quadrilateral formed by the bitangents. VII pass through the same four points.
The
conies
U=
0.]
M
Ex. If the conic
Ax* + By 1 + Ca2 .
IV pass through the same four
T=
0.
The conies
U=
0.
. 12. The diagonals of the quadrilateral formed by the bitangents are the polars of the nodes with respect to the conic IV.
W. = 0.bz ) + Gy (az2 .k) A3
=
{(8
. If the points of contact of the tangents from the locus of P is the cubic
1
P
lie
on a conic. civ.
9. A conic can be drawn touching the lines joining the nodes and the common tangents of the conies III and IV. C]
.
S2 =
0.
IV pass through the same four
points. [The intersections of U = with Obtain theorems by projecting these points into the circular points. 13.
I.Gzx .
Ex.ex*) + Hz (bx2 .
Ex.
[If
first
4. V. II. 1.8.ay*) =
[The polar cubic of
0. Use the tangential equation of the quartic]
Ex.
Ex. 4/3.
VII. III.
Deduce the equations of the conies III. A conic can be drawn having. 308. Ex.
P touches
a conic at A. 14. Conies are drawn touching the four bitangents and one side of the triangle ABC.]
Ex.
points. &c.
l
TRINODAL QUARTIC
The conies
0. 11.9i) +
.Fyz .x V
.
Fx {cy* .
Gross
see Stahl. the nodes as vertices of a selfconjugate triangle and touching the common tangents of the conies IV and V.
The
The
conies
conic's
{7=0.
Ss = we have
= 3^. 10. 15. VIII.
Ex. Show that the points of contact with these sides are
collinear. Any conic touching the bitangents {(bg^-eh^W-lAff + cAS + lcGvX-lbHXn} + k{-aB\* + (af -ch2 )iJ! + cBp + 2cFViJL-2amiJ}
2
. The chord of contact passes through the intersection of a common chord of V = and VII.4
280
Ex. 16.
[On AFx + BGy+CHz = 0. VI.
..
AX
X
X
F
. Crelle.
are the point-equations of the conies 'in the This result is due to S3 -2S2
S =
. III
have four
common
tangents.Hxy
meets the sides of the triangle of reference in
=
x
Z
and t 1\ and 2 the lines 1 and AX. The conic passes through the intersections of the conies V and IX.]
Ex.3/fc)t} \. The conies II and IV have double contact.
IX
pass through the same four points.. 17. Ex. inverting with respect to a node. and generalizing by projection.
I.
7. p.l
is
=
0.
1
and
Z
2.
. P2 and B'.]
Ex.
[Conic
I is
a line-pair. Blt P2 are Ax2 = By 2. if A 2
'
node with the
= 4FGH.]
Ex. 21.
[See §
3.
9.
Illustrate
by tracing
2
x2 y 2 -2xy(x-2y)-Bx 2 + 8xy + 3y
Ex. the points of contact of tangents from and C are collinear. = x 2 + y 2 + z" = + 2Fx2 yz + .]
threes in two
The tangents from the nodes cannot meet by
Ex. If the tangents at A are harmonic conjugates with respect to and AC. The intersections of the tangents at each node with the line joining the other two nodes cannot lie on two lines..
the polar.
.^23.
AB are harmonically conjugate with respect to the tangents is cx+gz = 0. 27. Show that AC and the harmonic conjugate of BC for BA.
A
= 4fgh.. the points of contact of tangents from C lie on a line through B. 22..
BC
Ex. Show also that three of the points Pt P2 Qx Q2 Et P2 (say P2 Q2 B2 ) are collinear. P1....]
Ex.. if A =. One diagonal AB the other two meet at and intersect AB in R1} P2 Similarly Show that AA'. 18.
Ex.
Similarly for conies III and IV.
AC
BC
=
Ex.. Conic II for ay 2 z 2 +
is
.
before attempting the
Ex. 28. §§ 1 and following examples..
if
[Conic III
is
a point-pair. ¥
Ex.
.
= 0.]
[AC and
at A. —c). XVIII. =
of conic
0.. Qlt @2 are concurrent..
AB
A
B
[P=0.
.
is
. if «A and C
are not both positive.
points.2fgh.
+ 2fxl yz + . The tangents at the nodes meet by threes in two points.
. and that APlt BQ lt CE 1 are concurrent. The points of contact of the four bitangents are the intersections of the quartic with two lines.. + . 19. The tangents at A and B form a quadrilateral.
.
. and the point of concurrency is fx = gy = hz.
4. If the tangents from are harmonic conjugates with respect to and AC. CC we obtain points A'..f.
+ +
.
The reader should read Ch. III.
V for
and vice
versa.
. 26. and so do the remaining intersections of the curve with the tangents at
C
[/-
o. 24.
. What conditions must hold in order that conies I. The intersections of the tangents from each line joining the other two nodes lie on two lines. 25. + . and similarly Hence 6" is (g.]
Ex. z 0. reciprocal with respect to
Aifz 2 +
. IV may be real? [The conic ax 1 * by 2 + cz i + 2fyz + 2gzx + 2hxy = is real.. Therefore is cy+fz = Q.. meet on a diagonal of the quadrilateral formed by the bitangents.. BB'.4
XVII AB
TRINODAL QUARTIC
281
Ex.
C
. Show that conies II and V are always real.]
.
. 20.
282
TRINODAL QUARTIC
AC,
XVII 4
touch a conic.
Ex. 29. The lines BA', CA', CB', AB',
[aX 2 +
...
BC
+....
~(f+bc/f)ixv+
...
+
...
= 0.]
Ex. 30. The conic through ABC touching and passes through the points of contact of the tangents from C to the quartic.
AC
BC C
]
[gyz+fzx + cxy
= 0.]
to the quartic
lie
Ex. 31. The points of contact of the tangents from on a conic through C.
[(fh
+ bg) xz + (gh + of) yz
The points of through A, C and the intersections of
•Ex. 32.
= Ax + By2 +2 (ch - 2fg) xy. contact of the tangents from A
2
lie
on a conic
Similarly
BC
with the quartic.
for the tangents
from B.
1
t
;•
[bgxz + ghyz
= By
+ (ch — 2fg) xy
fhxz + afyz
= Ax2 + (ch — 2fg) xy.]
Ex. 33. The conies of Ex. 31, 32 pass through the same four points.
Ex. 34. The tangents at C meet the quartic again in C1 and C2 and Cy C% meets the curve again at C\ and 2 Show that the points of contact of the tangents from C lie on a conic touching CC\ and CC\ at C\ and C'j and passing through A and B. Show also that through the intersections of the conic U = with this conic a conic can be drawn touching CA and CB at A and B.
,
C
.
[(/F+gG + hH)f-H(hz2 + cxy)
=
cU.]
Ex. 35. Through two pairs of vertices of the quadrilateral formed by the bitangents can be drawn (i) a conic touching at at A and B (ii) a conic touching conic I at its intersections with (iii) two conies each touching at A and at B and each passing through the points of contact of two bitangents.
CA
CB
;
CA
CB
AB
;
[(i)
uz + cxy
=
;
(ii)
u2 — bcx 2 — cay* + abz
1
=
;
(iii)
wz + cxy+\/abz2
=
0.]
Ex. 36.
Show
v
that, if
u lEfx+gy + hz,
= cxy + uz — tz w = bcx + acy — u
2
,
2
2
2
+ 2t (cxy + uz) —
t
'z
2
,
touches the quartic at its four intersections with than A and B. Show that v = touches and CB.
iv
=
v
= 0,
other
CA
[The quartic
is
v
2
+z2 w =
is
0.]
Ex. 37. Show that w = that in the former case w that in the latter case to
= =
a line-pair, if t = h or if t + 'Jab. Show is the tangents from C to the quartic, and is two bitangents. ,
=
is
Ex. 38. Show that the locus of the pole of AB with respect to v> = a conic through C, C, R lt R$; and that this and the two similar conies pass through the same four, points.
[Ax2 -By 2 + Fyz- Gxz
Ex. 39.
Show
that
= 0.] w — divides ^Bj harmonically.
XVII
5
TRICUSPIDAL QUARTIC
§ 5.
283
Tricuspidal Quartic.
The results of the preceding section require modification if the three double points of the quartic are not all ordinary nodes. Suppose the double points are all cusps. Since the tangents at each double point are coincident, the quartic takes the form
a —
-"
ar
H
— —
b
o
c
^
i
-I
y
zL
+
2V(bc) yz
2V(ca)
H —
zx
+ —
2V{ab) xy
—
u.
loss of generality.
all be taken as minus without For unless the quartic is a pair of coincident conies, three plus signs or one plus and two minus signs are impossible while if (say) the first sign is minus and the other two plus, we can make all signs minus by replacing Va by — Va. Finally, replacing x, y, z by -/a. x, Vby, -/cz we obtain the equation of the tricuspidal quartic in its canonical form
;
The ambiguous signs may
x2
y
2
s2
.
yz
zx
z
xy x
The cuspidal tangents y
= z,
=
x,
=y
are concurrent
at the point (1, 1, 1). Since the equation of the quartic
(yz + zx
may be written + xy) 2 = 4xyz(x + y + z),
is the bitangent, its points of contact being its x+y+z 0. intersections with yz + zx + xy The equation of the quartic may be also written
=
=
5+2/ The tangential equation is
a;
-
-2
+z
-2
=
0.
(A
+ n + vf =
27\pv.
§ 2
In the equations of the three-cusped hypocycloid given in (ii) put 2Z=a + 2x; </3y, 2X=a-x-</3y,
2Y=a-x+
=
are the sides of the triangle 0, Z 0, so that whose vertices are the three cusps.
X=
Y=
This gives
6Z =
a{l + 2cos(0 + §7r)} 2
*
67=c6{l + 2cos(0 + |7r)} 2 6£ = a{l+2cos0} 2
,
,
,
from which
is
readily deduced
any three-cusped quartic is projected so the cusps and the point of concurrency of the cuspidal that
It follows that, if
5
284
TRICUSPIDAL QUARTIC
XVII
triangle, the quartic
tangents become th» vertices and centroid of an equilateral becomes a three-cusped hypocycloid.
Ex. 1. The coordinates of any point on the tricuspidal quartic in canonical form are expressed rationally in terms of a parameter t by
{t-iy>x
Ex.
2.
=
(t-hlfy
=
4z.
conic is inscribed in a given triangle and passes through a fixed point. The locus of the point of concurrency of the lines joining the vertices of the triangle to the points of contact of the opposite sides is a tricuspidal quartic.
[If the point is (X, Y,
A
Z) and the conic
is
(*/fl*
+ (yA)*+(*/fl* =
+ (r/y)* + (Z/*)*
o,
the quartic Ex.
3.
is
(X/*)l
=
0.]
variable cubic touches three fixed lines at fixed collinear points and has inflexions at its other intersections with the fixed lines. Show that .the line of these inflexions envelops a tricuspidal quartic, having the line of the fixed points as bitangent.
A
[Writing
inflexions of
down the
conditions that
(0,
—v,y.)
and
0,
(
— v,
0, X)
are
xyz+(x + y + zy(\x + /j.y + i'z)
=
we
obtain
is
which
of the third class
and fourth degree since
it
has
(1, 1, 1)
as
a bitangent.]
Ex. 4. The locus of the centre of a conic having a given equilateral triangle as self-conjugate triangle and having its asymptotes inclined at
-1 an angle tan J */3
is
a three-cusped hypocycloid.
Ex. 5. A fixed radius of a circle and' the tangent at its extremity intercept on a moving line a segment which is bisected by the circle. Show that the envelope of the moving line is a three-cusped hypocycloid. Ex. 6. The envelope of the Simson (pedal) line of a given triangle is a three-cusped hypocycloid. The tangents from a vertex of the triangle are the two sides and altitude of the triangle through that vertex.
[Writing down the condition in trilinear coordinates that the lines perpendicular to the sides of the triangle at their intersections with
are concurrent,
\a + rt3 + vy= we get the tangential equation
.
of the envelope
fiv(ii
2
(1
+
cos .4
cos
.
B
.
cos C) \/iv
.
=
2 sin 2 A
.
.
cos
C + v cos B)
which touches sin A Oi + sin B See Converse, Annals of Math.,
this result.]
+ sin C
II. iii
at the circular points. (1904), p. 105, for an extension of
y
=
may
The
Ex. 7. The equation of a quartic with one real and two unreal cusps be put in the form
(a?
Ex. 1. If two of the nodes of a trinodal quartic are biflecnodes, so the third.
[/=</
Ex.
2.
= =
fc
0.]
a quartic has biflecnodes A, B, C, the tangents at A are harmonic conjugates for AB, AC. The tangents at A, B, C touch a conic for which ABC is a selfconjugate triangle.
If
Ex. 3. If a quartic has three real biflecnodes, one is an acnode, and the others are crunodes. Its equation cap be put in the form
-2
a?
+ jr 2
= s -2
.
If a quartic has three real biflecnodes, the intersection of two diagonals of the quadrilateral formed
Ex.
4.
acnode is the by the tangents
at the crunodes.
Ex. the
'
5.
Show
that the
'
cross-curve
x*f
carbon-point-curve
'
=
a 2 y1 + b i x*,
and the hour-glass-curve
are curves of this type. Trace the curves, and show that the two former are the loci of intersection of lines parallel to the axes of an ellipse or hyperbola through the intersections of any tangent with the axes.
to a quartic with Ex. 6. The points of contact of the tangents from three biflecnodes A, B, C lie on a conic which meets the quartic again with respect to the quartic. on the polar line of _2 Then use and as (£, /?, £). [Take the quartic as ar 2 + ir 2 +.z =
the identity
{£W+e)x+
•••
, + -}{f*(y +« )+
1,
•••
+
...
•••}
=
Hyz + £&x + £r,xy)
2
{( v
+ i 1)x>+
+
...
+v(ye+
...
+
...}]
t
1
286
Ex. 7. If in Ex. 6 a line whose envelope
OTHER QUA.RTICS
lies
is
XVII
6
[The conic
on the quartic, the points of contact lie on as a self-conjugate triangle. a conic having and the of Ex. 6 degenerates into x/t + y/n+z/C
Ex. 12. Through the crunodes A, B of the quartic of Ex. 8 four conies can be drawn having four-point contact with the curve. Their points, of
contact are the real sextactic points. The conic through A, B and three of the sextactic points touches the quartic at one of them.
Ex. 13. If in Ex. 9 the conic ABPQR meets the line joining S to the acnode on the quartic, the conic ABP'Q'R' meets the line joining S" to the acnode on the quartic.
Ex.
14.
cross-ratio
Obtain the envelope of a line divided in a range of given by a quartic with three biflecnodes.
in §
3.
[Put
The envelope becomes the conic touching the
biflecnodes, if the range is
f=g = h"=0
tangents
equianharmonic]
at the *
Ex. 15. The equation of a real quartic with one real and two unreal 1 (x 1 + # 2 ) 2 biflecnodes can be put in the form xyz The quartic can be projected into the lemniscate of Bemouilli
Now use the properties of invariants of conies to obtain the condition that polygons of 4, 5, 6 sides can be inscribed in the lattericonic and circumscribed to the former. We find
— 2 the cuspidal tangents pass that in the case through the points of contact of the bitangent, and that any two lines harmonically conjugate with respect to the tangents at the node meet the curve again. at four points such that the tangents at these points form a quadrilateral inscribed in and circumscribed to the quartic.
Ex.
4.
two tangents from any point of the bitangent of a quartic with a node and two cusps envelops a unicursal curve of the third class. [The tangents from the point (x, y, z) are given by 2 2 s (t + 2)x + kt (2kt+l)y-2(kt + t + l) z=0. If the point lies on the bitangent z = — 4k(x + y), this becomes 2 2 {2kt 2 + t + 2} {(4/fc 2 * 2 + 6fc*+4&+l)x+fc(47c* + « + 6* + 4)y} = 0. Hence if tx t3 are the parameters of the points of contact, while 6kv + (8k + l)u + 6 = 0. and t 1 t2 = v, t1 + t2 = u But the line joining the points with parameters t lt t2 involves u and v
,
in the third degree.]
* See Eoberts, Proc. London Math. Soc, xxiii (1892), pp. 202-211, for this and the following examples.
:
288
,
OTHER QUARTICS
B
and a node
C.
XVII
Show
6
Ex. 6. A quartio has cusps A, following points lie on a conic
:
that the
(i) A, B, the points of contact of tangents from points of contact of the bitangent.
A
and B, and the
(ii) A, B, the inflexions, the remaining intersections of the curve with the tangents at C.
[These and many similar results are proved by noticing that the curve can be projected so as to be symmetrical.]
and a node C. Show that the quartic has two cusps A, of the tangonts at A and B is a C to the intersection double line of the involution determined by the tangents at C and the
Ex.
7.
A
B
line joining
lines
CA, CB. Given A, B,
C, 0, find the locus of
the inflexions and of the points of
contact of the bitangent.
[A quartic with nodes at A, B and a cusp at a conic through ABO. See Ch. V, § 4, Ex. 8.]
Ex.
8.
C
touching OA,
OB
;
Reciprocate Ex.
III.
5, 6, 7.
Quartics with two Flecnodes and a Node.
real flecnodes
Ex.
1.
The equation of a quartic with two
the form
{x*
and a node
can be
putm
+ y 2) z^-x^f + Zhxy
{z
+ x){z + y)
= 0.
Show
that the
C.
Ex. 2. A quartic has two flecnodes A, B and a node C. following sets of six points lie on a conic
(i)
(ii)
A,
A,
B and the points of contact of the tangents from A, B, B and the intersections (other than A, B, C) with the
curve of
the tangents at A, B, C. [These and many similar results follow at once from the fact that the quartic can be projected into a quartis with symmetry.]
Ex. 3. unicursal quartic with two unreal flecnodes can be projected into the inverse of a conic with respect to the reflection of a focus in the corresponding directrix.
IV.
A
Quartics with a Biflecnode and two Nodes.
A to a quartic with nodes A, B, C form a harmonic pencil with AB, AC, and similarly for the tangents from B. Show that C is a biflecnode and that the tangents at A form a harmonic pencil with AB, AC, and similarly for B. Show that the equation of the quartic can be put in one of the forms
Ex.
1.
The tangents from
;
z* (a?
±y i + 2mxy) ±x"-yl =
0,
0.
[F = G =
gives
/=g=
if
the curve
is
not degenerate.]
Ex. 2. Show that, if a quartic has nodes A, B and a biflecnode C, it can be projected so as to have two axes of symmetry ; unless A is a crunode and B an acnode, or vice versa, when it can be projected so as to have a centre of symmetry at C. Trace the projected curves in the various cases which can occur. Prove that, if A and B are acnodes, C is a crunode.
[Putz
=
l in
Ex.
1.]
; ;
XVII
7
UNICURSAL QUARTICS
289
Ex. 3. A quartic has crunodes A, B and a biflecnode G. Show that the tangents at A and B meet at the vertices of a quadrangle having A, B, C as diagonal points; and that their remaining intersections with the curve are the vertices of a quadrangle whose diagonal points are C and
two points on AB. [This and similar theorems follow from symmetry, or by putting /=<7 = in§§ 3 and 4.]
(ii)
Ex. 4. A unicursal quartic cannot have (i) a biflecnode and a cusp, a biflecnode and a flecnode, (iii) a flecnode and two cusps, (iv) three
flecnodes.
§ 7.
Unicursal Quartics with
Two
Distinct Double Points.
Unicursal quartics with only two distinct double points include quartics with a tacnode or rhanrphoid cusp and another double point. The reader will readily obtain their properties by modification of the results of Ch. XVIII, §§ 14 and 15. He may illustrate his argument by tracing the curves of Ch. Ill, § 6 (xi) to (xv), § 8 (v) to (vii).
I.
Unicursal Quartics with a Tacnode.*
A quartic has a tacnode at C and another double point at B the tangent at C and BA the harmonic conjugate of BC with respect to the tangents at B. Show that its equation is
Ex.
1.
Ex. 7. The points of contact of \ and its intersections with the tangents from B form an involution with a double point on CA.
Ex. 8. The points of contact of the tangents from touching the quartic at C.
A
lie
on a conic
[Write down the polar cubic of A.]
*
22Ifi
For other examples see Messenger Math.,
xlvii (191V), p. 95,
U
7
[J? !
Ft
is
2x + 9>(a — l)y + z
=
.
meets & at L lines joining
J.
quartic with a triple point is converted into a unicursal cubic by projecting two points of the quartic into the circular points and then inverting with respect to the triple point. b. 1-a 2 ). if
B is
a flecnode
. a2 ). CA.
Ex.
[yz + x*
and Cx
lie
on a conic osculating
=
2y(x + ay). CB.
4. 1. 7. CC1 are concurrent.6.s y (x
+ ay).
A quartic has a rhaniphoid cusp at C and another double point at B CA is the tangent at C and BA the harmonic conjugate of BO
Ex.
Ex.
Ex.
Quartics with a Triple Point.
is
a cusp
a
=
1. B. The conic osculating the quartic at C and passing through the remaining intersections the curve passes 2 of lt 2 with x through Bj.
Show that a = 0. 12.
9.
2. The tangent from C is x + ay = 0.
AB.a) y\]
. 1 is
x + ay *=z touching
where x* — xy — ay* = 0.
A
. touch the conic of Ex. The conic osculating the quartic at C and passing through the intersections of the curve with (other than B) passes through Cv
AB
[yz + x*
= 4:y(x + ay). 12 X harmonically by CA.
F F
[yz + x 1
+ 2 xy = 4
(1
. and BK.
CH
as
C
to the points of contact of 6 has
CA
Ex.
& lie
Ex.290
II. The tangent from B is z = (l + a)x. if B cannot be a biflecnode. 9 at B passes through Bt
[yz + x*
C
and touching the conic
=
2xy.
0.
F F
is
.
with respect to the tangents at B. ..
on a conic
for
If
BB
and by the
a double ray.
A
lie
on a conic
[Write down the polar cubic of A. lt 3
Ex.]
Ex. touching at B x (\-a. 10.
XVII
7
Ehamphoid Cusp. The points of contact of the tangents from osculating the quartic at C.
6. &c.
.
that
its
equation
is
=
4a. The points of contact of b the quartic at C. B 1 C1 are concurrent.
UNICURSAL QUARTICS
Unicursal Quartics with a
1. 13.
and the conic
x i —4 xy + 2yz
= 0.]
Ex.
3.] Ex.
If the
B
CA
is
FF
F
:
F
. touching at Cj (a. The conic osculating the quartic at of Ex. 8.
BK
and
KK
X
2
is
divided
§ 8.
(yz + a?)*
Show
. C. [z(2x + ay) = (2 + a) a? + 2axy. 11. the involution pencil formed by CB. B lt Clt and the points of contact of which AC and BC^ are conjugate lines.
K K
.
Ex.]
Ex.1.]
Ex. 5.
and that
B
The bitangent
6 of the quartic of Ex.]
line
tangents at meet the quartic again at 1 and 2 the divided harmonically by 1 3 and the tangent at B to the conic touching CA at C and passing through B.
1). replacing z by z +px + qy and choosing p and q so as to make the coefficients of x3 y and
xy 3
zero..
bitangents*
I. we may suppose u = xy (x + y). the equation of the quartic is
An
zwj-v
=
. We may simplify this equation by a suitable choice
other
of the
two vertices of the triangle of reference. B.. 18. IX.
where
u = a x3 + 3a 1 x 2 y + 3a 2 xy 2 + a 3 y 3 v = A x* + 4:A x x 3 y + 6A 2 x 2 y 2 + 4>A 3 xy 3 + A^y*. the three
the quartic
O meet
[Project
A.y
XVII 8
QUARTICS WITH A TRIPLE POINT
291
Of course quadratic transformation with the triple point and two other points on the curve as the points C. If we take the triple point as (0.
V2
.
Ex. collinear points of a cubic are
collinear.
(
(iii)
meet the curve
(
where
)
Va x2 + txy + Vb y2 2 =
Va x2 + Txy .
=
(ii). O osculating the quartic at again on a conic through A. (T0.
If
is
A
and B.
a triple point of a quartic through conies through A. Hence from each property of a unicursal cubic can be
deduced a property of a quartic with a
triple point. The coordinates of any point can be expressed rationally in terms of a parameter by considering the intersection of the curve with a line through the triple point.
r
=
Therefore the lines (iii) are the four bitangents of (ii). and is a diagonal of the quadrilateral formed by these
Quartics with a Triple Point. B of Ch.
The reader
then
z
will easily verify that. 0.. 295. Then. comes to the same thing. O.
+ 2t(</ax+Vby) =
(ii)
z+2T(Vax. A. p.B
Then
:
'
The tangentials of three
*
to the circular points and invert with respect to O. For instance.']
An
exception arises
if
h2
=
ab. B. if all three tangents at (0.
Va+
Vb) 2
a + 2h +
b. 0.
alternative is to reduce the equation of the quartic to a simple form by a suitable choice of axes or triaDgle of reference.
See Ex.Vby) =
0.
(i).
.
1. 1) are real. § 1.Vb
2 2
)
= 0.
19.
we may
reduce
(i)
to the
form
zxy(x + y)
=
ax
i
+ 2hx2 y 2 + by i
. if
(t-Va-Vb) 2 =
a + 2h + b.
Ex. s 5= are four lines. In Ex.
3. whose reciprocal would be a quintic with
6.
r
=
p =
-s
0. 11. B. and also with respect to the lines joining to the other intersections of with the quartic. A triangle is inscribed in a quartic with a triple point 0. Show that OQ and OR are harmonically conjugate with respect to OP and 0. Q.
of a quartic meets the Ex.
a triple tangent. For the quartic § 8 (ii) the equation of the conic is
z2 + 4z (ax + by)
[It is
.]
Ex.8 (ab + btyy*} + 64-ab (a + 2h + b) {zxy (x + y). § 4. and OR trace out an involution pencil. Any diagonal of the quadrilateral formed by the bitangents of a quartic with a triple point O meets the curve and the conic through the points of contact of the bitangents in an involution whose double points lie on two tangents at O.
Ex. XIII. 12. OQ curve again in A.]
q=
s. The points of contact of the four bitangents of a quartic with a triple point lie on a conic. Q. r q «are concurrent. R.]
Ex. Show is a cubic.8 (ab
+ bh)
y*
=
0. B.
Ex. CA. is a quartic with
a triple point and with these lines as bitangents.292
QUAETICS WITH A TRIPLE POINT
XVII
8
which osculate Ex.
No
and circumscribed
to
triple point
[The quartic could be transformed quadratically into a quintic with a and three cusps. 0. R vary. If
§ 3. Ex.
The
line y
=
tx joining the triple point of § 8
(ii)
to
the
remaining intersections of the curve with the line joining the sections of y = tx x and y = t 1 x with the curve is given by 2 bt + {b + av-(a + 2h + b)u} t + av = 0. The conic OABQR meets the quartic again at S. B.] Ex.
Ex. P. 7. + */r + a/s = Vp +
-x/ff
such that p = s. B. XV.
or
we may
use Ex. the quartic at P. A quartic with a triple point passes through three fixed points A.ax* -2hxi yi . 11. AB again at six fixed points.
Ex. at A. §
4. {See §3.
[See Ch. 3.
6. 2.
9. R. A conic through the triple point If A. The lines joining to the remaining intersections of the quartic with the sides of the triangle form an involution.
13.]
triangle can be both inscribed in a quartic with a triple point. Q. XIII. C and meets BC.
Ex. 10.
Two
P on the
R
AB
[See Ch.)
EE
1. 1 there are three conies through A.] conies through the triple point O of a quartic meet again curve.
.8 (ab + ah)
2
x* . B. 4.
=
= 0. Show that A.
0. 9. that the locus of
[Use Ex. R lie on a conic. B. Ex.
inter-
where
91^=1. 8.by*} =
*
. P.
5.9. and they both touch the quartic at Q and respectively. P are fixed and Q.
[Use Ex.]
Ex.
•»
(l
+
sfj)
(1
+
fc.
readily seen that these bitangents are
1
{z"
+ 4z (ax + by)~8 (ab + ah) x . 7 and Ch.
.
The curve
16(x-y)(2x + y)y
with bitangents
=
a.
t
+ 4(. Let Ea 2 3 3 be the three pairs of vertices of 1 the quadrilateral formed by the bitangents of a quartic with a triple point 0. 2
l!
3(a a s -a 1 a )X/i-3(« a
i!
-«
2
l
)n
2
=
. Show that. . ts and l s form an involution.
.
Ex. Add and subtract these equations.
. 10. The conies of closest contact at
O with
§ 8
(ii)
l
are
i
)
zx-b{x — xy + y 2 -2hx = zy-a(x 2 -xy + y 2 ) — 2hy1 = 0.
(vi)
and
l lt lit l3
with the quartic.3 a2 Ai + 3 a^A % . . and subtract each from the equation
s2
of the conic of Ex.
be the conies of closest contact with the quartic the lines joining to their remaining intersections Then prove that tt and l x divide E1 1 harmonically. and the conic
52ar-52jBy-27 y .
F
and OFlt t a and \. (vii) A conic passes through E1 Ft and the four intersections of S2 and S3 It touches at the harmonic conjugate of tx for t2 and ts (viii) The conic through the intersections of St and S3 touching tx at meets t 2 and t3 in points collinear with the intersection of £./by) = 0. 2 z (x + tj) + (3a + 6)a.
four
(v)
Let
Slt S2 S3
.
The conic 6x 2 + 4xy + 3y' -24:X + 8y-96
i
=
passes through the points
of contact of the bitangents.16a.
contact with the conic through the points of contact of the bitangents at its intersections with tx
.
F E F
.
. . + (3a + 36 -2A) xy + (a + 3b)if = 0.V b) (>/ax-/by)z -&(h+ ^ab) (a/ox. 12. if the tangents t lt t 2 t3 at are taken in a certain order.^/byi2 = 0.4 (os A .]
0.152^-48 = touches the six inflexional tangents (only two are real). Through the four real intersections of these conies passes another conic through the
1
inflexions./a. For the curves § 8 (i) and (ii) the tangential equation of the conic is
4(A
A
i
-4:A 1 A 3 + BA i2)v'i -4:(a . then
t
E F
.A i -Sal A s + Ba2 A 2 -a A 1 ) vX .]
Ex.
=
0.
at O. 10 to 15.
The
lines
OE1
.
12x*
-Wx*y -ix^y* + 8xy + Syi
s
Sx + 2y + l2 =
0.a A s ) v/i — 3 (a x a3 — a 22
<
3
)
X2
X 2 + Al 2
.
will illustrate Ex.
. A line divided equianharmonically by a quartic with a triple point O envelops a conic touching the six inflexional tangents and touching the Hessian of the tangents at O.
+2
=
0.
:
(i) t2 t3 are the pair common to the two involutions subtended at by the points of contact of the pair of bitangents which meets at Ex and the pair which meets at F1 (ii) The two conies through and the points of contact of these two pairs of bitangents both touch t 1 at and have four-point contact.
.
.
-A|ii
+ + 4^(\ + F)v + 4(^+3o6)j.
1
.
(iii)
(iv)
The conic OE2 F2 E3 F3 touches tt at 0. Through E2 F2 E3 F3 can be drawn a conic having double
.XVII
8
(ii)
QUARTIGS WITH A TRIPLE POINT
the diagonal
is
293
[In § 8
z
=
0.
.
x-4 =
0.
y-3 =
0. 13. and E F1 [Two pairs of bitangents are 1 z" + 4(*/a + Vb) {-/ax + */by)z-8 (h.a/oB) (*/ax + .
&c. The conic through the points of contact of the bitangents the triple point O touches the linear branch at O. 3. 18.]
Quartics having a Triple J*oint with two Coincident Tangents. triple point at which two Ex. +
=
4 2/4 -84 2/
0.
2x + y + 2 = 0.
Ex.
p-iq.
Determine whether
bitangents in this case. one tangent at number of bitangents and inflexions for each type. Obtain the conies through its inflexions.
its
Hessian.
9x-8y+36 =
II.
and
is
find the
[For
§
8
(ii)
the condition
h?
= ab.
whose bitangents
0.
By replacing x. 19. For § 8 (ii) this becomes a + h + b = 0.
and the point-pair
unreal.Ex. [C41 — 0. 2. If one tangent at is real. &c]
%
. The equation of a quartic with a 2 tangents coincide may be put in the form zx y = y"(2 xt+ y) + ax*..
!i
. x-iy. Each of them is divided harmonically by two of the tangents at the triple point 0. 10.
[y(y + z)
=
2ax\]
.
0.]
Ex.+ y i )
0. 16 should degenerate into a
point-pair. b. three bitangents are concurrent. Sketch roughly the different types of quartic with a triple being the line at infinity. 22.]
. x + iy. The bitangents are z + y = ± 2 </ a (x 4 y)
Ex. The bitangents of a quartic with a triple point three tangents are real are all real or all unreal. 14 should degenerate into line-pairs.
a. Determine the condition that the conies of Ex. Their points of contact
2x-y+S =
lie
2x-y-6 =
on 2
(a:
-2. one is illustrated by
S6y(y + l)x
with bitangents
9a.
and .
!
+ 81
10y + 45
=
0. two bitangents are real.*+ (y-lf
=
21. 17. Find the condition that the conic touching the inflexional tangents of the quartics of § 8 (ii) or Ex. and determine the point 0. when the conies become respectively unreal and real line-pairs.
h in § 8
(ii)
respectively
by
p + iq. If the conic touching the inflexional tangents degenerates.
Ex.
Sx + 4y + 18 =
Sx-2y + 9 =
0.
[Illustrate by tracing 2ixy (x + y) are the line of infinity and
=
(2x i.
Ex.
this point-pair is real
^
is
or unreal. y. 20. The fourth bitangent is divided harmonically by the Hessian of the tangents at 0.
r-\(a + b)
obtain the equation of a quartic with a triple point at which one tangent is real.]
Ex. [There are four such types . Determine whether the points of contact of the bitangents are real
or unreal. 1.
XVII
8
QUARTICS WITH A TRIPLE POINT
at
295
which the
Ex. The inflexions lie on the conic Eys+2xz = 2ax [Eliminating a between the equations of the curve and wegef 2f + x 2 z = 0. 21.
a
<
0.]
Ex. the line joining the inflexions.
Ex. 1. distinguishing the cases
a
4. The tangent to the superlinear branch.
=
6X». The inflexions are real or unreal according as the points of contact of the bitangent are real or unreal.
1.
2
.
Any
Ex.]
Quartica with a Superlinear Branch. A conic passes through the inflexions and the remaining intersections of the inflexional tangents with the curve. § 8. and touching the inflexional
tangents. t three can be put in the form zy 3 The bitangent is z == 0.
[yz
= 4aa. and the line joining the remaining intersections of the inflexional tangents with the curve. The line divided harmonically by the quartic envelops a conic touching the curve at the singular point. We find that then I — m = h = 0. 1. the bitangent. are all concurrent. A quartic has two linear branches having three-point contact with one another.]
Ex. 3. The inflexions are (±ai.
the equation
=
(yz + x 2 f
+ 2 y (yz + x>)
+ my) + y
(ax 1 + 2 hxy + by 2 )
Choose a triangle of reference ABC such that C is the double point.
2.
[It is
X 2 +16a/i» +
128aV =
0. 1.]
TJnieursal Quartics of Class Six with one distinct
Double Point.
3.
Ill. of Order Three.]
Ex. 1 are
(p^ + iq) y
=
4^2
and
y"
=
0.
IV.
ry + 12a»»
III.
6.
is
1
[By Ch. which touches the bitangent and touches the quartic at the singular point.]
line met by the quartic in an equianharmonic range passes through the intersection of the inflexional tangents. 6.
The bitangents
of the quartic of Ex..
(yz + x*y
=y
(Ix
1
(px 1 + qy % ). 5.
36a 2 ).
Ex.
Show how
to transform the quartic into a conic
by quadratic
transformation. 2.
see that the curve may be projected into one with Trace this projection. B the other intersection of the osculating conies at C. 4. Show that its equation can be put in the form
.
we
an axis of symmetry.
1.
[Putting z
=
1.
a
= 0.
. 4a 2 ) and the inflexional tangents are
Ex.
Ex.
296
QUAETICS WITH A TKIPLE POINT
XVII 8
Ex.
Ex.
=
z+
8aix — 12a'y
at
(
=
0. A the intersection of the tangent at C with the other common tangent of the conies.
[Put
z-x /y
2
for z in Ex.
*
They meet the curve again
+ 3ai.
Ex.
.
> 0.
The equation of a quartic with a superlinear branch of order (Bay^ — x 2 ) 1. The envelope of the line divided equianharmonically by the quartic is a conic touching the inflexional tangents and touching the superlinear branch at 0.
2. ABC
of conies meet the curve at seven points coinciding with (0. 6. £w 2 Conies are yz + 'dx2 = 0.
If in IV. 8yz + 35x2 0. Show that the quartic can be projected into one with an axis of symmetry. 'A conic through two inflexions and the points of
contact of a bitangent touches the curve at &']
Ex.
Put 2=1.
1. Properties of the curve can be written down from the symmetry.
Ex. Indicate roughly its shape in the cases
p>0.
a
=
2
I
.]
Unicursal Quartics of Class Five with one distinct
Double Point. 1). such conies. Ex. if any such conic meets the curve at P.
3. More generally.
t2
t3
ti
Ex. h ts tt + <x ts tt + tt t% tt + tx tt ts + 1 = The tangent at any point is (l+4?)x + P(l-2ts )y + 2tz = 0.]
.
2
>0.]
The curve is of class 5.
the equation can be put in the form
(yz + x2 ) 2
{yz
= xy
s
. 2 [Putting z-x 2 /y for z we obtain z
=
xy.
(t
.
p<0.XVII
8
QUAETICS WITH A TRIPLE POINT
1
297
Ex.
1.
at the points with parameters
and meets the curve again
+ (l/2£)z — t.
p>0.
are collinear. 0. Q. the tangent at to that conic of closest contact at C which goes through B.
=
R
Ex.
1.g.
[Replacing z by z — lx — my we reduce the equation to
+ x2
2
)
= y'(px + qy).
[Put 2tpix
V. 4. Q.
— t* — t).
0. q<0. Express the coordinates of any point of the curve rationally in terms of a parameter.
They lie on a is real and two unreal. but no conic meets it eight times there.]
infinite
An
number
B
AB
B
Ex. One of the inflexions
B
B
=
. 4. through The inflexional tangents meet the curve again at points lying on such a conic. E.
5. e.
. the tangents at meet the quartic again in six points lying by threes on two P. Ex. Any point on the quartic If the points with parameters t t
is
.
Ex. 5. the point of contact of the tangent from C.
The bitangents coincide with y
=
0.
Derive properties of the quartic from those of the conic by
quadratic transformation. CA the tangent at C.
h + h + h + h = 0. Choose a new triangle of reference such that C is the singularity. and C with that conic which passes conic having double contact at and has closest contact at C with the quartic.
2
. [t h i<o. q>0.
=
(l-?)qly.
1) is a node.
. 0.CHAPTEK
XVIII
QUARTICS OF DEFICIENCY ONE OR TWO
§
1.
where u2 .
Nodal Quartics. Also.
Fig. and take the line IJ as z Since (0.
call tangents at meets the curve at another point.
In this chapter we shall consider quartics with deficiency one or two. the equation of the quartic is of the form
We
=
u4 + 2us z + u^z 2 =
. 1).u3 ui are homogeneous of degree 2. and u 2 = is the equation of the tangents at the node. 1. we take this node as (0. 4 in x and y. 3. A quartic with deficiency/two has a single node or cusp. 0. If Each of the it has a node 0. these points / and J. and w 2 = since z = intersect at two points on the curve.
. 0.
i00{2v +
i
ij
1
-lGx-ltj)(x'i + 3yi -Sx-l2y) + (ix + y)(2x + 3y)
«
=
0.
B'. OD Show that 0.
1.
The six points of contact of the tangents to the quartic from the node-0 lie on a conic touching OH and OK at and K.
A
D
meet the quartic again in
[If the line is z
= %.
Ex.
the quartic at H. and OA. C. B. C".
4. 1 and H
=
being the equation of the lines joining
to the points
2).
and the required conic
Ex. Putting z = 1. B.
u2 v2 + zu3 z + u2 z
.]
Ex. OB. are collinear. Eight foci of the projected 4-ic lie on j.
[Project
H and K into the circular points. OB. and OA. For the polar cubic of is u. lie on a conic.
2. and that the lines HABC. D'. are the points of contact withj of the common tangents of j and the polar
Ex. C. are collinear. C. + u 2 z = 0.
the lines OA. the points of contact of tangents
from
Ex. which meets the 6 quartic six times at and at the six intersections of the quartic with z 2 = v2 as is seen by writing (i) in the form
. we see that the curve can be projected so as to have as a centre of symmetry.
OD are
« 2 vi + '2v u 3 + 1« 2 1>! 2
l
= 0.
is
»2
=
v i z -]
meets the quartic again in A.
Ex.XVIII
1
NODAL QUARTICS
299
of the quartic
u2
v%
is a factor of be written
%
.
Show
can be put in the The equation of a quartic with a biflecnode form xyz 2 + u = Q or (xl + y 2 )z ! + u = 0.
Ex.
5.
H
K
its
own
cubic of 0. and the 4-ic is the envelope of a circle cuttings orthogonally whose centre lies on the polar reciprocal with respect to j of the .
line meets the quartic in A. K. the envelope of a circle cutting any given circle orthogonally whose centre lies on a tricuspidal 4-ic with an infinite bitangent is a nodal 4-ic.
line
A
through
H
If and are unreal. B'.
3.
.
3. OC.
Hence the equation
may
. KA'B'C meet on the conic through the points of contact of the tangents from 0.
Combine the equations of the curve and its Hessian. XI. OC.polar cubic of 0. C. The foci of the 4-ic on . H.
§ 11. OC meet the quartic again in A'. Prom this fact other properties of a biflecnodal quartic may be written down. C.
H and K at which IJ meets the quartic again (Figs. B'. or as x^ + y*
if
the node
is
an acnode. A'. B'. D' A'. If lie
6. A'. where u is homogeneous of degree 4 in x and y.
(i)
on two lines through H. (iii) The sixteen inflexions other that is divided harmonically by the quartic and (iv) Any line through
a fixed line.]
[(iii)
(iv)
. (ii) The points of contact of the tangents from lie on another quartic. the quartic can be projected so as to be inverse with respect to a circle j whose centre is the node 0..
IJ touches
that
:
See also Ch. Conversely. OB.. Show that K..
as
2
xy or x l —y i
if
the
node
is
a crunode.
u 2 (v2 ~ z2 + % z (us +u2 z) = We may without loss of generality take u
)
0.
"
" 2
=
_
(i).
0. p. where u and v are homogeneous of degree 2 in a: and y.-2
s2
+ 2 ». Light emanating from a fixed point is reflected at the wires to another fixed point.
ui + 2u3 z + u 2 z i =
where in
this case
0.
H.0 to the other. 8.
cubic
*
is
U = zu + u3 =
2
0.
J
H
We
.
meets the quartic in A. we may take in § 1 the equation of the-quartic is
O
as
(0. C. 10. it is the envelope of a circle cutting j orthogonally whose centre lies on the polar reciprocal with respect to j of the first polar curve of O.
if
the line
is
z
=
w. D' lie on a conic touching the non-inflexional branch at 0. D'. This polar reciprocal passes through 4 {n — 2) foci of the «-ic lying on j. Show that The equation of a quartio with a flecnode can be put in the form xyz 2 + 2yuz + xi v = 0. 1).. Find the locus of the point of
Ex. 5-26..
.
2 ««-i «V-i «»-» »r + «» «V-« See Bateman. shall leave the verification of these facts as an exercise to the reader. 48. OB. OC. 4 can be generalized as follows If an n-io with an (n — 2)-ple point O is self-inverse with respect to a circle j with centre O. 9. Archiv der Math.
•
=
are the curves.
Cuspidal Quartic s. we can find an »*-ic with an (>--2)-ple point at such that any line through Ois divided harmonically by the curves r being any given number ~^\n.
Most of the results of § 1 hold with slight modifications.
(iii)
If a line
meet the quartic again
[(ii)
xz + u
=
0. Richmond.
Quarterly Journal Math. A.
[If
m"_ 2 z*
+ 2 uB_i z + un
= 0. (ii) The tangents from to the curve lie on a conic touching the
non-inflexional branch at 0.
(i)
NODAL QUARTICS
XVIII
1
7.
:
large number of polished wires in the form of concentric on a table. Show that Ex.]
=
Ex.
xxvi (1893). xiii (1908). B'. Each curve passes through the points of contact of the tangents from . B'.
(iii)
x {z + w) + 2 u
=
0. The points /.
A
circles lie
reflexion.
[A nodal circular quartic]
§ 2. and give here only the method of finding the bitangents of a cuspidal quartic* Tfle polar cubic of the cusp O has also a cusp at O and the cuspidal tangents of the cubic and quartic coincide for this"
is
u2
a perfect square. OD in A'. and the tangent at this intersection meets the quartic again in and K.. D and OA. C. B. Show that the theorem of § 1 can be generalized as follows Given an «-ic with an (n — 2)-ple point 0. coincide at the intersection of the quartic with the cuspidal tangent.\
Ex. then A'._! z + ».
j'. und Physik.. C. pp.]
.:
:
300
Ex.
As
If a quartic has a single cusp 0.
/are real, ten bitangents are real. four bitangents are real. If two of
e,
them
none of are real, two
bitangents are real.]
Ex. 2. No two of a, a second double point.
[If
b, c, d, e,
f
can be equal unless the quartic has
a
=
6,
(1,
—a, a 3 )
is
a node.]
Through the points of contact of any bitangent can be drawn two conies touching the quartic at the cusp and each passing through
Ex.
3.
the points of contact of three tangents from the cusp.
\xz
= abc x* + (be + ca + ab) xy + (a + b + c) y
+ fl
l
,
&c]
Ex. 4. Show that the points of contact of e = 0, f = and the bitangents (abe, cdf) and (abf, cde) all lie on a conic through the cusp.
[0(«
= (O/3 + y8K.]
(abf, cde)
Ex. 5. The point of contact of a = 0, the intersection of /3 = with the bitangent (acd, bef), and the intersections of the bitangents
(abe, cdf)
and
are collinear.
/3
[They
Ex.
6.
lie
on 2
U+ a
2
=
The reader may
refer to
0ty8 + ape + a/3f, which is a straight line. Richmond, loc. cit., for further examples.]
What
modifications must be
made
in §
1,
Ex.
4,
in the case of
a cuspidal cubic ?
[Six foci lie oiij. The centre locus is a cuspidal cubic having threepoint contact with the line at infinity.]
Ex. 7. A quartic with deficiency two is referred to a triangle both inscribed and circumscribed to the curve, the double point being taken as Show that its equation takes the form (1, 1, 1).
ayz.y? + bzx » a
+ cxy w 1
=
0,
the tangents at the double point being an? + 6»2 + civ 2
=
;
where
(l
u
= (l + A)x-y-Az,
Show
also that
= -Bx + (l+B)y-z, iv=-x-Cy + + C)z. and that the quartic has a cusp if au+fiv+.yiv =
v
0,
bc(X'
i
+ caP i + aby*=
is
0;
where
a,
/3,
y denote
a
(y
BC+B+1, CA + C+1, AB + A + 1.
(z
Show
that the polar cubic of the node
+z) u 2 + 6
+ x) » 2 + c (x + y) w
1
=
0.
—
304
BICIRCULAR QUARTICS
§ 3.
XVIII 3
Bicircular Quartics.
If a real quartic has a pair of unreal nodes, they may be projected into the circular points at infinity. The quartic then
becomes
c (x
2
bicircular,
)
and
its
equation
)
is
of the form
+ y2 2 + 2
(Ix
+ my) (x2 + y 2 + ax2 + 2 hxy + by 2 + 2 gx + 2fy =
0,
if
the origin is taken on the curve. If we now invert with respect to a circle with centre the origin and unit radius, the quartic becomes the circular cubic
c
+ 2 (Ix + my) + ax2 + 2 hxy + by 2 + 2 (gx +fy) (x2 + y 2 =
)
0.
Since foci invert into foci and a circle and two inverse points into a circle and two inverse points, the, properties of the foci of a circular cubic proved in Ch. XIV, §§ 2, 3, hold for
a bicircular quartic, namely
:
bicircular quartic is self-inverse with respect to each of four mutually orthogonal circles each of which passes through four foci. If the four real foci are ccmcyclic, the quartic consists of two ovals and three of the four circles are real. If the four real foci are not concyclic, the quartic consists of a single oval and two of the circles are real, while each passes through two real foci.
A
The reader
will at once verify that, if the four lines
2 2 y +(x±<x)
=
each meet a bicircular quartic at only one finite point, the coefficients of x 3 , x 2 y, xy 2 y 3 xy in the equation of the quartic are all zero. Hence the equation of a bicircular quartic becomes
,
,
(x 2
+ y 2) 2 + ax2 + by2 + 2gx + 2fy+c =
.
.
.
(i)
the line joining a pair of singular foci is taken as y 0, and the middle point of the line is taken as origin. The four singular foci of this quartic are readily shown to be
the conic (ii) is a line-pair. It is therefore a pair of bitangents in this case. may exhibit the bicircular quartic (i) as an envelope in another manner. In fact it is the envelope of a circle whose centre lies on the conic
making use of (iv). The variable circle has double contact with its envelope (i).* The foci of (v) are the singular foci of (i). The curve (i) is self-inverse with respect to the circle (vi),
since the variable circle is self-inverse with respect to (vi), being orthogonal to it. Hence the four circles with respect to which (i) is self-inverse are obtained by putting into equation (vi) any value of t derived from (iv). The ordinary foci of the quartic (i) ai e the intersections of (v) and (vi), each value of t giving four such foci. is any intersection, the circular lines through For, if
-
P
P
form a degenerate
circle orthogonal to (vi)
and touching the
'
envelope of the circle
The conies
or
'
'
(v),
(viii), which is (i). where iisa root of (iv), are called the
focal'
deferent conies of the bicircular quartic. The reader will notice that (ii) and (vi) have the same centre. It follows that a pair of bitangents passes through the centre of each circle for which the quartic is self-inverse, as is geometrically obvious. In Fig. 3 is shown the bicircular quartic, whose real foci (0, 1), (-M, -ft), (¥. -I), (hi) lie on the focal conic The centres of the circles for which the quartic 9. «2 + 92/ 2 42 3 is self-inverse are (f, 3), (- T -, A), (M, t o), (If, -*!). being the harmonic points of the quadrangle formed by the real foci
=
* As is the case in general with any singly infinite family of the similar result for circular cubics see Ch. XIV, § 3.
951R
-^*-
circles.
For
306
BICIKCULAR QUARTICS
XVIII 3
X
E
:
XVIII 3
BICIECULAR QUARTICS
307
and the centre of the circle through the foci. The quartic is drawn by the method described in Ch. XI, § 11, Ex. 3. The real foci are denoted by O, the singular foci by x and
,
the centres
by
•
in the figure.
Pig. 4.
(x 2
+ y*) 2 -22x*-2y 2 -i
=
(
0.
Singular foei
(
+
-\/5,
0); ordinary foci
+
1, 0),
(0,
+2).
Ex. 1. The inverse of a bicircular quartic with respect to a point not on the curve is a bicircular quartic the inverse with respect to a point on the curve is a circular cubic.
;
Ex. 2. Through any point O of a bicircular quartic three real circles of curvature pass besides the circle of curvature at 0, and the three points of osculation lie on a circle through O.
[Invert with respect to O. may derive other theorems' by inverting properties of a circular cubic, e.g. Ch. XIV, § 3, Ex. 10, 11.]
We
Ex. 3. The circles of curvature of a bicircular quartic at its four intersections with any circle for which it is self- inverse have four- point contact. Ex. 4. If O is the point half-way between the real singular foci of a bicircular quartic, show that
(i) O is half-way between the unreal singular foci, and the sum of the squares of the distances of O from the singular foci is zero. (ii) The points of contact of the tangents from O lie on a conic.
(iii)
O
C,
is
A, B,
D are
The
equidistant from the middle points of and CD, where the intersections of any line with the quartic.
(i)
AB
[Use equation
of §
3,
and write down the
first
polar of O.]
bisectors of the angle between any pair of 'bitangents of a bicircular quartic are parallel and perpendicular to the line joining the real singular foci.
Ex.
5.
[Equation
Ex.
6.
(ii)
of § 3 has no term in xyi\
intersections of two pairs of bitangents lie on a circle half-way between the real singular foci. The sum of the squares of the radii of the circles obtained by taking two pairs of bitangents and also the other two pairs is the same for each way of dividing into pairs. X 2
The four
is
whose centre O
1
308
Ex.
lie
7.
BICIRCULAR QUARTICS
The eight points of contact of any two of the conies
Find the centres of the circles for which the quartic (i) of § 3 and show that they lie on a rectangular hyperbola is sell-Inverse through whose asymptotes are parallel and perpendicular to the line
Ex.
9.
;
joining the real singular
foci.
(iv).
-f/(b42t)), where t is given by The centre of (ii) lies on b — a = g/x—f/y-]
[(-g/(a + 2t),
Ex. 10. The centroid of the four finite intersections of a bicircular is half-way between the real quartic with any circle whose centre singular foci of the quartic is a fixed point V. It is also the centroid of any four concyclic foci, and the centroid of the four points of contact of any pair of bitangents. It is the intersection of the axes of the two parabolas of the family (ii) and the centre of the rectangular hyperbola of Ex. 9.
;
[Eliminate x or y between (i) and](iii) or between the point (g/(b-a), f/(a-b)).]
of
(v)
and
(vi).
V is
Ex. 11. In Ex. 10 the ratio of the distances from x = (ii) or (vi) and of Fis the square of the eccentricity of
of the centre
(v).
circle for
[This enables us to determine the focal conic corresponding to any which the quartic is self-inverse ]
Ex. 12. The centres of three circles for which a bicircular quartic is self-inverse are the vertices of the common self-conjugate triangle of the fourth circle and its focal conic.
[They are the harmonic points of the quadrangle whose vertices are the four foci on the fourth circle.]
Ex. 13. If a bicircular quartic consists of two ovals one inside the other, the focal conic through the real foci is a hyperbola and the other three focal conies are ellipses. If the qaartic consists of two ovals external to each other, the focal conic through the real foci is an ellipse and, the other three focal conies
are hyperbolas. If the quartic consists of one oval, one focal conic a hyperbola, and two are unreal.
is
an
ellipse,
one
is
[Use Ex. 11 and'Ch. XI, § 11, Ex. 3, 4]
Ex. 14. The directrices corresponding to four concyclic foci pass through the centre of the circle and form a pencil of the same crossratio as that subtended by the foci at 0. [The line joining to the point (vii) forms a pencil homographic with that traced out by the chord of contact of (viii) with its envelope.]
XVIII 3
Ex.
15.
BICIRCULAR QUARTICS
The normals
to a bicircular quartic at the points
309
where
it
meets a
circle for which it is self-inverse singular foci of the quartic.
touch a conic whose
foci are the
[The corresponding focal conic. It follows that these normals form a quadrangle whose opposite sides subtend supplementary angles at either
singular focus,
The pair of bitangents through the centre of any one of 'the for which a bicircular quartic is self-inverse are perpendicular to
Ex. 17. The distances of any point on a two-circuited bicircular quartic from three real foci A, B, C are connected by a linear relation C.PA + + n.PC=0, where I, m, n are constants. [Invert Ch. XIV, § 3, Ex. 20, with respect to any point, and we find that the result of Ch. XIV, § 3, Ex. 23, holds for the bicircular quartic. As in Ch. XIV, § 3, Ex. 26, we obtain results for the one-circuited
Ex. 19. Any bicircular quartic can be inverted into a bicircular quartic symmetrical about two perpendicular lines on which lie the real foci
and singular
foci.
[Invert with respect to the intersection of two real circles for which the quartic is self-inverse and on which lie the real foci. See Fig. 4.]
Ex. 20. Obtain the equations of the bicircular quartic symmetrical about the axes of reference whose real singular and ordinary foci are given and are on these axes. [There are three types. We may take the real singular foci as Then the curve is ( + m, 0).
{x 1
point of the curve from three foci (Ex. 17). 2 = a 2 cosh 2 f— j3 a sinh 2 e in type (1).]
For
this
purpose he
may
m
Ex. 21. Find the locus of the singular foci of a two-circuited bicircular quartic, given the real ordinary foci.
[The circular cubics with the given
foci.
See Ch. XIV,
§ 3,
Ex. 27.]
310
BICIRCULAR QUARTICS
XVIII
3
Ex. 22. Show that the equation of a two-circuited bicircular quartic can be put in the form
(x 2
+y
2 2 )
-2{lx + my)(x* + y' -l) + ax i + 2lmxy + bif +
l
l
=
0.
as the unreal circle with respect to which the quartic is self-inverse, and the axes of reference parallel and perpendicular to the line joining the real singular foci.]
[Take x^ + y' +
l
=
Ex. 23. Obtain the equations of the conies touching the quartic of Ex. 22 in four points, the equations of the bitangents, and the equations of the circles with respect to which the quartic is self-inverse.
+ (2 am- + 2W-ab-BP ni?) =*0. have their centres at the centre of v = where and are mutually orthogonal. Their equations are
Ex. 24. Show that the squares of the tangents from any point of a twe-circuited bicircular quartic to the three real circles with respect to which the quartic is self-inverse are connected by a homogeneous linear
relation.
Conversely, if the squares of the tangents from P to three given circles are connected by a homogeneous linear relation, the locus of P is a bicircular quartic which is self-inverse with respect to each of the given circles, if these circles are mutually orthogonal.
;
Ex. 26. Given the circles with respect to which a bicircular quartic is self-inve\'se and one point on the curve, find the locus of the singular foci.
Ex. 27. Find the locus of P,
(i) (ii)
if
PA. PB x PC. PD, PA PB PC,
.
<r.
where A, B,
[(i)
C,
D
are fixed points.
can be inscribed in
bicircular quartic such that an infinite number of quadrilaterals it whose sides pass alternately through the two circular points.
A
(ii) bicircular quartic with and as singular foci. inverse in respect to A, C or any point ?]
A
A
B
What
is its
Under what circumstances can a given bicircular quartic be generated by one of the methods of Ex.
.
K
K
.
B
are not ordinary nodes. 28.
. a slight modification If A and Such modification will of the above notation is necessary.
St
= 0. S2 =
t.
In our discussions of quartics with two real double points.]
Ex. find the intersections of <St = S. [A bicircular quartic with singular foci at the centres of the two
_
circles. 29. L X L2 ) are harmonic. double points are A and B. Trace the quartic
[It'is-
as the intersection of circles.
.
where SL
—
0.
. S =
Now
easy to put the equation in the form are circles. 27-29 ? For instance.
[The reader may consider the question.
.'
XVIII 4 QUARTICS
two fixed
circles
(i)
WITH TWO REAL NODES
311
Ex. 31.
if
varies as the distance of varies as the product of the tangents from to
circles
(i)
'
two fixed
the product of the tangents from P to P from a fixed line.
Quartics with
Two
Real Nodes. if we allow this circle to be unreal. The locus of the intersection of two orthogonal circles one belonging to a given coaxial family and the other belonging to another given coaxial family is a bicircular quartic through the common points and the limiting points of the two families. 5). any bicircular quartic can be generated by a point P the product of whose distances from the singular foci varies as the tangent from P to a fixed circle.]
Ex.
EE
. 2) X (BV. Find the locus of P.
MM
and
B2 and intersect at C (Fig. (ii) varies as the square root of the from a fixed line.
for different values of
the parameter
§ 4. The real shall adopt throughout the following notation.
E F
E
.
allow these alternatives. at A meets the curve again in 1 The harmonic conjugate of BA with respect to the and 2 The lines tangents at B meets the curve again in Zj and L 2 BL X L 2 meet at C.
.
.]
Sl S2 =
t
S.
is
distance of
P
if the product of the tangents from to constant. The other diagonals of the quadrilateral formed by the tangents at A and B meet AB at R x
we
M
H
M
H
t
. The tangents at A meet the curve t 2 again at x and 2 the tangents at B meet the curve again The line 1 2 meets the curve again at and 2 at x t The and the line F1 F2 meets it again at and x 2 2 tangents from A touch at A 1} A 2 A 3 A t and the tangents and from B touch at Bv B2 B3 B^ The ranges {ATI.
AM M
F
. usually be quite obvious without explanation. if we refuse to
'
. Find the locus of P. (ii) P two other fixed circles. or allow P to be inside the circle if real but not always.
[Invert with respect to a
common
(i)
point. (iii) varies as the taDgent from to
P
P
another fixed circle. The harmonic conjugate of AB with respect to the tangents. 30.]
Ex.
. . 5. and the other will be denoted by F. whose points of contact are A v A 2 A 3 Similarly in other cases. if B is a flecnode. Replacing x and y by suitable multiples of x and y.
of contact are lt 3 2 and % coincide at is a cusp. Taking ABC as triangle of reference. the coefficients of 4 4 a. . we may simplify the equation (i). one of the points F^ and F2 The points will be at B. now the intersection of the curve The points 1 and 2 are the other intersections called F.
(i). x6 y 3 x2 yz. the points Or again.312
QUARTICS WITH TWO REAL NODES XVIII 4
For instance. and 2 do not exist in this case. + qf) + 2m xyz 2 + 2 (fx + gy + hz) z 3 = and x2 +pz2 = where y 2 + qz2 = are the tangents at A and B. we may make/ and g
(x2
+pz 2)
2
(y
.
B B B B
F
F
K
K
Pig. xy 2 z are zero in the equation of the y quartic. which takes the form
. For instance. . There are only three 1 tangents from B (other than the tangents at B) their points
K
K
. .
.
tangents from A. if x with the cuspidal tangent. There are only three of the curve with the tangent at F. . .
. All quartics with two given nodes and passing through seven other fixed points.
case
One root
of
(iv)
= — 1.
(iv).]
= hm.]
jp
we may take
= g = — 1.
In
§
.
. and a corresponding
. ma 2 — 3 a).
[The quartic has either two crunodes or two acnodes. 26. as explained in § 4.
Ex. = 6 we get the tangent z + bx
and the corresponding bitangent qbx + 2b 2 y + (2b-q)z = 0. If one polygon of 2n sides can be inscribed in a binodal quartic so that the sides go alternately through the nodes. g=f.
t
i
-2mt 3 +(m 2 -2h)t 2 + 2t-q =
(iv). (iii).
Quartics with a
Node and
4 A and B were any two real double points of a quartic. Discuss the quartic for which f*/g* = q/p. and (iv) become qx2 + 2taiy + 2xz + 2yz + (2h. we obtain a tangent from the cusp z + ax = touching at the point B 1 (— 1.
.
[The conies of Ex. pass through an eighth fixed point. b. Ex. * Taking t = a. [Project the nodes into the circular points and invert with respect to one of the fixed points ] Ex. —1. we have p = and we may also take /= 1. d of a line-pair which must evidently be
If
t is
(ii)
represents
(z
+ tx)
{qtx + 2t 2 y+{2t-q)z}
=0.
bitangent b x with equation
qax + 2a2 y + (2a — q)z
touching where
Similarly taking t the cusp touching at
=
=
from
b2 . The quartic can be projected so as to be symmetrical. c. an infinite number of such polygons can be inscribed. 25.
a Cusp.
(i)
=
ez meets the curve
again in coincident
+ (2h-m2)e 2 + 4:(q~m) e + 4(2qh-l)
= 0. if
e
3
c.[m .
2
B
and so for t = The line 2y
points.a
.
. xy + (m — t)z 2 = (iii).
. Discuss the quartic for which fg 1 and 5 degenerate.
In the
is
t
first
the corresponding pair of bitangents meeting at (1.
2a2 y 2 +(2a-q)yz + qa(a — on) z 2 = 0.
and by § 4 this touches the quartic where it meets (iii). 27.
Ex.
.
§ 5. g = 1. Similarly in the second case. If A is a node and B is a cusp.
[Project the Steiner points of a polygon inscribed in a cubic (Ch. 24. 01.]
XVI.
any one of the roots a. The equation (i) of the quartic becomes x 2 (y2 + qz 2) + 2mxyz2 + 2(x + y + hz)z3 = (i) while equations (ii).
316
QUARTICS WITH TWO REAL NODES XVIII
4
Ex.
§ 6. 1) into the circular points and invert. The locus of the pole of AB with respect to the conies (ii) is x = y.
t
= d.1] 2) z 2 = (ii).
Let us take Then y ±ix are asymptotes twice over. Also y = i(x — d) which gives p touches the curve (i).
The locus of a point P whose distances from two fixed points B are connected by a linear relation is a Cartesian curve. but
it
consists of two ovals.:
XVIII 6
CARTESIAN CURVES
§ 6. and three real ordinary foci. which is the perpendicular bisector of the line
joining the real ordinary foci E.
0.
(i). the intersection of the tangents at co and co'.
(iii). Hence the quartic consists of two ovals (§ 3). 0). y now takes the form
'.. 0).
d has the roots ex. B. (y.
Or 'Cartesian oval'. which is selfinverse for a real circle j through Q.
The curve is called a Cartesian curve We now show that
A. 0).. When we invert with respect to Q.
319
Cartesian Curves!
If a quavtic has a cusp at each circular point co. B. These lie on the real circle.
Suppose the relation
is
.PA±l. G.PB = n. Let us consider case (1).
G
on the axis of
(2)
on the axis of
and one real ordinary focus G lie The singular focus x. (|8.*
from which
it
+ y 2 -^y-y0i-txBf + 4cx^y{2x-a-^-y) =
'
. a>\ its inverse with respect to any point Q has Q as a focus (Ch. and therefore there are three tangents from co other than the tangent at co. as origin and A. /3. The quartic with co and co' as cusps is therefore symmetrical about the line which is the inverse ofj. Considerations of symmetry show that there are only two
cases to consider. F. and a = b in (i). B.
±m. 1). the equation of the quartic takes the form
(x 2
+ y 2) 2 +p(x 2 + y 2)x + ax2 + by 2 + 2gx + c =
. Taking the line as axis of x. if
=
=
4gd 3 + (4c-a 2 )d 2 -2agd-g 2
This equation in readily follows that
(x 2
(i)
=
.AB
*
. Ex. G as the points (a..
and the
real ordinary foci A.
(ii).
(1)
lie
The
singular focus
x. V^ § 4) and is a bicircular quartic (§ 3. the real foci are Q and the inverses of A.
The curve is of class 6.
. Hence the curve has one distinct singular focus 0./.
F. &c. 0). 5.). B. when and A. G may
.
is
. 1.
from
(2).6
320
CARTESIAN CURVES
on
XVIII
Take a point
and
let
AB
such that
. 7.
+ y*-6iy +
(0.PA±0A3. 0)
. 0.
Case
is
OABC at points whose distances We have + 3PA ± 2PB = 10. G as (oc.PB = 0Cs. (8. the distances of any point on a Cartesian curve from the foci A. 0). 0). (j8. (f .
(iv).
Conversely. 0). (y. B
is
The curve meets are -13.BA
where
focus.
0). 0).=
Singular focus
ordinary foci
(2.
(x*
7. not so interesting geometrically.
= a + /3 + y.
the singular focus/and
G
the third real ordinary
A quartic with
of
(ii) is
two Cusps has one bitangent. The real foci E.
is
Then the equation of the locus P. B are connected by the linear relation
P
±0Bz.
(ii).
which when rationalized proves to be the same as
Pig.
.
.
G be
a point
0A:0B = P:m 2 on A B such that
V:rn?:<n?
= 0A:0B:0C. in which the three real foci of (i) are not collinear. 0) is
taken as origin
± p s{(x-uf+y
1
2
}i±a.i{(x-l3f + y2 }i=ys(p-a. 0).
UH4x-29).
The bitangent
evidently
2x
In Fig. A is (2.
(f
. 0). 7
is
(0. 0).
C
is
(8.
]
Ex.
2216
Y
.c 2 ) du + M (it2 . Py+yix + otp. if in
(iii)
I.]
Ex. and the singular focus (| ±it].
4.
9.
4. then
the ortho0. It is
? + y*-2£y-e-T.
What
relation connecting the distances of a point
on the curve
?
from two real foci replaces (iv) in the case of the quartic [Use Ch. if
PA + PB =
where
2u.
The curve
Soc. for diagrams of this case.
8. 10.y + y* = ((X-P)(0t-y). and P a variable point.
A
B are fixed points
is
PB-PA = 2v.
gonal trajectory of the family of curves with differential equation
M du + Ndv =
[Use
to obtain
N
(«
2
. or n
zero ?
Ex.
is
1.6
XVIII
be taken as
CARTESIAN CURVES
321
Then
(a
as ovigin.
Ex.
m.
Ex.
iii.
(v).
3. 25. § 3.
Show
and
that. becomes any circle with at two points only.
in § 3
Discuss the modifications which must be bicircular quartic is a Cartesian curve. Find the orthogonal trajectories of Cartesian curves (i) with given singular focus and two given ordinary foci.
(v)
By
inversion of
(iv)
obtain the results of
§ 3. [Invert with respect to an intersection and use Ch. Two Cartesian curves with the same real ordinary foci cut orthogonally.] Ex. The distances from a given focus of the intersections of a Cartesian curve with a variable line have a constant sum. and which cuts orthogonally a fixed circle with its centre at a The other two foci are the limiting points of the fixed circles. XIV. (0. 115. /3y—j-a + aft (3y+ 7 a-a/3. 2. Ex.
7.c 2 ) dv =
Ex. the other three make become the circular lines through a focus. 6. The roots of § 3 (iv) are
=
touching the curve
-07 +ya + a/3.] focus. Ex. 0) are also foci.
(vi)
become the circles &c.)
does the locus of
London Math.
What
P become. @ in (ii) by £±ir]. x i + y i = ^y.
made
when
the
centre on y [§ 3 (ii). y). we see that any two bicircular quartics with the same four real concyclic foci cut orthogonally. § 3. 17. Inverting with respect to any point. p.
§
(ii)
3 (v) becomes the circular lines through
or else
§ 3 (v).
[Use polar coordinates.
(iv)
M and N in Ex. AB = 2c.
the bitangent being
%x=2£ + y. (See Proc. Hence the equation of the curve is obtained by replacing a.
consists of a single oval. XIV.
Ex.
Ex. + 77). 5. (ii) with two given ordinary foci and passing through a fixed point. (x-a.
The points of contact of
2
(/3
(ii)
with
its
bitangent are real
if
y + yoc + a/3) >
(a 2 + /3 2 + /). Cartesian curve is in three different ways the envelope of Hence the a circle whose centre lies on a fixed circle with centre at the singular focus. The first root makes (ii) become the bitangent.*)* + 4y(? + V *)(2x-2i-y)
= 0. (£. taking the focus as pole.]
Ex.
Find the locus of a point fixed circles have a constant ratio.
.
if rays of light
Ex. (iv). emanating from a point in one medium are brought accurately to a focus in the other. a point Ex. the points of contact
'
x + y + kz=0 In fact from equation
W
1
and
W
2
lying on
xy + mz 2
If
=
0.
A man
B
Find the surface separating two homogeneous isotropic media.
take the root thebitangent
If
we
t
=
of
becomes z
=
and
of the quartic (i). As explained in § 4.
mz2 2 + 2(x + y + kz)z 3 =
)
.b.
g
= =1
=
/=
.
txy + xz+yz
+ (k + mt — \t 2 )z 2 = xy+(m — t)z 2 — s t{t -2mt 2 -2kt + 2} =
z(x-y)
.
Ex.
(i)
(iii).
(ii)
we
take the other roots
z
t=
a.
. c of
gives the
tangents
from
from
— 0.
.
is in a pond at A.6
.
14.
. and runs to on the land.
. find
foci.
CARTESIAN CURVES
XVIII
(i)
10. Then and B. swims to the bank at P.
322
Ex. 11.
The fourth power of the tangent from
P to
a circle varies as the
distance of
P from a line.
(iv).
13. z + cy = A A 2 A 3 and the tangents z + ax — Q.
and we
shall
put k
= h — -|m
and
The equation
while equations
(i)
of the quartic becomes
(xy +
(ii). We have the intersection of the tangents at p 0. 12.
Quartics with.
(ii)
(v).
Two
Real Cusps.
}
. we may suppose
G
is
A
2
.
the locus of the other § 7.
(v)
become
.
now
1.
whose normal distances from two
Ex.
(iii).
(ii)
The square of the tangent from
P to
one
circle varies as the
tangent from
(iii)
P to
another
circle.
(iv).
.
Given one focus and three points of a Cartesian curve.
. Show that the locus of P is a quartic with cusps at a and &>'. z + by = 0. q 0. If his time from A to B is the same for all positions of P. z + bx = 0.
3
.
(ii).
»
=
(iv).
. find the shape of the pond. if The tangents from P to two fixed circles are connected by a
linear relation.
Suppose that in § 4 both A and B were cusps. (i) it is obvious that this line is the bitangent of the quartic. z + cx = B touching at B B2 B
+ ay
A
touching at
x
.
the
Or Cassinian Oval
'.
with respect
Cassinian Curves.
But
it
consists of two ovals if
c
> a.
Y2
.
PF .
.
.
3
.
0).
2. TT2 lie on a conic. co
are x
—
2. + 1/) + 496 = 0.
)
(0.
A A A L lie
t
. and expressing the fact that the four lines
a>. Ex.
on a conic. The reader can write down a large number of theorems which follow from this fact. x [Putting z = 1 we see that the quartic can be projected so as to have an axis of symmetry. 0). E. B. The points A.. 3.
A
—
The bitangent
is
3x+3y+5 =
while the tangents from the cusp x = — 3 and so for the cusp (go
.
Ex.A 3 .A 2 .
meet the curve only at a> or a/.
.
.
x
= —§
.
W
.]
t
.
4. Sec.
called a ' Cassinian curve is a lemniscate of Bernoulli.
2
.
A. and so for the other points of contact.
Hence.
0.B1 . -a). y) from ( + c. For instance. F.*
If c
=
a. the quartic has the equation
§ 3
(i))
that
The left-hand side of (i) is at once proved to be the square of the product of the distances of the point (oo.
3
Ex. 0).
touches the quartic at A. z') on x + y+kz = to (i) meets (i) on the conic 2z(y'x + x'y + 2mz'z) = 3s' (xy + mz 2 ).
3. A 1 . oo'.
if
P
is
any point on
1
(i). B.
1.
XVIII
8
CASSINIAN CURVES
323
The coordinates of t are (a 3 ma 2 1.
Consider a quartic with a biflecnode at each circular point We may choose It has two real singular foci F1 F% axes of reference such that these singular foci are ( + c.
The
conic
[It is
xy + mz1
Wlt W A lt A A = 2y (x + y + kz). Writing down the general equation of a bicircular quartic.
y
= + * (& ± c)
we
see readily
(cf.]
Ex.]
§ 8. &c]
Ex.PF = a
2
2
.
[See § 4.Bi B3 lie on a conic.
The points of contact
drawn
to the curve
from any point
t
of the other four tangents which can be of the bitangent lie on a conic
through A and B.
The curve
*
'
is
Curve
'. [The polar cubic of the point (x' y'. The reader may illustrate the properties of the curve by tracing 9 cc 2 2/ 2 + 96x?/ + 1 44 (a.
t
.
The most interesting properties of the Cassinian curve are
given in the following examples. 8.
(i)
P is
the difference
4.e. V2.
any point of (i) and middle point of F1 F2 and of S^S.
are
l
. interested will find in the examples their more important
properties. 1.
Fig.
.
P
is
the centre of
i.
rectangular hyperbolas with
FF
1
2
(ii)
(iii)
The locus of the inflexions is a lemniscate of Bernouilli.
the circle on
FF
l
l
as diameter
and the perpendicular bisector of
(i)
FF
1
l
[The polar and pedal equations of
»-
when
is
pole are
j-* 4
. 3. A. V|. the locus of a point whose distances from the vertices of a regular polygon have a constant product.
(i).
are
is
(0.
±d) where
= c -a c d? = a -c
2 4
if
c<a.
2. The angle between OP and the normal at of the angles OPF1 and OPF2 Ex.
Ex. we shall confine ourselves to a very brief account of the other The reader who is varieties of quartic with unit deficiency. V3.
(
+
d.
. A.
The
real ordinary foci
c
2
Su S2
<2
of
4
(i)
4
. 184. 8 for the cases = •*.
A
family of Cassinian curves has given singular foci
trajectories are
Fu F
2
.. 8. See Messenger Math.
the
PS1 PS2 =
.* In order to keep this chapter within reasonable compass. The locus of the points of contact of tangents parallel to i^Fu
is
. i.
is
r
2
+c
2
cos 2 8
=
Fig. and
Ex. p.
PO\
Ex..324
CASSINIAN CURVES
XVIII
8
a/c
The shape of the curve is shown on Fig. then
If
c
2
. xlviii (1919). 0)
where
2
c>a.
0.e.
a2
. A.
if
1.]
* The reader may extend these results to the ' M-poled Cassinoid '.
giving their locus
2?Vcos20 = 4 + c4 -a4 and 2a?pr = + a4 -c The inflexions are shown in (Sr^ + c 1 — a?)p = 2aV.
Show that
Their orthogonal
as diameter.
PD
constant.
Llt Lt
. and so for other examples.]
Ex. (a-A) (0-.]
Ex.
PS. xy = H. 9.
Ex.]
6.QS=PS.
Ex.B) = (Ot + k-A) (P + k-B) = H\ Now eliminate A. The intersections of the curve with any line through C form two pairs of an involution with one double point at Cand the other on AB.
Quartics with
Two
Eeal Biflecnodes. 5.]
symmetry
. Find the locus of P. y) is a focus of A\* + 2H\p + Bfi 2 = 1 which has double contact with the confocals a\ 2 + 0/i2 = 1.
§
Ex. A conic touches the quartic at self-conjugate triangle. 1.
2.
having
. 3. Aliter. a Cassinian curve. 5. Ex.]
D
cut at a given angle. H.
Ex. 8.
Any
line
through A
is
divided harmonically by
BC and the curve. A quartic has real 2 4 its equation is (a: 2 +pz ) (f
biflecnodes A
With the notation
of
+ qz1 ) + 2 hz*
=
0.
[The curves are the inverses of the family in Fig. (OL + k)\ 2 + (B+k) M 2 = 1. C and the
Cassinian curve through
[Each locus
is
P with singular foci B. curve ?
[§
What do
gives
£
the results of § 3 become in the case of a Cassinian
2
or ±(c i — a i )i. (iii) The Cassinian curve through P with singular foci A.PC-. [If C is the centre and r the radius of the director circle.
The sum of the angles made by PA and PC with a fixed line less the sum of the angles made by PB and PD with the line is constant. The tangents from A to the quartic are and tangents from B are 2 X
BM
BM
AL
t
and AL^. 10.
The
as
Ex. if the polar conic of P with respect to a given Cassinian curve has a constant eccentricity.PS' = r*-CP\]
Ex.
If
ABCD is a parallelogram. P the fixed point and C bisects PQ.
Ex.PB. the envelope of a circle which cuts orthogonally the director circle of a hyperbola and whose centre lies on the curve.
(i) (ii)
to 0. The locus of the foci of a variable conic concentric with and having double contact with two given confocal conies is a Cassinian curve with singular foci at the foci of the given confocals. Find the orthogonal trajectories and the locus of the inflexions of Cassinian curves with given ordinary foci. project the quartic into the curve with double obtained by putting 1 for z .
and B.
M M
1)
2
ABC
[qx* + py* + (pq + 2h)z*=0.] 3
(iv)
. while S and S' are the foci of the ellipse.
= ±c
The Cassinian curve
is
§ 9. 11. B. 4. The locus of the intersections of two circles having their centres at the foci of a given conic and touching any tangent of the conic is a Cassinian curve. 7. we have x*-y*=*A-B. 8 with respect
Ex. [A Cassinian curve with the singular foci of the given curve as ordinary foci. [If (x.
is
find the locus of
P when
PA.XVIII 9
QUARTICS WITH TWO BIFLECNODES
4
325
Ex. The locus of the foci of an ellipse with a given director circle and passing through a given point is a Cassinian curve.
The tangents at A belong to the involution.
2 g <c. BC.]
=
B
Ex.
London Math.]
Ex.
[Putting
p=
q
= — 1.
2
lie
on a conic touching
AT at A.
2. The non-inflexional tangents at A and meet the curve again in and F.g*>c. Soc.
Quartics with a
Node and
a Biflecnode. 1 can be generated as the locus of of three ways.
5.
9. XIV. 5. Ex.
A A 1 A A s B touches AE. Bj.
xii.
C form an
involution and their points of
The tangents from the
intersection of
AE
CT
and
BF
form an
involution.
Ex. [For the following examples seS § 4 or use the symmetry in x and -y. and the inflexional tangents meet at 2'. from A form an involution pencil whose Ex. Ex.
PO
and the tangent from
P
to a fixed circle with
(3)
PO PS oc PO PS oc
the tangent from
the chord through
P to a fixed circle with centre 0.
will be found in Proc.
may
1.
With the
notation
E
2 i (x -zt ){y 2 ~z*) + 2mz (xz + yz + xy) + 2hz i 0. 4.
Ex.
A. §
Ex.
s
.
8.
s
2
.g >c.
Quartics with
Two
Real Mecnodes. 3.
+ 2(x + hz) z 3
=
[The quartic by putting 2 =
Ex.
2
3
. 0).
Ex.
4.
A it A t A Llt L
.]
§ 12.
Ch. B. f=g = m
in § 4
(iv).
Any
line through
The two tangents from
1
.on a conic touching AB at A. of § 4
1.
WiEh the notation
0.
Ex.
t
.]
§ 11.
Discuss the case in which the real foci are
3. Ex. Plie on a conic touching AT. 1. 6.
327
(y.
3.
one value of
— 1.). The four tangents double rays are AB.
.
[Cf. 7.
2.]
be projected into the symmetrical curve obtained
B is divided harmonically by the curve and AC. L lt L M1 M2 lie on a conic. F. B touch at the points M M Ex. The quartic of Ex. +>. E.
(1)
P
in one
The product
centre
(2)
S is
.
of constant.
A
its
equation
quartic has two real flecnodes may be written
A
and B. The points of contact lie on a pair of lines through B and . 4. A quartic has a node of § 4 its equation may be written 2 (x 2 +pz ! ) (v* + qz )
A
and a biflecnode B.XVIII 12
Ex.
.
p
22.
NODAL-BIFLECNODAL QUAETICS
(£.
(3)&>0. P 2 B lie on a conic.
.
Ex.
. [The three cases are given by
(l)k>0. The tangents from contact lie on a conic. AC. BT.
Three pairs of bitangents intersect on
and the other pair
t is
on AB.
A.
Diagrams
{2)k<0.
A lt A it A
E. P perpendicular to PS of a fixed
2
circle
with centre S.
25. The conic Ex.
B. Transform the curve of Ex.
NODAL-BIFLECNODAL QUARTICS XVIII
B.
The tangents from
tangents meet
AB
form an involution with BA. 1 into a cubic by quadratic transformation. The conic tangent at B. We take the taenode G (0.
B B .
With the notation
0.
7.
of § 4
The quartic has a node
equation
(x*
and a flecnode B.
24
.
B
B B A
2
s
touches
AC at A and the non-inflexional
Ex.
5.
[Putting
g
on
AC
= 0.
i
Es.
lie
Ex. The intersections of the bitangents lines through B.
The quartic
and a
flecnode. 1) and the tangent at C as y = O.
Hence. and investigate properties 8f the curve by this means. BC.
Ex.
We now consider
at as
quartics with a taenode. z 3 y.
The points B. » takes the form 2 2 2/ 2 + 202/w 2 + m4 O
Hence the equation
(i). + 2(fx + y + %z) 2s =
1
Ex. 0. 6.
B
§ 14.
readily seen that the coefficients of
.
the
The points of contact of a taenode C.
/=1
in
§4
(iv).
four tangents
to
the tangents at G.
1.
its
may
be written
. other than
touching the quartic at
C.
t
Ex.
z 3 x.e. 5.
[Replace
Ex.
Quartics with a Taenode.
6. if
(ad — be) (bd
— ca) (cd — ah)
= 0. or otherwise. y.] A
Quartics with a Flecnode and another Double Point.B
328
Ex.
3.
or
by pairs on
=
0.Mlt M
lt t
a
2
lie
on a conic touching the
inflexional tangent at
B. Then by Newton's
it is
diagram.
u2
does not contain y as a factor. lie
a quartic from on a conic
conies of closest contact to the two branches of the quartic at G meet in two points at G and in two other points
The
.
t
2
— pq + h± (h?+p)§.
The conic
ABE E F touches the inflexional tangent at B.z2) (f + qz*) + 2xyz. The biin two pairs of points forming an involution with A.
12
Elt E
m=
2
are collinear.]
quartic cannot have a cusp and a biflecnode. a double point which two linear branches touch. i.
the bitangents are
z*
qx 1 +py* + 2txy + 2xz + [pq + 2h-f)
where
§ 13.
of § 5 has a cusp
A
Ex. 4.
=
where u2 and ui are homogeneous of degree 2 and 4 in The polar cubic of G with respect to the quartic is y(zy + u2) = 0.
if
x.B . zx 3
»
in the equation of the quartic are zero.
x by z (z—x)/x. 2. z 2 x 2
z 2 xy.
See
Ch v III.
the points of contact
Cu C
2
C3
of the tangents from the cusp
on a conic osculating the
§ 8.&!.]
In Fig.axf/y.
properties
its
7. 10. 1 at B.XVIII
15
QUARTICS WITH RHAMPHOID CUSP
331
Ex. are represented by their
. &c
is
[CA
*
(3. oo ) and GA the line at infinity.
Clt Ca C. XVI.
Ex. 1 to
Ex. and
AB touches the
conic of Ex.
curve at C. b b.
. The equation of a quartic with a rhamphoid cusp can be put in the form 4 y (x + ay) (x + by) (x + cy).
2. are 2x + 5y = 5._8.
[Write
down the polar
cubic of C. In this case the points of contact of the four tangents from C are collinear. (0.*
rhamphoid cusp
lie
Ex.]
of
the
tacnodal
quartic by quadratic
[We can transform § 14 (iii) into the quartic of Ex.
=
2
. 3x = 5j/ + The equation of the curve is (6» + ^' = 4*(x-8) (a + 6). suffixes only. is bitangents b\.
[The quartic can be projected into a symmetrical curve. -5) &!. 0). and properties deduced from this symmetry or by modification of Ex. by replacing a. &c. -i). 9. (?. &c. Or we can transform it into the cubic of Ch.
Quarties with a Rhamphoid Cusp.ax)/y.
15. If in § 14 « 2 has y as a factor. .
.
Ex. zby z-(y. 10 G is (0.]
Fig.
If a quartic has a
C.]
The
(a.
. (yZ + x *y C. Obtain transformation. The equation of the quartic may be put in the form
My^z* + (x + ay) (x + by) (x + cy) (x + dy) = 0. and the equation of the bitangent 6 12 in the form 2a/Mz = (a + b-c-d)x + (ab-cd)y = 0. (5. y by x*/y. 6 a & 8 are 6x = 5y + 3. In the figure
.
(5y + x*y
= ix {x -3)
(as
+ 5). 2. are (6 + c + l) x + (a + bc)y-z = 0. the curvatures of the branches touching at C are equal and opposite.
the tangent at C..
§ 15. § 8.a2).
1. by x (y .
-l. 8 by replacing z by z — x^/y.
Obtain properties of the curve by transforming it quadratically into a cubic in this case.y.
(ii)
Replace y by x'jy and
.]
x
on the conic of Ex.332
QUARTICS WITH RHAMPHOID CUSP XVIII
.
[(i)
Take
B as C
z by z ..
The
and
and
6.
=
2y(x + ay).
Ex. There is no loss of generality in taking a = in Ex. The points of contact of quartic at C. 2.
C
and CL
lie
on a conic osculating the
[yz+x*
Ex.
AC passes through C
63 to the intersection
Ex.
of
6j 5.
Ex. 4. 7.
.
The points of contact of 6 2 and
6 3 lie
on
a conic
through
C.
1. 6 2 6 3 form together with a range of the same cross-ratio as the pencil C(Ci C1 C3 A). 3. The intersections of AC with 6.
C2 Cs
.
6.]
line joining the intersection of 6 2
x
.
15
Ex.
NON-SINGULAR QUARTICS
Bitangents of Non-singular Quartics.
s4 +
xyu + 2az 3 x + (a 2 + 2c)z 2 x2 + 2aczx 3 + c 2 xi + 2bz 3 y + (b 2 + 2d)z 2 y 2 + 2bdzy 3 + d2 yl
is
=
0. y.
XIX
2
Suppose that any two of the twenty-eight bitangents of a non-singular quartic are taken as x = 0.
. the left-hand side reduces to a perfect square..
. q
W=
=
=
=
=
=
=
Hence
Through the four points of contact of two bitangents of a non-singular quartic pass jive conies each of which passes through the points of contact of two more bitangents. 0. Let its equation arranged in descending powers of z be
z*
+ 2z 3 (ax + by) +
.
of factors p.
When we put 0. y = 0. y 0..
2
V= z
so that
+ (ax + by)z + (cx2 + dy 2).
where
u= This may
a conic. may be chosen in general in five ways.
of bitangents. so that k. z. it reduces 2 to a perfect square. Now these six pairs have the property that the eight points of contact of any two of these pairs lie on
0. say (z 2 + azx + ex 2 ) 2 and when we put x 0.
=
0. q 2 linear in x. p 0.
where
U=
2abz2 + 2 (bcx + ady) z + 2cdxy — u.
Now
may
be written
xy(U+2kV+k 2 xy) =
Choose k so that
(V+Jcxy) 2
. say (z 2 + ~bzy + dy 2 ) The equation must therefore evidently be of the form
y =
=
.
namely x
=
y
=
a conic.
we
obtain thus six pairs
and five pairs such as p = 0. and that 0. The equation of the quartic
H+2kV+k
2
xy has a pair
U+2kV+k
then takes the form
where
W = V+
xypq
Jcxy. q the conic passes through the eight points of contact of the four bitangents x 0.
Since k can be chosen in five ways.
=
W
2
(ii)»
It is evident that p are also bitangents.
be written
xyU =V 2
(i).:
334
§ 2. q = 0.
U=
(i)
and
V=
are conies. The condition that xy should factorize is readily seen to be of degree 5 in k.
or
W = i(W+Z) + %(W-Z)
will vanish
when x
.
p=
q
=
0. s= lie on a conic.
.
=
and
/ = xypq —W'2 =
xyrs — Z 2
.
is
a constant. 3/ =
and pq-rs
only meet the curve in their points of contact. since x = and. q Then the quartLe is / = 0.XIX
r
2
NON-SINGULAR QUAETICS
335
=
For suppose that any two such pairs are ^ = 0. Consider now the pair of bitangents x There 0. Let another 0.
deduce
xy(pq-rs)
= (W-£)
(W+Z). according as their six points of contact do or do not lie on a conic.
2
W
is
of the
form xy/fi + fi(pq-rs) similarly it is of the form xp/v + v(yq — tu). 1)
= y = 0. For otherwise
Now we
W—Z and y of W+iJ. Then we have
Now
— 4/= — 4xypq + 4tW 2 = —4xypq + (xy+pq — rs) = x2 y +p2 q 2 + r 2 s 2 — 2 xy pq — 2 xy rs — 2pq rs. Three bitangents of a quartic are called syzygetic or asyzygetic.
We
have shown
that. One such pair is y and p
=
=
=
=
=
=
pair be
t
=
and u
= 0.
W=
y
is
any conic through the
0.
2 2
The equation
of the quartic
(xy)i
is
therefore
+ (pq)? + (rs)i
=
(iii). and the quartic will go through which is impossible.
the point
(0.
. which
0.
W=
We
and
Z—
being conies.
cannot have x a factor of vice versa. Hence there is no loss of generality in supposing fi = 1 and the upper signs taken in the ambiguity. where 0. p are five pairs of bitangents such that the points of contact of any pair lie on a conic through the points of contact of x and q 0. if
points of contact of x
= 0.
=
0. Hence we must have
xy
where
ft
= n (W +Z)
=
1 n~ (W±Z). q = is the required result.
'
a bitangent is not altered by multiplying its equation through by a constant.
=
shows that the eight points of and r = 0.
The symmetry of
this result
contact of p 0. s = 0. 0. Similarly four bitangents will be called syzygetic or asyzygetic according as their eight points of contact do or do not lie on a conic.
B.AR2 =
.]
Two
bitangents. Ex.
. D.]
is the triangle formed Ex.
quartic are
syzygetic. D. If points of contact are P. C. QiQi).] Ex. If four bitangents. 19]
Ex.
. u = passes through the intersection of either x = 0.]
.) are overlapping in the two cases
respectively. [By a proper choice of coordinates the bitangents become and y = 0. t = and s = 0. (AB.
.:
336
NON-SINGULAR QUARTICS
for 2
XIX
2
Comparing these expressions
W we get
Therefore the line through the intersections of r = 0.
CQ1 CQ2 AR1 .
. limiting case of this. then all four bitangents are syzygetic. See also Ch. 3. the required [Taking the conies as V+k. AQ t AQ2 CPt CP2
Qs
.
2. (CA.
Ex. taking x = The four common tangents to two ovals of a
Ex.
Ex.
Ex.
.
.
.
.
[If
the quartic
is
conic
is xyU = V* U+2kV+ k^xy = 0.
. £> = 0. § 6.
§ 6. (PC. Four concurrent bitangents are syzygetic.
P
. Three concurrent bitangents are syzygetic. Show that a conic can be drawn touching the quartic at A. 5. C.
and the conic is V+kxy = 0.
Compare Ch. and V+k t xy = 0.
we
find that
will
17+ 2 k V+ k 2 {x 2 + 2 hxy + y 2 ) have y as a factor for a suitable value of k.
II.
lie
2.
2
. x 2 + 2hxy + y 2 = The equation of the quartic is (a.
[Use Ex. 1 A conic through the points of contact of two bitangents meets the curve again in A.
[Project the points of contact of one of the bitangents into the ftircular and remember that a pair of common chords of a circle and conic are equally inclined to the axes of the conic]
points.xy = The preceding example is a conic is D + {k 1 + k 2 ) V+k-Jc^xy = 0.
the upper or lower sign being taken according as the bitangents are An even or odd number of the involutions syzygetic or asyzygetic. P1 P2 ).
[Use Ch. 4. the required For the quartic may be written
xy (U+ 2kV+k 2 xy)
Ex.
if
y
=
is
a bitangent. BR. B. not all concurrent.or j/ = 0. 6.
Ex.
conies are drawn through the points of contact of two Show that their eight other intersections with the quartic
on a conic.
5.
[As in Ex. £.£.
BP PP
1
. § 3.
by three real bitangents whose Rx and P2 then + BR.
7.
=
{V+kxy)*. and 2 Qx and
ABC
.] 5. Hence The 378 intersections of the bitangents lie by threes on
straight lines.
as the fourth bitangent. are such that any three are syzygetic.
I. q = 0.
3. XII.2 + 2 hxy + y 2 ) U = V 2 and.
c may be changed without loss of respectively. the bitangents are the diagonals of a quadrilateral whose vertices are their
points of contact.f The twelve bitangents of a complex touch a curve of the third
tion. . y = 0. If the sides of the triangle of reference are three bitangents of a quartic.
See Humbert.
f=g = h=0. 423... But it is at once proved that these the conies
U+2kV+
conies cut x
=
in an involution. r.
exact differential.
=
as a e b e .
syzygetic
where p. a x (3 x a 2 /32) is an involution.
If the lower signs are taken.
second case
If in the
q = —hx + by+fz.
f Another proof is the following
§
Take x =
0.
'
x
.
. P and Q.
A
Either bitangent of
five pairs
'
any pair of a complex is met by the other in an involution. The signs of a. 8.
an involution. y.
We
. The lines ax and b x the x conic A^'B^A^A/.
A lt
B B
in
. p. z 1 2 i 1 (by* + 2jyz + cz' ) 1 . and similarly (PQ.
. b.
to
+ ax i + 2hxy + by i f. the equation of the quartic can be put in one of the two forms
2 i i 1 p x + q^y + r z' ±2qryz±2>pzx±2pqxy
1
=
xyzw. ( + cz + 2gzx + ax' )
. q= hx + by +fz.
Let a 6 be the bitangent. and similarly for the intersections of a 6 with a3 and b 3 &c.
vi (1890). . If pdx + qdy + rdz is an exact differential in this latter case. and meets a x b x a t b 2 in oc x /3lt a 2 /?2 Then
A-[. the bitangents are asyzygetic. pdx + qdy + rdz is an
r
= gx—fy + cz. and the conic A X A-[B X B{A 2 A£B 2 B 2 Suppose a e meets this last conic in are three such conies.
an involution. o 5 6 5 are included h?xy = 0. signs are taken.
(PQ. the line-pairs a^. The points of contact of the given
It
bitangent also belong to the involution.
the equation reduces
or
p
= ax + hy — gz.
is
a^. IV. the bitangents are
and
pdx + qdy + rdz
an exact
differential.
Six pairs of bitangents such that any two pairs are syzygetic are said to form a Steiner's Complex* shall denote the pairs by a x b v a 2 b 2 as b s a 4 6 4 a s b 5 a 6 b 6 The points of contact of a.
*
Or
(i)
Steiner's
Group or
Steiner's Set
:
or Double-six. A A
6
')
6
..
.
meets
all
conies
through
.
If the
q.
. &c. Liouville's Journal. will be denoted by 1 and A(.
.
Then taking
2
among
as the equation of the quartic. p = ax + hy + gz. z.
Steiner's Complex.
class.
.]
§ 3.
We shall say
y
'
a complex
'.
w are
is
upper
linear functions of x.
.
.
.
.
.
—
(
0.XIX
3
STEINER'S COMPLEX
337
Ex.
.
[Suppose that when we put x = 0. a? /3 2 A 6 A/) is an involuHence (A 6 A 6'. generality and we readily obtain in the two possible cases r = gx +fy + cz.
.. as the condition tliat the intersections should lie on a conic.. This is equivalent to proving that the six centres of the six
line-pairs included in the family of conies
S+21cS' + k 2 S"=0
(i). The conditions that the centres of the line-pairs lie on a conic
.
on a conic.
Taking the bitangents as the sides of the triangle of reference.. z = a 3 xi + b s y'i + 2yxy = lie on a. B..
S'
= a'x + . . when we replace S by S + 2^8" + h^ S" (i = 1.
by
Three bitangents of a complex.
&A i + bBi + cCi + 2fF + 2gG + 2hH =
i i i
0.
.
. 8'.
2
S"=a"x2 +. a line-pair. the points x = b 1 y 2 + c 1 z i + 2atyz = 0. are
asyzygetic. conic. V.
. or otherwise. For the sake of symmetry we have replaced the U. which is excluded. y = c 2 s 2 + a 2 a.
A — bc—f
2
.
&c.
F=
gh — af
Suppose now (i) is a line-pair when h = k lt k2 . become A t Bit .
gx+fy + cz
= 0. 2 6).
where
8
lie
= ax + by
2
2
+ cz2 + 2fyz + 2gzx + 2hxy.
.. In this case we get ..
ax2 + by2 + cz 2 + 2fyz + 2gzx + 2hxy
are
= =
1. = 0. its centre (x. and can readily be shown not to
lie
by § 2 The
(iii). . z) is
given by the
equations
ax + hy + gz = 0.. or 7c6 and that the quantities A..
(i
6).. if a^t^ = a 2 6 s c. the equation of the quartic takes the form
(l
t
x2 +
m
x
xy + n^zfi +
(l
2
yx +
my
2
+ n2 yz)~* + (l3 zx + m3 zy + n3 z 2)? =
2
intersections of this with the sides of the triangle of reference are at once found.
2. hx + by+fz — which give x2/A =y/B = z 2/G = yz/F
where
0.
t By Carnot's theorem.. .
*
See Ch.338 For
STEINER'S COMPLEX
#
XIX
3
it is readily proved that the envelope of a line divided three line-pairs in an involution is a eurve of the third class* touching the lines in question..
on a conic. y. 1 m 2 » 3 = 0...
Ex. 8".
= zx/0 = xy/H. XV..
§ 5.f
The six intersections of each pair of a complex lie on a conic.z + 2/3sa. and
If
xy of
S=
§ 2
by
is
S. no two of which belong
to the
same pair.
b^. We call three such complexes a complex-triple..
it
C.*
There are sixty-three complexes of bitangents. [The quartic is the envelope of the conic U+ 2kV+ l?xy = for varying h.
London Math.
.6 = 63.
P is
The
the corresponding Cayleyan.
§ 4. of the sixth order whose rows are
the determinant
A
At B
.
.
.
[See Ch. Between them they contain all the twenty. V= 0.. 3. to deduce from Ex. Use Ch. For by the result that three bitangents of a complex are asyzygetic..
. § 6.
pairs of bitangents of a complex are the polar conies of at an intersection of the fixed conic with the Hessian. say 04 and a2 with the pairs ct^a^.
t
. . The remaining bitangents of this new complex are not in the original complex.
.6)
vanishes. no one of a s b 3 . Ex. a 2 b 2 a 3 b3 a4 &4 a 6 b6 a 6 b 6 take any two bitangents not These determine a complex forming a pair. Soc.
.
XIX
4
RELATIONS BETWEEN COMPLEXES
if
339
Therefore the six centres lie on a conic. There are 28 (7 = 378 such pairs..
all
three
This proof
is
due to Baker.
F Q
t. Hence the number of complexes is 378 -J.
.
Ex. .
But
is
this is the case.
We' have seen in § 2 that from any two bitangents a single complex may be obtained. 2 that the intersections of pairs of bitangents of a complex lie on a conic. if no two belong to the same pair.
The
if
P
in Ex.
z2
. p.
four of the bitangents lying in
Proc. for A considered as of the fourth degree.
1.
.
H.
(i
=
l. curve of the third class touched by the bitangents of the complex is
2.
XV.]
Ex. ix (1910).
Relations between Complexes..
.
.2.
Hence A vanishes
identically. Any one of the six pairs of a complex 2 determines the same complex. xy =
U=
is
the Hessian. a 6 b6 can be syzygetic to both ax a 2 and'&j&g. In the complex whose pairs are %&!. .
.] Ex. 148. 25. .
.
II.
§ 4.
1. Similarly a x and b 2 determine a complex containing the pair a2 bx and no two of the three complexes thus determined by aj) x a x a 2 a 1 b 2 respectively have a bitangent in common other than a lt b x a 2 b 2 which are common to all three.
Any
non-singular quartic
is
the envelope of the polar conic of
any point P on a certain fixed conic with respect to a fixed cubic.. The cubic is a curve of which the Jacobian of 0.
. XVI.
For any two bitangents determine a complex.
an equation
in
/Cj
=
k 2 k3 kt k6 and k 6
. and has five roots
/Cj
=
.
eight bitangents
*
..
. their pairs are of the form
We
b^
a i C l! a2 C 2> a3 C 3> ai Ci> a6 C 5'
OjCj.
Two pairs of the given complex such as a. Suppose then that ax a2 ..
. we may without" loss of generality suppose a3 to be Let in the second (and b 3 in the third) complex of the triple. whereas a v a 3 b 3 are syzygetic.
For the bitangent Cj must lie in the complex determined There is no loss of by a x a2 or in that determined by a x b 2
. which with the tbiity-one complexes just mentioned
make up
the total of sixty-three.
.
a 4 6 4 a b b.a2 . i..
tt
6
C 6'
b3 c3 . For they are not a pair of this complex. &. then c^Cj. Hence the complex with the pairs b^. b 5 c 5
.
. b i ci .
to be in the former . Neither of them is in the first complex of the triple.
. Returning now to the given complex with pairs
=
a^. Hence each complex is a member of fifteen complex-triples including in all thirty-one complexes. we get thus thirty-two more complexes..
.
.
O-y
2
J
C1 C2
.
take any bitangent c 1 not in the complex.
o6 c6 .& 2 can be selected in e 2 15 ways..
. a 2 c 2 a 3 c 3 be three pairs of the second complex. b 2 c 2
.. and consider the three complexes with pairs
generality in supposing
it
<X-yOj
2.
These form a complex-triple and therefore contain all the bitangents. They cannot both be in the second (or both in the third) complex. Now the complex with the pairs a 2 a3 c 2 c3 contains the pair b 2 b3 since a 2 a 3 b 2 b 3 are syzygetic. and three bitangents of the complex no*two of which are a pair are asyzygetic. Since a 3 and b 3 are interchangeable in the given complex.
340
RELATIONS BETWEEN COMPLEXES
XIX
4
complexes.
. and the result proved. e.
either four syzygetic bitangents in another complex containing the same four bitangents a complex-triple including all the twenty-eight
Summing up we have Two complexes have common and form with
.a 6 6 6
. b 2 c2 contains the pair b3 c 3 This argument is at once extended. chosen in sixteen ways.. and the other twenty-four lying in one of the
complexes only. have six bitangents in common.
.
. a2 b 2 a3 b3>
. shall'show that the complexes determined by a^.
a2 c 2
.
. c x c 2 are pairs of a complex. and in particular a 3 and b 3 . We get two new Since c x can be complexes determined by c^Cj and b^.
.
5
ttjCj.. for the given complex determined by 04 & x is not altered by interchange of a2 and 62 .
.
.
b3
.
.
.. 2. 6.
oe ce
are such that any two have six asyzygetic bitangents in common.
therefore
ai a r
is
.XIX
5
THE HESSIAN NOTATION
341
bitangents.
.
.
is
cij.
. c. The bitangent paired to it in this complex denoted by ij or ji.
is
a{ bj. . .
\. 72. and form a complex-pair of
eighteen bitangents.
.
*
Seven
is
b1 . 4.
. and the bib v b 5 b 6 by 17.
. b2
.
b2 c 2
.
c6
. b s .
. 75.
. if they were. 84.
§ 5.
'>
j
j
I
of the digits 1.
OjCj. 71.
i
andj being any two
occur in the third. The seven bitangents ax a2 a3 at as c 6 be are such that no three of them are syzygetic. .. 74.
a complex
.
18.
o B c 5!
. For. 37. b 6 . 73.
Denote the bitangents a lt a2 a3 ait ab a6 by the symbols 68 or 81.
c is
ae c.
and
so
is b x . 76. 48.
.
«A.
The Hessian Notation. There are 8 C2 = 28 symbols such as ij = ji. 83. o 4 c .
The complexes formed by a iK a n°2> asK «A> a5&5 a 6 & 8 a i C l> *2 C 2' a3^3> ai®l> ab C5> a i C 6'
. 28. b 2 c 2 .
.
«2 & 2> a 3°3' a J>i> a b°s. 27. 57. 6 c 3 . b t .
ajbi. b 2 . tangents b 1 b2 b 3
.
b i ci
Then ax a 2 a 3 a4 a5 ae
. 5. Such a set of bitangents is called an Aronhold Seven '.. 38.
bjbs aja B
. . contrary to the supposition that ai} a it ar are
asyzygetic. 3.
b 6>
<x
6
.
three of
or else they have six bitangents in common no which are asyzygetic.
by interchanging a6 and become
In the three complexes at the end of § 4 alter the notation b 6 and by writing c for c 6 They
. o 6 c 5 3 4
.>
^2^2'
a6 & e!
^6 C
'
a3 CA> ^4 C4! a6 C63
b3 c 3
. 86.
. and cannot occur Hence it must in the second as it is asyzygetic to a^-. It does not occur in the first by definition. 67 or Denote c by 78 or 87.
ai Cl!
OjCj. b 3 . 82.
a complex.
a r bs a s b r
. Another Aronhold
. Hence we have a notation which will include all the twenty-eight bitangents. 47. 85. 58. an Aronhold Seven
.
. Now c must occur in one complex of the triple ai hj> a b i> «A> a bj> •'•> a i aj> hi hy
. We note that the bitangents ij and rs cannot be the same.
..
.
.
g.342
THE HESSIAN NOTATION
XIX
5
The notation is not symmetrical. 7. . 4. 6. see Fig.
their
from § 2. Ex. 78
*
or 12. e. 34. (ii) their symbols involve two digits twice and two digits once. that four bitangents are syzygetic. p. e. 4. It follows
if
The criterion of § 2. 1. but it is perhaps as convenient as any we can find.. The bitangents are now denoted by the symbols 12 or 21. 56.
. 8 in the figure. Ex. applied to show that three bitangents are syzygetic.
Buil. g = fg + e equal ellipses of small eccentricity and nearly concentric. &c.
. These portions are narrow ovals each touched by one bitangent. 12. the major axis of one being parallel to the minor axis of the other. symbols are of one of the two types 12. 3. and / 0. 2. 2). de France.
Fig.
proceed to the limit in which the ovals are indefinitely we may consider the bitangents as the twenty-eight lines joining in pairs the eight points 1. 34.. xxvii. 1 will readily if (i) their symbols in Fig. 23.
this section. 34.
12.
If
l. 3. 5.
=
=
Kg. We shall show that this is the same notation for the bitangents as that before derived at the beginning of
we
narrow. 229. 34. where e is a small constant. We may visualize * the notation by considering the quartic are 0. 23. 13 or 31. de la Soc. 8 (Fig. 56.
FontenS.g. 2 involve six distinct digits. 2. 41. Math. The quartic consists of four portions each lying inside one ellipse and outside the other. while any two ovals have four tangents in common. Hence the quartic has twenty-eight real bitangents each touching the curve in points which lie very close to two of the points labelled 1.
have the possible types f
We
We
We
j7. since we have just shown that i8j7. rs. 1 * begin with the type pq. 78 56.
a complex in Fig. retracing our steps. 58 67.
Fig. tu are syzygetic. 3. r. qr. 6. k.ji. It follows at once that we have complexes of the two types
or
18 17.. 68 75. 68 67 14 23. 2.. Hence we have thus all possible h ^4 sixty-three complexes and then. 28 27. 5. 58 57. we see that we have given above every possible set of three or four s
•
The
=
=
. 34 12.
.. 8i and 7j.
* The figure only gives a upon for a proof. s are 7 and 8. 48 47. 1. 38 37. qr. rs or three of the type pq. 24 31. q. 2.
XIX
5
THE HESSIAN NOTATION
343
In Fig.
denote any of the integers
. First take the case in which two of p. 8i
and
87.i7j8. rs. 2 the four bitangenta either join four distinct pairs of points or else are the sides of a quadrilateral.ji
and
7j.
chapter.ij78. we see that the notation for the bitangents obtained at the beginning of the section agrees with that obtained from Fig. 4. though it
. 7j.
fost type contains
S C2 28 complexes and the second 35 complexes. 78. ..
Moreover.
f
i. 2.
j. and therefore must not be relied assists us to visualize the results obtained in this
1..
special quartic.
is
can deduce the fact that three bitangents of' the type pq.
syzygetic bi tangents. from the original definition without the use of Fig.
j8.
I. 18.
Hence
. qr. 37 38. 5.. 4. 6.. 28 2k. jk.
8p
8q.. rs..
It is not in the first.
Repetition of this argument shows that
is
lZlj. for j7.
interchange 7 and
8.
a complex by definition.
.
. 78 7k is a complex. It follows that the other type 8i.
p and
q are any of the integers
1.
i8j7. i7 j7.
.
j8jk.
. 81 8j..jk
7 or
8."
are complexes... 3. ik. Therefore it is in the third which makes i8.
58 57. 2.. It is not in the second .
and 17
18. 7.
_
j 8 jk.. i7j8. 8j..
. We have now shown that there are complexes of the type
.
a complex... 38 37. 8j. 68 67. 21 2j.
ij
.
..
%
Now
We
take the type ij. 48 47. ..
is
ij 78.j8jk.
A
similar process holds if
we
kl. Therefore for any value of k 18 Ik.
.
-where
. kl are syzygetic. It follows that i8 is in the complex j8 jk But...344
These are
THE HESSIAN NOTATION
all syzygetic..
Iplq.
since this
complex
is
%8
ik. ik. for otherwise we should have three bitangents in different pairs of a complex syzygetic. kj. changing the notation.. 8.
78.
q... 2p2q.
18
78 7k. 81 8j a complex and ij. ik are syzygetic. 78. i7 j8.
.
. 8i. 71 7j.jk syzygetic. for
XIX
5
i8j7. jk are syzygetic. i8j8.
whence
klkj. This exhausts all trios of the type pq..
is
a complex. r...
is
For the first type we note that jk complexes
in one of the triple of
18 17.
We
and
8i. i8
is
ik. . s is
Now take the cases in which one of p. have shown that
Ik.
27 28. jk. j8.. have the two possible types
i8. 28 27. kj is also syzygetic...
72 74.
The points of contact of this bitangent e.
where u r is homogeneous of degree r in x.
u2 —
0..]
Ex. 61 63. tu.
.
c5
of § 5 are 16. Conversely. 62 64. 52 54.
. 56.
it is
in the third. 56. rs. 41 43.
and
is not in the are syzygetic.ws + 2u 2 w + u3 =
0. for the equation of any quartic
* The proof is similar to that which Shows that the points of contact of the tangents from to a plane curve lie on the' first polar curve of 0.
take any point on a cubic surface as the point the equation of the surface in homogeneous 1).
. 12.
Hence
12. coordinates is
If
(0. tu are syzygetic. 46.
i.
/ = «.
Obtain
all
the 288 Aronhold Sevens of bitangents. 12 34. 14 32.
lie
on
w.= 0.j
=
0. 82 84. 2 joining one point to the other seven or lines forming a triangle together with the lines joining one of the remaining points to
the other four. we have the equation u u3 = w2 for the intersection of the tangent-lines with the plane w = 0.
cs
.
3.
1. 32 34. 26. 34.XIX
6
LINES ON A CUBIC SURFACE
345
consider three bitangents of the type pq. y. 0.
c4
. on the inflexional tangents of the
cubic at 0. Since 56 is in one complex of the triple
Now
12 14.
Ex.. 71 73. 36.
2. qr.
How many
sets of four syzygetic bitangents exist ?
[105
+ 210. 51 53.*
w between / = and -^. This is a plane quartic with the interof w = and the tangent-plane at to the section Mj =
surface as a bitangent.
we
0.]
The bitangents c u
§ 6.
Ex. 21 23.
c2
.
56
This completes the proof that three bitangents of the type Then the proof that pq. complexes are of two types and that there are only two types of three or four syzygetic bitangents follows as before.
first
two. The tangent-lines from O to the surface have of contact on
their points
^=2(^u.+w dW
Eliminating
1
2)
=0. rs. z. . for instance.
Lines on a Cubic Surface.. 81 83. rs or pq. 34. any non-singular quartic can be derived from a cubic surface in this manner .
[Lines in Kg.
Hence either 3. Hence the curve of intersection of the plane with the surface has and Q for nodes. or 27 of the lines on a non-singular cubic surface are
real.
z(U+2kV+k2 u
Hence the conies
in
1
z)
=
0.w = u 1 = Q make up a Steiner complex (§ 3).
p. we obtain five pairs of bitangents of the quartic which together with the pair w z = 0. 129. we shall show that either 4.
which planes through the line w = z — meet the surface again project from into the family of conies
U+ 2kV + k u
.
. z.
homogeneous of degree 2 in The surface meets the plane w = Icz where
U and V are
x.
*
The method
is
due
to Geiser. Q one on each of the lines joining to the points of contact of the bitangent.
Through any line
m
drawn
m
i.
evidently no loss of generality in supposing that through a line of the surface. This result enables us to deduce properties of the plane quartic from those of the cubic surface and vice versa.
of the lines on on to Hence the projections from the surface are the bitangents of the quartic other than I. Hence
=
w
of a cubic surface five planes can be each touching the surface in three points and meeting it again in a line-pair. by I. = kz for such a value of k meets the surface in The plane three lines which form a degenerate cubic with three nodes. 15.
where
ut =
any
bitangent. Math.
w=
In Ch. u x For the present let us denote the bitangent to any bitangent of the quartic. y.
»
By choosing k to make such a conic a line-pair. The plane joining other than I.
i
1
z
=
0. The projections of these five linepairs form with I and the projection of a complex of bitangents of the quartic curve.
and must therefore degenerate into the
line
PQ
and
a conic.
Annalm. touches the surface in two points P. 16.* 0. or 28 of the bitangents of a non-singular quartic are real. But the curve is of
w=
=
P
degree
3. 7. It follows that there are twenty-seven lines on a cubic
surface. at each of which the plane touches the surface. 8.:
346
LINES ON A CUBIC SURFACE
XIX
is
6
can be put in the form u-^u^
=u
2 2
.
XX
There
is
the plane the line as
w = passes w — z = 0. Take The equation of the surface is
u 1 w 2 + 2Vw+Uz
where
=
0.
.planes of the
ci
Ne two
of c lt c 2 c 3 can meet. The plane of a 6 cB bB meets the surface in the three lines a e> c5> "6.
§ 7.
Z>
6
Cj.
.
tangent.
hyperboloid with e lt c 2 c 3 as generators in two points.
meet c 6 Let the planes
.
.
Sehlafli's Double-six. 3
(diagrammatic only).
<x
Suppose that the
line-pairs (§ 6)
line
e
of a cubic surface
is
met by the
five
Vl.
b B c3 b 6 cit 6 6 c s meet the surface
. . behig
. .
.
. It cannot meet a 6 .Hence b 6 meets one of these lines.
& 2 C 2> &3 C 3.
.
the planes a 6 6 1 c 1
surface. .
Since b 6 meets c x .
triple
&c.c 2 c 3 c 4 in four points lying on the surface. We now proceed to find the other thirty-six
complexes.
.
Suppose that
meets the
Oa
cs
Fig.
be c 2
. there is no loss of generality in supposing it to
.XIX
7
SCHLAFLI'S DOUBLE-SIX
of such triple tangent-planes
is
347
evidently
The number
27 X 5
-j-
3
=
45. b i c t> 6 5 c 5. 6 6 is a line of the surface.
Then it is at once seen again in the lines a lt a% a3 at> as that the configuration of twelve lines on the surface
(
ax
a2 a 3 a4 a 6 a 6
b2 b3
l
J
t
h
64
b5
b6
.
The above result gives us all the twenty-seven complexes which contain I. Through each of them passes a generator of the opposite family to c x c 2 c3 One of these is ae let the other be b a
' .
and 6 4 b 5 b 6 which meet them.
.
.
.
. the number of double. We showed that each line on a cubic surface meets ten other
and therefore does not meet sixteen lines.planes through a x are
'
«A> »A> «A> a A> a A>
follows readily that we might have constructed the doublesix starting from the pair a-fi^ instead of a 6 b e In other words. These lines can be chosen out of a 1> a 2 . hut does meet the other five lines...sixes is 216 -r 6 = 36. Hence all the complexes of bitangents are projections of
lines. as each double-six may be determined by any one of its pairs.
.
.
.
.
a configuration is called a Schlafli's double-six'. namely the conic in which the tangent-lines from to the hyperboloid meet the plane. The projections of the generators of a hyperboloid from any point on to a plane all touch a conic.
. Such
.
sets
of bitangents of a quartic there are twenty of six bitangents touching a conic. a 1. the double-six is symmetrical as regards its six pairs
it
. Since the triple tangent.
«A.
.
are 27
—
double-sixes or else contain I. For the plane through and a generator touches the hyperboloid.
A double-six projects into a complex of bitangents.
... Now the lines a x a 2 a z 6 4 b & b6 of a double-six lie on a hyperboloid. a2 a3 being generators of one family. Hence the corresponding bitangents of the quartic touch a conic. No two of these six bitangents belong to the same pair in the complex. being generators of the other family. Hence there x 16 2 = 216 non-intersecting pairs of lines on the surface.
.
In any complex
From
a complex
the theorem that the intersections of each pair of lie on a conic we obtain
li/nes
The six
drawn through any point
of
each intersecting a pair of a given double-six the second degree.
.
.
. a x meets b2 b3 bt bs b6 only. Hence:
.
. «A> as&fi> a ehDenoting the bitangents into which the lines a x b x c15 . or ax bx is a pair of the complex determined by ae b e Similarly for a2 b 2 a 3 b it &c. project by the same symbols. «A> «A.:
:
'348
is
SCHLAFLI'S DOUBLE-SIX
XIX
7
such that any line does not meet each of the lines in the same row or column in the array { }. For instance.
.
.
a cubic surface lie on a cone of
.
. a e Hence in twenty ways. Therefore. we showed in § 6 that a x be a B b x are pairs of the complex determined by I and c l Hence ai> ae> &i> ^6 are sy z ygetic.
These cusps are the intersections of a conic with a cubic toughing the sextic at the cusps and the twenty-seven bitangents of the sextic are the projections from of the twenty-seven lines on the surface. 2. By taking at the intersection of two lines of the surface obtain properties of a binodal quartic. 1.
Ex.
. By taking of § 6 on a line of the cubic surface obtain properties of the bitangents of a quartic with a node. 3.
Ex.XIX
7
SCHLAFLI'S DOUBLE-SIX
349
Ex. The intersection of any plane with the tangent-lines from a point to a cubic surface not passing through is a sextic with six cusps.
.
CHAPTER XX
CIRCUITS
§ 1. &c. unless the contrary is stated.
Circuit
and Branch. so that the direction of the tangent
'point'.
'
oo
Pig. Such a statement as the number of inflexions
curve. 'inflexion'
P
. and travels along the curve. Suppose a point starts from any point on a curve.
Heretofore. when we have discussed singularities on a we have not always distinguished real from unreal singularities. and by
—
mean 'real point'.
2)' is only true if we include of a non-singular n-ic is 3n(n unreal inflexions.
1. 'real inflexion'. In the present chapter we shall make the distinction between real and unreal points on a curve.
.. i. 'tripartite'. e. The part of the curve thus traced out by P is called a circuit of the curve. if an oval shrinks
into a point. number of its intersections with the Circuit alters. 'bipartite'.
§ 2. The reader must distinguish between the use of the word branch as curve at a multiple defined above and its use in such a phrase as branch of a
'
=
=
. all Suppose. that it does not Such cross itself.. it
next travels inwards from infinity to a very distant point B such that A and B are neighbouring points on any projection of the curve. Suppose that. though we shall be mainly concerned with algebraic
curves. Zug circuit. one three branches. * In German.
If line is moving in the plane of a circuit.
§ 3. has no crunode..* These definitions will apply to any continuous curve with continuously varying tangent.
Odd and Even
Circuits.XX
3
ODD AND' EVEN CIRCUITS
351
at P varies continuously. e.
Ovals. A curve loses one of its circuits. if in its journey P travels outwards to infinity from a very distant point A. two. Thus a hyperbola has one circuit composed of two branches. . The curve need not be algebraic.
its
Suppose that a circuit is closed. the the number is increased or decreased by two (or some even
Suppose that a
number) as
is
obvious from Fig. i. unipartite '. Ast circuits as 'part' for 'circuit'. as shown in Fig. 1 shows a curve with' two circuits. It is to be noticed that an acnode is not counted as a circuit. Fig. The portion of a circuit traced out by P between two infinitely distant positions of P is sometimes called a branch of the curve. An a circuit will be called an oval.
'
point
'-
. and speak of curves with one. oval can be reduced to a point by continuous deformation. .
3. three. and the other closed and crossing
composed of
itself. This means that in general P jumps from one end of an asymptote to the other.
Some English writers use branch. moreover.
points are finite. 2 which gives successive positions of the oval as it shrinks to a point.
which shows three conIt follows at once that
secutive positions of the
moving
line. Eventually P returns to 0.
In general an n-ia has an odd or even number of odd circuits. two an odd and an even circuit. as n is odd or even.
in particular any oval. In fact tRe argument of Fig. and odd circuits which are met by every straight line in an odd
number of points. The least number of points in which a circuit is met by any
The greatest number of points is The index and order are both odd or both called its order. The theorem will therefore be established if we show how to deform the first circuit so that all its
odd
circuit in
lines
points are distant.
3.
Two odd
circuits meet
even circuits.
line is called its index.
Fig. or
in an odd number of points.352
ODD AND EVEN CIRCUITS
XX
3
circuits can be divided into even circuits which are met by every straight line in an even number of points. and the number of its intersections with the other circuit alters. even according as the circuit is odd or even. Its intersections with the second (fixed) circuit will then be practically identical with its intersections with the asymptotes of the second circuit. First of all deform the circuit so that its crunodes are cut '. The oddness or evenness of a circuit is evidently unaltered
by
projection. meet in an even
number of points. is even. This is done by replacing the part near the node by the dotted
'
. if one of the circuits is continuously deformed. Any. and any number of lines meets an even circuit in an even number of points. for it meets the line at infinity in an even (zero) number of points. closed circuit. To prove this we
first notice that. the number is increased or diminished by two (or other even number). We now show that the first circuit can be deformed so that all its points are very distant. an even number of meets an odd or even circuit in an even number of points. But an odd number of lines meets an
an odd number of points. 3 still applies.
4 (i). but for our present purpose we
shall think of these distinct circuits as a single circuit
whose
Kg. 4 (ii) is would necessitate the alteration of the direction which a moving point traces out the circuit. 5. a 2216
. 4
(i). we can evidently withdraw such a branch continuously to an indefinitely great distance as shown in
The
ovals. as shown by
the arrow-heads. Keeping the two distant ends of. A nodeless curve cannot have more than one odd circuit. 4
(ii). for it
in
shown in Fig. The cutting of the nodes may really resolve the circuit into two or more distinct circuits. The method of Fig.
is
evidently unaltered by cutting the
circuit now consists of distinct nodeless branches and no two of which cut.XX
3
ODD AND EVEN CIRCUITS
353
line as
inferior.
-A-
.
oddness or evenness
nodes. each branch fixed.
Pig.
Fig.
Also each oval can be deformed into a point and we have achieved our purpose.
On the sphere an even circuit is deformable into a point-pair and an odd circuit into a great circle in such a way that each pair of diametrically opposite points remains diametrically opposite.
Use Ex.
2 to prove the theorem of §
3. Ex. 4. In particular. it cannot meet an even
[If it
met
in two. 2. The difference between the order and index of a and greater than zero. If an even circuit is projected similarly.
would meet
in one or
more nodes of the
A
71 is
non-singular n-io has therefore one or no odd circuit as
odd or even.
3.
Pig.354
ODD AND EVEN CIRCUITS
circuits
XX
3
For two odd
curve. Ex.
1. 5. the line joining
them would meet the quartic
in
five points.
Ex.
1.]
Ex.
more than one
circuit of a (singular) quartic cannot meet any other point.
5.
circuit is even
An odd
circuit in
circuit. whose points can be divided into diametrically opposite pairs. that the projection is a closed circuit on the sphere met by any great circle in ir+2 points.
Show
Ex. the projection Is a pair of diametrically opposite circuits each met by any great circle in an even number of points.]
. An odd circuit is projected on to any sphere from its centre.
[Project after cutting the crunodes.
the cutting of a node either divides the circuit into two even circuits or into two odd In the latter case it is possible. 4.
circuit
'
We call now
'
an odd
point-even number of points. if a node is cut as the circuit is divided into two closed circuits. tangent-even.
6.
on a circuit id As a point P travels from a fixed point (§ 1). tangent-even. the tangent at a fixed direction. What happens if a node
is
cut the
wrong way
as in Fig. by cutting a second node. tangent-odd. quintic with
A
two intersecting even
circuits cannot
have more than
one odd
Ex. 4
(i).
A
'
point-odd
To
its
and an even circuit meets any line in an odd reciprocal with respect to any base
'
point-odd
'
circuit
Point-even. 4
(ii)
?
[See Landsberg. reconvert the figure into a single even circuit with two nodes less than the original.
conic
any given
draw an odd number of tangents from '.XX 4
Ex.
355
circuits of a quintic
cannot meet
in
more than one
point. 6 shows the four possible varieties
we may
therefore
point.
Point-even.
A
by an even number. Show that. reciprocal will be called tangent-odd Similarly we obtain tangent-even circuits as the reciprocals of point-even circuits. Fig. but not closed.
6. if the circuit is even.
RECIPROCATION' OF CIRCUITS
Two odd
circuit. Annalen. if all the nodes are cut. the numher of circuits obtained differs from n + 1 in Fig. p.]
§ 4. Also that.
7.
Beciproeation of Circuits.
[See Pig.
Point-odd. to circuits.]
Ex. 8. lxx (1911). Show that.
The
'
'
'
of circuit. 563.
'. if the circuit is tangenthas passed an odd number of times odd. For the tangent at
P
P
Aa2
.
Fig. Math.
Point-odd. finally returning to turns through an odd multiple of tt.
closed circuit has n crunodes. tangent-odd.
which proves the result. arriving at
A
A
left
A.
point-odd or point-even circuit contains respectively an A tangent-odd or tangenteven circuit contains respectively an odd or even number of
A
odd or even number of inflexions. As starts from 0.
cusps. Now a is the number of tangents from A to the circuit. through an even multiple of n. according as the circuit is tangent-odd or tangent-even. and b is the number of tangents parallel to I.
A
§ 5.356
RECIPROCATION OF CIRCUITS
XX
4
through any fixed point A. since the number of tangents from Similarly the tangent at P turns to the circuit is odd. Q starts from and in the same direction as it finally returns to A. travelling along the Suppose that the tangent at a point Let the tangent at a fixed circuit meets a fixed line I in Q. of these statements is the reciprocal of the other. a being the number of times Q passes through A* b the number of times Q jumps from one end of I to the other. We confine our attention to the first. if and only if is an inflexion or an intersection of the circuit with I.
7. point of the circuit meet I in
One
may
P
A
P
Fig. Therefore a + b is even so that c is even.
Hence a + b + c must be even.
. 7).
. But Q reaches a limiting position on I and then moves back again.
P
A
*
The
initial starting
from
A and
the final arriving at
A
taken together
count as passing through
once. (Fig. Hence a and 6 are
both odd or both even. and c the number of times Q travels up to a certain point of I and then begins to move back again.
Inflexions and Cusps of Circuits. and returns to 0.
traverses the circuit. if the circuit is tangent-even.
.
9. since the circuit is of order 3.. The inverse of a circuit of r branches with respect to a point not on the circuit is an even closed circuit with an r-ple point at 0. [Project one inflexion to infinity. 2. Let the tangent at P meet the circuit again at B.
and
[State whether point. 4.. and notice that during
Ex. [(i) Invert with respect to 0.
. An odd circuit has 2r+3 inflexional tangents no one of which meets the circuit at more than three points (coinciding with the point of contact).
circuit of Ex. If the tangent at does the product PQ 1 PQ. cannot lie in this order on the circuit.
3. 4.
[Project one inflexional tangent to infinity.
7. [The curve and inflexional tangents divide up the plane into these
various portions. An odd circuit of order three without node or cusp has exactly three inflexions (Mobius's theorem). Describe the circuits of the curves in the diagrams contained
in Ch.or two-circuited cubic and show the portions of its plane from which 0. and therefore coincides with P twice.
Ex. 6. 8.
C
Ex. The reader may enunciate a theorem concerning the positions of the points of contact similar to that of Ex. or 6 tangents can be drawn. [Deform the circuit into some standard position. curve ? Deduce the theorem of §
.
PQa
meets the circuit at Q" Q2 Qa . Treat similarly some non-singular quartic curves.. the inverse circuit is odd or even as r is odd or even. Draw a one. Show also that 2r+l tangents can be drawn from any inflexion. (ii) cut orthogonally a given circle.]
Ex. 5. 10.
Ex. If is an r-ple point of the circuit.
An
odd
circuit without
node or cusp has at least three inflexions. give their order index. (ii) Deform the circle continuously
into its centre.
.
2. 1. change sign as P travels along the
. B'. Ill to
XIX. 11.
5. but has 2p (2p + 3) and q bitangents.
An
the deformation
p—q
is
never increased or diminished by an odd number.or tangent-even or odd.
. when . A'B'C A.
[Project one inflexion to infinity. 4
. Show that 4) — q is even. C.
5. R travels along it twice in the opposite direction. Hence there are two finite inflexions.+ 3) bitangents.
357
Ex.
.
Ex. .. &c]
Ex. B. An even number of circles of curvature of a circuit (i) pass through a given point 0.]
Ex.
A
non-singular curve of odd degree has at least three real
inflexions.]
Ex.]
.]
inflexions
even (odd) circuit has no node or cusp. and 2r + 2 tangents t'rom any other point of the circuit.' Show that it has r(2».]
Denoting the three parts into which the inflexions divide the by llt l2 l 3j show that there are two tangents from any point of ?i whose points of contact lie on l 2 and l 3 and that the tangent from the inflexion at which l 3 and ls meet has its point of contact on lt Show that if two lines meet the circuit in ABC.XX
5
INFLEXIONS AND CUSPS OF CIRCUITS
P
. A'. Ex.. There is only one other tangent through the infinite inflexion. As P travels along the circuit.
2. 14 has no inflexion. 15. Similarly for a cusp.]
true for each circuit of this quartic. where the curve C is shown with a crunode.
Method of Variation
of Coefficients. This is illustrated in Fig.
§ 6. 8. represents a curve of the same degree C and lying very close to it.
Another example
is
The theorem
as a whole.
5
it
Every tangent to a circuit without node or bitangent meets
in the same
number of points. An oval without double point has at least four vertices '.
and
for the quartic
Ex.
[E. 8. 14.g.
INFLEXIONS AND CUSPS OF CIRCUITS XX
12. a cusp. In general also an adjacent curve on one side of C has
as
. 13.]
Ex. 12 cannot be true for a circuit with a bitangent. vP
— yL
is
=
x l yi
. an acnode. Show that the result of Ex. points at which the circle of curvature has four-point contact. and' two adjacent curves also. remembering that the circles of curvature at two points of the oval lying between two consecutive vertices cannot both pass through 0. if has a crunode while an adjacent curve has none.
Ex. If the oval of Ex.
We
method enabling us to obtain the equaan assigned degree. number of circuits. the sum of the maximum radii of curvature less the sum of the minimum radii of curvature is half the perimeter of its evolute.
equation thus obtained.358
Ex. •
'
[Invert with respect to a point of the oval and use Ex.
Such a method
consists in giving small increments to the
coefficients in the equation of
some given curve
0. It will be noticed that in general.
The new
Fig. the adjacent curve has an inflexion close to each inflexion of C and two more inflexions close to the crunode of G. but may be true for a circuit with a node. i. &c.
shall require a tion of a curve with
inflexions. and the radius of curvature is a maximum or minimum. e.
the adjacent curve has an equation of = a + bx + cy i px1 + 2 qxy + ry1 + sx3 + .
&
The
from
results for the case of a cusp are at once written
down
Fig. The number of ideal bitangents of an adjacent curve is the same as for G and these bitangents are adjacent to those of G. 424. Annalen. bitangents whose points of contact are unreal or lie on the same circuit. F are conjugate
*
the form
.. c are small.
Two
. of these must be real bitangents in their final position just before becoming unreal. Clifford. F as nodes consists of one circuit and has two real bitangents t lt t as 2 may be readily proved by projecting E. where q2 < pr and a.
shall call a real bitangent with unreal points of contact The name is due to W. approximates to the curve G near EF. For take the case in which E. F. Let Q be the quartic with a real cusp which approximates most closely to C at E. The quartic in the position in which it has E. F. vii..
F
of these bitangents are consecutive to t x t 2 The other Now since the quartic Q to EF. In the neighbouring position Q will have four bitangents of the first sort.
. e. pro-
We
an ideal bitangent. which is in general a convex oval without inflexions. i. or no real bitangent. reasons as follows The line joining f two nodes counts as four bitangents.
Two
. as C varies. F and as C passes into a consecutive position let Q pass into a consecutive quartic still approximating to G near E.XX 6
VARIATION OP COEFFICIENTS
359
a small oval enclosing the acnode of G. the line EF counts as the limiting position of four bitangents.* The adjacent curve on the other side of G has no circuit near the acnode. G has a pair of conjugate imaginary nodes or cusps E. and the other two must be unreal becoming real.f similar argument holds if E.
vided
If.
G has no imaginary nodes or cusps. an adjacent curve has respectively two or three ideal bitangents adjacent to EF. This will be proved in § 8 by reasoning which does not involve the use of the result we are here establishing.
. To each line joining two nodes of corresponds four real bitangents of an adjacent curve having real points of contact. are cusps. p.
.
. 6. F are nodes.
:
imaginary. however. Math. into the circular points and inverting with resp*ecfc to the cusp. or no real bitangent. K. 8.
two must be consecutive
A
F
Taking the acnode of C as origin. If the nodes E. a bitangent of C must approximate to each bitangent of Q near EF. The approximation near the origin is the ellipse = a + bx + cy + px 1 + 2 qxy + ry2 Zeuthen. To each tangent from a node of G to G corresponds either two bitangents of an adjacent curve which are both real with
real points of contact.
2. Then if / = 0. If h is even.
e is
any small constant. is an (n+2)-io meeting the ellipse in the 2 (n + 2) real intersections of It has also (n + 2)n inflexions and no ideal e = 0. g = 0.-ic
and n 2 -ic.
i
+ n^-ic with n (ni — 2) + n2 (% — 2) + 2%! %2 = (%! + to 2 (nt + 2 — 2)
a non-singular (n^
inflexions
and no
ideal bitangent.
If a non-singular n x -ic and n^-ic exist having respectively ^(TOj — 2) and n 2 (n 2 — 2) inflexions and no ideal bitangent. r may have any of the values 0.
For
yj/2
if
=
is
/x = 0..
71.
[(i) x =f(t). prove that one Ex. which. g = and lines. The reader will readily establish the truth of this statement in the cases n — 2 and n = 3. Jc.
%K-2) +%(%. Take n + 2 lines each meeting the ellipse in real points. 2 Obtain the equations of (i) a unicursal curve. there exists a non-singular (n l + n i )-ic with (n 1 + n 2 )(n l + n 2 — 2) inflexions and no ideal
bitangent.
where
*n. where /.
is
inflexions. where e is a small constant. But the result is only required in the second part of § 7.2)
and no ideal bitangent. (ii) a non-singular curve.
/=
Ex.]
/=
is
any unreal non-
For we shall see in § 7 that the number of real bitangents of any non-singular «-ie adjacent to an n-io with nodes at E and F is the same. where singular («— 2)-ic or (» — l)-ic with real equation.
i (ii) (x + y*—l)f=f or (x-\)f=e. are the equations of -n-ic. ef = eg. and does not affect our discussion of the circuits of a non-singular curve in §§
7. bitangents as in the proof of the preceding theorem. 1./2 = are the equations of the an (n t + « 2 )-ic with n x n 2 crunodes. 0.
For any given value of n
with
it
\here exists
a non-singular
n-ic
n(n — £)
inflexions
and no
ideal bitangent. of degree n consisting of one circuit only which is of index or 1 as n is even or odd.. Now use induction.
. if h is odd. y degree n and \jr =
=
<f)
(t). ellipse. If (where q> = of the adjacent curves /+ e$ = does not go through and e is a small constant) has 2r inflexions near and that the other has 2 (Ic — r). let us take
above remarks which will be useful one or two theorems.
has ~k real branches through a point 0.
Let us assume that such a curve exists and that moreover meets an ellipse in the 2n (real) points. Perhaps neither line of argument will appear quite convincing. e = 0. .. 1.
Then /j/g
)
=
e.360
VARIATION OF COEFFICIENTS
illustrations of the
XX
6
As
later on.. z = yjr{t). r = \lc but. meet in n^n^ real points. 8. \jr are polynomials of has not more than one real root.
.
(The straight line
is
a bitangent in each case.. Now in the process no two ideal bitangents can coincide.* Hence the only way in which i + 2 1 can alter is (i) by the coincidence of two inflexions at a point of undulation and their subsequent disappearance (ii) by the points of contact of a bitangent becoming real instead of unreal. Then the curve is continuously changing its shape.
[/ =
ci
where
/=
is
the curve in but not if p
> 3. or vice versa. and we may thus continuously deform the n-ic till it coincides with some standard non-siugular n-ic. Show that an «-ic exists cutting the p-io in np real points. First consider the case of a non-singular n-ic.
* This would not apply to two bitangents with real points of contact.
for
.]
p
real points. and d is the number of real acnodes.
XX
7
KLEIN'S
THEOREM
361
Ex. n being any given integer. A p-ie is cut by a straight line I in p real points. For instance. but not a quadruple tangent. for this would imply the existence of a quadruple tangent. Suppose the coefficients in the equation of the curve to vary continuously. or one node.
I. no two of which are consecutive.
§ 7. k is the number of real cusps. where i is the number of real inflexions. (iii) by the appearance of a node.
In any curve of degree n and class m n + i + 2-t = m + k + 2d. &c. 9. t
Fig. or vice versa.
Klein's Theorem.
Now Fig. or pair of
unreal nodes. &c. During the deformation the curve may have one triple tangent. 9 shows that cases (i) and (ii) are really the same and that such an event decreases i by two and increases t by one. they might coincide to form a triple tangent.
is the number of ideal bitangents (real. or flecnode. Also we have shown in § 6 that in case (iii) the numbers i and t are not permanently altered.)
We may suppose
that during the deformation there
is
never
more than one relation between the coefficients.
the equation of n straight lines each meeting The line I can always be found if p = 2 or 3. with unreal points of contact). 3.
i is diminished by two. But we have shown in § 6 that an n-ic exists for which
i
= n {n.
as required.2t
by Klein's theorem. \d' pairs of conjugate imaginary nodes.
Ex. besides its k real cusps. For n = 2r + S.:
:
362
if
KLEIN'S
THEOREM
XX
7
a crunode appears. Klein (Math. Verify this.
The theorem of this section is due to F. the following: For n = 2r. and d 1 crunodes.
n + i + 2t = n(n-l)-2(d 1 + d + d')-3(k + k') + 2d + k = n(n-l)-2$-3 K + 2d + k = m + k+2d. d acnodes. XI.
5.' As a corollary from Klein's theorem
we deduce
No
n-ic has
more
i
than n(n — 2)
real inflexion.
n+(i + 2d1 + 2k) + 2(t + d' + ilc')
Hence
= n(n-l).
t
=
0.]
Prove
i-Si + 6d
= *-3k + 6t. which proves the result. &c. and is increased by two again as the node disappears.
Ex.
k
=d=
0. 6. Let it pairs of conjugate imaginary cusps. §
3. 199). i real inflexions. a curve adjacent to the curve consisting of r equal concentric ellipses with major axes parallel to the r sides of a regular polygon.
is
established for a non-singular curve. For an n-ic with double points it is sufficient to notice that we can find an adjacent non-singular n-ic for which v-may be
greater.
m = n (n—l). but cannot be less.
We have already shown that for any assigned value of n an n-ic with n(n—2) real inflexions exists.
What
property of a curve do we obtain by applying Klein's
2.
theorem
Ex.
2. Annalen.
Hence the theorem
II.
to its evolute?
[Use Ch. Klein takes as standard n-io with n(n — 2) inflexions. Ex.
Now
have
•§/«'
An
flexions and of the proof
adjacent non-singular ?i-ic has by §6 i + 2d x + 2k inTherefore by Part I t + d' + %k' ideal bitangents.
suppose the n-ic has nodes and cusps. x (1876). add to the ellipses a cubic whose odd circuit cuts all the ellipses in six points. It may be put in the form 'The quantity n + i + 2t is the same for a curve and its
polar reciprocal. 7.2). 1.
. p. and t ideal bitangents.
For a non-singular n-ic
= m — n— 2t = n(n — 2)~. Hence n + i+2t is the same for all non-singular n-ics.
For if it had five. There is some line meeting the curve in no (real) point. which consists of one. For if a quartie had three ovals A. Each oval is of order two or four. 6 for the case of quartics with double points.
the curve
inflexions. if
Ex. for an inflexional tangent to the inner oval would meet the inner oval in four points and the outer oval in at least two. a line cutting B and C would meet A. The reader may illustrate on r = a cos 35.
3. Klein's relation (§ 7) becomes i + It 8 for the non-singular quartie.
Circuits of a Quartie.
*
f There
. 6. If there is an inflexion. and G in at least two points each.
a cubic. there are four ideal bitangents each meeting the curve in no real point. XIV.]
Ex.
infinity as ideal bitangent.
result.
[Apply Klein's theorem to an adjacent non-singular curve.
Ex * A curve has no acnode and not more than a third Show that not more than a third of its inflexions
[Use Ex. It cannot have more than four even circuits.XX
8
-
CIRCUITS OF A QUARTIC
363
of its cusps are real. the conic through a point on each circuit would meet the curve in ten points. which has the line at 1. is then a circuit approximating at infinity to a semi-cubical parabola (ay 2 = x s ) and with one asymptote also.f If there is no inflexion. it will be readily seen that they have four and only four common tangents
circuits of
We
have already discussed (Ch. Cf. B. no oval can lie inside another. § 6. B. If there are three or four ovals. they may lie one inside the other or be external to each other.
We
=
See Ex. This will be seen to have two branches with a common tangent.
are real. or four ovals.
5. and discuss in detail the non-singular quartie* Such a quartie cannot have an odd circuit (§ 3). In the former case the inner oval has no inflexion. The quartie can therefore be projected into a closed curve. and a line adjacent to this tangent will meet the curve in no real point. G with B inside A. the reader will easily convince himself of the truth of this statement by projecting the inflexional tangent to infinity. If there are two ovals. three.
An
w-ic
with
^
crunodes cannot have more than
»(»-2)-2d. What modification must be made in Klein's has a multiple point with distinct tangents ?
.
Ex.
§ 1) the possible consider here the case of the quartie. If two ovals are external to each other.]
§ 8. which is impossible for a curve of degree four. two.
AB
=
Fig. of course.bitangents of the second sort. where r is 1. 3. it has also |)'(r-l)x4 bitangents of the second sort. Imagine a string wrapped round an oval of order four. 10.
2
+ 4'i/ 2 -36)(4a. If it has r ovals external to each other.
* If the
curve
such
common
is of degree higher than tangents. 8. The equation
(9a. We give an example to show how an equation may be found which shall represent a quartic belonging to an assigned class. is a bitangent subtending a 'bay' AGB of the Hence the quartic has quartic with two inflexions (Fig. 2. twice as many inflexions as it has bitangents of tho first sort with real points of contact. be also eight real inflexions and twenty-four . Hence it has 0. 16. Bitangents of the first sort are those whose points of contact are either unreal or lie on the same oval. or 4. which proves the theorem.
Non-singular quartics may be classified by means of their ovals and the nature of their bitangents.
or 28 real bitangents. 4.
It has four bitangents of the first sort.
=
e. Suppose we require a quartic with four ovals each with a bitangent. 10).
A
non-singular quartic has
4. There will. or 24 bitangents of the second sort and four of the first sort. 2 -l-92/ 2 -36)
five. The relation i + 2t 8 now proves the result. 12.
A
non-singular quartic has four bitangents of the first sort.
there might be
more than
four
.364
CIRCUITS OF A QUARTIC
XX
8
which are bitangents of the quartic*
Such bitangents are
called bitangents of the second sort. If
AB is A and
a straight portion of the string touching the oval at B.
will represent a quartic
adjacent to the pair of ellipses
9a 2 + 42/ 2
and lying inside one
=
36.
Ex. 3.
(y-l)(y-2)(y-3)(*-4).
Ex.
Ex.
sd>.
x 2 + §y" + 2x
i
x*-y2 = 9.]
Ex.
1
x2 .
. + 3
=
0. Ex.
0. 2
+ 9f =
36.
(y-1).
4^-^ = 4.y2
= = 2 x.
(x-l)(y-l)(x + y). XIX.
Fig. Each of the involutions is non-overlapping. § 2. Describe the nature of the ovals and bitangents of the quartics are the fg = ±e.
(i)
2 2 2 Describe the nature of the quartic (x + y —5y) = +
is
(ii)
(y-l)(y-2)(y-S).
first sort
can
[The argument of Ex. and / = 0. XIX. 11.
we take we
bitangents.
^+
x*
+ 4a.
4a. 11.
outside the other. Fig. as shown by the narrow line in Fig. g = curves given below
(i)a»«
(ii)
+ y'
2
«/
= 1. 3. § 2. where 2 2 as* + 4^-5. for the case of four concurrent bitangents. 6. where t is a small positive constant. 2.
1. 11 by the dotted line.
x'--y\
(ii)
(x + y-2){x-y
+ 2). It will evidently be a quartic of the
and as shown in
ellipse
kind
If
stated.
a small positive obtain a quartic with two ovals.
4.
The eight points
of contact of the four bitangents of the
first
sort lie
on a conic. The outer oval has eight inflexions and four
e
constant.
where
rf>
Ex. (y-l)(y-2).
(iii)
(iv)
(v)
(vi)
(y-2x){y-4){y + x). See Ch. Ex. 1 would show that the conic through the points of contact of the three bitangents passes through the points of contact of one of the common tangents to the two ovals which is impossible.:
XX
8
e is
CIRCUITS OF A QUARTIC
365
*
where
a small negative constant.
5.
(iii)
(iv)
(x-y)(x + y-2)(x + y + 2).
Describe the nature of the quartic fg = +e#. g = 4x + y -5
/=
and
<j>
is
(i)
x+y-2.
2x2 + y*
=
(iv)
(v)
(vi)
(vii)
x2 + y*±4:X-5 2 2 =4.
[Use Ch. j/ + 4y
= 0. No triangle formed by three bitangents of the enclose two ovals external to one another. x + 4y' = 4. one inside the other.
8.
(iii)
+ y 2 =4.
.
2*s + y» +
l
=
l.]
.
.
0.
ideal. due to Harnack. They lie on 1/x + l/y+l/z = 0.
x (1876).
Ex. — \(1 — t 2 ). and similarly
for §§ il.
common
No
oval of a quintic can have
more than twelve
inflexions or
six bitangents. For §§ 5. 7.e. 12.
and no quintic has more
§ 9. A quartic with one real and two unreal cusps has one bitangent with real points of contact. d are unreal. The real inflexions lie on a conic. 7. Show that x'y + y'z + z^x = has exactly six real inflexions and four real bitangents. suppose the curve to be of § 4). we see that the quartic
consists of a single oval touched in real points
by three bitangents. relative to the tangents at A and B. or four of a.
A
curve of deficiency
B has at most D + l
This result. (1 + t.366
*
CIRCUITS OF
nature
-A
QUARTIC
circuits
XX
the
quartics
8
in
Ex. degree n with r odd circuits.
A
A
quartic with three real cusps has one bitangent which is conic meeting the curve in eight real points intersects the
bitangent. a bitangent such that the ovals lie on the same side of it) and two internal bitangents (the ovals lying on opposite sides).] Ex. In the latter case all the six common tangents of curve and conic are real.
.* is familiar if (Ch.
circuits. unreal.
Maximum Number
of Circuits. 1— *). since two odd circuits meet in one or more crunodes. 15 we may classify the various types of curve by considering the cases in which none. §§ 3-15. after putting z = 1. or lying on odd or even circuits. To ovals of a quartic external to one another have two external bitangents (i. X. c. Let the curve have s other double points. For § 9 project the curve into x^y 1 + ax 1 + by 1 + c = 0. 189.
&c. 14. For § 4 take A and B at infinity and consider the possible positions of the conic of § 4.
p. A conic meeting the curve in eight real points does or does not intersect the bitangent. § 3. Then
-D
D=
= i(w-l)(TO-2)-fr(r-l)-8.
Ex.
The fourth bitangent is x + y + z = 0. To two ovals of a quintic external to one another four tangents can be drawn. These odd circuits meet in at least fr(r — 1) crunodes. 1. 10. where
Ex.
[The quartic can be projected into a three-cusped hypocycloid or a cardioid.]
circuit.
8.0). b. 9.] fi + 1 = t' + 9 1. Armalen. To prove it in general. 6.
* Math.
of the
of
[The nature of the bicircular quartics may be obtained by inversion The nature of their bitangents is given by from the circular cubic. 11.]
Ex. Ex. 13 and 16. [Drawing the curve. Describe the Ch. The real inflexions are (1. two.
[Three real inflexions lie on the odd than fifteen real inflexions. Ex. XVIII. acnodes.
four being on one side and side of the line of inflexions.
3.
ovals.
= n(n-2) + l
which
is
impossible. First consider the case in which n is even.
=
Ex.] the cases in which
[To find
is
n-ic
=
=
'
:
'
. Then r is even.XX
9
MAXIMUM NUMBER OF CIRCUITS
It will
367
that.
least
Hence the n-io and (n — 2)-ic meet in
at
2{%r(r-l) + s}+r + 2(D-r+l) + (n + r-2) =n(n-2) + 2
which is impossible. with twenty-four inflexions and eleven ovals. For the case of n and r odd. e. 6. another point on each of the r odd circuits and n — 4 more points on an even
real points. Show that not every non-singular n-ic can be projected into a closed curve. If a sextic has eleven ovals. protectable into such a sextic).
Suppose
—
D—
—
(D-r+2)+%r(r-l)+8 + {n + r-4) = \(n + l) {n-2).
Let
(j>
+ 8^ + 32^ + 28)
(a. and all meeting the odd Then / 2 + ?0 are (1) a sextic circuit of the cubic in three points. the (n — 2)-ic meets each odd circuit in three points at least.
=
2. take one point on each of the even circuits. and n + r 4 more points on one of the even circuits. the Jr(r— l)+s double points. Take one point on each of the r + 2 even circuits.
(tf-tf-V.
. (2) a sextic with eighteen inflexions and nine ovals (i.
=
two on the other
=
all possible arrangements of the circuits of a non-singular a difficult problem. The total number of these points is
circuits. the curve had D-f 2 or more have at least JD r + 2 even circuits. which ha9 been solved only for re 3. is a sextic with eleven
!
!
O.
circuit. for 6 take
(a. for meets them each once at a crunode. 1. if n is a given number greater than 4.
2
it
has an odd circuit.
odd. if possible. the |r(r-l)+s double points.
[If
n
is
For n
Ex. they consist For n 6 we have the result of ten ovals all external to each other and an oval enclosing either nine as in the above example. The reader may discuss of them or only one represent six or less lines in various positions. the curve can never be closed.]
=
be the equation of
six lines parallel to the line of
inflexions of a two-circuited cubic / 0. This
each odd circuit of the n-ic at least twice.
2
+ 8y2 .
Since n — 2 is odd. so that it meets one of them in at least w+r — 2 points and the
it
(n— 2)-ic meets
others at least twice.
Prove that
where
f
{2^sin3(0 + S) + 3^-l}{8r sin30 + llr }+<r = and 8 are small and positive. Hence a (n — 2)-ic can be drawn through them. Also it meets each even circuit in au even number of points-. and n — 2 is even.32^ + 28)
=
e. 4. and the n-ic and (n — 2)-ic meet in at
least
2{%r(r-l)+s} + 2r + 2(D-r+l)+(n-3)
real points.
Ex.
ovals. (ii) n odd. a line cutting this other circuit
Fig. In this diagram we have the three nests abed.
Then. ?*<£ (n — 2).
a>
2
2
inside
co^. the nest and one odd circuit being the ivhole n-ic.
and the innermost oval would cut the
-n-ic
in
mofe than
n
points.
. as in Fig. we have : (i) n even. Then. abf. 12.
Suppose n odd.
if
the curve had
%n
nested ovals
and another
circuit as well. a line cutting this even circuit (or two odd circuits) and the innermost oval would cut the n-io in more than n points. In the following section we prove the converse of this theorem. the nest being the whole n-w.
o>
3
. 12. ae.368
NESTED OVALS
§ 10.
XX
10
Nested Ovals. or r = \(n — 1).
Suppose
n
even.
A
o>
set of ovals o^. showing that for an assigned value of n it is always possible to find a non-singular n-ic with its maximum number of nested ovals and its maximum number of circuits and real
inflexions. if the curve had %(n— 1) nested ovals and also an even or two odd circuits in addition. or r = \n. It will be noticed that a nest may contain ovals not belonging to the nest.j
inside
w r such that ooj lies inside <o 2 w r is called a family of r nested
. r^^(n — 3).
If an n-ic has r nested ovals.
number ^(n — 2)
%(n — 3)
q/
nested ovals.
.
(iii)
It has the
maximum. Suppose we have also a thick line) with equation e curve (shown by the thin line) with equation f=0..
The theorem may be considered to hold even in the cases 2 or 3. 13...:
XX
11
HILBERT'S
THEOREM
is
369
even. if we adopt the convention that a single oval shall count as a nest of one oval when n > 3 and shall not count as a nest when n = 2 or 3.
Fig. (a?s + y»-|{»-l})(3!-») = «=.
Ex. y"-i") = «.. Suppose that in Fig.]
§ 11.» + (a? + tf-l)(a* + y*-2) .as the
maximum number n(n — 2)
of real inflexions. such as is shown at the bottom of B b 2216
=
. which possesses a nest of p ovals.
0. Find the equation of an «-ic with \n nested ovals when n and | (n — 1) nested ovals and an odd circuit when n is odd. has the maximum
number
\n{n% — 3% + 4)
or
of
circuits.
(iv)
ii A.For every assigned value of n greater than 3 there exists n-ic tulth the following properties
(i)
an
It is
(ii)
It
non-singular.. IX^ + ^-l) (a!» + y»-2) .
Hilbert's
Theorem. 13 we have a circle (shown by the
n=
. (a.
.
circle. if e is a small constant of suitable sign. If we
take r
= n + 2. 13 is
purely diagrammatic. now only to establish the existence of the
n-\e
=
eg. meeting the circle in 2r points passed through in the same order
whether we traverse the circle or the oval..
the derived curve
an (% + 2)-ic with a nest of ^n or ^(n — 1) ovals and an oval such as 12 meeting the circle in 2 (n + 2) points.
e
=
is
a
and g
=
a straight line not meeting the three straight lines all meeting the circle. 2 the lines B2 Bs Bi . since each intersection of the circle and original curve gives rise to
is
two extra
It
inflexions
by
§ 6.
. with §(« 2 — 3n + 4>) circuits of which |(ti — 2) or § (n — 3) form a nest as in Fig. B2r _ 1 B2r be g 0.
. and an oval 12 meeting the circle in 2q points (only) passed through in the same order whether we traverse the circle or the oval 12. 13 and one is an in 2n points (q = n). On this arc of the B2r and let the equation of circle take 2r points B t . 13. and inflexions are shown in the figure
exist.
(n + 2)n
its circuits is
S!
<
and the number of
!(7i 2 -37i
with the nest. There is a portion of the plane not containing the nest.
. which is bounded by an arc of the circle.
=
=
=
=
a curve such as is showrj by the broken line in Fig. It has a nest of p + 1 ovals and has in all 2^—1 more ovals than the original curve. q Then. For n = 2 it is sufficient to take any ellipse meeting the For n — 3 we may take the curve circle in four real points. and the n(n — 2) inflexions for the cases n = 2 and n = 3.
.
h
=
* Fig.*
has also an oval of the same nature as the oval 12.
eh
We
have
+ 4) + 2%-l = -§{(% + 2) -3(7i + 2) + 4}.. the oval 12..
. It has too 4g more inflexions. r 2.
B
. and an arc of 12 lying inside the circle.
where
circle.370
HILBEET'S
THEOREM
XX
11
the diagram. oval such as 12 meeting a fixed circle e = Suppose also the two has n(n— 2) real inflexions. The result will then follow by induction. The derived curve has also
is
i
n(n— 2) + 4% =
real inflexions. 3. Supposing now we have an n-io / = satisfying the conditions of the theorem.
B
.
which do not really
. 1 In the diagram p 2.
319 Rendiconti del Reale Istituto Lombardo. xlix (1916). p. III. lxxvii.
where
a.
(y-lx). p.
2. having its of circuits. cxiv. .
lxiii. p. The r lines g meet e in »• points. p. + y'i)(y-ax)(y-cx)
1. Berichte der K.
and use
induction. xxii.]
.. x. 349. and has a circuit meeting e in r points only which are passed through in the same order whether we traverse the line or the circuit. pp.:
XX
_
11
HILBERT'S
may
consult
THEOREM
371
The reader who wishes
to pursue further the subject of
•
circuits
American Journal Math. xxxviii. II. xlviii (1915). 177. Annali di Mat. 245 xxix. Soc.
i. New York Math. p. k.. 189. Crelle.
. Pura ed Applicata. . Annalen. p.
p. 489 and 797. such that the two segments containing the q points and the r points on e do not overlap. p.
A
/=
=
=
=
=
=
=
/=
non-singular ra-ic exists for any given value of n.
xli.. p.
Math. zu Leipzig. 495 and 577.
540.
Bb2
.
Ex. vii. der Wiss. p. 182 Trans. American Math. r = n + l in Ex.
. I
is
magnitude. b.. p. lxvii. xliii (1910). has q — 1 more circuits than 0. xlvii (1914). Soc. iii. 218. p. e. pp. 416. 48 and 143. Sachsischen Gesell. 197. with a circuit meeting a given line in n points passed through in the same order whether we traverse the line or the circuit. p. 305. p.
A
maximum number
[Take
Ex. 126. 117. Bull..
3.
q=
(x i
n. pp. tg.. d. p. p. xiv.
are 2n — 4 constants in ascending order of an n-ic of zero deficiency with a single circuit of index n — 2.. 170.
Ex. 1. 388.(tj-kx)
= y {y~bx)(y-dx)
... where e is a small constant of Prove that ef suitable sign. lxxiv. 410. lxxiii. lxix. p. 115. curve has a circuit meeting a line e in q points (only) which are passed through in the same order whether we traverse the line or the circuit.
.CHAPTER XXI
CORRESPONDING RANGES AND PENCILS
§
1. and that the curve lias no multiple point other than A and B. if certain relations hold between the coefficients of equation (i).
—
A
^
m
P
AB
* Assuming that the tangents at A and B are all distinct. VIII.
Correspondence of
Two
Pencils. D = (p-1)(?-1) The tangents from A to the curve
+ q 2 -p-q).
Suppose we take two lines PA.
To every
PB PA
PA
B
PB
A
Eliminating t and T between (i) and y tz.l + . remembering that a &-ple point counts as %h(Jc— 1) nodes (Ch. x Tz we It is a (p + q)-ic with multiple points obtain the locus of P. Pliicker's numbers are at once written down. The to every position of are said to trace out 2^>encils with vertices and lines and having a p:q correspondence.
0. This assumption may not he valid.+u p + n-' (b fP + b tP. + (Jc tP + k tP. These facts are obvious either from the equation of the curve or from simple geometrical considerations. If a = 0.
.. In (i) we have supposed a 0. We have *
=
=
A
A
..
. We verify easily from this that = 2pq. The locus of is now
= 2 (#V..
S
=
i (p
c
2
are the p tangents at each counted twice and the 2(q l)p lines given by those values of t which make (i).
position of correspond p positions of PA. have equal roots. tz. at A and B of orders p and q respectively. § 3). the line corresponds to itself in the two pencils.+kp ) =
1
)
1
1
.. considered as an equation in T.+bp ) + . such that t and Tare connected by a relation of the form
PB
=
—
B
Ti(a
tP
+ a htP..
(i).
n=p +
r
q..l + . k — =3(2pq-p-q). x Tz respecy of the triangle of reference tively through the points A and ABC. and correspond q positions of PB.1 + . The p tangents are the p lines of the pencil with vertex at which correspond to the line BA with vertex B and similarly for the tangents at B.
m=2pq..5z>q + 2p + 2q).
o
.
I)
is
=
(p-1) (q-1)
\
very well known. any %-ic with an r-ple point and an s-ple point B * may be considered as the locus of P._ s _ 2
. The case p = q = 1
).
We
have then
that the intersection of corresponding rays of two homographic pencils is a conic.
.. the lines and joining any point on a (p + q)-ic to multiple points and B of orders p and "q respectively have a p q correspondence.
.
. the locus of P is similarly a (p + q — 2)-ic with (p — 2)-ple and (2 — 2)-ple points at A and B respectively
p
=
q
1
and so on.
o n _.
The polar reciprocals of two pencils with a p:q correspondence are ranges of points on two lines with a p q correspondence.] T=2(p 2 q*-7pq + 2p + 2q + 4 [.
!
n=p + q-l.
o1
.
We readily show that.
..
a1
.]
Correspondence of
Two
Ranges. conversely. This is evident on putting t for y/z. To any point of the first range correspond 2 points of the second. r
=
1 if
A
is
an ordinary point
of the
curve
and
so for B. when we get a relation of the form (i) with
PA
P
PB
A
:
=
A
PA
PB
BA
AB
a
all zero.. = S(2pq-p-q-l). We have a relation such
:
* r
=
.
.
if
A
is
not on the curve.. T for x/z in the equation of the curve.. If a — a = b = 0..
Bb3
. More generally. In fact. when the pencils traced out by and have a (n — s):(n — ?•) correspondence such that n — r — s of the lines through A corresponding to coincide with AB. and n — r — s of the lines through B corresponding to coincide with BA. or is a straight line if the line joining the vertices of the pencils corresponds to itself. and to any point of the second range correspond p points of the first. putting y = tz and x Tz in the equation of the curve we have a relation of the form (i)..
Ex.
XXI
2
CORRESPONDENCE OE TWO RANGES
1
373
the line point at
AB together with a (p + q — l)-ic having a (p — l)-ple A and a (q — )-ple point at B.
ci n
_ r_ s _ 1
. For this cuvve
m=2(pq-l).
K
i
= Q.. The quadratic transform of the (p + q)-ic of § 1 with respect to a conic touching CA and CB at A and B is the intersection of pencils through A and B having & q:p correspondence. The reader may also verify the results obtained by taking p = 2.. q 1 or
= = 2.
A
is
a line through B. 5 = i (p 2 + g 2 -3p-3g + 4).
.
[The quadratic transform of a line through
§ 2.
.
a node at B ]
:
The
locus
is
Ex. B. The line joining corresponding points of two ranges with a p:q correspondence in general envelops a curve of class p + q having the lines on which the ranges lie as p-ple and g-ple tangents respectively. 5.]
:
AB
Ex. § 2.
BC
being
self-
. to a given family of confocal conies. Two pencils with &p:q correspondence meet any two lines
p q correspondence. if t and T are the distances of corresponding points measured from fixed origins on the two lines.
[The pencils
Ex. Two ranges on the same line with p q correspondence have p + q self corresponding points. conic through intersection of
C
are fixed points. (ii) is.
and B have a 2 2 correspondence in which The locus is a cubic through A and B. one of the
given lines.]
Through A and C are drawn conEx. A.
A
A
[The pencils
Ex. The tangent at locus of the intersection of BQ and CP.374
CORRESPONDENCE OF TWO RANGES XXI
2
as (i) of § 1. (ii) A (2m — 2)-ic with (m — l)-ple points at the circular points. B. 1.
AP and BQ have
a
1
:
2 correspondence. B. Find the locus of the foci of ftrves of the m-th class touching given lines when the line at infinity (i) is not. Find the locus of the intersection of BP and CQ. Find the locus of the intersection of tangents from two fixed points A. 3.
BQ
and
is
CP have
a
1
:
2 correspondence. C are fixed points.
*
Ex. A. jugate chords of a fixed conic through A and B meeting the conic in P and Q respectively. Two pencils with a common -vertex and p q correspondence have
: :
p+q
self-corresponding rays.
The [(i) A (2w — l)-ic with (m — l)-ple points at the circular points. Consider the correspondence of the tangents to the curve from the circular points.
B
[The tangents from
is
A
self-corresponding. C to touch meets I at Q. and the pencils formed by joining any two vertices respectively to the points of ranges with a p:q correspondence have themselves a p q
in ranges with a
: :
correspondence. C.
7.]
conic is drawn through three fixed points A. 5. Find the a fixed line I at any point P. B. V.]
drawn through four fixed points A. D and meets a fixed line through A in P and another fixed line in Q. 2.
through A meets a fixed Find the locus of the
have a
therefore a cubic through
C with
1 2 correspondence.
Any
line
[The
B in P and BP and CQ.]
BP
and CQ have a
1
:
2 correspondence. singular foci of the locus lie on the locus. Ex.
A
conic
[The pencils corresponding. pencils CQ and BP
a fixed line in Q. See also Ch. Find the locus of the intersection of AP and BQ.
Ex.
6.
m— 1
4.
H
[The tangent and
12. B. Ex. 9. C. <p (x) are polynomials of degree n.
=
k<t>(x)
and
are said to be projective. D. How many circles can be drawn through a fixed point A touching a given line at P such that P and the intersection Q of another ? fixed line with the tangent at A are collinear with a fixed point
[OQ and OP have a
1
:
2 correspondence. F are fixed points. Find the locus of the intersection of CP and DQ. C. Any conic through A. J are fixed points. B. variable conic touches fixed lines at A and B. 13. that the involution-range has 2(» — 1) double points. 11. H. D.]
Ex. 8. 10. B.
Discuss
Ex.
IP have a
1
:
3 correspondence.' A similar definition holds for pencils.]
:
Ex.ndEQ. F. 14.]
:
is
6. G. 0.]
: '
Draw a circle Ex.
Therefore by Ex. F.
[The pencils
EP and FQ
have a 2 2 correspondence. C. A.]
Ex. D meets two fixed lines in P and Q respectively. How many solutions are there to the problem through two fixed points A. Show that two projective involution-ranges self-corresponding points and that the line on the same line have n + joining corresponding points of two projective involutions along different having the lines containing the lines envelops a curve of class n + Show also that involutions as «-ple and IV-ple tangents respectively. A. Q collinear
with a given point
'
?
[OP and OQ have a answer is 2 + 2=4. E.
The two involutions
f{x)
of degree
n and
F(y)
N given by the equations
=
&'*
{y).
E
A
[The pencils
DP and EQ have
a 2
:
4 correspondence. fixed conic passes through C.
Therefore the answer
is
1+2 =
3. How many solutions are there to the problem Draw a circle through two given points to meet two given lines in points P.
N
. Any cubic through A. D. C. G. Find the locus of the intersection of the tangent at A and the line IP. A common tangent of the two conies meets this tangent at C in Q. Find the locus of the intersection of EP and FQ.
D
A and B P and Q
[The pencils
CPand DQ
have a 2 :2 correspondence.]
' :
Ex.XXI
2
CORRESPONDENCE OF TWO RANGES
375
Ex. If x is the distance of a point on a given line from a fixed origin and f(x).
1
the
Ex. and meets the tangent at C to the fixed conic in P. A. E. A. B meeting a given conic at P and a given ? line at Q so that PQ meets the conic again at a given point
'
[OP and OQ have a 4
the case in which
The answer 2 correspondence. Find the locus of the intersection of DPa. are fixed points Tangents are drawn from to any conic of a given confocal family to meet a given line in respectively. D. if k and k' are connected by a relation of the form akk' + bk + ck' + d = 0. meets a fixed line at P. 15. are fixed points. a corresponding result holds for pencils. E. B. C. B.]
2
:
2 correspondence.
16. the group of n points given by f(x) — k<p(x) is said to trace out'an involution-range of Show degree n as k varies. the given line passes through A.
N
. B. A.
Suppose Q. (p 2 '»)• Show that. If and classes m.)
m
:
. Now by a series of quadratic transformations we may transform a given curve successively into curves with less and less complex singularities.
AQ. The are found by eliminating k from intersections of the locus with y = and /2 (a:. IV. k) = 0. till we obtain a curve with no singularities other than ordinary multiple points with distinct tangents. § 8.376
COKRESPONDENCE OF TWO RANGES XXI
.
may use our knowledge of the locus of the intersection of corresponding rays of two pencils to establish the theorem
:
We
If the points of two curves have «1:1 correspondence.
R
and
in PThe locus of is an algebraic eliminate the coordinates of Q and from the rational equations connecting them.
2
Ex. 0. while the coordinates of P' can be expressed rationally in terms of P" and vice versa. k) =
§ 3. p x and p 2 in k. y. then the curves have the same deficiency. characteristic n x and n 2 . guch that the coordinates of P can be expressed rationally in terms of P' and vice versa. y' may be expressed rationally in terms of x. the curves have the same deficiency. IX. Zj) families have a 1 1 correspondence. fc) = 0. the locus of the intersection of corresponding curves is in general of degree n l p 1 + n l p 1 and of class mjZj + t l 1 . Now we have proved (Ch. thus getting an algebraic eliminant which is the equation
let
. if to each point P with Cartesian coordinates (x. y and vice versa. Hence it suffices to prove the theorem for two curves with ordinary multiple * points with distinct tangents only. then the coordinates of P can be expressed rationally in terms of P" and vice versa. Two families of curves have respectively degree. k) = where/! and/2 are of degrees » x and n 2 in x and y. (x. It will suffice therefore to prove the result for any two curves into which the given curves may be transformed by a series
:
of quadratic transformations. This means that. §§ 1. It is at once seen that. P'.
m
[Corresponding curves have equations
and /z (x. /x (x. class. B. m^. 0. Take two fixed vertices A. M. from the equations of the given curves.] /. Bit meet
P
curve
for
we may
R
BR
. and from the equations of the lines and AQ. if the lt (p 1 . (See Ch. P" are taken on three curves.
Curves with a One-to-One Correspondence. if points P. y. 8) that two curves derived from each other by quadratic transformation have a 1 1 correspondence and have also the same deficiency. 17. corresponding points on two such curves of degrees n. y) on one curve corresponds a point P' with coordinates* (x\ y') on the other so that x'.
For two of the lines corresponding to such a position of coincide. xvi (1888). 132.y) = 0.
Ex. Similarly for y. Hence and have an correspondence. we express x rationally in terms of f. the two pencils of planes in two pencils of lines with a p q correspondence. y) of a given curve f(x. we pointed out that the coordinates (f.) of a point on the evolute corresponding to a point (x. making use oif(x.
A
curve and
its
reciprocal have the
same
deficiency.
[See Ch. so we may have two pencils of planes with p q correspondence. 3. 1).
and q are t t
a* +
Steinerian.
M—
(Ch. § 2.
:
:
. and have therefore the same deficiency. de la Soc.y from these equations. Hence the class of the locus of Similarly it is 2n + M. §3. 77 by eliminating. Ex. de France.VIII. the
establishes a 1 1 correspondence [The transformation x = £ 9 y = between the given curve and xP-i-yp +l = Q which is non-singular of degree p. ?j. Hence a curve and its evolute have a 1 1 correspondence. For example. Conversely. VIII.
tf+\
=
Q
is
\{p-l){p-2). Bull. § 1
2.
same
deficiency.
. For otherwise (£.]
§ 4.
:
^
p. is Therefore in—2n + 2 and 2iV+2 are equal. Jamet.
Ex. and Cayleyan of a curve have the
relatively prime integers.
:
Ex. The tangents from to the locus of are the lY tangents at each counted twice. ?.
:
Just as in § 1 we had two pencils of lines with p q correspondence. or it may be proved independently as in §1. together with the lines through touching the locus of Q. with and B as if-ple and %-ple points. XI. y) = are expressible rationally in terms of x and y by means of equations (i) of Ch. Math. To eacli position of correspond n positions of BR. when we solve for x in terms of £.
Correspondence in Three Dimensions. the planes of each pencil passing through a given line called the axis of the pencil. and the locus of is an (n + iV)-ic.
(vi).y) = 0. See V. If p deficiency of
p
being positive.
1.]
The Hessian. But these are twice the deficiencies of the curves since they have no cusps
AQ
BR
AQ
BR
P
N:n A
A
P
A
m
A
BR
AQ
P
2N+m. Corresponding planes meet on a ruled surface of degree p + q having the axes of the pencil as p-ple and g-ple lines respecThis is evident from the fact that any plane meets tively. which is not in general the case. and to each position of correspond impositions of AQ. 77) would be the centre of curvature at more than one point of f(x.XXI 4 CORRESPONDENCE IN THREE DIMENSIONS
377
of the locus of P.
and let any plane II meet the generators through in A and B. But these tangent planes are the planes joining I to the generators through the intersections of I with 2.'. one on each of two given twisted curves.
Reciprocating.
The (p + q)-ie of § 1 may be employed to study algebraic curves on a conicoid. these two generators not being counted as part of the curve of inter-. If the coordinates of two points P and P. and a generator of the same family as OB as an (n— £>)-ple line. 1 is well known. Suppose that the curve on j we are investigating is the intersection oi' j with a surface of degree n having a generator of the same family as OA as an (n — <7)-ple line.
P
Curves on a Coniooid. the line joiniDg corresponding points of two ranges on non-intersecting lines with a p:q correspondence generates a ruled surface 2 such that p + q tangent planes can be drawn to 2 through any line I. and the p + q selt-corresponding planes give the generators of the surface which meet I. P and spondence.* Let be a point of the conicoid. the whole or partial intersection of the conicoid with any algebraic surface. which does not pass through 0.4 378
CORRESPONDENCE IN THREE DIMENSIONS XXI
If the axes of the pencil intersect at 0.
for
. that is. then the line traces out an algebraic ruled surface of degree p + q. Then OA meets the curve in p points all projecting into the point A.
* Every twisted curve of the third or fourth degree is of this nature a conicoid through nine points of the curve must contain the curve.]
§ 5. Moreover.
. and OB meets the curve in q points all projecting into B.. Similarly the other family of generators project into straight lines through B. Consider the projection from on to IT of any curve on j. Any generator of the same family as OA
. section. are connected by rational relations so that p points P correspond to each position of and q points to each position of P. p:q corre[The planes joining any given line I to.
PP
P
P
have a. the reciprocation shows that the lines on which the ranges lie are g-ple and p-ple lines of 2. The case p q
=
=
Ex. All generators of one family on j intersect OA they therefore project into straight lines through A. so that these intersections are p + q in number. We see then that the line joining corresponding points of two ranges on non-intersecting lines with a p q correspondence generates a ruled surface of degree p + q with the given lines
:
as g-ple and p-ip\e lines respectively. the ruled surface
becomes a cone with vertex 0.
we obtain a (p + q — l)-ic with J.
through 0.
= =
V
OA
OB
P
VP
P
VP
A
B
AB
V
A
B
AB
A
B
envelope. and the tangent at P tangent at to the projection lies in the osculating plane of the curve at 0.e. The harmonic envelope is the envelope of a line divided harmonically by the conies.XXI
5
CURVES ON A CONICOID
379
meets the curve in p points. the into the tangents at pole of the plane of a conic onj projects into the pole of with respect to the projected conic. Hence the projection of the curve is a (p + q)-ic with . if the planes of two conies onj are conjugate. If OA and OB meet the curve respectively in p — 1 and q — 1 points other than 0. Hence the lines and VQ project planes at and to the projected conic.*
We have so far supposed that the point does not lie on the does lie on this curve. This is clear on putting^ 1.A as £>-ple point and B as (/-pie point. q Let be the pole of the plane of a conic on j. in other words. and the harmonic locus is the locus of a point from which a harmonic pencil of tangents can be drawn to
the conies. the projections are two conies with degenerate harmonic locus and and through
A
A
B
.
vice versa. If we project on to the plane II. the generators given twisted curve. as (p — l)-ple point and B as (q — l)-ple again at a point P on the The projection meets point. If the plane of any other conic on j passes through V. which meets and in and Q. Each inflexion of the plane curve is the projection of a point of the twisted curve at which the osculating plane passes
. or otherwise. Projecting we see that. Then and VQ are the tangent and Q to j. The tangents at A to the projection lie in the planes through A touching the twisted curve where it meets OA and so for B. conic on j projects into a conic through and and conversely. the polar of with respect to it is the line of intersection of its plane with the polar plane of V. when the conies become orthogonal circles. to the given twisted curve.
. A twisted curve on j meeting all generators of one family in points and all generators of the other family in q points
AB
p
* This is obvious on projecting A and B to the circular points. and projects into a line meeting the projection of the curve q times at B and p times elsewhere. Similarly for a generator of the other family. Every property of the plane curve gives a property of the twisted curve on projecting back on to the conicoid j and
. i. their projections for and such that the pole of are two conies through one conic is the pole of their other common chord for the second conic .
Q. For since a twisted cubic meets any conicoid not containing it in six points. 9.
3. 5.
[Projecting from
we huve
Ch. Projecting from A.]^
Ex. XIII. otherwise of type 3 + 1. With any point as vertex three cones of the secon°ree can be drawn having six-point contact with a given twisted cubic. The points of contact of the three osculating planes of a twisted cubic which pass through any point lie on a plane through 0. The tangent planes to the cone at the points of contact pass through the points of contact of the osculating planes through 0.second degree.
may
Projecting from [The cubic lies on a conicoid through 0. 5 the pencil is harmonic.
Ex.
Ex. we have the fact that the pencils of tangent planes through and have the same Similarly project from C for the chords CA and CD. and the cross-ratio of the pencil of planes
is
constant. it is of type 2 + 2 . Similarly a twisted quartic lies on the conicoid through any nine points of the curve. 8. the pencils AB{PQRS) and AC(PQRS) have the same cross-ratio. Any twisted cubic curve is the partial intersection of two conicoids with a common generator and is of the type 2 + 1.
[Project from 0.
[Project from a point
on the
line. The planes joining four given points P. If in Ex. Nine osculating planes of a curve of type 2 + 2 pass through a given point on the curve. we see that [Let AB and CD be two chords. 4.
Four tangent lines of a given twisted cubic meet any given
line in space.380
CURVES ON A CONICOID
XXI
5
be called a curve of type p + q on the conicoid. § 4. it must lie wholly on all conicoids through any eight points of the curve. have The three inflexions of a plane nodal cubic are collinear. §
4.]
Ex. XV11I.
S
of a twisted
Projecting from A. The tangent planes through any point to a curve of type 2 + 2 at the points where it meets the generators through of a conicoid containing the curve meet the curve again in four coplanar points.
[See Ch. 7. the tangent at osculating plane at in a point on the plane RQS.
P
meets the
R
[Project from R. R. 3 + 1 are well known. Ex. Through the line joining a fixed point to any point of a twisted cubic two tangent planes are drawn to the curve touching at Q and R.] cross-ratio.
[Let and CD be two chords. The plane through and any two of the points of contact passes through a third point of contact. Similarly project from C for the pencils joining CA and CD to P. 2.
AB
AB
AC
Ex. S.']
:
'
we
Ex. R. 6. 2 + 2. 13
(iii). Ex.
l.]
Ex.]
Ex. Through any variable chord of a curve of type 2 + 2 four tangent planes can be drawn. Q.]
Ex.J
. cubic to any chord form a pencil of constant cross-ratio. If it lies on a second conicoid. Show that the plane OQR envelops a cone of the. The curves of type 2+1. 1.
The properties of conies on j proved in § 5 now become the well-known theorems that a cirqje on j and the pole of its plane project into a circle and its centre. If we take n as the diametral plane parallel to the tangent plane
.
m
m
. The a curve on j. it has S nodes and k cusps curve is of degree n and class besides a ^Ti-ple point at each circular point. The line of striction of one family of generators of a hyperboloid is a curve of type 3 + 1. r great circles are bitangent to the Then the projected curve.
If the conicoid j of § 5 is a sphere. 10. and that two circles onj whose planes are conjugate project into orthogonal circles.]
Ex. all great circles project into circles orthogonal to a fixed (unreal) circle which
OB
A
B
the projection of the circle at infinity. The generators OA. while it has t bitangent circles and i osculating circles which are orthogonal
.
§ 6. the projection becomes the well-known stereographic projection. become the circular lines through in the tangent plane at 0. Suppose a spherical curve df even degree n has 8 nodes
is
We
F
F
F
great circles can be drawn through any and k cusps while point to touch the curve. generators through Hence the stereographic projections of the foci of a spherical curve are the foci of the projected curve. 11. These results also follow from the fact that a spherical curve are inverses of each and its stereographic projection from other with respect to 0. Show that the plane OQR envelops a cone of the second degree. Ex. and Pis any other point on the curve. project into lines through <o and to'. while and become the circular points a> and a/ in IT./ touch If the generators through a point may be called a focus of the curve. on the sphere. is a fixed point on a curve of type 3 + 1. 12.
. we must have p = q when the curve onj is real and the degree of the curve is even.
at 0. § 8. Ex. XVII.
to
any
fixed circle. The tangent planes through OP to the curve touch at Q and R.
Curves on a Sphere. and l great circles osculate it.XXI
6
CURVES ON A SPHERE
1
381
by projecting
Ex. In particular. The three points of contact of the osculating planes of a curve of type 3 + 1 which pass through a point of the curve lie on a plane through 0. Obtain properties of a twisted curve of type 3 + from any point of the conicoid containing the curve. may show similarly that arjgles are unaltered by stereographic projection. 13.
[See Ch.
B.
Ex. 5. C. § 11. A chord PQ of a circular cubic subtends a right angle at a fixed point of the curve.
See also Ch. C.
3. Ex. if the line traces out a locus on the sphere. (iv) A self-inverse spherical 2re-ic is the intersection of the sphere with a cone of degree n.
whence we get
(ii)
and
(iii).
2
-2<5-3k.
A
AB
[The tangent-arcs to a sphero-conic from a point P make equal angles Project from the point diametrically with the focal distances of P. If are inverse with respect to the circle. BPD. Show Ex. so pole of the plane of.
Show that
foci are concyclic
:
may be
bicircular quartic has real concyclic foci A.
Ex. AQC. passes through the respect to a circle on the sphere.
2. general a spherical 2ra-ic of the same type (with the same Pliicker's numbers). See also Ch.
Obtain properties of the spherical
4-ic
from those of
tb^e
plane
bicircular 4-ic.]
Ex.the circle.
.382
CURVES ON A SPHERE
results of Ch. Two bicircular quartics with the same four real concyclic foci cut orthogonally. D. P and Q are two points on a bicircular quartic with real concyclic foci A. does P.]
Ex. 5.
Ex. It follows that the stereographic projection of a circle and two inverse curves on the sphere is a circle and two inverse
curves
. (ii) The angle between two spherical curves is equal to the angle between their inverses with respect (iii) The inverse of a spherical 2m-ic is in to any circle of the sphere. opposite to P.
§ 6. § 3. Ex. Show that the circle through PQ bisecting the circumference of a fixed circle with centre passes through two fixed
points.
'
P
'
PP
P
P
'
'
[For (i) use the fact that the circle is cut orthogonally by any circle through the points. n = m (m — 1) — 2t — 3i.
as Pliicker's equations for a spherical
may be regarded
on a sphere are said to be inverse with Ex. BQD are all touched by the same two circles.
properties of a plane bicircular quartic whose real obtained from the properties of a conic as follows Project the conic on to a sphere from the centre. 1.
second degree.
Ex. 7. Two points P.
[Two confocal sphero-conics cut orthogonally. that the bitangents to the curve from the intersection of and CD make equal angles with OAB. i = §n(n-2)-68-$K. 6. D.
6.]
The foci of a spherical 4-ic lie by fours on four circles with respect to which the 4-ic is self-inverse. 8. Show that the four circles APC. XVIII. XI. Then project the spherical curve thus obtained (a sphero-conic) stereographically into a bicircular quartic.
XXI
we
get
6
Looking at the
m = f™
These
curve. B.
4. and the loci of jP and Show that (i) the stereographic projection of a circle and two inverse points is a circle and two inverse points. The 4-ic is the intersection of the sphere with four cones of the
Ex. XVIII. OCD. 1.
and that their points of contact lie on a circle through O. section is a circle. The locus of a point on the Earth's surface at which a given place has a given azimuth is a spherical quartic.
(sin
and
foci S. 16.. The spherical quartic has a node point on the curve. Q in
.
cosec \ OS') cosec \ OP
is
is
constant. 4
applicable to any curve
which
is
self-inverse with respect to a circle.
sin
1ST'
. VII.
cosec |
OS± sin ±S'P
. find the locus of P.. Through a given point If drawn meeting a given spherical 2»-ic in Q lt Q2 .XXI
6
CURVES ON A SPHERE
383
Ex.
[Project stereographically from the point diametrically opposite to
and use Ch. pp.
.
sin|O02
sin
\OQr
.
Show
that.
Show
that the method of Ex. 1873).. 15.
cosec \ PQi
. S'.
if
P is any
Ex.
[Projecting stereographically from collinear inflexions. the plus or minus sign being taken according as acnode or crunode.
%SP
.]
2n C. A circle spherical quartic has a node touching the curve is met again by the orthogonal circles Show that Y and Y' lie on a fixed and 8. S'.
in Darboux's
[The reader will find a very interesting discussion of spherical quartics Sur une classe remarquable de courbes et de surfaces alg4briques
(Paris.
cosec
\OY cosec|Or
.
[Projecting stereographically from is a hyperbola.']
we have
is
:
'A cubic has three real
Ex. 13.
cosec
|P§2
cosec \PQ r
summation extending over
§ 1.. Show that the two properties are obtainable from each other by ordinary inversion. Two properties of a plane curve are derived by projecting a given spherical curve stereographically from two different points of the sphere.
is
constant. 17.
drawn through the node The locus of their second
of a
inter-
Ex. Y'
A
and that sinJSr.
Ex. The locus of their second intersection is a spherical quartic with a node at 0. 18. Two circles are drilwn through the cusp of a spherical quartic touching the curve and cutting each other at a constant angle.
and loci S. A spherical sextic has a triple point 0. of a sphere any great circle OP is Ex. and S' at Y.]
Ex. terms).
O = 2sinJ0&
(the
. Two orthogonal circles are spherical quartic touching the curve. a parabola
Ex. 9. 1-60.
Two
spherical quartics with a
common node and
foci cut
orthogonally. 12.
. 11.
Ex.']
we have
' :
The
isoptic locus of
through through
circle. 14. 10. Show that three real circles of curvature of the sextic pass through 0.
an
[Projecting stereographically from
we
project the curve into a conic]
Ex. | 677.169 | 1 |
Summary and Info
Minimal prerequisites make this text ideal for a first course in number theory. Written in a lively, engaging style by the author of popular mathematics books, it features nearly 1,000 imaginative exercises and problems. Solutions to many of the problems are included, and a teacher's guide is available. 1978 edition.
More About the Author
Underwood Dudley (born January 6, 1937) is a mathematician, formerly of DePauw University, who has written a number of research works and textbooks but is best known for his popular writing. | 677.169 | 1 |
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Unformatted text preview: UNIT 5 VECTOR SPACES INTRO: This Unit is on vector spaces and on the dot product of a pair of vectors in the vector spaces R 2 and R 3 . First, though, some preliminary remarks. 1. Abstract Vector Spaces Nearly all students will have already been exposed to the notion of a vector, perhaps defined as a quantity with magnitude and direction, perhaps defined as a pair of numbers ( a,b ) or a triple of numbers ( a,b,c ) with rules for adding and multiplying by constants. These are called the vector spaces R 2 and R 3 . Because these two vector spaces can be closely idenified with physical two dimensional and three dimensional space, we are comfortable saying that these vectors have magnitude (length) and direction. Actually, the definition of a vector is much more encompassing. To begin it is necessary to define the entire set of vectors V , called the vector space, and to define how to add and to multiply by constants the elements of the set. Any element in the set V defined as the vector space is called a vector ~v V . The small arrow above the vector v is unnecessary; it is used simply as a device to remind us that ~v is a vector. In general, a vector space may have no concept of direction and no concept of magnitude. For that reason, defining a vector as an object having magnitude and direction would be misleading. For example, in many technology areas complex systems are described using a vector space of functions, in which the functions will be the vectors and there may be no connection with a notion of direction. We, however, are going to be working exclusively with the two vector spaces R 2 and R 3 , so there will be no drawback to thinking of vectors as objects with magnitude and direction, at least in this course. For further information on vectors in more general settings, please refer to the Appendix of this Unit. 2. Vectors in R 2 and R 3 In many elementary courses in linear algebra, vectors are written either as a row ( a 1 ,a 2 ,...,a n ) or as a column a 1 a 2 a n of n real numbers, and the set of all such vectors of fixed length n is called the vector space R n . We shall be interested only in the cases n = 2 and n = 3, ie, only in R 2 and R 3 . Editors (and we) prefer to have such vectors written as rows in textbook explanations, since it takes up substantially less room in printing a paragraph, although if the vectors are to be multiplied in the normal way by a matrix, then it is important to write them as columns. Since we will not be carrying out matrix multiplication, we will generally display our vectors as rows. Except for operations with matrices, the distinction between row vector and column vector is of no importance. For additional clarity, many textbooks surround the vector with brackets < > and, if given a letter name, place an arrow above. So, ~v = < a,b > is in the vector space R 2 , while ~v = < a,b,c > is in the vector space R 3 . This enables the author to distinguish between a....
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CliffsNotes Basic Math & Pre-Algebra Quick Review, 2nd Edition
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Inside the Book: Preliminaries Whole numbers Decimals Fractions Percents Integers and rationals Powers, exponents, and roots Powers of ten and scientific notation Measurements Graphs Probability and statistics Number series Variables, algebraic expressions, and simple equations Word problems Review questions Resource center Glossary Why CliffsNotes? Go with the name you know and trust Get the information you need-fast! Master the Basicsa€"Fast Complete coverage of core concepts Easy topic-by-topic organization Access hundreds of practice problems at CliffsNotes.comExample 11: What is the area of the triangle shown in Figure 9-2? Figure 9-2
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Title
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CliffsNotes Basic Math & Pre-Algebra Quick Review, 2nd Edition
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Summary
Reinforces student understanding and aids in test preparation with detailed explanations, worked-out examples, and practice problems. Lists key ideas to master and builds problem-solving skills. Includes worked solutions to the odd-numbered problems in the text. | 677.169 | 1 |
Intended Outcomes for the course
Upon successful completion students should be able to:
Analyze real world scenarios to recognize when derivatives or integrals are appropriate, formulate problems about the scenarios, creatively model these scenarios (using technology, if appropriate) in order to solve the problems using multiple approaches, judge if the results are reasonable, and then interpret and clearly communicate the results.
Appreciate derivative and integral concepts that are encountered in the real world, understand and be able to communicate the underlying mathematics involved to help another person gain insight into the situation.
Work with derivatives and integrals in various situations and use correct mathematical terminology, notation, and symbolic processes in order to engage in work, study, and conversation on topics involving derivatives and integrals with colleagues in the field of mathematics, science or engineering.
Enjoy a life enriched by exposure to Calculus.
Outcome Assessment Strategies
Demonstrate an understanding of the concepts of derivatives and integrals and their application to real world problems in:
at least two proctored exams, one of which is a comprehensive final that is worth at least 25% of the overall grade
Consistently demonstrate proper notation, documentation, and use of language throughout all assessments and assignments. For proper documentation standards see Addendum.
Demonstrate an ability to work and communicate with colleagues, on the topics of derivatives and integrals, in at least two of the following:
A team project with a written report and/or in-class presentation
Participation in discussions
In-class group activities
Course Content (Themes, Concepts, Issues and Skills)
Context Specific Skills
Students will learn to use the first and second derivatives of a function to find extreme function values and to solve applied maximum/minimum problems.
Students will learn the formal definition of the definite integral and several estimation techniques rooted in this definition.
Students will learn to antidifferentiate function formulas and use the Fundamental Theorem of Calculus to evaluate definite integrals.
Students will learn to model and solve several types of applications using definitive integrals.
Students will learn to evaluate indeterminate form limits using L'Hospital's Rule.
Learning Process Skills
Classroom activities will include lecture/discussion and group work.
Students will communicate their results in oral and written form.
Students will apply concepts to real world problems.
The use of calculators and/or computers will be demonstrated and encouraged by the instructor where appropriate. Technology will be used (at least) when graphing curves, evaluating derivatives, and evaluating definite integrals.
Competencies and Skills
Applications of the Derivative
The goal is to use the first and second derivatives to analyze the behavior of families of functions.
Use the first derivative to help find absolute extreme function values over a closed interval.
Use the Second Derivative Test to classify the behavior of a function at appropriate critical points.
Solve applied problems involving optimization.
Use L'Hospital's Rule to evaluate limits.
The Antiderivative
The goal is to find the antiderivative(s) of a function expressed in graphical or symbolic form.
Draw the family of antiderivative curves given the graph of the derivative.
Estimate values of an antiderivative given the graph of the derivative and initial conditions for the antiderivative.
Utilize the rules of antidifferentiation.
Antidifferentiate power, exponential, and trigonometric functions.
Antidifferentiate a function using substitution.
Antidifferentiate a function using integration by parts.
Evaluate integrals using a computer algebra system
The Definite Integral
The goal is to develop a practical understanding of the definite integral, to make connections between the derivative and the definite integral, and to understand multiple techniques for definite integral evaluation and estimation.
Determine/estimate the totals change in a function when the derivative of the function is presented in graphical, tabular, and/or symbolic form.
Express definite integrals as the sum and/or difference of the areas of the regions between the integrand€™s curve and the horizontal axis.
Evaluate definite integrals using the Fundamental Theorem of Calculus.
Apply the properties of the definite integral.
Approximate definite integrals numerically.
Construct and evaluate a Riemann Sum using left-hand endpoints, right-hand endpoints, or midpoints.
Utilize the trapezoid rule to approximate a definite integral.
Utilize Simpson's rule to approximate a definite integral.
Estimate the errors accrued in the approximation techniques listed above.
Evaluate improper integrals.
Determine divergence or convergence utilizing the Fundamental Theorem of Calculus.
Determine divergence or convergence by comparison.
Using the Integral
The goal is to use the definite integral to solve application problems.
Study applications to geometry.
Find areas of planar regions using definite integrals.
Calculate volumes of a given cross-section by slicing.
Calculate volumes of a given cross-section by revolving a region in a plane around a line.
Calculate arc length of a curve over a closed interval.
Study applications to physics, engineering, and other sciences
Discuss unit analysis and emphasize the difference between mass and weight.
Calculate work done by a force applied to an object to move it a given distance.
Find the average value of a continuous function over a given interval.
Documentation Standards for Mathematics
All work in this course will be evaluated for your ability to meet the following writing objectives as well as for mathematical content.
Every solution must be written in such a way that the question that was asked is clear simply by reading the submitted solution.
Any table or graph that appears in the original problem must also appear somewhere in your solution.
All graphs that appear in your solution must contain axis names and scales. All graphs must be accompanied by a figure number and caption. When the graph is referenced in your written work, the reference must be by figure number. Additionally, graphs for applied problems must have units on each axis and the explicit meaning of each axis must be self-apparent either by the axis names or by the figure caption.
All tables that appear in your solution must have well defined column headings as well as an assigned table number accompanied by a brief caption (description). When the table is referenced in your written work, the reference must be by table number.
A brief introduction to the problem is almost always appropriate.
In applied problems, all variables and constants must be defined.
If you used the graph or table feature of your calculator in the problem solving process, you must include the graph or table in your written solution.
If you used some other non-trivial feature of your calculator (e.g., SOLVER), you must state this in your solution.
All (relevant) information given in the problem must be stated somewhere in your solution.
A sentence that orients the reader to the purpose of the mathematics should usually precede symbol pushing.
Your conclusion shall not be encased in a box, but rather stated at the end of your solution in complete sentence form.
Remember to line up your equal signs.
If work is word-processed, all mathematical symbols must be generated with a math equation editor. | 677.169 | 1 |
This book is a comprehensive introduction to the mathematical theory of vorticity and incompressible flow ranging from elementary introductory material to current research topics. While the contents center on mathematical theory, many parts of the book showcase the interaction between rigorous mathematical theory, numerical, asymptotic, and qualitative simplified modeling, and physical phenomena.In the rapidly advancing field of flight aerodynamics, it is important for students to completely master the fundamentals. This text, written by renowned experts, clearly presents the basic concepts of underlying aerodynamic prediction methodology. These concepts are closely linked to physical principles so that they may be more readily retained and their limits of applicability are fully appreciated | 677.169 | 1 |
About this product
Description
Description
First published in 1942, as the second edition of a 1941 original, this book was intended to supply a course of mathematical instruction for the Air Training Corps Cadets during the Second World War. The text provides a cursory review of mathematical basics, then goes on to demonstrate the practical applications of those basics to 'the many and varied problems and interests of the Services'. This book will be of value to anyone with an interest in the history of education in the military during the Second World War. | 677.169 | 1 |
About this product
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Description
New and Updated for the LEAP Mathematics Assessments for 2015-2016! This practice test book is the perfect preparation tool for the new LEAP Mathematics assessments. It covers all the skills assessed on the test, and will provide practice with the types of questions and tasks found on the real assessments. Complete Preparation for the New LEAP Assessments - Begins with two short practice sets to introduce students to testing - Continues with eight 60-minute practice sets to give students the experience they need to perform well on assessments - Each practice test mimics the content and style of the real assessments - Prepares students for all the question types and tasks found on the LEAP tests - Covers all the mathematics skills assessed on the real tests Developed Specifically for the New LEAP Mathematics Assessments - Covers the same skills assessed on the LEAP assessments - Provides practice completing a wide range of question types - Questions mimic the techlogy-enhanced questions and will prepare students to take the tests online - Includes more advanced questions to reflect more rigorous LEAP tasks - Provides practice with tasks that require students to show their work or explain their thinking - Includes practice for the new complex tasks that require expressing mathematical reasoning and modeling or applying skills Key Benefits - Develop and build on all the mathematics skills that students need - Build confidence by using the tests to improve student performance - Reduce test anxiety by allowing low-stress practice - Ensure students are comfortable with a range of question formats - Introduce students to the more complex tasks found on the new assessments - Help students prepare for the more rigorous assessments - Use the full answer key and skills list to identify gaps in kwledge and target revision accordingly - Use the practice sets for testing, revision, and retesting | 677.169 | 1 |
Strictly as per the new CBSE course structure and NCERT guidelines, this thoroughly revised and updated textbook is meant for class XII of senior secondary schools (under the 10 + 2 pattern of education). The subject matter of this book is presented in a very systematic and logical manner. Every effort has been made to make the contents as lucid as possible so that the beginners will grasp the fundamental concepts in an unambiguous manner. KEY FEATURES ?Large number of solved examples to understand the subject. ?Categorization of problems under: ? Level of Difficulty A (Cover the needs of the students preparing for CBSE exams) ? Level of Difficulty B (Guide the students for engineering entrance examinations). ?A Smart Table at the beginning of each chapter to decide the relative importance of topics in the CBSE exam. ?Problem Solving Trick(s) to enhance the problem solving skills. ?A list of Important Formulae at the beginning of the book. Besides this, each chapter is followed by a Chapter Test and an exercise in which the questions from the CBSE papers of previous years are provided. Working hints to a large number of problems are given at the end of each and every exercise. In a nut shell, this book will help the students score high marks in CBSE, and at the same time build a strong foundation for success in any competitive examination. | 677.169 | 1 |
Mathematical sciences
Although much of its discovery process is descriptive and qualitative, chemistry is fundamentally a
quantitative science. It serves a wide range of human needs, activities, and concerns. The mathematical sciences
provide the language for quantitative science, and this language is growing in many directions as computational
science in general continues its rapid expansion. A timely opportunity now exists to strengthen and increase the
beneficial impacts of chemistry by enhancing the interaction between chemistry and the mathematical sciences.
The 2002 Clay School on Geometry and String Theory was held at the Isaac
Newton Institute for Mathematical Sciences, Cambridge, UK from 24 March - 20
April 2002. It was organized jointly by the organizers of two concurrent workshops
at the Newton Institute, one on Higher Dimensional Complex Geometry organized
by Alessio Corti, Mark Gross and Miles Reid, and the other on M-theory organized
by Robbert Dijkgraaf, Michael Douglas, Jerome Gauntlett and Chris Hull, in
collaboration with Arthur Jaffe, then president of the Clay Mathematics Institute....Learning mathematics in the middle
grades is a critical component in the
education of our nation's youth. The
mathematics foundation laid during
these years provides students with the
skills and knowledge to study higher
level mathematics during high school,
provides the necessary mathematical
base for success in other disciplines such
as science, and lays the groundwork for
mathematically literate citizens. A
variety of evidence suggests that the
mathematics education landscape is
shifting and evolving rapidly....
The Institute for Mathematical Sciences at the National University of
Singapore was established on 1 July 2000 with funding from the Ministry
of Education and the University. Its mission is to provide an international
center of excellence in mathematical research and, in particular, to promote
within Singapore and the region active research in the mathematical
sciences and their applications. It seeks to serve as a focal point for scientists
of diverse backgrounds to interact and collaborate in research through
tutorials, workshops, seminars and informal discussions....
The last decade has seen significant reform in the South African mathematics curriculum and the mathematics education research community has also grown markedly. Drawing on the proceedings from nearly a decade of the South African Association for Research in Mathematics, Science and Technology Education (SAARMSTE) conferences, this book reflects on the theoretical and ideological work that has contributed to the growth of mathematics education research in South Africa.
Mathematics and science are key areas of knowledge for the development of individuals and for the social and economic development of South Africa. In November 2002, about 9000 Grade 8 learners from South African public schools participated in the Trends in International Mathematics...Environmental Engineer's Mathematics Handbook
brings together and integrates in a single text
the more practical math operations of environmental engineering for air, water, wastewater, biosolids
and stormwater. Taking an unusual approach to the overall concept of environmental engineering
math concepts, this offers the reader an approach that emphasizes the relationship between the
principles in natural processes and those employed in engineered processes.
This report is based on a process of the same name is given at Stanford University in
fall quarter, 1987. Below is a description of stores:
CS 209. Write the text matter of mathematical techniques and effective
presentation of mathematics and computer science. A term paper on a topic of your choice, this paper may be used for credit in another course.
Over half of the students who enrol on economics degree courses have not studied mathematics
beyond GCSE or an equivalent level. These include many mature students whose last
encounter with algebra, or any other mathematics beyond basic arithmetic, is now a dim and
distant memory. It is mainly for these students that this book is intended. It aims to develop
their mathematical ability up to the level required for a general economics degree course (i.e.
one not specializing in mathematical economics) or for a modular degree course in economics
and related subjects, such as business studies.
Clearly and elegantly presented, Mathematical Methods in Science and Engineering provides a coherent treatment of mathematical methods, bringing advanced mathematical tools to a multidisciplinary audience. The growing interest in interdisciplinary studies has brought scientists from many disciplines such as physics, mathematics, chemistry, biology, economics, and finance together, which has increased the demand for courses in upper-level mathematical techniques.
Sincemany excellent treatises on the history ofmathemat-
ics are available, there may seem little reason for writing
still another. But most current works are severely techni-
cal, written by mathematicians for other mathematicians
or for historians of science. Despite the admirable schol-
arship and often clear presentation of these works, they are not especially well adapted
to the undergraduate classroom. (Perhaps the most notable exception is Howard Eves's
popular account, An Introduction to the History of Mathematics.
At first glance mathematics and persuasive communication – writing, and particularly
public speaking - would seem to have little in common. After all, mathematics is an
objective science, whilst speaking involves voice quality, inflection, eye contact,
personality, body language, and other subjective components.
Earth science (also known as geoscience, the geosciences or the Earth sciences) is an all-embracing term for the sciences related to the planet Earth.[1] It is arguably a special case in planetary science, the Earth being the only known life-bearing planet. There are both reductionist and holistic approaches to Earth sciences. The formal discipline of Earth sciences may include the study of the atmosphere, hydrosphere, oceans and biosphere, as well as the solid earth.This is an intermediate level post-calculus text on mathematical and statistical
methods, directed toward the needs of chemists. It has developed out of a
course that I teach at the University of Massachusetts Dartmouth for thirdyear
undergraduate chemistry majors and, with additional assignments, for
chemistry graduate students.
The long-awaited sequel to the "Concepts and Practice of Mathematical Finance" has now arrived. Taking up where the first volume left off, a range of topics is covered in depth. Extensive sections include portfolio credit derivatives, quasi-Monte Carlo, the calibration and implementation of the LIBOR market model, the acceleration of binomial trees, the Fourier transform in option pricing and much more. Throughout Mark Joshi brings his unique blend of theory, lucidity, practicality and experience to bear on issues relevant to the working quantitative analyst.... | 677.169 | 1 |
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Unformatted text preview: About the Midterm There will be a mix of problems on the midterm; some will be similar to written homework exercises, and some will be similar to WebWork exercises. There will be some definitions and maybe some short proofs. However, you dont need to memorize the 10 axioms of a vector space, I will give those to you. There will be no notes on the midterm, and you wont need a calculator. Anything presented in lecture up until Wednesday, January 27, is fair game for the midterm. I will not try to write an exhaustive list of everything that will appear on the midterm. However, off the top of my head, some things that I think are important: You should know how to solve a system of linear equations and write the solution set in vector form. Related topics: augmented matrices, row equivalence, matrix equations, geometric interpretation as lines, planes and hyperplanes, particular and homogeneous solutions, pivot variables, free variables, ......
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Math 073 Study Guide for Test 2,
Wednesday, March 18.
Vector Functions and Their Derivatives and Integrals (Section 13.1 and 13.2)
What is a vector valued function? What can it be used to model? How do you sketch its graph? How do you
compute its derivati | 677.169 | 1 |
9780321760111
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Summary
This package consists of the textbook plus an access kit for MyMathLab/MyStatLab. #xA0; Elayn Martin-Gay firmly believes that every student can succeed, and her developmental math textbooks and video resources are motivated by this belief. Basic College Mathematics with Early Integers, Second Editionwas written to help students effectively make the transition from arithmetic to algebra. The new edition offers new resources like the Student Organizerand now includes Student Resourcesin the back of the book to help students on their quest for success. #xA0; MyMathLabprovides a wide range of homework, tutorial, and assessment tools that make it easy to manage your course online. #xA0;
Table of Contents
1. The Whole Numbers
1.1 Tips for Success in Mathematics
1.2 Place Value, Names for Numbers, and Reading Tables
1.3 Adding Whole Numbers and Perimeter
1.4 Subtracting Whole Numbers
1.5 Rounding and Estimating
1.6 Multiplying Whole Numbers and Area
1.7 Dividing Whole Numbers
Integrated Review-Operations on Whole Numbers
1.8 An Introduction to Problem Solving
1.9 Exponents, Square Roots, and Order of operations
Group Activity
Vocabulary Check
Highlights
Review
Test
2. Integers and Introduction to Variables
2.1 Introduction to Variables and Algebraic Expressions
2.2 Introduction to Integers
2.3 Adding Integers
2.4 Subtracting Integers
Integrated Review-Integers
2.5 Multiplying and Dividing Integers
2.6 Order of Operations
Group Activity
Vocabulary Check
Chapter Highlights
Chapter Review
Chapter Test
Cumulative Review
3. Fractions
3.1 Introduction to Fractions and Mixed Numbers
3.2 Factors and Simplest Form
3.3 Multiplying and Dividing Fractions
3.4 Adding and Subtracting Like Fractions and Least Common Denominator | 677.169 | 1 |
Elementary And Intermediate Algebra For College Students - new book
This. This intermediate algebra topics such as systems of equations, inequalities in one and two variables, roots, radicals, quadratic functions, logarithmic functions, conic sections, and the binomial theorem. Books, Science and Geography~~Mathematics~~Algebra, Elementary And Intermediate Algebra For College Students~~Book~~9780130139801~~Allen R. Angel, , , , , , , , , ,, [PU: Prentice Hall]
An emphasis on the practical applications of algebra motivates readers and encourages them to see algebra as an important part of their daily lives. Strongly emphasizes good problem-solving skills, uses real-world applications. For anyone interested in Algebra. algebra and trigonometry,college and university,education,education and reference,mathematics,science and math,science and math,textbooks College & University, Prentice Hall | 677.169 | 1 |
Euclidean and Transformational Geometry: A Deductive Inquiry
Euclidean and Transformational Geometry: A Deductive Inquiry by Shlomo Libeskind
Book Description
Ideal for mathematics majors and prospective secondary school teachers, Euclidean and Transformational Geometry provides a complete and solid presentation of Euclidean geometry with an emphasis on solving challenging problems. The author examines various strategies and heuristics for approaching proofs and discusses the process students should follow to determine how to proceed from one step to the next through numerous problem solving techniques. A large collection of problems, varying in level of difficulty, are integrated throughout the text and suggested hints for the more challenging problems appear in the instructor's solutions manual and can be used at the instructor's discretion | 677.169 | 1 |
An Introduction To Differential Equations And Its Applications
An Introduction To Differential Equations and Its Applications
This text is intended for a one-term course in introductory differential equations and is designed for students in pure and applied mathematics who have had a course in calculus. The text presents a balance of mathematical rigour and intuitive thinking. The illustrations aim to enhance the conceptual material and allow students to visualize the mathematics. The treatment of chaotic dynamical systems introduces students to the basic ideas surrounding chaotic motion. Problem sets, which contain computer applications, are carefully graduated from the routine to the more challenging and extension exercises asking students to expand on the material are included to pique student interest. Brief historical notes place topics in their proper historical and cultural context.
Specifications of An Introduction To Differential Equations and Its Applications | 677.169 | 1 |
9780534371630
0534371600
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The basic concepts of algebra are presented in a simple, straightforward manner. Algebraic ideas are developed in a logical sequence, through examples, continuously reinforced through additional examples, and then applied in a variety of problem-solving situations. In the examples, students are guided to organize their work and to decide when a meaningful shortcut might be used.
Table of Contents
Basic Concepts and Properties
1
(43)
Sets, Real Numbers, and Numerical Expressions
2
(9)
Operations with Real Numbers
11
(10)
Properties of Real Numbers and the Use of Exponents
21
(8)
Algebraic Expressions
29
(15)
Summary
39
(2)
Review Problem Set
41
(2)
Test
43
(1)
Equations, Inequalities, and Problem Solving
44
(61)
Solving First-Degree Equations
45
(7)
Equations Involving Fractional Forms
52
(8)
Equations Involving Decimals and Problem Solving
60
(7)
Formulas
67
(10)
Inequalities
77
(7)
More on Inequalities and Problem Solving
84
(9)
Equations and Inequalities Involving Absolute Value
93
(12)
Summary
100
(2)
Review Problem Set
102
(2)
Test
104
(1)
Linear Equations and Inequalities in Two Variables
105
(55)
Rectangular Coordinate System
106
(13)
Linear Equations in Two Variables
119
(9)
Linear Inequalities in Two Variables
128
(5)
Distance and Slope
133
(10)
Determining the Equation of a Line
143
(17)
Summary
155
(2)
Review Problem Set
157
(1)
Test
158
(2)
Polynomials
160
(58)
Polynomials: Sums and Differences
161
(6)
Products and Quotients of Monomials
167
(6)
Multiplying Polynomials
173
(9)
Factoring: Use of the Distributive Property
182
(7)
Factoring: Difference of Two Squares and Sum or Difference of Two Cubes | 677.169 | 1 |
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Odyssey Algebra
04/01/05
CompassLearning ( has expanded its entire suite of Odyssey products, including Odyssey Algebra for middle schools and secondary education. The browser-based curriculum will help teachers offer a comprehensive approach to math education, while providing a platform that supports a variety of instructional strategies and learning styles. Odyssey Algebra has 13 chapters and 131 objectives to cover in an entire school year. The curriculum's online features include interactive tutorials that are woven throughout the program and aids such as online calculators, graph paper, number lines, protractors, spreadsheets and rulers. The program also provides additional offline materials for students that are designed to extend learning beyond the classroom.
This article originally appeared in the 04 | 677.169 | 1 |
...
Show More important ideas in each, usually through the lens of important problems that involve differential equations. Written at a sophomore level, the text is accessible to students who have completed multivariable calculus. With a systems-first approach, the book is appropriate for courses for majors in mathematics, science, and engineering that study systems of differential equations. Because of its emphasis on linearity, the text opens with a full chapter devoted to essential ideas in linear algebra. Motivated by future problems in systems of differential equations, the chapter on linear algebra introduces such key ideas as systems of algebraic equations, linear combinations, the eigenvalue problem, and bases and dimension of vector spaces. This chapter enables students to quickly learn enough linear algebra to appreciate the structure of solutions to linear differential equations and systems thereof in subsequent study and to apply these ideas regularly. The book offers an example-driven approach, beginning each chapter with one or two motivating problems that are applied in nature. The following chapter develops the mathematics necessary to solve these problems and explores related topics further. Even in more theoretical developments, we use an example-first style to build intuition and understanding before stating or proving general results. Over 100 figures provide visual demonstration of key ideas; the use of the computer algebra system Maple and Microsoft Excel are presented in detail throughout to provide further perspective and support students' use of technology in solving problems. Each chapter closes with several substantial projects for further study, many of which are based in applications. Errata sheet available at: | 677.169 | 1 |
McCaulay's GED* Math contains sample questions of the type that are most likely to appear in the mathematics section of the General Education Development Tests designed to measure the math skills and knowledge equivalent to a high school course of study. These practice questions will help improve scores on the GED*. The categories of questions include Number Operations and Number Sense; Measurement and Geometry; Data Analysis, Statistics, and Probability; and Algebra, Functions, and Patterns. *GED is a trademark registered by the GED Testing Service of the American Council of Education, which was not involved in the production of, and does not endorse, this product | 677.169 | 1 |
MATLAB: An Introduction with Applications – Amos Gilat – 3rd Edition
f you want a clear, easy-to-use introduction to MATLAB®, this book is for you! The Third Edition of Amos Gilat's popular MATLAB®, An Introduction with Applications requires no previous knowledge of MATLAB and computer programming as it helps you understand and apply this incredibly useful and powerful mathematical tool.
Thoroughly updated to match MATLAB®'s newest release, MATLAB® 7.3 (R2007b), the text takes you step by step through MATLAB®'s basic features—from simple arithmetic operations with scalars, to creating and using arrays, to three-dimensional plots and solving differentialequations. You'll appreciate the many features that make it easy to grasp the material and become proficient in using MATLAB®, including: | 677.169 | 1 |
9780321279088
032127908500
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Summary
Gary Rockswold focuses on teaching algebra in context, answering the question, "Why am I learning this?" and ultimately motivating the reader to succeed. Introduction to Functions and Graphs. Linear Functions and Equations. Quadratic Functions and Equations. Nonlinear Functions and Equations. Exponential and Logarithmic Functions. Systems of Equations and Inequalities. Conic Sections. Further Topics in Algebra. Basic Concepts From Algebra and Geometry. For all readers interested in college algebra. | 677.169 | 1 |
Omtale:
An Introduction to Computation and Modeling for Differential Equations
An introduction to scientific computing for differential equations Introduction to Computation and Modeling for Differential Equations provides a unified and integrated view of numerical analysis, mathematical modeling in applications, and programming to solve differential equations, which is essential in problem-solving across many disciplines, such as engineering, physics, and economics. This book successfully introduces readers to the subject through a unique "Five-M" approach: Modeling, Mathematics, Methods, MATLAB, and Multiphysics. This approach facilitates a thorough understanding of how models are created and preprocessed mathematically with scaling, classification, and approximation, and it also illustrates how a problem is solved numerically using the appropriate mathematical methods. The book's approach of solving a problem with mathematical, numerical, and programming tools is unique and covers a wide array of topics, from mathematical modeling to implementing a working computer program.
The author utilizes the principles and applications of scientific computing to solve problems involving:* Ordinary differential equations* Numerical methods for Initial Value Problems (IVPs)* Numerical methods for Boundary Value Problems (BVPs)* Partial Differential Equations (PDEs)* Numerical methods for parabolic, elliptic, and hyperbolic PDEs* Mathematical modeling with differential equations* Numerical solution* Finite difference and finite element methods Real-world examples from scientific and engineering applications including mechanics, fluid dynamics, solid mechanics, chemical engineering, electromagnetic field theory, and control theory are solved through the use of MATLAB and the interactive scientific computing program Comsol Multiphysics. Numerous illustrations aid in the visualization of the solutions, and a related Web site features demonstrations, solutions to problems, MATLAB programs, and additional data. Introduction to Computation and Modeling for Differential Equations is an ideal text for courses in differential equations, ordinary differential equations, partial differential equations, and numerical methods at the upper-undergraduate and graduate levels.
The book also serves as a valuable reference for researchers and practitioners in the fields of mathematics, engineering, and computer science who would like to refresh and revive their knowledge of the mathematical and numerical aspects as well as the applications of scientific computation. | 677.169 | 1 |
MATLAB is a powerful programme, which naturally lends itself to the rapid implementation of most numerical algorithms. This text, which uses MATLAB, gives a detailed overview of structured programming and numerical methods for the undergraduate student.
The book covers numerical methods for solving a wide range of problems, from integration to the numerical solution of differential equations or the stimulation of random processes. Examples of programmes that solve problems directly, as well as those that use MATLAB's high-level commands are given.
Each chapter includes extensive examples and tasks, at varying levels of complexity. For practice, the early chapters include programmes that require debugging by the reader, while full solutions are given for all the tasks. The book also includes:
a glossary of MATLAB commands
appendices of mathematical techniques used in numerical methods.
Designed as a text for a first course in programming and algorithm design, as well as in numerical methods courses, the book will be of benefit to a wide range of students from mathematics and engineering, to commerce.
Steve Otto was formerly a lecturer at Birmingham University but now heads a research unit at St. Andrews, UK. This book is based on courses taught at Birmingham and at Adelaide to students in mathematics, engineering and economics.
"This book is an introduction to the numerical methods that are frequently used in science and engineering undergraduate courses and it is based on the MATLAB programming environment. … The text is easy to read even for those who have little experience with computer practice. … There are three appendices and the most useful of them, about 70 pages, gives solutions to the tasks at the end of chapters." (Matti Vuorinen, Zentralblatt MATH, Vol. 1076, 2006) | 677.169 | 1 |
How is Math X different from other modes?
Mastery Learning The Math X Program is a Mastery Learning program. It requires you prove that you have mastered the material by passing each chapter exam and/or quiz with a score of 80% or better, before proceeding to the next chapter. Students are given three attempts to pass each exam and/or quiz, with guided remediation in-between attempts.
Math X Mode
Lecture Mode
Students choose the dates and times of their exams
Exam dates and times are fixed
All homework must be completed before attempting a test
Usually, homework does not have to be completed prior to taking a test
Three attempts to pass a test
One attempt to pass a test
Personalized review and support between attempts
Little revisiting of the concepts
Why is this important?
In Math X students:
take tests when they are ready
continue on to new material prepared to be successful
build on their knowledge instead of struggling just to get by
develop confidence in their mathematical skills
Math X mode vs. Lecture mode
In Math X mode there is no lecture. Students are supported in their learning by an instructor, an instructional assistant and, sometimes, a Math X tutor. In addition, students use web-based instructional software which includes a number of built-in support features.
Math X Mode
Lecture Mode
Flexible pacing
Instructor determines the pacing
Students determine how much time they need to spend on a concept before moving on
Students may have to move forward before fully grasping a concept
Why is this important?
In Math X students:
can move at a slower or faster pace and adjust their work pace when work or family needs intervene
can spend more time on difficult concepts and work quickly through concepts they understand well
are more apt to be successful when they do not have to move on before understanding basics
Committing to Success
We commend you for your commitment to enroll in our Math X Program. Commitment to learning is vital to your success in this program. A learning brain requires focus, attention and motivation, according to psychologists.
To be successful in the Math X program students should:
attend every class
work diligently on homework and lab assignments
construct a good working relationship with their instructor and instructional assistant
take full advantage of all support resources
commit to learning
For more information about LPC campus support services for students in mathematics classes, click here.
"Learning is not attained by chance, it must be sought for with ardor and attended to with diligence." Abigail Adams, 1780 | 677.169 | 1 |
Limits Involving Radicals
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0.07 MB | 2 pages
PRODUCT DESCRIPTION
This worksheet asks students to evaluate several limits involving radicals. The last problem asks the student to use a calculator to complete a table of values. The values are then used to estimate a limit involving a radical. Finally, the student is asked to evaluate the limit algebraically and compare it to the answer found using the table of values | 677.169 | 1 |
Description
Create a review
Contents
was developed over a period of two years at the University of Utah Department of Computer Science in conjunction with the U.S. Department of Energy-funded Undergraduate Computation in Engineering Science (UCES) program. To encourage the integration of computation into the science and engineering curricula, an introductory course in computing was designed expressly for science and engineering students. This course was created to satisfy the standard programming requirement, while preparing students to immediatly expliot the broad power of modern computing in their sicence and engineering courses. The course has at least four distinguishing features: - It uses a sumbolic mathematics package (Mathematica) in combination with a conventional programming language (C). - It teaches programming concepts in parallel with a scientific problem-solving methodology. - It draws upon a variety of computational problems from the breadth of science and engineering to interest students and establish the relevance of the computational problem-solving approach. - The author has developed an extensive suite of interactive, on-line laboratory materials that students can use via any HTML viewer. All of the "Introduction to Scientific Programming" notes are available on -line and contain a number of embedded interactive URLs. | 677.169 | 1 |
Omar Al-Khayyam's famous book on algebra and equations is considered to be perhaps his most important contribution to mathematics - the cream of his work. Scholars of mathematics regard the book as an essential item in both their personal libraries and institutional collections. This book deals with the solution of quadratic and cubic equations. Al-Khayyam solved all possible cases of such equations by using geometrical approaches, sometimes involving conic sections such as parabolas and hyperbolas. The proofs presented are precise and very deep. The author made due acknowledgment and referral to the work and contributions of others who came before him. A modern reader may well be astonished by the high quality of the work, linguistically as well as mathematically. Historians of science, teachers of mathematics and mathematicians themselves will find the book both interesting and informative.
Author Biography - Omar Al-Khayyam
'Abel Fath Omer Bin Al-Khayyam was an eleventh-century Persian mathematician, astronomer, philosopher, physician and poet, most famous for his Rubaiyat, one of the classics of world literature. His work greatly influenced the development of mathematics, particularly analytical geometry, and was unsurpassed for many centuries. Although he is generally better known as a poet, his work as a philosopher, scientist and mathematician was a major contribution to the growth of human | 677.169 | 1 |
The Pleasures of Probability (Undergraduate Texts in Mathematics)
Delivery: 10-20 Working Days
(5 reviews)
The ideas of probability are all around us. Lotteries, casino gambling, the al most non-stop polling which seems to mold public policy more and more these are a few of the areas where principles of probability impinge in a direct way on the lives and fortunes of the general public. At a more re moved level there is modern science which uses probability and its offshoots like statistics and the theory of random processes to build mathematical descriptions of the real world. In fact, twentieth-century physics, in embrac ing quantum mechanics, has a world view that is at its core probabilistic in nature, contrary to the deterministic one of classical physics. In addition to all this muscular evidence of the importance of probability ideas it should also be said that probability can be lots of fun. It is a subject where you can start thinking about amusing, interesting, and often difficult problems with very little mathematical background. In this book, I wanted to introduce a reader with at least a fairly decent mathematical background in elementary algebra to this world of probabil ity, to the way of thinking typical of probability, and the kinds of problems to which probability can be applied. I have used examples from a wide variety of fields to motivate the discussion of concepts.
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Specifications
Country
USA
Author
Richard Isaac
Binding
Hardcover
EAN
9780387944159
Edition
1st ed. 1995. Corr. 2nd printing 1996
IsAdultProduct
ISBN
038794415X
IsEligibleForTradeIn
1
Label
Springer
Manufacturer
Springer
NumberOfItems
1
NumberOfPages
244
PublicationDate
1996-10-30
Publisher
Springer
Studio
Springer
Most Helpful Customer Reviews
Professor Isaac has written a book for those interested in learning about probability. It is at a high school algebra level although knowledge of calculus could be helpful at times. He starts with the now famous Monte Hall problem and provides the most lucid explanation I have seen to date. This is a great way to introduce important probability notions such as sample space and probability models for the sample outcomes. Deals mainly with discrete probability which is easiest to understand and yet rich with applications in gambling and other areas. Important theory is presented but without the detailed mathematical proofs. Covers the gambler's ruin, geometric probability, Monte Carlo methods and some statistical decision theory. He also presents both the frequentist (throughout the text)and the Bayesian paradigms (Chapter 4) for statistical inference. Examples of the application of probability to statistical inference is nicely treated in Chapter 15. The deeper material on Markov...
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This introduction to probability and statistics teaches you about important concepts, theorems and applications without going into proving most of them. It's easily accessible to amateur mathematicians with a bit of persistance, and it illuminates many of its concepts using famous problems. I'm going to take a statistics course next year, and I found this to be a delightful introduction to the topic.
This is a wonderful book for both the serious and amateur mathematician. For the serious student and practitioner it will provide an excellent background. For the amateur who enjoy dabbling and reading about math, this is a wonderful book - just read the opening section which explains clearly and an illustrated fashion the Monty Hall | 677.169 | 1 |
Aleks Presentation How can using Aleks™ daily help you?
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Aleks Presentation What is Algebra? Why do I need to take it? Why do schools make such a big deal about getting ready for it? Why do my parents care if I pass, or get a high grade? Isn't it just another Math class?
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Aleks Presentation What is Algebra? Algebra, for most students, can be thought of as a system of brain puzzles that are common to all students around the world. Solving the problems involves using rules that you need to learn, memorize, and put to use when you take quizzes and tests.
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Aleks Presentation Why do I need to take it? Taking and passing Algebra 1 with a "C" or better, along with other Math classes, is a graduation requirement for all LAUSD high schools. It is also a requirement for admission to California State Universities.
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Aleks Presentation Why do schools make such a big deal about getting ready for it? The reason your teachers make such a big deal about learning your math skills in earlier grades, and then passing Algebra 1, is this: Students who pass Algebra in Middle School or High School almost always go to college, but students who don't almost never get to college.
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Aleks Presentation Why do my parents care if I pass, or get a high grade? Your parents want you to be successful in school. Parents want their children to someday be smarter in school, and possibly more successful in careers, than they were. Your parents want to be proud of you. You want to be proud of yourself.
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Aleks Presentation Isn't Algebra just another Math class? No. Passing Algebra 1, more than almost anything else you do in school, is an indicator of the choices you'll have in your future. In The United States of America, you can decide what future you'll have, but you have to work hard to get to college where you'll have the chance to make your choices.
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Aleks Presentation What if I don't pass Algebra 1 the first time I take it? You can re-take Algebra 1 if you get less than a C the first time. Many students don't pass Algebra 1 the first time. Ideally, you'll pass the first time. It doesn't matter as much when you pass it, but that you ultimately do pass.
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Aleks Presentation Algebra is like a mountain to climb…. Using Aleks, along with carefully doing your classwork and homework, will help you get to the top.
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Mt. Algebra You may be asking yourself, Can I get to the top? You can, but it will take a lot of hard work on your part. Teachers can teach you how to climb, but you have to do the climbing!
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Aleks Presentation You can teach yourself how to climb the Algebra mountain by using Aleks….
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You can log in to Aleks™ from anywhere you can access the Internet, including at home, or the library.
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The first time you log in, you are given a short test. This helps Aleks find out what you already know, and what you need to learn.
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You will need to use paper and pencil to answer most of the questions.
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When you enter your answer and click Next, you will not be told if you are right or wrong.
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If you don't know how to answer a question, click "I haven't learned this yet."
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The progress bar tells you how far along in the test you are.
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Some questions will require you to control a tool like a ruler or protractor.
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Some questions require the use of special math vocabulary.
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Many Aleks tools are well designed to help you locate a correct answer.
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Geometry is a part of all Aleks courses.
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Critical thinking and logic are also tested in Aleks, as with this Venn Diagram.
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Aleks tasks often take far less time than doing the work using a textbook.
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Some Aleks tasks, called Topics, can be completed in just a few minutes.
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When you have completed an Aleks assessment, you will be given a report of how you did. You'll click Next to see it.
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The report comes in the form of a pie chart. Your goal in Aleks, over time, is to fill in the entire pie.
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Aleks explains how to use the pie chart.
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Choose a slice to work on. Choose the first topic in the list. You can only work on topics with an arrow next to them.
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Completing 5 Topics in 30 minutes is a good daily goal.
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Make note of where you are each day you start to work on Aleks. Try to increase this number by 5 or more each day.
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A goal for the end of this session would be 93. (That's )
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Use paper and pencil, if necessary. Enter your answer, then click Next.
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If you answer correctly, Aleks will tell you immediately.
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You only have to answer a few questions correctly to complete a topic. This Aleks message says that.
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Now you can click Done, or choose to continue to practice this skill.
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The Topic just completed is removed from the list, and the dark part of the pie grows. You can choose another topic from the same slice, or a different one.
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Here, a new Topic has been chosen. Again, only three or four correct answers may be needed to complete the Topic.
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That's a lot better than having to do a whole page of questions from a workbook.
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If you're not sure how to do a problem, click on the Explain button.
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Aleks shows how to work out the problem, all the way to the correct answer.
44
If you answer incorrectly, Aleks will let you know with red letters. You can try again and change your answer.
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If you're still not right, Aleks may prompt you to use the Explain button, and give you more help in highlighted colors.
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If you're still having trouble, Aleks will tell you to "…try something else." Just return to the pie and pick another Topic.
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Choose another Topic from the pie that you can do by yourself. Ask for help with especially difficult ones when a teacher is available.
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Some Topics are not hard, but you do need to be careful. Notice the red stops between two numbers.
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Taking good notes in Aleks is very important when working on Topics.
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You can also use your notes when you take an assessment.
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Aleks assessments mostly contain questions you've practiced. If you've taken good notes while working with your pie, the questions should be easy.
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If you do well, your pie will likely stay the same, but if you do poorly, Aleks will take Topics you had previously passed off the pie, and you'll have to do them again.
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When working on Aleks with a parent who speaks Spanish, you can change the language to Spanish.
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Everything will work exactly the same way. You can switch between English and Spanish any time.
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QuickTables is a part of Aleks that will help you become faster in multiplication.
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You can use QuickTables for a few minutes each day. Time in QuickTables is limited.
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Remember, your goal is to complete 5 or more Topics each day you work on Aleks
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3 Topics completed so far…
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Your teacher can view your progress.
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5 Topics completed!
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5 Topics per day is 25 topics in one week.
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25 topics per week ≈ 100 Topics in one month.
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Completing a pie is like making a year's worth of math growth.
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Many students are able to complete more than one pie, or one year's growth in a school year.
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The teacher is notified when a student has completed an Aleks course (pie).
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A student could begin the year as a D student who hates Math, but finish the year as an A or B student who gets it. Hard Work in Math Class + Hard Work in Aleks
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A student could begin the year as BB (Below Basic) and finish the year as a P (Proficient) student. There are direct connections between Aleks and your State Tests you'll take at the end of the year. | 677.169 | 1 |
by Todd Keene Timberlake (Author), J. Wilson Mixon.Jr. (Author) Introduces students to the use of a leading open-source computer algebra system The text groups input and output so that students see precisely how Maxima commands are implemented Accompanied by a set of Maxima files that allow students to replicate all of the text's material and to build on that analysis
Differential equations with "maxima"—differential equations that contain the maximum of the unknown function over a previous interval—adequately model real-world processes whose present state significantly depends on the maximum value of the state on a past time interval.
Economists can use computer algebra systems to manipulate symbolic models, derive numerical computations, and analyze empirical relationships among variables. Maxima is an open-source multi-platform computer algebra system that rivals proprietary software. Maxima's symbolic and computational capabilities enable economists and financial analysts to develop a deeper understanding of models by allowing them to explore the implications of differences in parameter values, providing numerical solutions to problems that would be otherwise intractable, and by providing graphical representations that can guide analysis. | 677.169 | 1 |
MATH 1B03 / 1H03: Project 2 Model for the Age Distribution of a Population Instructor: Dr. D. Lozinski SOLUTIONS
Instructions: Completed projects are to be handed in to the homework lockers on the first floor of Hamilton Hall (near HH 105) by 2:00 pm on t
MATH 1B03 / MATH 1H03 McMaster University Project: Mendelian Genetics Due: 7pm June 9, 2008 SOLUTIONS
Instructions: Completed projects are to be handed in at the beginning of class on Monday, June 9. LATE ASSIGNMENTS WILL RECEIVE A GRADE OF ZERO. On the f
TA: Geoff Williams Tutorials: Thursday 1:30-2:20 in TSH-120 E-mail:williagg@math.mcmaster.ca Office hours: Wednesday 1:30-3:30 at the Math Help Centre
Tutorial #3 Transformations An Example of a (Linear) Transformation: We have a machine that transforms L
Triangular prism
Definition
A triangular prism is
aprismcomposed of two triangular
bases and three rectangular sides. It
is apentahedron. It is implemented
inMathematicasPolyhedron
Data["Triangular Prism"].
The triangular prism has nine distinct nets, a | 677.169 | 1 |
Elementary Geometry, extensively revised, this new edition continues in the fine tradition of its predecessor. Major changes include: a notation that formalizes the distinction between equality and congruence and between line, ray and line segment; a completely rewritten chapter on mathematical logic with inclusion of truth tables and the logical basis for the discovery of non-Euclidean geometries; expanded coverage of analytic geometry with more theorems discussed and proved with coordinate geometry; two distinct chapters on parallel lines and parallelograms; a condensed chapter on numerical trigonometry; more problems; expansion of the section on surface areas and volume; and additional review exercises at the end of each chapter. Concise and logical, it will serve as an excellent review of high school geometry. | 677.169 | 1 |
Exams Philosophy:
Exams areawonderful opportunity for students to broaden their mathematical techniques in the application process. I believe problems should be doable but help to stretch the student to higher levels. My exams are based on the material from my lecture.
In summary, my teaching technique is to go from the known to the unknown through examples and analogies. Central to my teaching philosophy is the notion that theory and practice are integrated activities.
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Pure mathematics is, in its way, the poetry of logical ideas. ~Albert Einstein | 677.169 | 1 |
Product Details
Math is an important part of everyday life and an integral part of the skills necessary to become certified in the safety profession. Many who pursue certification have long since completed their college math courses and have not actively pursued the math skills they once had. Background Math provides the basics necessary to successfully negotiate the math included on the certification exams, as well as a handy primer for those who already have their credentials.
Topics include:
Calculator selection and use, including BCSP rules for calculators, strategies for examinations and hierarchy for operations
Fractions, reciprocals, proportions, rounding and absolute value
Exponents, roots and logarithms and antilogs
Systems of measurement, including English, metric, conversions and dimensional analysis
Notation, both scientific and engineering
Algebraic properties and simple equations, including variables, commutation, associative and distributive properties, order of operations, rules of equations, multiplying polynomials, and solving equations | 677.169 | 1 |
As the field of economics becomes ever more specialized and complicated, so does the mathematics required of economists. With Mathematics for Economists, expert mathematician Viatcheslav V. Vinogradov offers a straightforward, practical textbook for students in economics—for whom mathematics is not a scientific or philosophical subject but a practical necessity. Focusing on the most important fields of economics, the book teaches apprentice economists to apply mathematical algorithms and methods to economic analysis, while abundant exercises and problem sets allow them to test what they've learned | 677.169 | 1 |
MATHEMATICS: A PRACTICAL ODYSSEY, 7E demonstrates mathematics' usefulness and relevance to students' daily lives through topics such as calculating interest and understanding voting systems. Well known for its clear writing and unique variety of topics, the text emphasizes problem-solving skills, practical applications, and the history of mathematics, and unveils the relevance of mathematics and its human aspect to students. To offer flexibility in content, the book contains more information than might be covered in a one-term course. The chapters are independent of each other so instructors can select the ideal topics for their courses.
Additional Product Information
Features and Benefits
Accessibility: Emphasizing problem-solving skills, the variety of applications-oriented examples and exercises connect the material to students' lives. "...in the Real World" feature incorporates specific chapter material into a real-world context. The authors have an accessible writing style that engages students and retains their interest.
Usability: This book is user friendly. The examples do not skip steps; key points are boxed for emphasis; procedures are given step-by-step, making them easier to follow and understand; and an abundance of exposition is provided.
History: Students see the human side of mathematics as the history of the subject matter is interwoven throughout most chapters. In addition, "Historical Notes" give in-depth biographies of the prominent people involved.
Flexibility: The text offers an assortment of content to choose from. The chapters are independent of each other so instructors can select the ideal topics for their courses.
Design: An eye-catching design, including color art and photos, livens the appearance of the text. Color is used pedagogically in figures and examples, helping to convey the mathematics and elevating students' understanding of the material.
This CD-ROM.
Solution Builder
(ISBN-10: 0495915564 | ISBN-13: 9780495915560)
This online instructor database offers complete worked solutions to all exercises in the text, allowing you to create customized, secure solutions printouts (in PDF format) matched exactly to the problems you assign in class.
This guide helps students navigate Enhanced WebAssign. It includes instructions on how to use the Assignment page and its Summary, tips on using MathPad for providing easy input of math notation and symbols, an overview of the Graphing Utility's drawing tools for completing graphing assignments, and information on how to access grades and scores summary MATHEMATICS: A PRACTICAL ODYSSEY, 7th Edition.
If your instructor has chosen to package Enhanced WebAssign with your text, this manual will help you get up and running quickly with the Enhanced WebAssign system so you can study smarter and improve your performance in classDavid B. JohnsonThomas A. Mowry
Thomas Mowry earned a
Thomas Mowry earned a | 677.169 | 1 |
New York Math Circle's goal is to constantly challenge your mind.
You'll get to solve unusual problems and invent your own, apply existing
knowledge in new situations, learn famous gems of mathematics, and explore the
unknown. The Math Circle will open your eyes and increase your sensitivity to
all the mathematics around us. Our main requirement is that you have an open
mind and a willingness to work.
NYMC is not a tutoring or test preparation program. Classes focus on
material you won't encounter in the regular curriculum. We'll help
you develop reasoning and problem solving skills, and along the way will help
you enjoy, appreciate, and expand your knowledge of mathematics.
To subscribe to announcements about student classes, see our contact page.
We generally recommend that new students start in level Middle School A for those below 9th
grade, and High School A for those in 9th grade and above. Check out sample problems listed
under every level to decide if you are ready for it. If still in doubt about the right level,
please ask, and we'll be happy to offer guidance.
Once you start the program, we expect you to come for the full semester.
Any course for which fewer than 10 students register may have to be cancelled.
Middle School A
When:
Saturdays and Sundays February 4 - May 20, 2017
Click on a course to see schedule details February 4 - May 21, 2017 C
When:
Saturdays February 4 - May 20, 2017
Click on a course to see schedule details February 4 - May 21, 2017 B
When:
Saturdays and Sundays February 4 - May 21, 2017 C
When:
Saturdays February 4 - May 20, 2017
Click on a course to see schedule detailsCollege Bridge
When:
Saturdays February 4 - May 20, 2017
Click on a course to see schedule details.
College Bridge is a new course designed to challenge our most advanced students who plan to take advanced mathematics classes in college. It will provide preparation for the formal theorem/proof approach taken in college by examining underlying structures and developing proof-writing skills. Three instructors will teach three topics (calculus, set theory and abstract algebra) after an introductory session by Larry Zimmerman.
Students will be required to take a placement test at home which they can submit by email; they will also be required to provide a reference either from their High School teacher or their NYMC instructor to complete registration Sep 17-Dec 18 Sep 17-Dec 18 this class 2 or more semesters in the past.
High School B
When:
Saturdays and Sundays Sep 17-Dec 18Past Courses
We had our first full course for students in Spring 2008. You can see
the listing of past courses here.
Registration Policy for Weekend Courses
When registering online for weekend courses, we expect you to make a payment of at least a 20%
non-refundable deposit upon registration to reserve your spot (we do encourage you to pay the full
registration fee, though). The balance is due two weeks before the class starts. If we don't
receive your full payment in time, we may cancel your registration and give your spot to
another student.
Cancellation Policy:
We'll refund the fee except the 20% non-refundable deposit if you cancel at least 2 weeks
before classes start.
We'll refund 50% if you cancel afterwards but before the 3rd class. No refunds are possible
after the 3rd class.
If on the first day of classes you find that the level isn't right for you, please speak with
the instructor about a more appropriate level, and email
[This e-mail address is being protected from spam bots, you need JavaScript enabled to view it].
We will do what we can to place you in the right class. But if we aren't successful, we'll refund
your registration fee, less the non-refundable deposit.
If you find you need to cancel, please let us know as soon as possible out of consideration for
students who may be waiting for a spot.
Frequently Asked Questions
Probably not. Our program assumes that participants are already comfortable with
regular school material, and introduces a lot of additional interesting topics
and problems. It is for children who like math, and would like to deepen their understanding
through exposure to new topics and greater challenges.
No. Our instructors provide a rich variety of material that varies each semester. We generally
assume that students attend the same circle for two semesters before moving on to the next level,
and occasionally a student enjoys one level so much that she or he will stay several semesters in a
row. All our courses vary the material offered each year. Important topics and concepts may be
repeated and reinforced, but there are always plenty of new topics and problems to keep everyone
engaged and challenged.
We generally ask parents to refrain from sitting in with their children. Students participate
better without their parents in the room, and they take more mathematical risks, which is
important.
We do recognize that there are situations where it makes sense for a parent to be present, and
special occasions (like the first and last days of class) where it's exciting for parents to sit
in. In these cases, parents may observe the class on the following conditions: there is enough
space in the classroom; they have the permission of their child; and they observe quietly from
the back row. In all cases, the teacher has the final word, and may ask parents to wait outside.
Please respect the teacher's requests.
We are committed to making our courses available to everyone. If your family is experiencing
financial hardship, you may request a fee reduction. The difference would come from contributions
from private individuals who value and support our work.
To request a fee reduction, register for the course you are interested in.
After submitting the registration form, there is a link to the Fee Reduction
Form, where we ask for basic information about your family's financial
situation. Once you submit that form, we will consider your situation, and get
back to you soon with further information.
If a student wants to learn mathematics at NYMC, but money stands in the way,
do not hesitate to apply.
Private contributions are vital to us, and we would be extremely grateful for a monetary donation.
Donations to NYMC are tax-deductible. You can make them here.
There are also some limited volunteer opportunities available, but mainly
involving special skills and non-trivial commitment. If you are interested
in volunteering, please email us at info@nymathcircle.org, and describe
your interests and skills. We'd love to hear from you!
Another great help would be if you can help us spread the word about NYMC.
Tell your friends, write about us on your blog, or become our fan on facebook. | 677.169 | 1 |
HO1%20Math%20review - GLY 4822/GLY 6934 Hydrogeology Math...
GLY 4822/GLY 6934 Hydrogeology Math Review Guidelines for solving quantitative problems One of my goals for this course is to help those students that are not comfortable with quantitative problems improve their problem solving skills and gain confidence. Basic algebra is all that is needed for this course. I will not ask you do complicated mathematical operations - just addition, subtraction, multiplication, and division. You are all more than capable of doing the math required in this course. However, depending on your background you might find it challenging to determine what you are being asked and to identify the equations necessary for finding the answer. One way to improve your problem solving skills is to develop a set of steps that you follow rigorously when solving problems. Therefore, if you explicitly follow the below steps for all quantitative problems on homeworks and exams, I will be generous with partial credit. I will be less generous with partial credit if your answer is incorrect and you did not approach the problem in the following stepwise manner. On in-class problems, I will not give any partial credit unless you complete the following steps. Problem solving steps (I suggest you write out the questions and answers to the questions as you work through the problems.). I will not give any credit for "magical answers" if you show no work. 1. What are you being asked to find in words and symbols? 2. What units and or dimensions should the answer be in? 3. What information are you given in the problem? List all parameters and values given. 4. What equations might apply to this problem? List them. 5. Solve. Include units all the way through your calculations. 6. Check your answer. Is the magnitude reasonable? Are the units correct?
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SIGNIFICANT FIGURES For measured numbers, significant figures relate the certainty of the measurement. As the number of significant figures increases, the more certain the measurement. The means for obtaining the measurement also becomes more sophisticated as the number of significant figures increase. Scientific notation is the most reliable way of expressing a number to a given number of significant figures. In scientific notation, the power of ten is insignificant. For instance, if one wishes to express the number 2000 to varying degrees of certainty: 2000 2 x 10 3 is expressed to one significant figure 2000 2.0 x 10 3 is expressed to two significant figures 2000 2.00 x 10 3 is expressed to three significant figures 2000 2.000 x 10 3 is expressed to four significant figures What do these numbers imply as to the certainty? Let's see what the number can be distinguished from: The number 2000 to one significant figure lies between: 1 x 10 3 = 1000 2 x 10 3 = 2000 3 x 10 3 = 3000 It is a number that lies between 1000 and 3000 -- not very certain, is it. The number 2000 to two significant figures lies between:
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Introduction to Quantum Computing
This concise, accessible text provides a thorough introduction to quantum computing - an exciting emergent field at the interface of the computer, engineering, mathematical and physical sciences. Aimed at advanced undergraduate and beginning graduate students in these disciplines, the text is technically detailed and is clearly illustrated throughout with diagrams and exercises. Some prior knowledge of linear algebra is assumed, including vector spaces and innerproducts. However, prior familiarity with topics such as quantum mechanics and computational complexity is not required | 677.169 | 1 |
This mathematics curriculum guide was developed in response to the need for an updated regional mathematics guide identifying common student performance expectations of the professional staff. The expectations, stated as instructional objectives, have been organized into two major components: (1) an overall "Mathematics Comprehensive Program Guide: K-8 and Algebra I," and (2) a separate "Mathematics Grade Level Guide" designed for each grade level K-8. Each of these two components organizes the objectives into six categories: numeration and numbers, operations, sets and set notation, geometry, measurement, and graphs. For each grade level, the two guide components indicate whether the objective is to be introduced, reinforced, mastered, or continued. Certain objectives are designated as optional. (Author/MK) | 677.169 | 1 |
Two different elements inform this talk. The first is my experience over the past year working on two NCTM books on geometry for secondary mathematics teachers, books that have involved creating four 'Big Ideas' (different ones
for grades 6-8 teachers and grades 9-12 ones). The second is a realization that it is now twenty years since David Tall edited the book Advanced Mathematical Thinking (Kluwer, 1991), whose blurb claims "This book is the first major study of advanced mathematical thinking as performed by mathematicians and taught to students in senior high school and university." I wondered what has happened in the intervening time in this area.
So in my talk I will attempt to explore what a 'big idea' is (whether in mathematics or in mathematics education at the undergraduate level), as proposed by mathematicians (ancient and modern) and by researchers in post-secondary mathematics education, before considering what it might mean to teach mathematics bearing one or more such guiding themes in mind.
There has been a great deal of discussion of late about the use of open-ended problems in the teaching and learning of mathematics. But what exactly are these problems and how can they be used? In this workshop we will look at different types of open-ended problems and discuss the limitations and affordances in using them in our teaching. Participants will leave with a greater understanding of the open-ended problems, resources for locating them, and strategies for using them.
Everybody likes games and puzzles, but few people realize the deep mathematical thinking involved in solving them. The three leaders will bring their favourite games and puzzles. The participants will explore them and their connections to mathematics, and discuss ways to use these puzzles to enrich our teaching and encourage our students to think mathematically.
13:30 Should we change what we do in the calculus classroom? Warren Code, Carl Wieman Science Education Initiative, UBC (1900, Fletcher Challenge Theatre)
In an attempt to answer this question, I will present results from various measurements of student learning in differential calculus taken over the last few years at UBC. There are indicators which suggest room for improvement.
To address the inevitable follow-up: "But how might we change?", I will also describe our recent, formal comparison of teaching methods similar to Deslauriers, Schelew & Wieman's "Improved Learning in a large-enrollment physics class" published in Science last year: each of two sections of the same calculus course were subject to an "intervention" week where a less-experienced instructor produced a much higher level of student engagement by design. Our instructional choices encouraged more active learning ("clicker" questions, small-group discussions, worksheets) during a significant amount of class time, building on assigned pre-class tasks. The lesson content and analysis of the assessments were informed by existing research on student learning of mathematics. Based on our initial analysis, we can report improved student performance - on conceptual items in particular - in the higher engagement section in both cases.
This represents the work of several people at UBC Mathematics, including fellows Costanza Piccolo and Joseph Lo of CWSEI, faculty member Mark MacLean (honoured at Changing The Culture 2012) and graduate student and calculus instructor David Kohler.
Abstract: It does not matter whether you are teaching elementary school students, high school students, or post-secondary students - it is the questions you ask that make the difference in your students' perceptions of what mathematics is and also their success.
I think we can all do a better job of asking questions that focus on deeper understandings in mathematics rather than on details. We can also do a better job of asking questions that are accessible to a broader range of our students and that better engage them. And we can ask more open questions that evoke richer and broader, rather than narrower, conversations about mathematics. I will share samples of some student responses to this approach to questioning and will also describe the impact that this has had on many teachers and their students. | 677.169 | 1 |
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To Use Calculus or Not to Use CalculusIn the past if you have studied Algebra and Trigonometry then your knowledge has prepared you to master the next step calculus Calculus is complicated but not quite as bad as everyone thinks It is the study of changing quantities Take for example the curve as a path of a rocket A tangent line at any point on the orbit displays the direction that the rocket is flying at that point If gravity disappears the rocket would move off on the tangent line Within changing quantities calculus is based on two fundamentals derivatives and integrals and Computing the derivative is easy but has a lot of algormathically The derivative finds the slope of a function Integration is the same as the anti-derivative in that it finds the original equation of the function and computes the area under the function These are two concepts that calculus is moldedDerivatives and integrals are all to do with limits The concept of limits is essential to all calculus The limit is the basis for all calculus problems A knowledgeable understanding of limits will help explain many theories in calculus In limits the only thing that matters is how a function is defined near the point a point a is the quantity that the limit gets close to but never reaches In taking the derivative of a height function the out come is a velocity function In taking the derivative of a velocity function the rate of change of a position you get a acceleration function the rate of change of velocity Or if given an acceleration function you can integrate to get a velocity function Newton realized that the pattern of velocities is simpler while the pattern of accelerations is simpler still The two known operations of calculus differentiation taking the derivative | 677.169 | 1 |
Affordable Access
Chapter 0 Introduction
Abstract
Publisher Summary This chapter presents the concepts of power geometry. The equations may be algebraic, ordinary differential or partial differential and systems may comprise the equations of one type, but may include equations of different types. The solutions to these equations and systems subdivide into regular and singular ones. The main concept of Power Geometry is to study the properties of solutions to an equation through the power exponents of its monomials. Power Geometry is based upon the three concepts—that is, Newton polyhedron, power transformation, and logarithmic transformation. The theory and algorithms presented considered only as the first steps on the way of using the concepts of Power Geometry. Algorithms of solution of systems of linear inequalities are set forth, as well as their modifications for the purposes of Power Geometry, and the corresponding computer programs are also presented in the chapter.
There are no comments yet on this publication. Be the first to share your thoughts. | 677.169 | 1 |
Sets, Sequences and Mappings: The Basic Concepts of Analysis
Overview systematic development of many of the important properties of the real number system, plus detailed treatment of such concepts as mappings, sequences, limits, and continuity. The sixth and final chapter discusses metric spaces and generalizes many of the earlier concepts and results involving arbitrary metric spaces. An index of axioms and key theorems appears at the end of the book, and more than 300 problems amplify and supplement the material within the text. Geared toward students who have taken several semesters of basic calculus, this volume is an ideal prerequisite for mathematics majors preparing for a two-semester course in advanced calculus. | 677.169 | 1 |
Hall and Mercer's text is intended for schools that want a single book covering the standard topics from elementary algebra through intermediate algebra. The text is fully integrated, rather than being simply the joining of two, separate texts. Topics are organized not following the historical pattern, but by using as the guiding principles, the AMATYC standards as outlined in Crossroads in Mathematics. BEGINNING AND INTERMEDIATE ALGEBRA: THE LANGUAGE AND SYMBOLISM OF MATHEMATICS is oriented toward recent reforms in college level mathematics curricula.
"synopsis" may belong to another edition of this title.
From the Publisher:
This book has been written from the ground up as an integrated, combined book. The authors have used the AMATYC Standards as a guiding document in writing the book. The use of the graphing calculator is assumed, and the authors use the graphing calculator to explain a large number of real-data applications. Concepts are presented using "Rule of Four" (multiple representations of mathematical solutions to problems, including graphical, algebraic, numerical, and verbal approaches). "Multiple Perspectives" text boxes feature two or more of the Rule of Four approaches (numerical, algebraic, graphical, verbal) to solving a given problem.
Book Description McGraw-Hill Science/Engineering/Math. PAPERBACK. Book Condition: New. 00725041610072504161 | 677.169 | 1 |
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Use these engaging real-world scenarios to help infuse your mathematics program with a problem-based approach to the kwledge and skills required by the Common Core State Standards for Mathematics.This collection of tasks addresses all of the Common Core State Standard categories for high school mathematics: Number and Quantity Algebra Functions Modeling Geometry Statistics and Probability Each Problem-Based Task is set in a meaningful context to engage student interest and reinforce the relevance of mathematics. Each is tightly aligned to one or more specific standards from the High School CCSS for Algebra I or Mathematics I. | 677.169 | 1 |
Popular computer algebra systems such as Maple, Macsyma, Mathematica, and REDUCE are now basic tools on most computers. Efficient algorithms for various algebraic operations underlie all these systems. Computer algebra, or algorithmic algebra, studies these algorithms and their properties and represents a rich intersection of theoretical computer science with classical mathematics
Bringing geometric algebra to the mainstream of physics pedagogy, Geometric Algebra and Applications to Physics not only presents geometric algebra as a discipline within mathematical physics, but the book also shows how geometric algebra can be applied to numerous fundamental problems in physics, especially in experimental situations. | 677.169 | 1 |
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Linear algebra is relatively easy for students during the early stages of the course, when the material is presented in a familiar, concrete setting. But when abstract concepts are introduced, students often hit a brick wall. Instructors seem to agree that certain concepts (such as linear independence, spanning, subspace, vector space, and linear transformations), are not easily understood, and require time to assimilate. Since they are fundamental to the study of linear algebra, students' understanding of these concepts is vital to their mastery of the subject. David Lay introduces these concepts early in a familiar, concrete "Rn" setting, develops them gradually, and returns to them again and again throughout the text so that when discussed in the abstract, these concepts are more accessible. | 677.169 | 1 |
A bestselling introductory course that covers all areas of trigonometry, including: the theory and equations of tangent, sine and cosine, dimensions and angles, radians, ratios, compound angles and circles related to triangles | 677.169 | 1 |
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Maths in Focus Author: Margaret Grove
ISBN: 9780170354622 Format: Book with Other Items Published: 28 August 2014 Country of Publication: AU Description: The bestselling Maths in Focus series is the trusted senior mathematics resource for students and teachers across New South Wales. The series is renowned for its clear structure and approach, and now comes in a smaller, more functional format supported by a range of digital solutions. Each book contains a wealth of fully-worked examples and explanations, as well as a logical progression of topics, graded exercises for continual revision and straightforward language. An interactive NelsonNetBook, which can be personalised and annotated, accompanies each student text, in addition to a NelsonNet student and teacher website. The NelsonNet teacher website includes teaching programs and notes, which support the use of the resources for assessment | 677.169 | 1 |
Publisher Description
Looking math students. The program is capable of plotting Cartesian, polar, table defined, as well as specialty graphs, such as trigonometric functions (sin, cos, tg, and others). Importantly, it features a simple data and formula input format, making it very practical for solving in-class and homework algebra or calculus problems. The program comes with customizable axis options (color, style, width, grid), and table data import/export options. The one feature that makes it very popular among both math teachers and students is that the graphs the program plots are fully interactive. This object-oriented approach lets the students zoom in and out, see the points of intercept and do much more. Also, the program allows multiple graphs plotting. GraphSight is much easier to use than most similar software titles. Programs user interface is well labeled and is extremely simple to follow.
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GraphSight Junior FREE GraphSight Junior is an easy to use freeware handy 2D math-graphing program. It was originally designed to help students and teachers satisfy their day-after-day math plotting needs. It makes it easy to plot and explore common Y(X) = F(X) Cartesian graphs, set text labels on the coordinate... Download | 677.169 | 1 |
MathBitsNotebook - Algebra 1 is a series of lesson and practice pages for students studying high school Algebra 1. These materials cover ALL standards stated in the Common Core State Standards for Mathematics (Appendix A, traditional pathway for Algebra 1), and coordinate with both SBAC and PARCC assessments. Differences reflected in the PARCC MCF are included and highlighted.
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Currently there is substantial exchange and communication between academic communities around the world as researchers endeavour to discover why so many children 'fail' at a subject that society deems crucial for future economic survival. This book charts current thinking and trends in teacher education around the world, and looks critically at the... more...
On the occasion of the celebration of ?Twenty Years of Didactique of Ma- ematics? in France, Jeremy Kilpatrick commented that though the works of Guy Brousseau are known through texts referring to them or mentioning their existence, the original texts are unknown, or known only with difficulty, in the non-Fren- speaking world. With very few exceptions,... more...
Do formulas exist for the solution to algebraical equations in one variable of any degree like the formulas for quadratic equations? The main aim of this book is to give new geometrical proof of Abel's theorem, as proposed by Professor V.I. Arnold. The theorem states that for general algebraical equations of a degree higher than 4, there are no formulas... more...
This book focuses on aspects of mathematical beliefs, from a variety of different perspectives. Current knowledge of the field is synthesized and existing boundaries are extended. The volume is intended for researchers in the field, as well as for mathematics educators teaching the next generation of students. more...
Didactics of Mathematics as a Scientific Discipline describes the state of the art in a new branch of science. Starting from a general perspective on the didactics of mathematics, the 30 original contributions to the book, drawn from 10 different countries, go on to identify certain subdisciplines and suggest an overall structure or `topology'... more...
Since abstract algebra is so important to the study of advanced mathematics, it is critical that students have a firm grasp of its principles and underlying theories before moving on to further study. To accomplish this, they require a concise, accessible, user-friendly textbook that is both challenging and stimulating. A First Graduate Course in Abstract... more...
A significantly revised and improved introduction to a critical aspect of scientific computation Matrix computations lie at the heart of most scientific computational tasks. For any scientist or engineer doing large-scale simulations, an understanding of the topic is essential. Fundamentals of Matrix Computations, Second Edition explains matrix computations... more...
The leading reference on probabilistic methods in combinatorics-now expanded and updated When it was first published in 1991, The Probabilistic Method became instantly the standard reference on one of the most powerful and widely used tools in combinatorics. Still without competition nearly a decade later, this new edition brings you up to speed... more...
The favorable reaction to the ?rst edition of this book con?rmed that the publication of such an application-oriented text on bifurcation theory of dynamical systems was well timed. The selected topics indeed cover - jor practical issues of applying the bifurcation theory to ?nite-dimensional problems. This new edition preserves the structure of the... more... | 677.169 | 1 |
Precise Calculator has arbitrary precision and can calculate with complex numbers, fractions, vectors and matrices. Has more than 150 mathematical functions and statistical functions and is programmable (if, goto, print, return, for). | 677.169 | 1 |
sixth edition of FUNDAMENTALS OF ALGEBRAIC MODELING strives to show the student connections between math and their daily lives. Algebraic modeling concepts and solutions are presented in non-threatening, easy-to-understand language with numerous step-by-step examples to illustrate ideas. Whether they are going on to study early childhood education, graphic arts, automotive technologies, criminal justice, or something else, students will discover that the practical applications of mathematical modeling will continue to be useful well after they have finished this course.
Available with InfoTrac® Student Collections
Features and Benefits
Laboratory Exercises at the end of each chapter get students involved in guided hands-on activities. Some are designed to be completed as individual assignments and others require group work.
Calculator Mini-Lessons encourage students to become more familiar with the functions of their calculators.
Each section ends with an extensive problem set allowing students to practice what they have learned. Each chapter has review problems and a chapter test with answers provided in the back of the book.
The Chapter Summary includes all important terms, equations and concepts in one page.
Chapter 8 Modeling with Statistics now includes a section on reading and interpreting graphical informationI give this text an A. I have looked at numerous other texts for the MAT 115 course, but none have been better suited than this book, in my opinion, to the Mathematical Modeling course."
— Lori Kiel
"I have also asked my students for their opinions on the textbook, and I heard the following things: They really like that the definitions and formulas they will need are highlighted in blue. They also said that mathematics textbooks either make the sections impossible to follow on your own, or so easy to follow on your own that you feel you do not need to attend class, so they stop going. They felt this textbook was right in the middle. They said they need to come to my lecture because it helps them to understand exactly what the examples are doing in the book, but they wouldn't be able to follow completely on their own. They like the way the book works directly with lecture and vise versa. "
— Erica M. Farelli
"We have not found any textbook that even runs a close second."
— Michael Bradshaw
Student Supplements | 677.169 | 1 |
Summary and Info
This book is wholeheartedly recommended to every student or user of mathematics. Although the author modestly describes his book as 'merely an attempt to talk about' algebra, he succeeds in writing an extremely original and highly informative essay on algebra and its place in modern mathematics and science. From the fields, commutative rings and groups studied in every university math course, through Lie groups and algebras to cohomology and category theory, the author shows how the origins of each algebraic concept can be related to attempts to model phenomena in physics or in other branches of mathematics. Comparable in style with Hermann Weyl's evergreen essay The Classical Groups, Shafarevich's new book is sure to become required reading for mathematicians, from beginners to experts.
More About the Author
Igor Rostislavovich Shafarevich (Russian: И́горь Ростисла́вович Шафаре́вич, born June 3, 1923) is a Russian mathematician who has contributed to algebraic number theory and algebraic geometry. | 677.169 | 1 |
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Unformatted text preview: Math 216: Differential Equations Lab 3: Higher-Order Numerical Methods, Linearity and Superposition Goals In this lab we have two goals: we will first examine some new numerical methods to find approximations to the solutions to differential equation: that is, we will test-drive some new numerical solvers. We will implement an improved Eulers method, and will use Matlabs built in numerical solver which is an implementation of the Runge-Kutta method. Our second goal is to use these solvers to explore solutions to a linear and a nonlinear differential equation. This will allow us to see how superposition works (for linear equations) and doesnt (for nonlinear ones). Application: a pendulum The angular position of a simple pendulum can be described by differential equations. Con- sider the pendulum shown below. If measures the angular displacement of the pendulum bob from the vertical, then 00 ( t ) + g L sin( ( t )) = f ( t ) , (1) where g is the acceleration due to gravity, L is the length of the pendulum, and f ( t ) is any forcing imposed on the system. For this lab we will consider f ( t ) = 0. Note that this equation is conspicuously nonlinear ! As a result, we are unable to solve it in terms of elementary functions. Because of this, we often use its linearization, 00 ( t ) + g L ( t ) = f ( t ) , (2) which is valid for small angles and is obviously much easier (possible!) to solve. For this lab, we will take f ( t ) = 0, g = 9 . 81 m/s 2 , and L = 1 m (quite a big pendulum). 1 Prelab assignment Before arriving in the lab, answer the following questions. You will need your answers in lab to work the problems, and your recitation instructor may check that you have brought them. These problems are to be handed in as part of your lab report. First, consider the linearized equation modeling the pendulums motion, (2), with the values indicated: 00 ( t ) + 9 . 81 ( t ) = 0 . 1. Suppose that we pull the pendulum back an angle of 4 and then release it. Write the initial conditions that correspond to this physical situation, and then solve the linearized pendulum equation with these initial conditions. 2. Next, suppose that instead of pulling the pendulum back and releasing it, we give it a tap so that it has an initial velocity, say 2 m/s. Write out the initial conditions for this physical situation, and then solve the linearized pendulum equation with these initial conditions. 3. Now consider the combination of these initial conditions: we pull the bob of the pen- dulum back an angle of 4 and also push it outwards with a speed of 2 m/s. Write down the initial conditions in this case. Then note that you can write down a solution to this initial problem using your work in the preceding two questions. Why are you able to do this? What is the solution to this initial value problem?...
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Solving 1-Step and 2-Step Equations Solution
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This MS PowerPoint presentation provides an easy-to-understand and classroom-tested technique for solving 1-Step and 2-Step Equations. I have used this presentation multiple times in my 7th and 8th grade math classes. The equation solution technique includes an easy-to-remember mnemonic that is appreciated by students. I recently reviewed this presentation with both of current Math I classes as a refresher for them in equation solving. It is fully animated and includes a lot of audio to keep the students engaged and focused. Please note that any of the audio or animations can be modified as desired by the teacher/end user | 677.169 | 1 |
MYTUTOR SUBJECT ANSWERS
How do I conquer Abstract Reasoning?
Abstract Reasoning (AR) is a new way of thinking which draws students to make links between shapes. Although it is a new and unfamiliar topic, the best way to tackle this section is using SCANS which should cover all the elements to look for in an AR questionAbout the author | 677.169 | 1 |
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