Commission on Higher Education
in collaboration with the Philippine Normal University
INITIAL RELEASE: 13 JUNE 2016
TEACHING GUIDE FOR SENIOR HIGH SCHOOL
Precalculus
CORE SUBJECT
This Teaching Guide was collaboratively developed and reviewed by educators from public
and private schools, colleges, and universities. We encourage teachers and other education
stakeholders to email their feedback, comments, and recommendations to the Commission on
Higher Education, K to 12 Transition Program Management Unit - Senior High School
Support Team at k12@ched.gov.ph. We value your feedback and recommendations.
Published by the Commission on Higher Education, 2016
Chairperson: Patricia B. Licuanan, Ph.D.
Commission on Higher Education
K to 12 Transition Program Management Unit
Office Address: 4th Floor, Commission on Higher Education,
C.P. Garcia Ave., Diliman, Quezon City
Telefax: (02) 441-1143 / E-mail Address: k12@ched.gov.ph
DEVELOPMENT TEAM
Team Leader: Dr. Ian June L. Garces
Writers: Dr. Jerico B. Bacani, Dr. Richard B. Eden,
Mr. Glenn Rey A. Estrada, Dr. Flordeliza F. Francisco,
Mr. Mark Anthony J. Vidallo
Technical Editors: Dr. Maria Alva Q. Aberin,
Dr. Flordeliza F. Francisco, Dr. Reginaldo M. Marcelo
Copy Reader: Naomi L. Tupas
Cover Artists: Paolo Kurtis N. Tan, Renan U. Ortiz
CONSULTANTS
THIS PROJECT WAS DEVELOPED WITH THE PHILIPPINE NORMAL UNIVERSITY.
University President: Ester B. Ogena, Ph.D.
VP for Academics: Ma. Antoinette C. Montealegre, Ph.D.
VP for University Relations & Advancement: Rosemarievic V. Diaz, Ph.D.
Ma. Cynthia Rose B. Bautista, Ph.D., CHED
Bienvenido F. Nebres, S.J., Ph.D., Ateneo de Manila University
Carmela C. Oracion, Ph.D., Ateneo de Manila University
Minella C. Alarcon, Ph.D., CHED
Gareth Price, Sheffield Hallam University
Stuart Bevins, Ph.D., Sheffield Hallam University
SENIOR HIGH SCHOOL SUPPORT TEAM
CHED K TO 12 TRANSITION PROGRAM MANAGEMENT UNIT
Program Director: Karol Mark R. Yee
Lead for Senior High School Support: Gerson M. Abesamis
Lead for Policy Advocacy and Communications: Averill M. Pizarro
Course Development Officers:
Danie Son D. Gonzalvo, John Carlo P. Fernando
Teacher Training Officers:
Ma. Theresa C. Carlos, Mylene E. Dones
Monitoring and Evaluation Officer: Robert Adrian N. Daulat
Administrative Officers: Ma. Leana Paula B. Bato,
Kevin Ross D. Nera, Allison A. Danao, Ayhen Loisse B. Dalena
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Introduction
As the Commission supports DepEd’s implementation of Senior High School (SHS), it upholds the vision
and mission of the K to 12 program, stated in Section 2 of Republic Act 10533, or the Enhanced Basic
Education Act of 2013, that “every graduate of basic education be an empowered individual, through a
program rooted on...the competence to engage in work and be productive, the ability to coexist in fruitful
harmony with local and global communities, the capability to engage in creative and critical thinking,
and the capacity and willingness to transform others and oneself.”
To accomplish this, the Commission partnered with the Philippine Normal University (PNU), the
National Center for Teacher Education, to develop Teaching Guides for Courses of SHS. Together with
PNU, this Teaching Guide was studied and reviewed by education and pedagogy experts, and was
enhanced with appropriate methodologies and strategies.
Furthermore, the Commission believes that teachers are the most important partners in attaining this
goal. Incorporated in this Teaching Guide is a framework that will guide them in creating lessons and
assessment tools, support them in facilitating activities and questions, and assist them towards deeper
content areas and competencies. Thus, the introduction of the SHS for SHS Framework.
The SHS for SHS Framework
The SHS for SHS Framework, which stands for “Saysay-Husay-Sarili for Senior High School,” is at the
core of this book. The lessons, which combine high-quality content with flexible elements to
accommodate diversity of teachers and environments, promote these three fundamental concepts:
SAYSAY: MEANING
HUSAY: MASTERY
SARILI: OWNERSHIP
Why is this important?
How will I deeply understand this?
What can I do with this?
Through this Teaching Guide,
teachers will be able to
facilitate an understanding of
the value of the lessons, for
each learner to fully engage in
the content on both the
cognitive and affective levels.
Given that developing mastery
goes beyond memorization,
teachers should also aim for deep
understanding of the subject
matter where they lead learners
to analyze and synthesize
knowledge.
When teachers empower
learners to take ownership of
their learning, they develop
independence and selfdirection, learning about both
the subject matter and
themselves.
About this Teaching Guide
The Pre-Calculus course bridges Basic Mathematics and Calculus. This course completes the
foundational knowledge on Algebra, Geometry, and Trigonometry of students who are
planning to take courses in the STEM track. It provides them with conceptual understanding
and computational skills that are crucial for Basic Calculus and future STEM courses.
Based on the Curriculum Guide for Pre-Calculus of the Department of Education, the primary
aim of this Teaching Guide is to give Math teachers adequate stand-alone material that can be
used for each session of the Grade 11 Pre-Calculus course.
The Guide is divided into three units: Analytic Geometry, Summation Notation and
Mathematical Induction, and Trigonometry. Each unit is composed of lessons that bring
together related learning competencies in the unit. Each lesson is further divided into sublessons that focus on one or two competencies for effective teaching and learning. Each sublesson is designed for a one-hour session, but the teachers have the option to extend the time
allotment to one-and-a-half hours for some sub-lessons.
Each sub-lesson ends with a Seatwork/ Homework, which consists of exercises related to the
topic being discussed in the sub-lesson. As the title suggests, these exercises can be done in
school (if time permits) or at home. Moreover, at the end of each lesson is a set of exercises
(simply tagged as Exercises) that can be used for short quizzes and long exams. Answers,
solutions, or hints to most items in Seatwork/ Homework and Exercises are provided to guide
the teachers as they solve them.
Some items in this Guide are marked with a star. A starred sub-lesson is optional and it is
suggested that these be taken only if time permits. A starred example or exercise requires the
use of a calculator.
To further guide the teachers, Teaching Notes are provided on the margins. These notes
include simple recall of basic definitions and theorems, suggested teaching methods,
alternative answers to some exercises, quick approaches and techniques in solving particular
problems, and common errors committed by students.
We hope that Pre-Calculus teachers will find this Teaching Guide helpful and convenient to
use. We encourage the teachers to study this Guide carefully and solve the exercises
themselves. Although great effort has been put to this Guide for technical correctness and
precision, any mistake found and reported to the Team is a gain for other teachers. Thank you
for your cooperation.
The Parts of the Teaching Guide
This Teaching Guide is mapped and aligned to the DepEd SHS Curriculum, designed to be highly usable
for teachers. It contains classroom activities and pedagogical notes, and integrated with innovative
pedagogies. All of these elements are presented in the following parts:
1. INTRODUCTION
•
Highlight key concepts and identify the essential questions
•
Show the big picture
•
Connect and/or review prerequisite knowledge
•
Clearly communicate learning competencies and objectives
•
Motivate through applications and connections to real-life
2. MOTIVATION
•
Give local examples and applications
•
Engage in a game or movement activity
•
Provide a hands-on/laboratory activity
•
Connect to a real-life problem
3. INSTRUCTION/DELIVERY
•
Give a demonstration/lecture/simulation/hands-on activity
•
Show step-by-step solutions to sample problems
•
Give applications of the theory
•
Connect to a real-life problem if applicable
4. PRACTICE
•
Provide easy-medium-hard questions
•
Give time for hands-on unguided classroom work and discovery
•
Use formative assessment to give feedback
5. ENRICHMENT
•
Provide additional examples and applications
•
Introduce extensions or generalisations of concepts
•
Engage in reflection questions
•
Encourage analysis through higher order thinking prompts
•
Allow pair/small group discussions
•
Summarize and synthesize the learnings
6. EVALUATION
•
Supply a diverse question bank for written work and exercises
•
Provide alternative formats for student work: written homework, journal, portfolio, group/
individual projects, student-directed research project
On DepEd Functional Skills and CHED’s College Readiness Standards
As Higher Education Institutions (HEIs) welcome the graduates of the Senior High School program, it is
of paramount importance to align Functional Skills set by DepEd with the College Readiness Standards
stated by CHED.
The DepEd articulated a set of 21st century skills that should be embedded in the SHS curriculum across
various subjects and tracks. These skills are desired outcomes that K to 12 graduates should possess in
order to proceed to either higher education, employment, entrepreneurship, or middle-level skills
development.
On the other hand, the Commission declared the College Readiness Standards that consist of the
combination of knowledge, skills, and reflective thinking necessary to participate and succeed - without
remediation - in entry-level undergraduate courses in college.
The alignment of both standards, shown below, is also presented in this Teaching Guide - prepares
Senior High School graduates to the revised college curriculum which will initially be implemented by
AY 2018-2019.
College Readiness Standards Foundational Skills
DepEd Functional Skills
Produce all forms of texts (written, oral, visual, digital) based on:
1. Solid grounding on Philippine experience and culture;
2. An understanding of the self, community, and nation;
3. Application of critical and creative thinking and doing processes;
4. Competency in formulating ideas/arguments logically, scientifically,
and creatively; and
5. Clear appreciation of one’s responsibility as a citizen of a multicultural
Philippines and a diverse world;
Visual and information literacies
Media literacy
Critical thinking and problem solving skills
Creativity
Initiative and self-direction
Systematically apply knowledge, understanding, theory, and skills
for the development of the self, local, and global communities using
prior learning, inquiry, and experimentation
Global awareness
Scientific and economic literacy
Curiosity
Critical thinking and problem solving skills
Risk taking
Flexibility and adaptability
Initiative and self-direction
Work comfortably with relevant technologies and develop
adaptations and innovations for significant use in local and global
communities;
Global awareness
Media literacy
Technological literacy
Creativity
Flexibility and adaptability
Productivity and accountability
Communicate with local and global communities with proficiency,
orally, in writing, and through new technologies of communication;
Global awareness
Multicultural literacy
Collaboration and interpersonal skills
Social and cross-cultural skills
Leadership and responsibility
Interact meaningfully in a social setting and contribute to the
fulfilment of individual and shared goals, respecting the
fundamental humanity of all persons and the diversity of groups
and communities
Media literacy
Multicultural literacy
Global awareness
Collaboration and interpersonal skills
Social and cross-cultural skills
Leadership and responsibility
Ethical, moral, and spiritual values
Table of Contents
About This Teaching Guide
1
DepEd Curriculum Guide for Precalculus
2
Unit 1:
6
Analytic Geometry (19 one-hour sessions)
Lesson 1.1: Introduction to Conic Sections and Circles . . . . . . . .
7
1.1.1: An Overview of Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.1.2: Definition and Equation of a Circle . . . . . . . . . . . . . . . . . . . . . . .
8
1.1.3: More Properties of Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.1.4: Situational Problems Involving Circles. . . . . . . . . . . . . . . . . . . . 12
Lesson 1.2: Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
1.2.1: Definition and Equation of a Parabola . . . . . . . . . . . . . . . . . . . . 19
1.2.2: More Properties of Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
1.2.3: Situational Problems Involving Parabolas . . . . . . . . . . . . . . . . 27
Lesson 1.3: Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
1.3.1: Definition and Equation of an Ellipse . . . . . . . . . . . . . . . . . . . . . 32
1.3.2: More Properties of Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
1.3.3: Situational Problems Involving Ellipses . . . . . . . . . . . . . . . . . . . 40
Lesson 1.4: Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
1.4.1: Definition and Equation of a Hyperbola . . . . . . . . . . . . . . . . . . 45
1.4.2: More Properties of Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
1.4.3: Situational Problems Involving Hyperbolas . . . . . . . . . . . . . . . 54
Lesson 1.5: More Problems on Conic Sections . . . . . . . . . . . . . . . . 59
1.5.1: Identifying the Conic Section by Inspection . . . . . . . . . . . . . . . 59
1.5.2: Problems Involving Di↵erent Conic Sections . . . . . . . . . . . . . . 61
iii
Lesson 1.6: Systems of Nonlinear Equations . . . . . . . . . . . . . . . . . . 66
1.6.1: Review of Techniques in Solving Systems of Linear
Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
1.6.2: Solving Systems of Equations Using Substitution . . . . . . . . . 68
1.6.3: Solving Systems of Equations Using Elimination . . . . . . . . . . 71
1.6.4: Applications of Systems of Nonlinear Equations . . . . . . . . . . 75
Unit 2:
Mathematical Induction (10 one-hour sessions)
80
Lesson 2.1: Review of Sequences and Series . . . . . . . . . . . . . . . . . . . 81
Lesson 2.2: Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
2.2.1: Writing and Evaluating Sums in Sigma Notation . . . . . . . . . 86
2.2.2: Properties of Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
Lesson 2.3: Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . . . . . 95
2.3.1: Proving Summation Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
2.3.2: Proving Divisibility Statements . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
? 2.3.3: Proving Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
Lesson 2.4: The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
2.4.1: Pascal’s Triangle and the Concept of Combination . . . . . . . . 108
2.4.2: The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
2.4.3: Terms of a Binomial Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . 114
? 2.4.4: Approximation and Combination Identities . . . . . . . . . . . . . . . 116
Unit 3:
Trigonometry (29 one-hour sessions)
121
Lesson 3.1: Angles in a Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
3.1.1: Angle Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
3.1.2: Coterminal Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
3.1.3: Arc Length and Area of a Sector . . . . . . . . . . . . . . . . . . . . . . . . . 128
Lesson 3.2: Circular Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
3.2.1: Circular Functions on Real Numbers . . . . . . . . . . . . . . . . . . . . . 135
3.2.2: Reference Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
Lesson 3.3: Graphs of Circular Functions and Situational
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
3.3.1: Graphs of y = sin x and y = cos x . . . . . . . . . . . . . . . . . . . . . . . . 144
3.3.2: Graphs of y = a sin bx and y = a cos bx . . . . . . . . . . . . . . . . . . . 146
3.3.3: Graphs of y = a sin b(x c) + d and
y = a cos b(x c) + d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
3.3.4: Graphs of Cosecant and Secant Functions . . . . . . . . . . . . . . . . 154
3.3.5: Graphs of Tangent and Cotangent Functions . . . . . . . . . . . . . 159
3.3.6: Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
Lesson 3.4: Fundamental Trigonometric Identities . . . . . . . . . . . . . 176
3.4.1: Domain of an Expression or Equation . . . . . . . . . . . . . . . . . . . . 176
3.4.2: Identity and Conditional Equation . . . . . . . . . . . . . . . . . . . . . . . 178
3.4.3: The Fundamental Trigonometric Identities . . . . . . . . . . . . . . . 180
3.4.4: Proving Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . 183
Lesson 3.5: Sum and Di↵erence Identities . . . . . . . . . . . . . . . . . . . . . 189
3.5.1: The Cosine Di↵erence and Sum Identities . . . . . . . . . . . . . . . . 189
3.5.2: The Cofunction Identities and the Sine Sum and
Di↵erence Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192
3.5.3: The Tangent Sum and Di↵erence Identities . . . . . . . . . . . . . . . 195
Lesson 3.6: Double-Angle and Half-Angle Identities . . . . . . . . . . . 199
3.6.1: Double-Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
3.6.2: Half-Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
Lesson 3.7: Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . 208
3.7.1: Inverse Sine Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
3.7.2: Inverse Cosine Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
3.7.3: Inverse Tangent Function and the Remaining Inverse
Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218
Lesson 3.8: Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 233
3.8.1: Solutions of a Trigonometric Equation . . . . . . . . . . . . . . . . . . . . 234
3.8.2: Equations with One Term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
3.8.3: Equations with Two or More Terms . . . . . . . . . . . . . . . . . . . . . . 241
Lesson 3.9: Polar Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . 251
3.9.1: Polar Coordinates of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252
3.9.2: From Polar to Rectangular, and Vice Versa . . . . . . . . . . . . . . . 257
3.9.3: Basic Polar Graphs and Applications . . . . . . . . . . . . . . . . . . . . . 260
References
273
2
Semester: First Semester
No. of Hours/ Semester: 80 hours/ semester
Pre-requisite (if needed):
key concepts of
conic sections and
systems of
nonlinear
equations
The learners
demonstrate an
understanding
of...
CONTENT
STANDARDS
model situations
appropriately and solve
problems accurately using
conic sections and systems
of nonlinear equations
The learners shall be able
to...
PERFORMANCE
STANDARDS
graph a circle in a rectangular coordinate system
4.
5.
6.
7.
8.
9.
10.
11.
12.
define a parabola
determine the standard form of equation of a parabola
graph a parabola in a rectangular coordinate system
define an ellipse
determine the standard form of equation of an ellipse
graph an ellipse in a rectangular coordinate system
define a hyperbola
determine the standard form of equation of a hyperbola
determine the standard form of equation of a circle
3.
2.
illustrate the different types of conic sections: parabola, ellipse,
circle, hyperbola, and degenerate cases.***
define a circle.
LEARNING COMPETENCIES
1.
The learners...
K to 12 Senior High School STEM Specialized Subject – Pre-Calculus December 2013
Analytic
Geometry
CONTENT
Page 1 of 4
STEM_PC11AG-Ia-5
STEM_PC11AG-Ib-1
STEM_PC11AG-Ib-2
STEM_PC11AG-Ic-1
STEM_PC11AG-Ic-2
STEM_PC11AG-Ic-3
STEM_PC11AG-Id-1
STEM_PC11AG-Id-2
STEM_PC11AG-Ia-4
STEM_PC11AG-Ia-3
STEM_PC11AG-Ia-2
STEM_PC11AG-Ia-1
CODE
Subject Description: At the end of the course, the students must be able to apply concepts and solve problems involving conic sections, systems of nonlinear equations,
series and mathematical induction, circular and trigonometric functions, trigonometric identities, and polar coordinate system.
Grade: 11
Core Subject Title: Pre-Calculus
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
3
key concepts of
series and
mathematical
induction and the
Binomial
Theorem.
CONTENT
STANDARDS
keenly observe and
investigate patterns, and
formulate appropriate
mathematical statements
and prove them using
mathematical induction
and/or Binomial Theorem.
PERFORMANCE
STANDARDS
recognize the equation and important characteristics of the
different types of conic sections
solves situational problems involving conic sections
illustrate systems of nonlinear equations
determine the solutions of systems of nonlinear equations using
techniques such as substitution, elimination, and graphing***
solve situational problems involving systems
of nonlinear equations
14.
15.
16.
17.
differentiate a series from a sequence
use the sigma notation to represent a series
illustrate the Principle of Mathematical Induction
apply mathematical induction in proving identities
illustrate Pascal’s Triangle in the expansion of 𝑥 + 𝑦 𝑛 for small
positive integral values of 𝑛
prove the Binomial Theorem
determine any term of 𝑥 + 𝑦 𝑛 , where 𝑛 is a positive integer,
without expanding
solve problems using mathematical induction and the Binomial
Theorem
2.
3.
4.
5.
6.
9.
7.
8.
illustrate a series
1.
18.
graph a hyperbola in a rectangular coordinate system
LEARNING COMPETENCIES
13.
K to 12 Senior High School STEM Specialized Subject – Pre-Calculus December 2013
Series and
Mathematical
Induction
CONTENT
STEM_PC11SMI-Ih-1
STEM_PC11AG-Ig-2
STEM_PC11AG-If-g-1
STEM_PC11AG-If-1
STEM_PC11AG-Ie-2
STEM_PC11AG-Ie-1
STEM_PC11AG-Id-3
CODE
Page 2 of 4
STEM_PC11SMI-Ij-2
STEM_PC11SMI-Ij-1
STEM_PC11SMI-Ii-3
STEM_PC11SMI-Ii-2
STEM_PC11SMI-Ih-2
STEM_PC11SMI-Ih-3
STEM_PC11SMI-Ih-4
STEM_PC11SMI-Ih-i-1
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
4
CONTENT
STANDARDS
key concepts of
circular functions,
trigonometric
identities, inverse
trigonometric
functions, and
the polar
coordinate
system
4. formulate and solve
accurately situational
problems involving the
polar coordinate system
3. formulate and solve
accurately situational
problems involving
appropriate trigonometric
functions
2. apply appropriate
trigonometric identities in
solving situational
problems
PERFORMANCE
STANDARDS
1. formulate and solve
accurately situational
problems involving
circular functions
illustrate the different circular functions
uses reference angles to find exact values of circular functions
determine the domain and range of the different circular functions
graph the six circular functions (a) amplitude, (b) period, and (c)
phase shift
solve problems involving circular functions
determine whether an equation is an identity or a conditional
equation
derive the fundamental trigonometric identities
derive trigonometric identities involving sum and difference of
angles
derive the double and half-angle formulas
simplify trigonometric expressions
prove other trigonometric identities
solve situational problems involving trigonometric identities
illustrate the domain and range of the inverse trigonometric
functions.
evaluate an inverse trigonometric expression.
solve trigonometric equations.
solve situational problems involving inverse trigonometric
functions and trigonometric equations
locate points in polar coordinate system
convert the coordinates of a point from rectangular to polar
systems and vice versa
solve situational problems involving polar coordinate system
3.
4.
5.
6.
7.
22.
20.
21.
17.
18.
19.
12.
13.
14.
15.
16.
10.
11.
8.
9.
illustrate angles in standard position and coterminal angles
2.
LEARNING COMPETENCIES
illustrate the unit circle and the relationship between the linear
and angular measures of a central angle in a unit circle
convert degree measure to radian measure and vice versa
1.
K to 12 Senior High School STEM Specialized Subject – Pre-Calculus December 2013
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Trigonometry
CONTENT
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Competency
Week
Quarter
Domain/Content/
Component/ Topic
Grade Level
illustrate the different types
of conic sections: parabola,
ellipse, circle, hyperbola,
and degenerate cases
Week one
First Quarter
Analytic Geometry
Grade 11
Science, Technology,
Engineering and Mathematics
Pre-Calculus
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LEGEND
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a
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Series and Mathematical Induction
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Unit 1
Analytic Geometry
https://commons.wikimedia.org/wiki/File%3ASan Juanico Bridge 2.JPG
By Morten Nærbøe (Own work)
[CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0)
or GFDL (http://www.gnu.org/copyleft/fdl.html)],
via Wikimedia Commons
Stretching from Samar to Leyte with a total length of 2.16 kilometers, the San
Juanico Bridge has served as one of the main thoroughfares of economic and social
development in the country since its completion in 1973. Adding picturesque
e↵ect on the whole architecture, geometric structures are subtly built to serve
other purposes. The arch-shaped support on the main span of the bridge helps
maximize its strength to withstand mechanical resonance and aeroelastic flutter
brought about by heavy vehicles and passing winds.
Lesson 1.1. Introduction to Conic Sections and Circles
Time Frame: 4 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate the di↵erent types of conic sections: parabola, ellipse, circle, hyperbola, and degenerate cases;
(2) define a circle;
(3) determine the standard form of equation of a circle;
(4) graph a circle in a rectangular coordinate system; and
(5) solve situational problems involving conic sections (circles).
Lesson Outline
(1) Introduction of the four conic sections, along with the degenerate conics
(2) Definition of a circle
(3) Derivation of the standard equation of a circle
(4) Graphing circles
(5) Solving situational problems involving circles
Introduction
We introduce the conic sections, a particular class of curves which sometimes
appear in nature and which have applications in other fields. In this lesson, we
discuss the first of their kind, circles. The other conic sections will be covered in
the next lessons.
1.1.1. An Overview of Conic Sections
We introduce the conic sections (or conics), a particular class of curves which
oftentimes appear in nature and which have applications in other fields. One
of the first shapes we learned, a circle, is a conic. When you throw a ball, the
trajectory it takes is a parabola. The orbit taken by each planet around the sun
is an ellipse. Properties of hyperbolas have been used in the design of certain
telescopes and navigation systems. We will discuss circles in this lesson, leaving
parabolas, ellipses, and hyperbolas for subsequent lessons.
• Circle (Figure 1.1) - when the plane is horizontal
• Ellipse (Figure 1.1) - when the (tilted) plane intersects only one cone to form
a bounded curve
7
• Parabola (Figure 1.2) - when the plane intersects only one cone to form an
unbounded curve
• Hyperbola (Figure 1.3) - when the plane (not necessarily vertical) intersects
both cones to form two unbounded curves (each called a branch of the hyperbola)
Figure 1.1
Figure 1.2
Figure 1.3
We can draw these conic sections (also called conics) on a rectangular coordinate plane and find their equations. To be able to do this, we will present
equivalent definitions of these conic sections in subsequent sections, and use these
to find the equations.
There are other ways for a plane and the cones to intersect, to form what are
referred to as degenerate conics: a point, one line, and two lines. See Figures 1.4,
1.5 and 1.6.
Figure 1.4
Figure 1.5
Figure 1.6
1.1.2. Definition and Equation of a Circle
A circle may also be considered a special kind of ellipse (for the special case when
the tilted plane is horizontal). For our purposes, we will distinguish between
these two conics.
See Figure 1.7, with the point C(3, 1) shown. From the figure, the distance
of A( 2, 1) from p
C is AC = 5. By the distance formula, the distance of B(6, 5)
from C is BC = (6 3)2 + (5 1)2 = 5. There are other points P such that
P C = 5. The collection of all such points which are 5 units away from C, forms
a circle.
8
Figure 1.7
Figure 1.8
Let C be a given point. The set of all points P having the same
distance from C is called a circle. The point C is called the center of
the circle, and the common distance its radius.
The term radius is both used to refer to a segment from the center C to a
point P on the circle, and the length of this segment.
See Figure 1.8, where a circle is drawn. It has center C(h, k) and radius r > 0.
A point P (x, y) is on the circle if and only if P C = r. For any such point then,
its coordinates should satisfy the following.
p
(x
(x
PC = r
h)2 + (y
h)2 + (y
k)2 = r
k)2 = r2
This is the standard equation of the circle with center C(h, k) and radius r. If
the center is the origin, then h = 0 and k = 0. The standard equation is then
x2 + y 2 = r 2 .
Example 1.1.1. In each item, give the
standard equation of the circle satisfying the given conditions.
(1) center at the origin, radius 4
p
(2) center ( 4, 3), radius 7
(3) circle in Figure 1.7
(4) circle A in Figure 1.9
(5) circle B in Figure 1.9
(6) center (5, 6), tangent to the yaxis
Figure 1.9
9
(7) center (5, 6), tangent to the x-axis
(8) It has a diameter with endpoints A( 1, 4) and B(4, 2).
Solution. (1) x2 + y 2 = 16
(2) (x + 4)2 + (y
3)2 = 7
(3) The center is (3, 1) and the radius is 5, so the equation is (x
25.
3)2 + (y
1)2 =
(4) By inspection, the center is ( 2, 1) and the radius is 4. The equation is
(x + 2)2 + (y + 1)2 = 16.
(5) Similarly by inspection, we have (x
3)2 + (y
2)2 = 9.
(6) The center is 5 units away from the y-axis, so the radius is r = 5 (you can
make a sketch to see why). The equation is (x 5)2 + (y + 6)2 = 25.
(7) Similarly, since the center is 6 units away from the x-axis, the equation is
(x 5)2 + (y + 6)2 = 36.
(8) The center C is the midpoint of A and B: C = 1+4
, 4+2 = 32 , 3 . The
2 q 2
q
2
29
radius is then r = AC =
. The circle has
1 32 + (4 3)2 =
4
equation x
3 2
2
+ (y
3)2 =
29
.
4
2
Seatwork/Homework 1.1.2
Find the standard equation of the circle being described in each item.
p
(1) Center at the origin, radius 11
Answer: x2 + y 2 = 11
Answer: (x + 6)2 + (y
(2) Center ( 6, 7), tangent to the y-axis
7)2 = 36
(3) It has a diameter with endpoints A( 3, 2) and B(7, 4).
Answer: (x
1.1.3. More Properties of Circles
After expanding, the standard equation
✓
◆2
3
x
+ (y
2
3)2 =
29
4
can be rewritten as
x2 + y 2
3x
6y
5 = 0,
an equation of the circle in general form.
If the equation of a circle is given in the general form
Ax2 + Ay 2 + Cx + Dy + E = 0,
10
A 6= 0,
2)2 + (y
3)2 = 26
or
x2 + y 2 + Cx + Dy + E = 0,
we can determine the standard form by completing the square in both variables.
Completing the square in an expression like x2 + 14x means determining
the term to be added that will produce a perfect polynomial square. Since the
coefficient of x2 is already 1, we take half the coefficient of x and square it, and
we get 49. Indeed, x2 + 14x + 49 = (x + 7)2 is a perfect square. To complete
the square in, say, 3x2 + 18x, we factor the coefficient of x2 from the expression:
3(x2 + 6x), then add 9 inside. When completing a square in an equation, any
extra term introduced on one side should also be added to the other side.
Example 1.1.2. Identify the center and radius of the circle with the given equation in each item. Sketch its graph, and indicate the center.
(1) x2 + y 2
2
(2) x + y
2
6x = 7
14x + 2y =
(3) 16x2 + 16y 2 + 96x
14
40y = 315
Solution. The first step is to rewrite each equation in standard form by completing the square in x and in y. From the standard equation, we can determine the
center and radius.
(1)
x2 6x + y 2 = 7
x2 6x + 9 + y 2 = 7 + 9
(x 3)2 + y 2 = 16
Center (3, 0), r = 4, Figure 1.10
(2)
x2
x2 14x + y 2 + 2y = 14
14x + 49 + y 2 + 2y + 1 = 14 + 49 + 1
(x 7)2 + (y + 1)2 = 36
Center (7, 1), r = 6, Figure 1.11
(3)
16x2 + 96x + 16y 2 40y = 315
✓
◆
5
2
2
16(x + 6x) + 16 y
y = 315
2
✓
◆
✓ ◆
5
25
25
2
2
16(x + 6x + 9) + 16 y
y+
= 315 + 16(9) + 16
2
16
16
11
Teaching Notes
Recall the
technique of
completing the
square. This was
introduced in
Grade 9.
✓
2
Teaching Notes
A common mistake
committed by
students is to add 9
and 25
only. They
16
16(x + 3) + 16 y
✓
2
(x + 3) + y
Center
3, 54 , r = 5.5, Figure 1.12.
5
4
5
4
◆2
◆2
= 484
484
121
=
=
=
16
4
✓
11
2
◆2
2
often forget the
multiplier outside
the parenthesis.
Figure 1.10
Figure 1.11
Figure 1.12
In the standard equation (x h)2 + (y k)2 = r2 , both the two squared
terms on the left side have coefficient 1. This is the reason why in the preceding
example, we divided by 16 at the last equation.
Seatwork/Homework 1.1.3
Identify the center and radius of the circle with the given equation in each item.
Sketch its graph, and indicate the center.
(1) x2 + y 2
5x + 4y = 46
Answer: center
(2) 4x2 + 4y 2 + 40x
5
,
2
2 , radius
15
2
= 7.5, Figure 1.13
32y = 5
Answer: center ( 5, 4), radius
13
2
= 6.5, Figure 1.14
Figure 1.13
Figure 1.14
1.1.4. Situational Problems Involving Circles
We now consider some situational problems involving circles.
12
? Example 1.1.3. A street with two lanes, each 10 ft wide, goes through a
semicircular tunnel with radius 12 ft. How high is the tunnel at the edge of each
lane? Round o↵ to 2 decimal places.
Figure 1.15
Solution. We draw a coordinate system with origin at the middle of the highway,
as shown in Figure 1.15. Because of the given radius, the tunnel’s boundary is
on the circle x2 + y 2 = 122 . Point P is the point on the arc just above the edge
of a lane, so its x-coordinate is 10. Wepneed its y-coordinate. We then solve
102 + y 2 = 122 for y > 0, giving us y = 2 11 ⇡ 6.63 ft.
2
Example 1.1.4. A piece of a broken plate was dug up in an archaeological site.
It was put on top of a grid, as shown in Figure 1.16, with the arc of the plate
passing through A( 7, 0), B(1, 4) and C(7, 2). Find its center, and the standard
equation of the circle describing the boundary of the plate.
Figure 1.16
Figure 1.17
Solution. We first determine the center. It is the intersection of the perpendicular
bisectors of AB and BC (see Figure 1.17). Recall that, in a circle, the perpendicular bisector of any chord passes through the center. Since the midpoint M
13
Teaching Notes
A perpendicular
bisector of a
segment is the line
that passes
through the
midpoint of the
segment and is
perpendicular to
the segment.
of AB is 7+1
, 0+4
= ( 3, 2), and mAB = 41+70 = 12 , the perpendicular bisector
2
2
of AB has equation y 2 = 2(x + 3), or equivalently, y = 2x 4.
Since the midpoint N of BC is 1+7
, 4+2
= (4, 3), and mBC = 27 41 = 13 ,
2
2
the perpendicular bisector of BC has equation y 3 = 3(x 4), or equivalently,
y = 3x 9.
The intersection of the two lines y = 2x 4 and y = 3x 9 is (1, 6) (by
solving a system of linear equations). We can take the radius as the distance of
this point from any of A, B or C (it’s most convenient to use B in this case). We
then get r = 10. The standard equation is thus (x 1)2 + (y + 6)2 = 100.
2
Seatwork/Homework 1.1.4
? 1. A single-lane street 10 ft wide goes through a semicircular tunnel with radius
9 ft. How high is the tunnel at the edge of each lane? Round o↵ to 2 decimal
places.
Answer: 7.48 ft
2. An archeologist found the remains of an ancient wheel, which she then placed
on a grid. If an arc of the wheel passes through A( 7, 0), B( 3, 4) and C(7, 0),
locate the center of the wheel, and the standard equation of the circle defining
its boundary.
Answer: (0, 3), x2 + (y + 3)2 = 58
Exercises 1.1
1. Identify the center and radius of the circle with the given equation in each
item. Sketch its graph, and indicate the center.
(a) x2 + y 2 = 49
Answer: center (0, 0), r = 7
(b) 4x2 + 4y 2 = 25
Answer: center (0, 0), r =
7 2
4
2
(c) x
+ y+
(d) x2 + y
12x
(e) x2 + y 2 + 8x
3 2
4
=
169
16
10y =
Answer: center
12
3
4
,r=
13
4
Answer: center (6, 5), r = 7
9y = 6
(f) x2 + y 2 + 10x + 12y =
7
,
4
5
2
Answer: center ( 4, 4.5), r =
13
2
12
Answer: center ( 5, 6), r = 7
(g) 2x2 + 2y 2
14x + 18y = 7
(h) 4x2 + 4y 2
20x + 40y =
Answer: center (3.5, 4.5), r = 6
p
Answer: center (2.5, 5), r = 30
p
7
14
Answer: center
,
,
r
=
2
5
3
3
2
5
2
(i) 9x + 9y + 42x + 84y + 65 = 0
14
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
2. Find the standard equation of the circle which satisfies the given conditions.
p
(a) center at the origin, radius 2 2
Answer: x2 + y 2 = 8
(b) center at (15, 20), radius 9
(c) center at (5, 6), through (9, 4)
Answer: (x
Answer: (x
15)2 + (y + 20)2 = 81
5)2 + (y
6)2 = 20
Solution. The radius is the distance from the center to (9, 4):
p
p
(5 9)2 + (6 4)2 = 20.
(d) center at ( 2, 3), tangent to the x-axis
Answer: (x + 2)2 + (y
3)2 = 9
(e) center at ( 2, 3), tangent to the y-axis
Answer: (x + 2)2 + (y
3)2 = 4
(f) center at ( 2, 3), tangent to the line y = 8
Answer: (x + 2)2 + (y
3)2 = 25
15
Teaching Notes
To determine the
equation of a
circle, we just need
to determine the
center and the
radius.
Solution. We need to determine the radius. This is best done by
sketching the center and line, to see that the center ( 2, 3) is 5 units
away from the nearest point on the line, ( 2, 8) (which is the point of
tangency).
(g) center at ( 2, 3), tangent to the line x = 10
Answer: (x + 2)2 + (y
3)2 = 64
(h) center in the third quadrant, tangent to both the x-axis and y-axis,
radius 7
Answer: (x + 7)2 + (y + 7)2 = 49
(i) a diameter with endpoints ( 9, 2) and (15, 12)
Answer: (x
(j) concentric with x2 + y 2 + 2x
3)2 + (y
4y = 5, radius is 7
Answer: (x + 1)2 + (y
7)2 = 169
2)2 = 49
Solution. Two circles are said to be concentric if they have the same
center.
The standard equation of the given circle is
(x + 1)2 + (y 2)2 = 10. Thus, the circle we’re looking for has center
( 1, 2) and radius 7.
(k) concentric with x2 + y 2
8x
10y = 16 and 4 times the area
Answer: (x 4)2 + (y 5)2 = 100
Solution. The given circle has standard equation
(x
4)2 + (y
5)2 = 52 .
Its radius is 5, so its area is 25⇡ sq. units. The circle we are looking
for should have area 100⇡ sq. units, so its radius is 10.
(l) concentric with x2 + y 2
x2 + y 2 14x + 6y = 33
10x
6y =
2, same radius as
Answer: (x 5)2 + (y 3)2 = 25
(m) center at C(3, 4), tangent to the line y = 13 x 13
Answer: (x
Teaching Notes
The radius drawn
to a point on the
circle is
perpendicular to
the line tangent to
the circle at that
point.
3)2 + (y
4)2 = 10
Solution. (A sketch will greatly help in understanding the argument.)
If P is the point of tangency, then line CP is perpendicular to the given
tangent line. Since the tangent line has slope 13 , line CP has slope 3.
Because it passes through C, line CP has equation y 4 = 3(x 3),
or y = 3x + 13. Solving the system {y = 13 x 13 , y = 3x + 13}
yields p
x = 4 and y = 1, the p
coordinates of P . The radius is then
2
2
CP = (3 4) + (4 1) = 10.
(n) center at ( 4, 3), tangent to the line y = 4x 30
Answer: (x + 4)2 + (y
16
3)2 = 17
Solution. (Similar to the previous problem) Let P be the point of
tangency, so line CP is perpendicular to the tangent line. The tangent
line has slope 4, so line CP has slope 14 . Line CP passes through C,
so it has equation y 3 = 14 (x + 4), or y = 14 x + 4. Solving the system
{y = 4x 30, y = 14 x + 4} yields x = 8 and y = 2, the coordinates
p
p
of P . The radius is then CP = ( 4 + 8)2 + (3 2)2 = 17.
? 3. A seismological station is located at (0, 3), 3 km away from a straight
shoreline where the x-axis runs through. The epicenter of an earthquake
was determined to be 6 km away from the station.
(a) Find the equation of the curve that contains the possible location of
the epicenter.
Answer: x2 + (y + 3)2 = 62
(b) If furthermore, the epicenter was determined to be 2 km away from
the shore, find its possible coordinates (rounded o↵ to two decimal
places).
Answer: (±3.32, 2)
Solution. Since the epicenter is 6 units away from (0, 3), it could be any
of the points of a circle with center (0, 3) and radius 6. The equation is
then x2 + (y + 3)2 = 62 . Next, we solve thispequation for x if y = 2, and we
get x2 = 62 (2 + 3)2 = 11, and so x = ± 11 ⇡ ±3.32.
4. A ferris wheel is elevated 1 m above ground. When a car reaches the highest
point on the ferris wheel, its altitude from ground level is 31 m. How far
away from the center, horizontally, is the car when it is at an altitude of
25 m?
Answer: 12 m
Solution. The ferris wheel, as shown
below, is drawn 1 unit above the xaxis (ground level), center on the yaxis, and highest point at y = 31.
The diameter is thus 30, and the radius 15. We locate the center at
(0, 16), and write the equation of the
circle as x2 + (y 16)2 = 152 .
If y = 25, we have x2 + (25 16)2 =
152 , so x2 = 152 92 = 144, and
x = ±12. (Clearly, there are two
points on the ferris wheel at an altitude of 25 m.) Thus, the car is 12 m
away horizontally from the center.
? 5. A window is to be constructed as shown, with its upper boundary the arc
of a circle having radius 4 ft and center at the midpoint of base AD. If the
17
vertical side is to be 34 as long as the base, find the dimensions (vertical side
and base) of this window. Round o↵ your final answer to 1 decimal place.
Answer: base 4.44 ft, side 3.33 ft
Solution. We put two lines corresponding to the x-axis and y-axis, as shown,
with the origin coinciding with the midpoint of the window’s base. This
origin is the center of the circle containing the arc. The equation of the circle
is then x2 + y 2 = 16. Let n be length of the base AD, so the side AD has
2
2
length 34 n. Point B then has coordinates n2 , 3n
. Therefore, n2 + 3n
=
4
4
16. Solving this for n > 0 yields n = p1613 . The base is then n ⇡ 4.44 ft and
the side 34 n ⇡ 3.33 ft.
4
Lesson 1.2. Parabolas
Time Frame: 3 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) define a parabola;
(2) determine the standard form of equation of a parabola;
(3) graph a parabola in a rectangular coordinate system; and
(4) solve situational problems involving conic sections (parabolas).
Lesson Outline
(1) Definition of a parabola
(2) Derivation of the standard equation of a parabola
18
(3) Graphing parabolas
(4) Solving situational problems involving parabolas
Introduction
A parabola is one of the conic sections. We have already seen parabolas which
open upward or downward, as graphs of quadratic functions. Here, we will see
parabolas opening to the left or right. Applications of parabolas are presented
at the end.
1.2.1. Definition and Equation of a Parabola
Consider the point F (0, 2) and the line ` having equation y = 2, as shown in
Figure 1.18. What are the distances of A(4, 2) from F and from `? (The latter
is taken as the distance of A from A` , the point on ` closest to A). How about
the distances of B( 8, 8) from F and from ` (from B` )?
AF = 4
BF =
p
( 8
0)2 + (8
and
AA` = 4
2)2 = 10
and
BB` = 10
There are other points P such that P F = P P` (where P` is the closest point on
line `). The collection of all such points forms a shape called a parabola.
Figure 1.18
Figure 1.19
Let F be a given point, and ` a given line not containing F . The set of
all points P such that its distances from F and from ` are the same, is
called a parabola. The point F is its focus and the line ` its directrix.
Consider a parabola with focus F (0, c) and directrix ` having equation x = c.
See Figure 1.19. The focus and directrix are c units above and below, respectively,
the origin. Let P (x, y) be a point on the parabola so P F = P P` , where P` is the
19
point on ` closest to P . The point P has to be on the same side of the directrix
as the focus (if P was below, it would be closer to ` than it is from F ).
PF
p
x2 + (y c)2
x2 + y 2 2cy + c2
x2
= P P`
= y ( c) = y + c
= y 2 + 2cy + c2
= 4cy
The vertex V is the point midway between the focus and the directrix. This
equation, x2 = 4cy, is then the standard equation of a parabola opening upward
with vertex V (0, 0).
Suppose the focus is F (0, c) and the directrix is y = c. In this case, a
point P on the resulting parabola would be below the directrix (just like the
focus).
Instead of opening upward, it will open downward. Consequently, P F =
p
2
x + (y + c)2 and P P` = c y (you may draw a version of Figure 1.19 for
this case). Computations similar to the one done above will lead to the equation
x2 = 4cy.
We collect here the features of the graph of a parabola with standard equation
x = 4cy or x2 = 4cy, where c > 0.
2
(1) vertex : origin V (0, 0)
• If the parabola opens upward, the vertex is the lowest point. If the
parabola opens downward, the vertex is the highest point.
(2) directrix : the line y =
c or y = c
• The directrix is c units below or above the vertex.
(3) focus: F (0, c) or F (0, c)
• The focus is c units above or below the vertex.
• Any point on the parabola has the same distance from the focus as it
has from the directrix.
(4) axis of symmetry: x = 0 (the y-axis)
20
• This line divides the parabola into two parts which are mirror images
of each other.
Example 1.2.1. Determine the focus and directrix of the parabola with the
given equation. Sketch the graph, and indicate the focus, directrix, vertex, and
axis of symmetry.
(1) x2 = 12y
(2) x2 = 6y
Solution. (1) The vertex is V (0, 0) and the parabola opens upward. From 4c =
12, c = 3. The focus, c = 3 units above the vertex, is F (0, 3). The directrix,
3 units below the vertex, is y = 3. The axis of symmetry is x = 0.
(2) The vertex is V (0, 0) and the parabola opens downward. From 4c = 6, c = 32 .
The focus, c = 32 units below the vertex, is F 0, 32 . The directrix, 32 units
above the vertex, is y = 32 . The axis of symmetry is x = 0.
Example 1.2.2. What is the standard equation of the parabola in Figure 1.18?
Solution. From the figure, we deduce that c = 2. The equation is thus x2 =
8y.
2
21
Seatwork/Homework 1.2.1
1. Give the focus and directrix of the parabola with equation x2 = 10y. Sketch
the graph, and indicate the focus, directrix, vertex, and axis of symmetry.
Answer: focus 0, 52 , directrix y = 52
2. Find the standard equation of the parabola with focus F (0, 3.5) and directrix
y = 3.5.
Answer: x2 = 14y
1.2.2. More Properties of Parabolas
The parabolas we considered so far are “vertical” and have their vertices at the
origin. Some parabolas open instead horizontally (to the left or right), and some
have vertices not at the origin. Their standard equations and properties are given
in the box. The corresponding computations are more involved, but are similar
to the one above, and so are not shown anymore.
In all four cases below, we assume that c > 0. The vertex is V (h, k), and it
lies between the focus F and the directrix `. The focus F is c units away from
the vertex V , and the directrix is c units away from the vertex. Recall that, for
any point on the parabola, its distance from the focus is the same as its distance
from the directrix.
22
(x
(x
h)2 = 4c(y
h)2 =
4c(y
k)
(y
k)
(y
k)2 = 4c(x
k)2 =
4c(x
h)
h)
directrix `: horizontal
directrix `: vertical
axis of symmetry: x=h, vertical
axis of symmetry: y=k, horizontal
The following observations are worth noting.
• The equations are in terms of x h and y k: the vertex coordinates are
subtracted from the corresponding variable. Thus, replacing both h and k
with 0 would yield the case where the vertex is the origin. For instance, this
replacement applied to (x h)2 = 4c(y k) (parabola opening upward) would
yield x2 = 4cy, the first standard equation we encountered (parabola opening
upward, vertex at the origin).
• If the x-part is squared, the parabola is “vertical”; if the y-part is squared,
the parabola is “horizontal.” In a horizontal parabola, the focus is on the left
or right of the vertex, and the directrix is vertical.
• If the coefficient of the linear (non-squared) part is positive, the parabola
opens upward or to the right; if negative, downward or to the left.
23
Teaching Notes
In finding the
equation of a
parabola, we just
need to determine
the vertex and the
value of c.
Example 1.2.3. Figure 1.20 shows the graph of parabola, with only its focus
and vertex indicated. Find its standard equation. What is its directrix and its
axis of symmetry?
Solution. The vertex is V (5, 4) and the focus is F (3, 4). From these, we
deduce the following: h = 5, k = 4, c = 2 (the distance of the focus from the
vertex). Since the parabola opens to the left, we use the template (y k)2 =
4c(x h). Our equation is
(y + 4)2 =
8(x
5).
Its directrix is c = 2 units to the right of V , which is x = 7. Its axis is the
horizontal line through V : y = 4.
Figure 1.20
The standard equation (y + 4)2 = 8(x 5) from the preceding example can
be rewritten as y 2 + 8x + 8y 24 = 0, an equation of the parabola in general
form.
If the equation is given in the general form Ax2 + Cx + Dy + E = 0 (A and C
are nonzero) or By 2 + Cx + Dy + E = 0 (B and C are nonzero), we can determine
the standard form by completing the square in both variables.
Example 1.2.4. Determine the vertex, focus, directrix, and axis of symmetry
of the parabola with the given equation. Sketch the parabola, and include these
points and lines.
24
(1) y 2
5x + 12y =
16
(2) 5x2 + 30x + 24y = 51
Solution. (1) We complete the square on y, and move x to the other side.
y 2 + 12y = 5x 16
y 2 + 12y + 36 = 5x 16 + 36 = 5x + 20
(y + 6)2 = 5(x + 4)
The parabola opens to the right. It has vertex V ( 4, 6). From 4c = 5, we
get c = 54 = 1.25. The focus is c = 1.25 units to the right of V : F ( 2.75, 6).
The (vertical) directrix is c = 1.25 units to the left of V : x = 5.25. The
(horizontal) axis is through V : y = 6.
(2) We complete the square on x, and move y to the other side.
5x2 + 30x =
5(x2 + 6x + 9) =
5(x + 3)2 =
(x + 3)2 =
24y + 51
24y + 51 + 5(9)
24y + 96 = 24(y
24
(y 4)
5
4)
In the last line, we divided by 5 for the squared part not to have any coefficient. The parabola opens downward. It has vertex V ( 3, 4).
25
From 4c = 24
, we get c = 65 = 1.2. The focus is c = 1.2 units below V :
5
F ( 3, 2.8). The (horizontal) directrix is c = 1.2 units above V : y = 5.2. The
(vertical) axis is through V : x = 3.
Example 1.2.5. A parabola has focus F (7, 9) and directrix y = 3. Find its
standard equation.
Solution. The directrix is horizontal, and the focus is above it. The parabola
then opens upward and its standard equation has the form (x h)2 = 4c(y k).
Since the distance from the focus to the directrix is 2c = 9 3 = 6, then c = 3.
Thus, the vertex is V (7, 6), the point 3 units below F . The standard equation is
then (x 7)2 = 12(y 6).
2
Seatwork/Homework 1.2.2
1. Determine the vertex, focus, directrix, and axis of symmetry of the parabola
with equation x2 6x + 5y = 34. Sketch the graph, and include these points
and lines.
Answer: vertex (3, 5), focus (3, 6.25), directrix y =
26
3.75, axis x = 3
2. A parabola has focus F ( 2, 5) and directrix x = 6. Find the standard
equation of the parabola.
Answer: (y + 5)2 = 16(x 2)
1.2.3. Situational Problems Involving Parabolas
We now solve some situational problems involving parabolas.
Example 1.2.6. A satellite dish has a shape called a paraboloid, where each
cross-section is a parabola. Since radio signals (parallel to the axis) will bounce
o↵ the surface of the dish to the focus, the receiver should be placed at the focus.
How far should the receiver be from the vertex, if the dish is 12 ft across, and 4.5
ft deep at the vertex?
27
Solution. The second figure above shows a cross-section of the satellite dish drawn
on a rectangular coordinate system, with the vertex at the origin. From the
problem, we deduce that (6, 4.5) is a point on the parabola. We need the distance
of the focus from the vertex, i.e., the value of c in x2 = 4cy.
x2 = 4cy
62 = 4c(4.5)
62
c=
=2
4 · 4.5
Thus, the receiver should be 2 ft away from the vertex.
2
Example 1.2.7. The cable of a suspension bridge hangs in the shape of a
parabola. The towers supporting the cable are 400 ft apart and 150 ft high.
If the cable, at its lowest, is 30 ft above the bridge at its midpoint, how high is
the cable 50 ft away (horizontally) from either tower?
Solution. Refer to the figure above, where the parabolic cable is drawn with
its vertex on the y-axis 30 ft above the origin. We may write its equation as
(x 0)2 = a(y 30); since we don’t need the focal distance, we use the simpler
variable a in place of 4c. Since the towers are 150 ft high and 400 ft apart, we
deduce from the figure that (200, 150) is a point on the parabola.
x2 = a(y 30)
2002 = a(150 30)
2002
1000
a=
=
120
3
The parabola has equation x2 = 1000
(y
30), or equivalently,
3
y = 0.003x2 + 30. For the two points on the parabola 50 ft away from the
towers, x = 150 or x = 150. If x = 150, then
y = 0.003(1502 ) + 30 = 97.5.
28
Thus, the cable is 97.5 ft high 50 ft away from either tower. (As expected, we
get the same answer from x = 150.)
2
Seatwork/Homework 1.2.3
? 1. A satellite dish in the shape of a paraboloid is 10 ft across, and 4 ft deep at
its vertex. How far is the receiver from the vertex, if it is placed at the focus?
Round o↵ your answer to 2 decimal places. (Refer to Example 1.2.6.)
Answer: 1.56 ft
Exercises 1.2
1. Determine the vertex, focus, directrix, and axis of symmetry of the parabola
with the given equation. Sketch the graph, and include these points and lines.
(a) x2 =
4y
(b) 3y 2 = 24x
(c) y +
5 2
2
=
9
2
5 x
(d) x2 + 6x + 8y = 7
(e) y 2
12x + 8y =
(f) 16x2 + 72x
40
112y =
221
Answer:
Item
Vertex
Focus
Directrix
Axis of Symmetry
(a)
(0, 0)
(0, 1)
y=1
x=0
(b)
(0, 0)
(2, 0)
(c)
(4.5, 2.5)
(3.25, 2.5)
x = 5.75
(d)
( 3, 2)
( 3, 0)
y=4
(e)
(2, 4)
(5, 4)
(f)
( 2.25, 1.25)
( 2.25, 3)
29
x=
x=
y=
2
1
0.5
y=0
y=
2.5
x=
3
y=
4
x=
2.25
Teaching Notes
It is helpful to
draw a diagram for
each item.
(a)
(b)
(c)
(d)
(e)
(f)
2. Find the standard equation of the parabola which satisfies the given conditions.
Answer: (y + 9)2 =
(a) vertex (1, 9), focus ( 3, 9)
(b) vertex ( 8, 3), directrix x =
10.5
Answer: (y
16(x
1)
2
3) = 10(x + 8)
(c) vertex ( 4, 2), focus ( 4, 1)
Answer: (x + 4)2 =
(d) focus (7, 11), directrix x = 1
(e) focus (7, 11), directrix y = 4
12(y
2)
Answer: (y
11)2 = 12(x
4)
Answer: (x
7)2 = 14(y
7.5)
(f) vertex ( 5, 7), vertical axis of symmetry, through the point P (7, 11)
Answer: (x + 5)2 = 8(y + 7)
Solution. Since the axis is vertical and P is above the vertex, then the
parabola opens upward and has equation of the form (x + 5)2 = 4c(y + 7).
We plug the coordinates of P : (7 + 5)2 = 4c(11 + 7). We then get c = 2.
Thus, we have (x + 5)2 = 8(y + 7).
(g) vertex ( 5, 7), horizontal axis of symmetry, through the point P (7, 11)
Answer: (y + 7)2 = 27(x + 5)
Solution. Since the axis is horizontal and P is to the right of the vertex,
then the parabola opens to the right and has equation of the form (y +
30
7)2 = 4c(x + 5). We plug the coordinates of P : (11 + 7)2 = 4c(7 + 5).
We then get c = 6.75. Thus, we have (y + 7)2 = 27(x + 5).
3. A satellite dish shaped like a paraboloid, has diameter 2.4 ft and depth 0.9 ft.
If the receiver is placed at the focus, how far should the receiver be from the
vertex?
Answer: 0.4 ft
4. If the diameter of the satellite dish from the previous problem is doubled, with
the depth kept the same, how far should the receiver be from the vertex?
Answer: 1.6 ft
? 5. A satellite dish is shaped like a paraboloid, with the receiver placed at the
focus. It is to have a depth of 0.44 m at the vertex, with the receiver placed
0.11 m away from the vertex. What should the diameter of the satellite dish
be?
Answer: 0.88 m
? 6. A flashlight is shaped like a paraboloid, so that if its light bulb is placed at
the focus, the light rays from the bulb will then bounce o↵ the surface in a
focused direction that is parallel to the axis. If the paraboloid has a depth of
1.8 in and the diameter on its surface is 6 in, how far should the light source
be placed from the vertex?
Answer: 1.25 in
7. The towers supporting the cable of a suspension bridge are 1200 m apart and
170 m above the bridge it supports. Suppose the cable hangs, following the
shape of a parabola, with its lowest point 20 m above the bridge. How high is
the cable 120 m away from a tower?
Answer: 116 m
4
Lesson 1.3. Ellipses
Time Frame: 3 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) define an ellipse;
(2) determine the standard form of equation of an ellipse;
(3) graph an ellipse in a rectangular coordinate system; and
(4) solve situational problems involving conic sections (ellipses).
Lesson Outline
(1) Definition of an ellipse
(2) Derivation of the standard equation of an ellipse
31
(3) Graphing ellipses
(4) Solving situational problems involving ellipses
Introduction
An ellipse is one of the conic sections that most students have not encountered
formally before, unlike circles and parabolas. Its shape is a bounded curve which
looks like a flattened circle. The orbits of the planets in our solar system around
the sun happen to be elliptical in shape. Also, just like parabolas, ellipses have
reflective properties that have been used in the construction of certain structures
(shown in some of the practice problems). We will see some properties and
applications of ellipses in this section.
1.3.1. Definition and Equation of an Ellipse
Teaching Notes
You may review
the distance
formula.
Consider the points F1 ( 3, 0) and F2 (3, 0), as shown in Figure 1.21. What is the
sum of the distances of A(4, 2.4) from F1 and from F2 ? How about the sum of
the distances of B (and C(0, 4)) from F1 and from F2 ?
AF1 + AF2 = 7.4 + 2.6 = 10
BF1 + BF2 = 3.8 + 6.2 = 10
CF1 + CF2 = 5 + 5 = 10
There are other points P such that P F1 + P F2 = 10. The collection of all such
points forms a shape called an ellipse.
Figure 1.21
Figure 1.22
Let F1 and F2 be two distinct points. The set of all points P , whose
distances from F1 and from F2 add up to a certain constant, is called
an ellipse. The points F1 and F2 are called the foci of the ellipse.
32
Given are two points on the x-axis, F1 ( c, 0) and F2 (c, 0), the foci, both c
units away from their center (0, 0). See Figure 1.22. Let P (x, y) be a point on
the ellipse. Let the common sum of the distances be 2a (the coefficient 2 will
make computations simpler). Thus, we have P F1 + P F2 = 2a.
p
P F1 = 2a
(x + c)2 + y 2 = 2a
P F2
p
(x c)2 + y 2
p
4a (x c)2 + y 2 + x2
x2 + 2cx + c2 + y 2 = 4a2
p
a (x c)2 + y 2 = a2 cx
⇥
⇤
a2 x2 2cx + c2 + y 2 = a4 2a2 cx + c2 x2
(a2 c2 )x2 + a2 y 2 = a4 a2 c2 = a2 (a2
b 2 x 2 + a2 y 2 = a2 b 2
x2 y 2
+ 2 =1
a2
b
c2 )
by letting b =
p
a2
2cx + c2 + y 2
c2 , so a > b
p
When we let b = a2 c2 , we assumed a > c. To see why this is true, look at
4P F1 F2 in Figure 1.22. By the Triangle Inequality, P F1 + P F2 > F1 F2 , which
implies 2a > 2c, so a > c.
We collect here the features of the graph of an ellipse with standard equation
p
x2 y 2
+
=
1,
where
a
>
b.
Let
c
=
a2 b 2 .
a2
b2
(1) center : origin (0, 0)
(2) foci : F1 ( c, 0) and F2 (c, 0)
• Each focus is c units away from the center.
• For any point on the ellipse, the sum of its distances from the foci is 2a.
(3) vertices: V1 ( a, 0) and V2 (a, 0)
• The vertices are points on the ellipse, collinear with the center and foci.
33
• If y = 0, then x = ±a. Each vertex is a units away from the center.
• The segment V1 V2 is called the major axis. Its length is 2a. It divides
the ellipse into two congruent parts.
(4) covertices: W1 (0, b) and W2 (0, b)
• The segment through the center, perpendicular to the major axis, is the
minor axis. It meets the ellipse at the covertices. It divides the ellipse
into two congruent parts.
• If x = 0, then y = ±b. Each covertex is b units away from the center.
• The minor axis W1 W2 is 2b units long. Since a > b, the major axis is
longer than the minor axis.
Example 1.3.1. Give the coordinates of the foci, vertices, and covertices of the
ellipse with equation
x2 y 2
+
= 1.
25
9
Sketch the graph, and include these points.
Solution. With a2 = 25 and b2 = 9, we have a = 5, b = 3, and c =
foci: F1 ( 4, 0), F2 (4, 0)
p
a2
b2 = 4.
vertices: V1 ( 5, 0), V2 (5, 0)
covertices: W1 (0, 3), W2 (0, 3)
Example 1.3.2. Find the (standard) equation of the ellipse whose foci are
F1 ( 3, 0) and F2 (3, 0), such that for any point on it, the sum of its distances
from the foci is 10. See Figure 1.21.
Solution. We have 2a = 10 and c = 3, so a = 5 and b =
equation is
x2 y 2
+
= 1.
25 16
34
p
a2
c2 = 4. The
2
Seatwork/Homework 1.3.1
1. Give the coordinates of the foci, vertices, and covertices of the ellipse with
x2
y2
equation
+
= 1. Sketch the graph, and include these points.
169 25
Answer: foci: F1 ( 12, 0) and F2 (12, 0), vertices: V1 ( 13, 0) and V2 (13, 0),
covertices: W1 (0, 5) and W2 (0, 5)
2. Find the equation in standard form of the ellipse whose foci are F1 ( 8, 0) and
F2 (8, 0), such that for any point on it, the sum of its distances from the foci
y2
x2
is 20.
Answer:
+
=1
100 36
1.3.2. More Properties of Ellipses
The ellipses we have considered so far are “horizontal” and have the origin as their
centers. Some ellipses have their foci aligned vertically, and some have centers
not at the origin. Their standard equations and properties are given in the box.
The derivations are more involved, but are similar to the one above, and so are
not shown anymore.
p
In all four cases below, a > b and c = a2 b2 . The foci F1 and F2 are c
units away from the center. The vertices V1 and V2 are a units away from the
center, the major axis has length 2a, the covertices W1 and W2 are b units away
from the center, and the minor axis has length 2b. Recall that, for any point on
the ellipse, the sum of its distances from the foci is 2a.
35
Center
Corresponding Graphs
(0, 0)
x2 y 2
+ 2 =1
a2
b
x2 y 2
+ 2 =1
b2
a
(h, k)
(x
h)2
a2
+
k)2
(y
b2
=1
h)2
(x
b2
+
(y
k)2
a2
=1
major axis: horizontal
major axis: vertical
minor axis: vertical
minor axis: horizontal
In the standard equation, if the x-part has the bigger denominator, the ellipse
is horizontal. If the y-part has the bigger denominator, the ellipse is vertical.
Example 1.3.3. Give the coordinates of the center, foci, vertices, and covertices
of the ellipse with the given equation. Sketch the graph, and include these points.
36
(x + 3)2 (y 5)2
+
=1
24
49
(2) 9x2 + 16y 2 126x + 64y = 71
(1)
p
Solution. p(1) From a2 = 49 and b2 = 24, we have a = 7, b = 2 6 ⇡ 4.9, and
c = a2 b2 = 5. The ellipse is vertical.
center:
foci:
vertices:
covertices:
( 3, 5)
F1 ( 3, 0), F2 ( 3, 10)
V1 ( 3, 2), V2 ( 3, 12)
p
W1 ( 3 2 6, 5) ⇡ ( 7.9, 5)
p
W2 ( 3 + 2 6, 5) ⇡ (1.9, 5)
(2) We first change the given equation to standard form.
9(x2
9(x2 14x) + 16(y 2 + 4y) = 71
14x + 49) + 16(y 2 + 4y + 4) = 71 + 9(49) + 16(4)
9(x 7)2 + 16(y + 2)2 = 576
(x 7)2 (y + 2)2
+
=1
64
36
37
We have a = 8 and b = 6. Thus, c =
horizontal.
center:
foci:
vertices:
covertices:
p
a2
p
b2 = 2 7 ⇡ 5.3. The ellipse is
(7, 2)
p
2 7, 2) ⇡ (1.7, 2)
p
F2 (7 + 2 7, 2) ⇡ (12.3, 2)
F1 (7
V1 ( 1, 2), V2 (15, 2)
W1 (7, 8), W2 (7, 4)
Example 1.3.4. The foci of an ellipse are ( 3, 6) and ( 3, 2). For any point
on the ellipse, the sum of its distances from the foci is 14. Find the standard
equation of the ellipse.
Solution. The midpoint ( 3, 2) of the foci is the center of the ellipse. The
ellipse is vertical (because the foci are vertically
p aligned)pand c = 4. From the
given sum, 2a = 14 so a = 7. Also, b = a2 c2 = 33. The equation is
(x + 3)2 (y + 2)2
+
= 1.
2
33
49
p
p
Example 1.3.5. An ellipse has vertices (2
61, 5) and (2 + 61, 5), and
its minor axis is 12 units long. Find its standard equation and its foci.
38
Solution. The midpoint (2, 5)pof the vertices is the center of the ellipse, which is
horizontal. Each vertex is a = 61 units away from the center. From the length of
(x 2)2 (y + 5)2
the minor axis, 2b = 12 so b = 6. The standard equation is
+
=
61
36
p
1. Each focus is c = a2 b2 = 5 units away from (2, 5), so their coordinates
are ( 3, 5) and (7, 5).
2
Seatwork/Homework 1.3.2
1. Give the coordinates of the center, foci, vertices, and covertices of the ellipse
with equation 41x2 + 16y 2 + 246x 192y + 289 = 0. Sketch the graph, and
include these points.
Answer:
center C( 3,
p
p6), foci F1 ( 3, 1) and F2 ( 3, 11), vertices V1 ( 3, 6
41) and V2 ( 3, 6 + 41), covertices W1 ( 7, 6) and W2 (1, 6)
2. An ellipse has vertices ( 10, 4) and (6, 4), and covertices ( 2, 9) and
( 2, 1). Find its standard equation and its foci.
p
p
(x + 2)2 (y + 4)2
Answer:
+
= 1, foci ( 2
39, 4) and ( 2 + 39, 4)
64
25
39
1.3.3. Situational Problems Involving Ellipses
We now apply the concept of ellipse to some situational problems.
? Example 1.3.6. A tunnel has the shape of a semiellipse that is 15 ft high at
the center, and 36 ft across at the base. At most how high should a passing truck
be, if it is 12 ft wide, for it to be able to fit through the tunnel? Round o↵ your
answer to two decimal places.
Solution. Refer to the figure above. If we draw the semiellipse on a rectangular
coordinate system, with its center at the origin, an equation of the ellipse which
contains it, is
x2
y2
+
= 1.
182 152
To maximize its height, the corners of the truck, as shown in the figure, would
have to just touch the ellipse. Since the truck is 12 ft wide, let the point (6, n)
be the corner of the truck in the first quadrant, where n > 0, is the (maximum)
height of the truck. Since this point is on the ellipse, it should fit the equation.
Thus, we have
62
n2
+
=1
182 152
2
n = 15
2
✓
1
62
182
◆
p
n = 10 2 ⇡ 14.14 ft
2
Example 1.3.7. The orbit of a planet has the shape of an ellipse, and on one
of the foci is the star around which it revolves. The planet is closest to the star
when it is at one vertex. It is farthest from the star when it is at the other vertex.
Suppose the closest and farthest distances of the planet from this star, are 420
million kilometers and 580 million kilometers, respectively. Find the equation of
the ellipse, in standard form, with center at the origin and the star at the x-axis.
Assume all units are in millions of kilometers.
40
Solution. In the figure above, the orbit is drawn as a horizontal ellipse with
center at the origin. From the planet’s distances from the star, at its closest
and farthest points, it follows that the major axis is 2a = 420 + 580 = 1000
(million kilometers), so a = 500. If we place the star at the positive x-axis,
then it is c = 500 420 = 80 units away from the center. Therefore, we get
b2 = a2 c2 = 5002 802 = 243600. The equation then is
x2
y2
+
= 1.
250000 243600
The star could have been placed on the negative x-axis, and the answer would
still be the same.
2
Seatwork/Homework 1.3.3
? 1. The arch of a bridge is in the shape of a semiellipse, with its major axis at the
water level. Suppose the arch is 20 ft high in the middle, and 120 ft across its
major axis. How high above the water level is the arch, at a point 20 ft from
the center (horizontally)? Round o↵ to 2 decimal places. Refer to Example
1.3.6.
Answer: 18.86 ft
Exercises 1.3
1. Give the coordinates of the center, vertices, covertices, and foci of the ellipse
with the given equation. Sketch the graph, and include these points.
(a)
x2
y2
+
=1
169 25
41
x2
y2
+
=1
144 169
(c) 4x2 + 13y 2 = 52
(x + 7)2 (y 4)2
(d)
+
=1
16
25
(e) 9x2 + 16y 2 + 72x 96y + 144 = 0
(b)
(f) 36x2 + 20y 2
Answer:
144x + 120y
396 = 0
Item
Center
Vertices
Covertices
Foci
(a)
(0, 0)
(±13, 0)
(0, ±5)
(±12, 0)
(b)
(0, 0)
(±12, 0)
(0, ±5)
(c)
(0, 0)
(0, ±13)
p
(± 13, 0)
(0, ±2)
(±3, 0)
(d)
( 7, 4)
( 7, 1)
( 11, 4)
( 7, 1)
( 7, 9)
( 3, 4)
( 8, 3)
( 4, 0)
( 7, 7)
p
( 4 ± 7, 3)
(0, 3)
( 4, 6)
p
(2 ± 2 5, 3)
(e)
(f)
( 4, 3)
(2, 3)
(2, 9)
(2, 3)
(2, 3)
(2, 7)
(2, 1)
(a)
(b)
(c)
(d)
(e)
(f)
42
2. Find the standard equation of the ellipse which satisfies the given conditions.
(a) foci ( 7, 6) and ( 1, 6), the sum of the distances of any point from the
(x + 4)2 (y 6)2
foci is 14
Answer:
+
=1
49
40
(b) center (5, 3), horizontal major axis of length 20, minor axis of length 16
(x 5)2 (y 3)2
Answer:
+
=1
100
64
(c) major axis of length 22, foci 9 units above and below the center (2, 4)
(x 2)2 (y 4)2
Answer:
+
=1
40
121
(d) covertices ( 4, 8) and (10, 8), a focus at (3, 12)
(x 3)2 (y 8)2
Answer:
+
=1
49
65
Solution. The midpoint of the covertices is the center, (3, 8). From this
point, the given focus is c = 4 units away. Since b = 7 (the distance from
the center to a covertex), then a2 = b2 + c2 = 65. The ellipse then has
(x 3)2 (y 8)2
equation
+
= 1.
49
65
(e) focus ( 6, 2), covertex ( 1, 5), horizontal major axis
(x + 1)2 (y + 2)2
Answer:
+
=1
74
49
Solution. Make a rough sketch of the points to see that the center is to
the right of the given focus, and below the given covertex. The center is
thus ( 1, 2). It follows that c = 5, b = 7, so a2 = b2 + c2 = 74. The
(x + 1)2 (y + 2)2
ellipse then has equation
+
= 1.
74
49
3. A semielliptical tunnel has height 9 ft and a width of 30 ft. A truck that is
about to pass through is 12 ft wide and 8.3 ft high. Will this truck be able to
pass through the tunnel?
Answer: No
4. A truck that is about to pass through the tunnel from the previous item is 10
ft wide and 8.3 ft high. Will this truck be able to pass through the tunnel?
Answer: Yes
5. An orbit of a satellite around a planet is an ellipse, with the planet at one
focus of this ellipse. The distance of the satellite from this star varies from
300, 000 km to 500, 000 km, attained when the satellite is at each of the two
vertices. Find the equation of this ellipse, if its center is at the origin, and the
vertices are on the x-axis. Assume all units are in 100, 000 km.
Answer:
43
x2
16
+
y2
15
=1
6. The orbit of a planet around a star is described by the equation
y2
x2
640,000
+
= 1, where the star is at one focus, and all units are in millions of
kilometers. The planet is closest and farthest from the star, when it is at the
vertices. How far is the planet when it is closest to the sun? How far is the
planet when it is farthest from the sun?
630,000
Answer: 700 million km, 900 million km
Solution. The ellipse has center at the origin, and major axis on the x-axis.
Since a2 = 640, 000, then a = 800, so p
the verticespare V1 ( 800, 0) and
2
V2 (800, 00). Since b = 630, 000, then c = a2 b2 = 10, 000 = 100. Suppose the star is at the focus at the right of the origin (this choice is arbitrary,
since we could have chosen instead the focus on the left). Its location is then
F (100, 0). The closest distance is then V2 F = 700 (million kilometers) and
the farthest distance is V1 F = 900 (million kilometers).
7. A big room is constructed so that the ceiling is a dome that is semielliptical
in shape. If a person stands at one focus and speaks, the sound that is made
bounces o↵ the ceiling and gets reflected to the other focus. Thus, if two
people stand at the foci (ignoring their heights), they will be able to hear each
other. If the room is 34 m long and 8 m high, how far from the center should
each of two people stand if they would like to whisper back and forth and hear
each other?
Answer: 15 m
Solution. We could put a coordinate system with the floor of the room on
the x-axis, and the center of the room at the origin, as shown in the figures.
The major axis has length 34, and the height of the room is half of the minor
y2
x2
axis. The ellipse that contains the ceiling then has equation 17
2 + 82 = 1. The
p
p
distance of a focus from the center is c = a2 b2 = 172 82 = 15. Thus,
the two people should stand 15 m away from the center.
8. A whispering gallery has a semielliptical ceiling that is 9 m high and 30 m
long. How high is the ceiling above the two foci?
Answer: 5.4 m
44
Solution. As in the previous problem, put a coordinate system with the floor
of the room on the x-axis, and the center of the room at the origin. The major
axis has length 30, and half the minor axis is 9. The ellipse that contains the
y2
x2
ceiling then has equation 15
2 + 92 = 1. The distance of a focus from the center
p
p
is c = a2 b2 = 152 92 = 12. If we put x = 12 in the equation of the
2
2
ellipse, we get 12
+ y92 = 1. Solving for y > 0 yields y = 27
= 5.4. The height
152
5
of the ceiling above each focus is 5.4 m.
4
Lesson 1.4. Hyperbolas
Time Frame: 3 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) define a hyperbola;
(2) determine the standard form of equation of a hyperbola;
(3) graph a hyperbola in a rectangular coordinate system; and
(4) solve situational problems involving conic sections (hyperbolas).
Lesson Outline
(1) Definition of a hyperbola
(2) Derivation of the standard equation of a hyperbola
(3) Graphing hyperbolas
(4) Solving situational problems involving hyperbolas
Introduction
A hyperbola is one of the conic sections that most students have not encountered formally before, unlike circles and parabolas. Its graph consists of two
unbounded branches which extend in opposite directions. It is a misconception
that each branch is a parabola. This is not true, as parabolas and hyperbolas
have very di↵erent features. An application of hyperbolas in basic location and
navigation schemes are presented in an example and some exercises.
1.4.1. Definition and Equation of a Hyperbola
Consider the points F1 ( 5, 0) and F2 (5, 0) as shown in Figure 1.23. What is the
absolute value of the di↵erence of the distances of A(3.75, 3) from F1 and from
45
F2 ? How about the absolute value of the di↵erence of the distances of B
from F1 and from F2 ?
|AF1
|BF1
5, 16
3
AF2 | = |9.25 3.25| = 6
16 34
BF2 | =
=6
3
3
There are other points P such that |P F1 P F2 | = 6. The collection of all such
points forms a shape called a hyperbola, which consists of two disjoint branches.
For points P on the left branch, P F2 P F1 = 6; for those on the right branch,
P F1 P F2 = 6.
Figure 1.23
Figure 1.24
Let F1 and F2 be two distinct points. The set of all points P , whose
distances from F1 and from F2 di↵er by a certain constant, is called a
hyperbola. The points F1 and F2 are called the foci of the hyperbola.
In Figure 1.24, given are two points on the x-axis, F1 ( c, 0) and F2 (c, 0), the
foci, both c units away from their midpoint (0, 0). This midpoint is the center
of the hyperbola. Let P (x, y) be a point on the hyperbola, and let the absolute
value of the di↵erence of the distances of P from F1 and F2 , be 2a (the coefficient
2 will make computations simpler). Thus, |P F1 P F2 | = 2a, and so
p
p
(x + c)2 + y 2
(x c)2 + y 2 = 2a.
Algebraic manipulations allow us to rewrite this into the much simpler
x2
a2
y2
= 1,
b2
where b =
p
c2
a2 .
p
When we let b = c2 a2 , we assumed c > a. To see why this is true, suppose
that P is closer to F2 , so P F1 P F2 = 2a. Refer to Figure 1.24. Suppose also
46
that P is not on the x-axis, so 4P F1 F2 is formed. From the triangle inequality,
F1 F2 + P F2 > P F1 . Thus, 2c > P F1 P F2 = 2a, so c > a.
Now we present a derivation. For now, assume P is closer to F2 so P F1 > P F2 ,
and P F1 P F2 = 2a.
P F1 = 2a + P F2
p
p
(x + c)2 + y 2 = 2a + (x c)2 + y 2
⇣p
⌘2 ⇣
⌘2
p
(x + c)2 + y 2 = 2a + (x c)2 + y 2
p
cx a2 = a (x c)2 + y 2
⇣ p
⌘2
(cx a2 )2 = a (x c)2 + y 2
(c2
a2 )x2
a2 y 2 = a2 (c2
a2 )
b 2 x 2 a2 y 2 = a2 b 2
x2 y 2
=1
a2
b2
by letting b =
p
c2
a2 > 0
We collect here the features of the graph of a hyperbola with standard equation
x2 y 2
= 1.
a2
b2
p
Let c = a2 + b2 .
Figure 1.25
Figure 1.26
(1) center : origin (0, 0)
(2) foci : F1 ( c, 0) and F2 (c, 0)
• Each focus is c units away from the center.
47
Teaching Notes
If it is assumed
that P is closer to
F1 , then the same
equation will be
obtained because
of symmetry.
• For any point on the hyperbola, the absolute value of the di↵erence of
its distances from the foci is 2a.
(3) vertices: V1 ( a, 0) and V2 (a, 0)
• The vertices are points on the hyperbola, collinear with the center and
foci.
• If y = 0, then x = ±a. Each vertex is a units away from the center.
• The segment V1 V2 is called the transverse axis. Its length is 2a.
(4) asymptotes: y = ab x and y =
b
x,
a
the lines `1 and `2 in Figure 1.26
• The asymptotes of the hyperbola are two lines passing through the center which serve as a guide in graphing the hyperbola: each branch of
the hyperbola gets closer and closer to the asymptotes, in the direction
towards which the branch extends. (We need the concept of limits from
calculus to explain this.)
• An aid in determining the equations of the asymptotes: in the standard
2
2
equation, replace 1 by 0, and in the resulting equation xa2 yb2 = 0, solve
for y.
• To help us sketch the asymptotes, we point out that the asymptotes
`1 and `2 are the extended diagonals of the auxiliary rectangle drawn
in Figure 1.26. This rectangle has sides 2a and 2b with its diagonals
intersecting at the center C. Two sides are congruent and parallel to
the transverse axis V1 V2 . The other two sides are congruent and parallel
to the conjugate axis, the segment shown which is perpendicular to the
transverse axis at the center, and has length 2b.
Example 1.4.1. Determine the foci, vertices, and asymptotes of the hyperbola
with equation
x2 y 2
= 1.
9
7
Sketch the graph, and include these points and lines, the transverse and conjugate
axes, and the auxiliary rectangle.
2
2
Solution. With
p a = 9 and
p b = 7, we have
a = 3, b = 7, and c = a2 + b2 = 4.
foci: F1 ( 4, 0) and F2 (4, 0)
vertices: V1 ( 3, 0) and V2 (3, 0)
p
p
asymptotes: y = 37 x and y = 37 x
The graph is shown at the right. The conjup
gate axis drawn has its endpoints b = 7 ⇡
2.7 units above and below the center.
2
48
Example 1.4.2. Find the (standard) equation of the hyperbola whose foci are
F1 ( 5, 0) and F2 (5, 0), such that for any point on it, the absolute value of the
di↵erence of its distances from the foci is 6. See Figure 1.23.
Solution. We have 2a = 6 and c = 5, so a = 3 and b =
x2 y 2
hyperbola then has equation
= 1.
9
16
p
c2
a2 = 4. The
2
Seatwork/Homework 1.4.1
1. Determine foci, vertices, and asymptotes of the hyperbola with equation
x2
16
y2
= 1.
20
Sketch the graph, and include these points and lines, along with the auxiliary
rectangle.
Answer: foci
F1 ( 6, 0) and
F2 (6, 0), vertices V1 ( 4, 0) and V2 (4, 0), asympp
p
5
5
totes y = 2 x and y = 2 x
p
2. Find the p
equation in standard form of the hyperbola whose foci are F1 ( 4 2, 0)
and F2 (4 2, 0), such that for any point on it, the absolute value of the di↵erence of its distances from the foci is 8.
Answer:
x2
y2
=1
16 16
1.4.2. More Properties of Hyperbolas
The hyperbolas we considered so far are “horizontal” and have the origin as their
centers. Some hyperbolas have their foci aligned vertically, and some have centers
49
not at the origin. Their standard equations and properties are given in the box.
The derivations are more involved, but are similar to the one above, and so are
not shown anymore.
Center
Corresponding Hyperbola
(0, 0)
x2
a2
y2
=1
b2
y2
a2
x2
=1
b2
(h, k)
(x
h)2
a2
k)2
(y
b2
=1
(y
k)2
a2
h)2
(x
b2
=1
transverse axis: horizontal
transverse axis: vertical
conjugate axis: vertical
conjugate axis: horizontal
p
In all four cases above, we let c = a2 + b2 . The foci F1 and F2 are c units
away from the center C. The vertices V1 and V2 are a units away from the center.
The transverse axis V1 V2 has length 2a. The conjugate axis has length 2b and is
50
perpendicular to the transverse axis. The transverse and conjugate axes bisect
each other at their intersection point, C. Each branch of a hyperbola gets closer
and closer to the asymptotes, in the direction towards which the branch extends.
The equations of the asymptotes can be determined by replacing 1 in the standard
equation by 0. The asymptotes can be drawn as the extended diagonals of the
auxiliary rectangle determined by the transverse and conjugate axes. Recall that,
for any point on the hyperbola, the absolute value of the di↵erence of its distances
from the foci is 2a.
In the standard equation, aside from being positive, there are no other restrictions on a and b. In fact, a and b can even be equal. The orientation of the
hyperbola is determined by the variable appearing in the first term (the positive
term): the corresponding axis is where the two branches will open. For example,
if the variable in the first term is x, the hyperbola is “horizontal”: the transverse
axis is horizontal, and the branches open to the left and right in the direction of
the x-axis.
Example 1.4.3. Give the coordinates of the center, foci, vertices, and asymptotes of the hyperbola with the given equation. Sketch the graph, and include
these points and lines, the transverse and conjugate axes, and the auxiliary rectangle.
(y + 2)2 (x 7)2
=1
25
9
(2) 4x2 5y 2 + 32x + 30y = 1
(1)
2
2
Solution.
(1) From
p a = 25 and b = 9, we have a = 5, b = 3, and c =
p
a2 + b2 = 34 ⇡ 5.8. The hyperbola is vertical. To determine the asymp2
(x 7)2
totes, we write (y+2)
= 0, which is equivalent to y + 2 = ± 53 (x 7).
25
9
We can then solve this for y.
center: C(7, 2)
foci: F1 (7, 2
p
34) ⇡ (7, 7.8) and F2 (7, 2 +
vertices: V1 (7, 7) and V2 (7, 3)
asymptotes: y = 53 x
41
3
and y =
5
x
3
+
p
34) ⇡ (7, 3.8)
29
3
The conjugate axis drawn has its endpoints b = 3 units to the left and right
of the center.
51
(2) We first change the given equation to standard form.
4(x2 + 8x) 5(y 2 6y) = 1
4(x2 + 8x + 16) 5(y 2 6y + 9) = 1 + 4(16) 5(9)
4(x + 4)2 5(y 3)2 = 20
(x + 4)2 (y 3)2
=1
5
4
p
p
We have a = 5 ⇡ 2.2 and b = 2. Thus, c = a2 + b2 = 3. The hyperbola
2
(y 3)2
is horizontal. To determine the asymptotes, we write (x+4)
= 0
5
4
2
p
which is equivalent to y 3 = ± 5 (x + 4), and solve for y.
center: C( 4, 3)
foci: F1 ( 7, 3) and F2 ( 1, 3)
p
p
vertices: V1 ( 4
5, 3) ⇡ ( 6.2, 3) and V2 ( 4 + 5, 3) ⇡ ( 1.8, 3)
asymptotes: y =
p2 x
5
+
p8
5
+ 3 and y =
p2 x
5
p8
5
+3
The conjugate axis drawn has its endpoints b = 2 units above and below
the center.
52
Example 1.4.4. The foci of a hyperbola are ( 5, 3) and (9, 3). For any point
on the hyperbola, the absolute value of the di↵erence of its of its distances from
the foci is 10. Find the standard equation of the hyperbola.
Solution. The midpoint (2, 3) of the foci is the center of the hyperbola. Each
focus is c = 7 units away from the center. From the given di↵erence, 2a = 10 so
a = 5. Also, b2 = c2 a2 = 24. The hyperbola is horizontal (because the foci are
horizontally aligned), so the equation is
(x
2)2
25
(y + 3)2
= 1.
24
2
Example 1.4.5.
p A hyperbola has vertices ( 4, 5) and ( 4, 9), and one of its
foci is ( 4, 2
65). Find its standard equation.
Solution. The midpoint (
which is vertical (because
a = 7 units away from the
the center. Thus, b2 = c2
4, 2) of the vertices is the center of the hyperbola,
the vertices are vertically aligned).
Each vertex is
p
center. The given focus is c = 65 units away from
a2 = 16, and the standard equation is
(y
2)2
49
(x + 4)2
= 1.
16
53
2
Seatwork/Homework 1.4.2
1. Give the coordinates of the center, foci, vertices, and asymptotes of the hyperbola with equation 9x2 4y 2 90x 32y = 305. Sketch the graph, and
include these points and lines, along with the auxiliary rectangle.
p
p
Answer: center C(5, 4), foci F1 (5, 4 2 13) and F2 (5, 4+2 13), vertices
V1 (5, 10) and V2 (5, 2), asymptotes y = 32 x + 72 and y = 32 x 23
2
2. A hyperbola has vertices (1, 9) and (13, 9), and one of its foci is ( 2, 9). Find
(x 7)2 (y 9)2
its standard equation.
Answer:
=1
36
45
1.4.3. Situational Problems Involving Hyperbolas
We now give an example on an application of hyperbolas.
Example 1.4.6. An explosion is heard by two stations 1200 m apart, located at
F1 ( 600, 0) and F2 (600, 0). If the explosion was heard in F1 two seconds before
it was heard in F2 , identify the possible locations of the explosion. Use 340 m/s
as the speed of sound.
Solution. Using the given speed of sound, we deduce that the sound traveled
340(2) = 680 m farther in reaching F2 than in reaching F1 . This is then the
di↵erence of the distances of the explosion from the two stations. Thus, the
explosion is on a hyperbola with foci are F1 and F2 , on the branch closer to F1 .
54
We have c = 600 and 2a = 680, so a = 340 and b2 = c2 a2 = 244400.
The explosion could therefore be anywhere on the left branch of the hyperbola
y2
x2
= 1.
2
115600
244400
Seatwork/Homework 1.4.3
? 1. Two stations, located at M ( 1.5, 0) and N (1.5, 0) (units are in km), simultaneously send sound signals to a ship, with the signal traveling at the speed of
0.33 km/s. If the signal from N was received by the ship four seconds before
the signal it received from M , find the equation of the curve containing the
y2
x2
possible location of the ship.
Answer: 0.4356
= 1 (right branch)
1.8144
Exercises 1.4
1. Give the coordinates of the center, foci, vertices, and the asymptotes of the
hyperbola with the given equation. Sketch the graph, and include these points
and lines.
x2
36
y2
(b)
25
(c) (x
(a)
y2
=1
64
x2
=1
16
1)2 y 2 = 4
(y + 2)2 (x + 3)2
=1
15
10
(e) 3x2 2y 2 42x 16y =
(d)
(f) 25x2
67
39y 2 + 150x + 390y =
225
55
Answer:
Item
Center
Vertices
Foci
(a)
(0, 0)
(±6, 0)
(b)
(0, 0)
(0, ±5)
(c)
(1, 0)
(d)
( 3, 2)
( 1, 0), (3, 0)
p
( 3, 2 ± 15)
(±10, 0)
p
(0, ± 41)
p
(1 ± 2 2, 0)
(e)
(7, 4)
(3, 4), (11, 4)
(f)
( 3, 5)
( 3, 0), ( 3, 10)
( 3, 3), ( 3, 13)
Item
Asymptotes
(a)
y = ± 43 x
y = ± 54 x
(b)
(c)
(d)
(e)
(f)
(a)
( 3, 7), ( 3, 3)
p
(7 ± 2 10, 4)
y=x
x+1
q
y = ± 32 x ± 3 32 2
q
q
3
y = ± 2 x ⌥ 7 32 4
q
1, y =
y = ± p539 x ±
(b)
56
p15
39
+5
(c)
(d)
(e)
(f)
2. Find the standard equation of the hyperbola which satisfies the given conditions.
(a) foci ( 4, 3) and ( 4, 13), the absolute value of the di↵erence of the
distances of any point from the foci is 14
(y 5)2 (x + 4)2
Answer:
=1
49
15
(b) vertices ( 2, 8) and (8, 8), a focus (12, 8)
(x 3)2 (y 8)2
Answer:
=1
25
56
(c) center ( 6, 9), a vertex ( 6, 15), conjugate axis of length 12
(y 9)2 (x + 6)2
Answer:
=1
25
36
(d) asymptotes y = 43 x + 13 and y = 43 x + 41
, a vertex ( 1, 7)
3
(x 5)2 (y 7)2
Answer:
=1
36
64
Solution. The asymptotes intersect at (5, 7). This is the center. The
distance of the given vertex from the center is a = 6. This vertex and
center are aligned horizontally, so the hyperbola has equation of the form
(x h)2
(y k)2
= 1. The asymptotes consequently have the form y k =
a2
b2
b
± a (x h), and thus, have slopes ± ab . From the given asymptotes, ab = 43 .
Since a = 6, then b = 8. The standard equation is then
(x
(e) asymptotes y = 13 x +
5
3
5)2
36
and y =
(y
1
x
3
7)2
= 1.
64
+ 73 , a focus (1, 12)
(y 2)2
Answer:
10
(x
1)2
=1
90
Solution. The asymptotes intersect at (1, 2). This is the center. The
distance of the given focus from the center is c = 10. This focus and
57
center are aligned vertically, so the hyperbola has equation of the form
(y k)2
(x h)2
= 1. The asymptotes consequently have the form y k =
a2
b2
a
± b (x h), and thus, have slopes ± ab . From the given asymptotes, ab = 13 ,
so b = 3a.
c2 = 100 = a2 + b2 = a2 + (3a)2 = 10a2
Thus, a2 = 10, and b2 = 9a2 = 90. The standard equation is
(y
2)2
10
(x
1)2
= 1.
90
3. Two control towers are located at points Q( 500, 0) and R(500, 0), on a
straight shore where the x-axis runs through (all distances are in meters).
At the same moment, both towers sent a radio signal to a ship out at sea, each
traveling at 300 m/µs. The ship received the signal from Q 3 µs (microseconds)
before the message from R.
(a) Find the equation of the curve containing the possible location of the
x2
y2
ship.
Answer:
= 1 (left branch)
202500 47500
(b) Find the coordinates (rounded o↵ to two decimal places) of the ship if it
is 200 m from the shore (y = 200).
Answer: ( 610.76, 200)
Solution. Since the time delay between the two signals is 3 µs, then the di↵erence between the distances traveled by the two signals is 300 · 3 = 900 m. The
ship is then on a hyperbola, consisting of points whose distances from Q and R
(the foci) di↵er by 2a = 900. With a = 450 and c = 500 (the distance of each
focus from the center, the origin), we have b2 = c2 a2 = 5002 4502 = 47500.
y2
x2
Since a2 = 202500, the hyperbola then has equation 202500
= 1. Since
47500
the signal from Q was received first, the ship is closer to Q than R, so the
ship is on the left branch of this hyperbola. Using y = 200, we then solve
x2
2002
= 1 for x < 0 (left branch), and we get x ⇡ 610.76.
202500
47500
4
58
References
[1] R.N. Aufmann, V.C. Barker, and R.D. Nation, College Trigonometry, Houghton
Mi✏in Company, 2008.
[2] E.A. Cabral, M.L.A.N. De Las Peñas, E.P. De Lara-Tuprio, F.F. Francisco,
I.J.L. Garces, R.M. Marcelo, and J.F. Sarmiento, Precalculus, Ateneo de
Manila University Press, 2010.
[3] R. Larson, Precalculus with Limits, Brooks/Cole, Cengage Learning, 2014.
[4] L. Leithold, College Algebra and Trigonometry, Addison Wesley Longman
Inc., 1989, reprinted by Pearson Education Asia Pte. Ltd., 2002.
[5] M.L. Lial, J. Hornsby, and D.I. Schneider, College Algebra and Trigonometry
and Precalculus, Addison-Wesley Educational Publisher, Inc., 2001.
[6] J. Stewart, L. Redlin, and S. Watson, Precalculus: Mathematics for Calculus,
Brooks/Cole, Cengage Learning, 2012.
[7] M. Sullivan, Algebra & Trigonometry, Pearson Education, Inc., 2012.
[8] C. Young, Algebra and Trigonometry, John Wiley & Sons, Inc., 2013.
273
Biographical Notes
Ian June L. Garces, Ph.D.
Richard B. Eden, Ph.D.
Team Leader
Writer
Dr. Ian June Garces is currently an Associate
Professor at the Department of Mathematics,
Ateneo de Manila University. He finished his
Bachelor of Science in Mathematics at Mindanao
State University in Marawi City, as Magna Cum
Laude. He was granted a straight program to the
Doctor of Philosophy in Mathematics degree at
the Ateneo de Manila University as a DOST
scholar. He was the leader and head coach of the
Philippine Team to the International
Mathematical Olympiad for three years, the
convenor of the Math Learning Area Team of the
K to 12 Program of the Department of Education,
and one of the test developers of the MetrobankMTAP-DepEd Math Challenge. He has published
articles and research papers in integration
theory, graph theory, mathematics education,
mathematical recreation, and mathematical
Olympiad. He is also the Team Leader of the
Grade 11 Precalculus Learning Manual
commissioned by the Department of Education.
Dr. Richard Eden is currently an Assistant
Professor at the Ateneo de Manila University,
where he has been teaching Probability and
Calculus at the undergraduate and graduate
levels. He finished his bachelor’s degree in
Mathematics at Ateneo de Manila, finishing Cum
Laude; and his master’s degree and doctorate
degree in Mathematics at Purdue University in
Indiana, USA. He also presented papers at
Purdue University, University of Kansas, and
Duke University - focusing on Statistics,
Malliavin Calculus, Stochastic Analysis, and
Probability. As a teacher, he was also involved in
various Mathematics competitions, such as the
International Mathematical Olympiad, and the
Philippine Mathematical Olympiad.
Jerico B. Bacani, Ph.D.
Writer
Dr. Jerico B. Bacani is currently the Department
Chairman of the Department of Mathematics and
Computer Science at University of the
Philippines Baguio, and the Regional
Coordinator of the Philippine Mathematical
Olympiad. He finished his doctorate degree at
Karl-Franzens Universität Graz in Austria,
through the Austria’s Science and Technology
Grant for Southeast Asia. He finished his
master’s degree in Mathematics at UP Baguio as
a CHED scholar, and his bachelor’s degree in
Mathematics at UP Diliman as a DOST SEI
scholar. Besides publishing numerous academic
articles both locally and internationally, he also
wrote learning resources for Grade 7 students,
and teaching modules for Grade 11 Mathematics
teachers.
Glenn Rey A. Estrada
Writer
Mr. Glenn Rey Estrada has been teaching
Mathematics at the Leyte Normal University
(LNU) for 10 years now. He finished his master’s
of Mathematics at the University of San Carlos
as a CHED scholar. He finished Cum Laude with
a degree of Elementary Education and Bachelor
in Secondary Education major in Mathematics at
the LNU. Mr. Estrada is a member of
international and national Mathematics
organization such as the Asian Qualitative
Research Association, Mathematical Society of
the Philippines, and the Philippine Council of
Mathematics Teacher Educators. An upholder of
the K to 12 education, Mr. Estrada has also
served as Trainer in several teacher trainings on
K to 12 Basic Education Curriculum.
Flordeliza F. Francisco, Ph.D.
Writer
Dr. Flordeliza Francisco is a current member of
the CHED Technical Panel for General
Education and the DepEd Mathematics Group of
Consultants for K to 12. She has been teaching
Advanced Calculus, Real Analysis, Finite
Mathematics, and related courses at the Ateneo
de Manila University for 26 years. She completed
her doctorate degree in Mathematics and
bachelor’s degree in Mathematics both at the
Ateneo. Dr. Francisco is also member of the
Philippine Council of Mathematics Teacher
Educators. She co-authored and edited books on
Mathematics, including Precalculus (Ateneo
Press) and Math Challenge Questions (Anvil
Publications).
Mark-Anthony J. Vidallo
Writer
Mr. Mark-Anthony Vidallo is a Mathematics
teacher and trainer at the Makati Science High
School. He graduated magna cum laude with a
degree in Secondary Education major in
Mathematics at the University of Makati and is
pursuing graduate studies at the Philippine
Normal University. He received citations
including the MTAP Exemplary Secondary
Mathematics Teacher Award and won first place
at the 1st Casio-DepEd NCR Mathematics
Teachers Olympics. Mr. Vidallo co-authored
textbooks on Calculus, Analytic Geometry, and
Math Ed magazines for junior high school. He is
a Trainer for Math competitions such as the
Philippine Math Olympiad, Metrobank-MTAPDepEd Math Challenge, and Australian
Mathematics Competition.
Maria Alva Q. Aberin, Ph.D.
Technical Editor
Dr. Maria Alva Aberin is an Assistant Professor
at the Ateneo de Manila University where she
has been teaching undergraduate and graduate
Mathematics courses for over 15 years. She has
also served as National Trainer for a number of
teacher trainings initiated by the DepEd. Dr.
Aberin was Vice President of the National Board
of Directors for the Philippine Council of
Mathematics Teacher Educators from 2015 to
2016. She completed her doctorate degree in
Mathematics Education, her master’s and
bachelor’s degrees in Mathematics at the
University of the Philippines Diliman.
Reginaldo M. Marcelo, Ph.D.
Technical Editor
Dr. Reginaldo Marcelo is currently a member of
the CHED Technical Panel for Science and
Mathematics. He is Assistant Professor at the
Ateneo de Manila University, where he has been
teaching for 31 years. He finished his doctorate
degree in Mathematics at the Sophia University
through a scholarship from the Japan’s Ministry
of Education and his master’s degree in
Mathematics at the University of the Philippines
Diliman. He graduated with a degree in
Mathematics at the Ateneo under a scholarship
grant from the Department of Science and
Technology. Dr. Marcelo is a member of the
Mathematical Society of the Philippines and the
Southeast Asian Mathematics Society, and coauthored and edited publications such as
Precalculus and College Algebra textbooks.
Naomi L. Tupas
Copyreader
Ms. Naomi Tupas holds various technical writing
and editing positions for both private and public
institutions. She is the technical writer at the
provincial government of Camarines Sur, an
editor at Operation Blessing Foundation
Philippines Inc., a copywriter at Coastal Training
Technologies, Inc., and the editor-in-chief for the
Marketing Communication Office at the De La
Salle University Manila. She is currently taking
her Master of Arts in Language Education at the
University of the Philippines Diliman, and she
finished her degree in Bachelor of Arts in
English, major in Creative Writing at the same
university.