Precalculus Learner's Guide

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Precalculus
Learner’s Material
Department of Education
Republic of the Philippines
This learning resource was collaboratively developed and
reviewed by educators from public and private schools, colleges, and/or
universities. We encourage teachers and other education stakeholders
to email their feedback, comments and recommendations to the
Department of Education at action@deped.gov.ph.
We value your feedback and recommendations.
All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

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Precalculus
Learner’s Material
First Edition 2016
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Table of Contents
To the Precalculus Learners 1
DepEd Curriculum Guide for Precalculus 2
Unit 1: Analytic Geometry 6
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.2.1: Definition and Equation of a Parabola. . . . . . . . . . . . . . . . . . . . 19
1.2.2: More Properties of Parabolas. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
1.2.3: Situational Problems Involving Parabolas . . . . . . . . . . . . . . . . 26
Lesson 1.3: Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
1.3.1: Definition and Equation of an Ellipse. . . . . . . . . . . . . . . . . . . . . 33
1.3.2: More Properties of Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
1.3.3: Situational Problems Involving Ellipses. . . . . . . . . . . . . . . . . . . 40
Lesson 1.4: Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
1.4.1: Definition and Equation of a Hyperbola . . . . . . . . . . . . . . . . . . 46
1.4.2: More Properties of Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
1.4.3: Situational Problems Involving Hyperbolas . . . . . . . . . . . . . . . 54
Lesson 1.5: More Problems on Conic Sections . . . . . . . . . . . . . . . . 60
1.5.1: Identifying the Conic Section by Inspection. . . . . . . . . . . . . . . 60
1.5.2: Problems Involving Different Conic Sections . . . . . . . . . . . . . . 62
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Lesson 1.6: Systems of Nonlinear Equations . . . . . . . . . . . . . . . . . . 67
1.6.1: Review of Techniques in Solving Systems of Linear
Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
1.6.2: Solving Systems of Equations Using Substitution . . . . . . . . . 69
1.6.3: Solving Systems of Equations Using Elimination. . . . . . . . . . 70
1.6.4: Applications of Systems of Nonlinear Equations . . . . . . . . . . 73
Unit 2: Mathematical Induction 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 . . . . . . . . . 87
2.2.2: Properties of Sigma Notation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
Lesson 2.3: Principle of Mathematical Induction . . . . . . . . . . . . . . 96
2.3.1: Proving Summation Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
2.3.2: Proving Divisibility Statements. . . . . . . . . . . . . . . . . . . . . . . . . . .101
2.3.3: Proving Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
Lesson 2.4: The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
2.4.1: Pascal’s Triangle and the Concept of Combination. . . . . . . . 109
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 123
Lesson 3.1: Angles in a Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
3.1.1: Angle Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
3.1.2: Coterminal Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
3.1.3: Arc Length and Area of a Sector . . . . . . . . . . . . . . . . . . . . . . . . . 129
Lesson 3.2: Circular Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
3.2.1: Circular Functions on Real Numbers . . . . . . . . . . . . . . . . . . . . . 136
3.2.2: Reference Angle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

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Lesson 3.3: Graphs of Circular Functions and Situational
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
3.3.1: Graphs of y = sin x and y = cos x . . . . . . . . . . . . . . . . . . . . . . . . 145
3.3.2: Graphs of y = a sin bx and y = a cos bx . . . . . . . . . . . . . . . . . . . 147
3.3.3: Graphs of y = a sin b(x − c) + d and
y = a cos b(x − c) + d. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
3.3.4: Graphs of Cosecant and Secant Functions . . . . . . . . . . . . . . . . 154
3.3.5: Graphs of Tangent and Cotangent Functions . . . . . . . . . . . . . 158
3.3.6: Simple Harmonic Motion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
Lesson 3.4: Fundamental Trigonometric Identities. . . . . . . . . . . . .171
3.4.1: Domain of an Expression or Equation . . . . . . . . . . . . . . . . . . . . 171
3.4.2: Identity and Conditional Equation . . . . . . . . . . . . . . . . . . . . . . . 173
3.4.3: The Fundamental Trigonometric Identities . . . . . . . . . . . . . . . 174
3.4.4: Proving Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . 176
Lesson 3.5: Sum and Difference Identities. . . . . . . . . . . . . . . . . . . . .181
3.5.1: The Cosine Difference and Sum Identities . . . . . . . . . . . . . . . . 181
3.5.2: The Cofunction Identities and the Sine Sum and
Difference Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
3.5.3: The Tangent Sum and Difference Identities . . . . . . . . . . . . . . . 186
Lesson 3.6: Double-Angle and Half-Angle Identities. . . . . . . . . . .192
3.6.1: Double-Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192
3.6.2: Half-Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
Lesson 3.7: Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . 201
3.7.1: Inverse Sine Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
3.7.2: Inverse Cosine Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
3.7.3: Inverse Tangent Function and the Remaining Inverse
Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
Lesson 3.8: Trigonometric Equations. . . . . . . . . . . . . . . . . . . . . . . . . .220
3.8.1: Solutions of a Trigonometric Equation. . . . . . . . . . . . . . . . . . . . 221
3.8.2: Equations with One Term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
3.8.3: Equations with Two or More Terms . . . . . . . . . . . . . . . . . . . . . . 227

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Lesson 3.9: Polar Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . 236
3.9.1: Polar Coordinates of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237
3.9.2: From Polar to Rectangular, and Vice Versa. . . . . . . . . . . . . . . 241
3.9.3: Basic Polar Graphs and Applications . . . . . . . . . . . . . . . . . . . . . 244
Answers to Odd-Numbered Exercises in Supplementary Problems
and All Exercises in Topic Tests 255
References 290

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To the Precalculus Learners
The Precalculus course bridges basic mathematics and calculus. This course
completes your foundational knowledge on algebra, geometry, and trigonometry.
It provides you with conceptual understanding and computational skills that are
prerequisites for Basic Calculus and future STEM courses.
Based on the Curriculum Guide for Precalculus of the Department of Edu-
cation (see pages 2-5), the primary aim of this Learning Manual is to give you
an adequate stand-alone material that can be used for the Grade 11 Precalculus
course.
The Manual is divided into three units: analytic geometry, summation no-
tation 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 sub-lessons that focus on one or two competencies
for effective learning.
At the end of each lesson, more examples are given in Solved Examples to
reinforce the ideas and skills being developed in the lesson. You have the oppor-
tunity to check your understanding of the lesson by solving the Supplementary
Problems. Finally, two sets of Topic Test are included to prepare you for the
exam.
Answers, solutions, or hints to odd-numbered items in the Supplementary
Problems and all items in the Topic Tests are provided at the end of the Manual
to guide you while solving them. We hope that you will use this feature of the
Manual responsibly.
Some items are marked with a star. A starred sub-lesson means the discussion
and accomplishment of the sub-lesson are optional. This will be decided by your
teacher. On the other hand, a starred example or exercise means the use of
calculator is required.
We hope that you will find this Learning Manual helpful and convenient to
use. We encourage you to carefully study this Manual and solve the exercises
yourselves with the guidance of your teacher. Although great effort has been
put into this Manual for technical correctness and precision, any mistake found
and reported to the Team is a gain for other students. Thank you for your
cooperation.
The Precalculus LM Team

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Kto12BASICEDUCATIONCURRICULUM
SENIORHIGHSCHOOL–SCIENCE,TECHNOLOGY,ENGINEERINGANDMATHEMATICS(STEM)SPECIALIZEDSUBJECT
Kto12SeniorHighSchoolSTEMSpecializedSubject–Pre-CalculusDecember2013Page1of4
Grade:11Semester:FirstSemester
CoreSubjectTitle:Pre-CalculusNo.ofHours/Semester:80hours/semester
Pre-requisite(ifneeded):
SubjectDescription:Attheendofthecourse,thestudentsmustbeabletoapplyconceptsandsolveproblemsinvolvingconicsections,systemsofnonlinearequations,
seriesandmathematicalinduction,circularandtrigonometricfunctions,trigonometricidentities,andpolarcoordinatesystem.
CONTENT
CONTENT
STANDARDS
PERFORMANCE
STANDARDS
LEARNINGCOMPETENCIESCODE
Analytic
Geometry
Thelearners
demonstratean
understanding
of...
keyconceptsof
conicsectionsand
systemsof
nonlinear
equations
Thelearnersshallbeable
to...
modelsituations
appropriatelyandsolve
problemsaccuratelyusing
conicsectionsandsystems
ofnonlinearequations
Thelearners...
1.illustratethedifferenttypesofconicsections:parabola,ellipse,
circle,hyperbola,anddegeneratecases.***
STEM_PC11AG-Ia-1
2.defineacircle.STEM_PC11AG-Ia-2
3.determinethestandardformofequationofacircleSTEM_PC11AG-Ia-3
4.graphacircleinarectangularcoordinatesystemSTEM_PC11AG-Ia-4
5.defineaparabolaSTEM_PC11AG-Ia-5
6.determinethestandardformofequationofaparabolaSTEM_PC11AG-Ib-1
7.graphaparabolainarectangularcoordinatesystemSTEM_PC11AG-Ib-2
8.defineanellipseSTEM_PC11AG-Ic-1
9.determinethestandardformofequationofanellipseSTEM_PC11AG-Ic-2
10.graphanellipseinarectangularcoordinatesystemSTEM_PC11AG-Ic-3
11.defineahyperbolaSTEM_PC11AG-Id-1
12.determinethestandardformofequationofahyperbolaSTEM_PC11AG-Id-2
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CONTENT
CONTENT
STANDARDS
PERFORMANCE
STANDARDS
13.graphahyperbolainarectangularcoordinatesystemSTEM_PC11AG-Id-3
14.recognizetheequationandimportantcharacteristicsofthe
differenttypesofconicsections
STEM_PC11AG-Ie-1
15.solvessituationalproblemsinvolvingconicsectionsSTEM_PC11AG-Ie-2
16.illustratesystemsofnonlinearequationsSTEM_PC11AG-If-1
17.determinethesolutionsofsystemsofnonlinearequationsusing
techniquessuchassubstitution,elimination,andgraphing***
STEM_PC11AG-If-g-1
18.solvesituationalproblemsinvolvingsystems
ofnonlinearequations
STEM_PC11AG-Ig-2
Seriesand
Mathematical
Induction
keyconceptsof
seriesand
mathematical
inductionandthe
Binomial
Theorem.
keenlyobserveand
investigatepatterns,and
formulateappropriate
mathematicalstatements
andprovethemusing
mathematicalinduction
and/orBinomialTheorem.
1.illustrateaseries
STEM_PC11SMI-Ih-1
2.differentiateaseriesfromasequenceSTEM_PC11SMI-Ih-2
3.usethesigmanotationtorepresentaseriesSTEM_PC11SMI-Ih-3
4.illustratethePrincipleofMathematicalInductionSTEM_PC11SMI-Ih-4
5.applymathematicalinductioninprovingidentitiesSTEM_PC11SMI-Ih-i-1
6.illustratePascal’sTriangleintheexpansionof𝑥+𝑦𝑛
forsmall
positiveintegralvaluesof𝑛
STEM_PC11SMI-Ii-2
7.provetheBinomialTheoremSTEM_PC11SMI-Ii-3
8.determineanytermof𝑥+𝑦𝑛
,where𝑛isapositiveinteger,
withoutexpanding
STEM_PC11SMI-Ij-1
9.solveproblemsusingmathematicalinductionandtheBinomial
Theorem
STEM_PC11SMI-Ij-2
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CONTENT
CONTENT
STANDARDS
PERFORMANCE
STANDARDS
Trigonometrykeyconceptsof
circularfunctions,
trigonometric
identities,inverse
trigonometric
functions,and
thepolar
coordinate
system
1.formulateandsolve
accuratelysituational
problemsinvolving
circularfunctions
1.illustratetheunitcircleandtherelationshipbetweenthelinear
andangularmeasuresofacentralangleinaunitcircleSTEM_PC11T-IIa-1
2.convertdegreemeasuretoradianmeasureandviceversaSTEM_PC11T-IIa-2
3.illustrateanglesinstandardpositionandcoterminalanglesSTEM_PC11T-IIa-3
4.illustratethedifferentcircularfunctionsSTEM_PC11T-IIb-1
5.usesreferenceanglestofindexactvaluesofcircularfunctionsSTEM_PC11T-IIb-2
6.determinethedomainandrangeofthedifferentcircularfunctionsSTEM_PC11T-IIc-1
7.graphthesixcircularfunctions(a)amplitude,(b)period,and(c)
phaseshift
STEM_PC11T-IIc-d-1
8.solveproblemsinvolvingcircularfunctionsSTEM_PC11T-IId-2
2.applyappropriate
trigonometricidentitiesin
solvingsituational
problems
9.determinewhetheranequationisanidentityoraconditional
equation
STEM_PC11T-IIe-1
10.derivethefundamentaltrigonometricidentitiesSTEM_PC11T-IIe-2
11.derivetrigonometricidentitiesinvolvingsumanddifferenceof
angles
STEM_PC11T-IIe-3
12.derivethedoubleandhalf-angleformulasSTEM_PC11T-IIf-1
13.simplifytrigonometricexpressionsSTEM_PC11T-IIf-2
14.proveothertrigonometricidentitiesSTEM_PC11T-IIf-g-1
15.solvesituationalproblemsinvolvingtrigonometricidentitiesSTEM_PC11T-IIg-2
3.formulateandsolve
problemsinvolving
appropriatetrigonometric
functions
16.illustratethedomainandrangeoftheinversetrigonometric
functions.
STEM_PC11T-IIh-1
17.evaluateaninversetrigonometricexpression.STEM_PC11T-IIh-2
18.solvetrigonometricequations.STEM_PC11T-IIh-i-1
19.solvesituationalproblemsinvolvinginversetrigonometric
functionsandtrigonometricequations
STEM_PC11T-IIi-2
4.formulateandsolve
problemsinvolvingthe
polarcoordinatesystem
20.locatepointsinpolarcoordinatesystemSTEM_PC11T-IIj-1
21.convertthecoordinatesofapointfromrectangulartopolar
systemsandviceversa
STEM_PC11T-IIj-2
22.solvesituationalproblemsinvolvingpolarcoordinatesystemSTEM_PC11T-IIj-3
***SuggestionforICT-enhancedlessonwhenavailableandwhereappropriate
4

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CodeBookLegend
Sample:STEM_PC11AG-Ia-1
DOMAIN/COMPONENTCODE
AnalyticGeometryAG
SeriesandMathematicalInductionSMI
TrigonometryT
LEGENDSAMPLE
FirstEntry
LearningAreaand
Strand/Subjector
Specialization
Science,Technology,
EngineeringandMathematics
Pre-Calculus
STEM_PC11AGGradeLevelGrade11
Uppercase
Letter/s
Domain/Content/
Component/Topic
AnalyticGeometry
-
RomanNumeral
*Zeroifnospecific
quarter
QuarterFirstQuarterI
Lowercase
Letter/s
*Putahyphen(-)in
betweenlettersto
indicatemorethana
specificweek
WeekWeekonea
-
ArabicNumberCompetency
illustratethedifferenttypes
ofconicsections:parabola,
ellipse,circle,hyperbola,
anddegeneratecases
1
5

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Unit 1
Analytic Geometry
San Juanico Bridge, by Morten Nærbøe, 21 June 2009,
https://commons.wikimedia.org/wiki/File%3ASan Juanico Bridge 2.JPG. Public Domain.
Stretching from Samar to Leyte with a total length of more than two kilome-
ters, the San Juanico Bridge has been serving as one of the main thoroughfares
of economic and social development in the country since its completion in 1973.
Adding picturesque eﬀect 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.
6

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Lesson 1.1. Introduction to Conic Sections and Circles
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate the different types of conic sections: parabola, ellipse, circle, hyper-
bola, 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 present the conic sections, a particular class of curves which sometimes
appear in nature and which have applications in other fields. In this lesson, we
first illustrate how each of these curves is obtained from the intersection of a
plane and a cone, and then 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
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• 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 hyper-
bola)
Figure 1.1 Figure 1.2 Figure 1.3
We can draw these conic sections (also called conics) on a rectangular co-
ordinate 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.
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). As we get to know more about a circle, we will
also be able to distinguish more between these two conics.
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See Figure 1.7, with the point C(3, 1) shown. From the ﬁgure, the distance
of A(−2, 1) from 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
PC = 5. The collection of all such points which are 5 units away from C, forms
a circle.
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 PC = r. For any such point then,
its coordinates should satisfy the following.
PC = r
(x − h)2 + (y − k)2 = r
(x − h)2
+ (y − 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
+ y2
= r2
.
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Example 1.1.1. In each item, give the
standard equation of the circle satisfy-
ing the given conditions.
(1) center at the origin, radius 4
(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 y-
axis Figure 1.9
(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
+ y2
= 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−3)2
+(y−1)2
=
25.
(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
2
, 4+2
2
= 3
2
, 3 . The
radius is then r = AC = −1 − 3
2
2
+ (4 − 3)2 = 29
4
. The circle has
equation x − 3
2
2
+ (y − 3)2
= 29
4
. 2
1.1.3. More Properties of Circles
After expanding, the standard equation
x −
3
2
2
+ (y − 3)2
=
29
4
can be rewritten as
x2
+ y2
− 3x − 6y + 4 = 0,
an equation of the circle in general form.
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If the equation of a circle is given in the general form
Ax2
+ Ay2
+ Cx + Dy + E = 0, A = 0,
or
x2
+ y2
+ 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 equa-
tion in each item. Sketch its graph, and indicate the center.
(1) x2
+ y2
− 6x = 7
(2) x2
+ y2
− 14x + 2y = −14
(3) 16x2
+ 16y2
+ 96x − 40y = 315
Solution. The first step is to rewrite each equation in standard form by complet-
ing the square in x and in y. From the standard equation, we can determine the
center and radius.
(1)
x2
− 6x + y2
= 7
x2
− 6x + 9 + y2
= 7 + 9
(x − 3)2
+ y2
= 16
Center (3, 0), r = 4, Figure 1.10
(2)
x2
− 14x + y2
+ 2y = −14
x2
− 14x + 49 + y2
+ 2y + 1 = −14 + 49 + 1
(x − 7)2
+ (y + 1)2
= 36
Center (7, −1), r = 6, Figure 1.11
(3)
16x2
+ 96x + 16y2
− 40y = 315
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16(x2
+ 6x) + 16 y2
−
5
2
y = 315
16(x2
+ 6x + 9) + 16 y2
−
5
2
y +
25
16
= 315 + 16(9) + 16
25
16
16(x + 3)2
+ 16 y −
5
4
2
= 484
(x + 3)2
+ y −
5
4
2
=
484
16
=
121
4
=
11
2
2
Center −3, 5
4
, r = 5.5, Figure 1.12. 2
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.
1.1.4. Situational Problems Involving Circles
Let us now take a look at some situational problems involving circles.
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 oﬀ to 2 decimal places.
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Solution. We draw a coordinate system with origin at the middle of the highway,
as shown. Because of the given radius, the tunnel’s boundary is on the circle
x2
+ y2
= 122
. Point P is the point on the arc just above the edge of a lane, so
its x-coordinate is 10. We need its y-coordinate. We then solve 102
+ y2
= 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 1.13, 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 1.13
Figure 1.14
Solution. We ﬁrst determine the center. It is the intersection of the perpendicular
bisectors of AB and BC (see Figure 1.14). Recall that, in a circle, the perpen-
dicular bisector of any chord passes through the center. Since the midpoint M
of AB is −7+1
2
, 0+4
2
= (−3, 2), and mAB = 4−0
1+7
= 1
2
, the perpendicular bisector
of AB has equation y − 2 = −2(x + 3), or equivalently, y = −2x − 4.
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Since the midpoint N of BC is 1+7
2
, 4+2
2
= (4, 3), and mBC = 2−4
7−1
= −1
3
,
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
More Solved Examples
1. In each item, give the standard equation of the circle satisying the given con-
ditions.
(a) center at the origin, contains (0, 3)
(b) center (1, 5), diameter 8
(c) circle A in Figure 1.15
(d) circle B in Figure 1.15
(e) circle C in Figure 1.15
(f) center (−2, −3), tangent to the y-
axis
(g) center (−2, −3), tangent to the x-
axis
(h) contains the points (−2, 0) and
(8, 0), radius 5
Figure 1.15
Solution:
(a) The radius is 3, so the equation is x2
+ y2
= 9.
(b) The radius is 8/2 = 4, so the equation is (x − 1)2
+ (y − 5)2
= 16.
(c) The center is (−2, 2) and the radius is 2,
so the equation is (x+2)2
+(y −2)2
= 4.
(d) The center is (2, 3) and the radius is 1,
so the equation is (x−2)2
+(y −3)2
= 1.
(e) The center is (1, −1) and by the
Pythagorean Theorem, the radius (see
Figure 1.16) is
√
22 + 22 =
√
8, so the
equation is (x − 1)2
+ (x + 1)2
= 8.
Figure 1.16
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(f) The radius is 3, so the equation is (x + 2)2
+ (y + 3)2
= 9.
(g) The radius is 2, so the equation is (x + 2)2
+ (y + 3)2
= 4.
(h) The distance between (−2, 0) and (8, 0) is 10; since the radius is 5, these
two points are endpoints of a diameter. Then the circle has center at
(3, 0) and radius 5, so its equation is (x − 3)2
+ y2
= 25.
2. Identify the center and radius of the circle with the given equation in each
item. Sketch its graph, and indicate the center.
(a) x2
+ y2
+ 8y = 33
(b) 4x2
+ 4y2
− 16x + 40y + 67 = 0
(c) 4x2
+ 12x + 4y2
+ 16y − 11 = 0
Solution:
(a)
x2
+ y2
+ 8y = 33
x2
+ y2
+ 8y + 16 = 33 + 16
x2
+ (y + 4)2
= 49
Center (0, −4), radius 7, see Figure 1.17.
(b)
4x2
+ 4y2
− 16x + 40y + 67 = 0
x2
− 4x + y2
+ 10y = −
67
4
x2
− 4x + 4 + y2
+ 10y + 25 = −
67
4
+ 4 + 25
(x − 2)2
+ (y + 5)2
=
49
4
=
7
2
2
Center (2, −5), radius 3.5, see Figure 1.18.
(c)
4x2
+ 12x + 4y2
+ 16y − 11 = 0
x2
+ 3x + y2
+ 4y =
11
4
x2
+ 3x +
9
4
+ y2
+ 4y + 4 =
11
4
+
9
4
+ 4
x +
3
2
2
+ (y + 2)2
= 9
Center −
3
2
, −2 , radius 3, see Figure 1.19.
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3. A circular play area with radius 3 m is
to be partitioned into two sections using
a straight fence as shown in Figure 1.20.
How long should the fence be?
Solution: To determine the length of the
fence, we need to determine the coordi-
nates of its endpoints. From Figure 1.20,
the endpoints have x coordinate −1 and
are on the circle x2
+ y2
= 9. Then
1 + y2
= 9, or y = ±2
√
2. Therefore,
the length of the fence is 4
√
2 ≈ 5.66 m. Figure 1.20
4. A Cartesian coordinate system was used to identify locations on a circu-
lar track. As shown in Figure 1.21, the circular track contains the points
A(−2, −4), B(−2, 3), C(5, 2). Find the total length of the track.
Solution: The segment AB is vertical and has midpoint (−2, −0.5), so its
perpendicular bisector has equation y = −0.5. On the other hand, the segment
BC has slope −1/7 and midpoint (1.5, 2.5), so its perpendicular bisector has
equation y − 2.5 = 7(x − 1.5), or 7x − y − 8 = 0.
The center of the circle is the intersection of y = −0.5 and 7x − y − 8 = 0;
that is, the center is at 15
14
, −1
2
.
The radius of the circle is the distance from the center to any of the points A,
B, or C; by the distance formula, the radius is
2125
98
=
5
14
√
170. Therefore,
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the total length of the track (its circumference), is
2 × π ×
5
14
√
170 ≈ 29.26 units.
Supplementary Problems 1.1
Identify the center and radius of the circle with the given equation in each item.
Sketch its graph, and indicate the center.
1. x2
+ y2
=
1
4
2. 5x2
+ 5y2
= 125
3. (x + 4)2
+ y −
3
4
2
= 1
4. x2
− 4x + y2
− 4y − 8 = 0
5. x2
+ y2
− 14x + 12y = 36
6. x2
+ 10x + y2
− 16y − 11 = 0
7. 9x2
+ 36x + 9y2
+ 72y + 155 = 0
8. 9x2
+ 9y2
− 6x + 24y = 19
9. 16x2
+ 80x + 16y2
− 112y + 247 = 0
Find the standard equation of the circle which satisﬁes the given conditions.
10. center at the origin, radius 5
√
3
11. center at (17, 5), radius 12
12. center at (−8, 4), contains (−4, 2)
13. center at (15, −7), tangent to the x-axis
14. center at (15, −7), tangent to the y-axis
15. center at (15, −7), tangent to the line y = −10
16. center at (15, −7), tangent to the line x = 8
17. has a diameter with endpoints (3, 1) and (−7, 6)
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18. has a diameter with endpoints
9
2
, 4 and −
3
2
, −2
19. concentric with x2
+ 20x + y2
− 14y + 145 = 0, diameter 12
− 2x + y2
− 2y − 23 = 0 and has 1/5 the area
+ 4x + y2
− 6y + 9 = 0 and has the same circumference as
x2
+ 14x + y2
+ 10y + 62 = 0
22. contains the points (3, 3), (7, 1), (0, 2)
23. contains the points (1, 4), (−1, 2), (4, −3)
24. center at (−3, 2) and tangent to the line 2x − 3y = 1
25. center at (−5, −1) and tangent to the line x + y + 10 = 0
26. has center with x-coordinate 4 and tangent to the line −x + 3y = 9 at (3, 4)
27. A stadium is shaped as in Figure 1.23, where its left and right ends are circular
arcs both with center at C. What is the length of the stadium 50 m from one
of the straight sides?
Figure 1.23
28. A waterway in a theme park has a
semicircular cross section with di-
ameter 11 ft. The boats that are
going to be used in this waterway
have rectangular cross sections and
are found to submerge 1 ft into the
water. If the waterway is to be
ﬁlled with water 4.5 ft deep, what is
the maximum possible width of the
boats?
Figure 1.24
4
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Lesson 1.2. Parabolas
(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
(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 1.25. 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 and AA = 4
BF = (−8 − 0)2 + (8 − 2)2 = 10 and BB = 10
There are other points P such that PF = PP (where P is the closest point on
line ). The collection of all such points forms a shape called a parabola.
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.
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Figure 1.25
Figure 1.26
Consider a parabola with focus F(0, c) and directrix having equation y = −c.
See Figure 1.26. The focus and directrix are c units above and below, respectively,
the origin. Let P(x, y) be a point on the parabola so PF = PP , where P is the
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 = PP
x2 + (y − c)2 = y − (−c) = y + c
x2
+ y2
− 2cy + c2
= y2
+ 2cy + c2
x2
= 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, PF =
x2 + (y + c)2 and PP = c − y (you may draw a version of Figure 1.26 for
this case). Computations similar to the one done above will lead to the equation
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x2
= −4cy.
We collect here the features of the graph of a parabola with standard equation
x2
= 4cy or x2
= −4cy, where c > 0.
(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)
• 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.
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(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 = 3
2
.
The focus, c = 3
2
units below the vertex, is F 0, −3
2
. The directrix, 3
2
units
above the vertex, is y = 3
2
. The axis of symmetry is x = 0.
Example 1.2.2. What is the standard equation of the parabola in Figure 1.25?
Solution. From the ﬁgure, we deduce that c = 2. The equation is thus x2
=
8y. 2
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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.
(x − h)2
= 4c(y − k) (y − k)2
= 4c(x − h)
(x − h)2
= −4c(y − k) (y − k)2
= −4c(x − h)
directrix : horizontal directrix : vertical
axis of symmetry: x=h, vertical axis of symmetry: y=k, horizontal
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Note the following observations:
• 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.
Example 1.2.3. Figure 1.27 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 1.27
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The standard equation (y + 4)2
= −8(x − 5) from the preceding example can
be rewritten as y2
+ 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 By2
+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.
(1) y2
− 5x + 12y = −16
(2) 5x2
+ 30x + 24y = 51
Solution. (1) We complete the square on y, and move x to the other side.
y2
+ 12y = 5x − 16
y2
+ 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 = 5
4
= 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.
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(2) We complete the square on x, and move y to the other side.
5x2
+ 30x = −24y + 51
5(x2
+ 6x + 9) = −24y + 51 + 5(9)
5(x + 3)2
= −24y + 96 = −24(y − 4)
(x + 3)2
= −
24
5
(y − 4)
In the last line, we divided by 5 for the squared part not to have any coeﬃ-
cient. The parabola opens downward. It has vertex V (−3, 4).
From 4c = 24
5
, we get c = 6
5
= 1.2. The focus is c = 1.2 units below V :
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
1.2.3. Situational Problems Involving Parabolas
Let us 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
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oﬀ 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?
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)
c =
62
4 · 4.5
= 2
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?
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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)
a =
2002
120
=
1000
3
The parabola has equation x2
= 1000
3
(y − 30), or equivalently,
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.
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
For Examples 1 and 2, determine the focus and directrix of the parabola with the
given equation. Sketch the graph, and indicate the focus, directrix, and vertex.
1. y2
= 20x
Solution:
Vertex: V (0, 0), opens to the right
4c = 20 ⇒ c = 5
Focus: F(5, 0), Directrix: x = −5
See Figure 1.28.
2. 3x2
= −12y
Solution: 3x2
= −12y ⇔ x2
= −4y
Vertex: V (0, 0), opens downward
4c = 4 ⇒ c = 1
Focus: F(0, −1), Directrix: y = 1
See Figure 1.29.
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3. Determine the standard equation of the
parabola in Figure 1.30 given only its
vertex and focus. Then determine its di-
rectix and axis of symmetry.
Solution:
V −
3
2
, 4 , F(−4, 4)
c =
5
2
⇒ 4c = 10
Parabola opens to the left
Equation: (y − 4)2
= −10 x +
3
2
Directrix: x = 1, Axis: y = 4 Figure 1.30
4. Determine the standard equation of the
parabola in Figure 1.31 given only its
vertex and diretrix. Then determine its
focus and axis of symmetry.
Solution:
V 5,
13
2
, directrix: y =
15
2
c = 1 ⇒ 4c = 4
Parabola opens downward
Equation: y −
13
2
2
= −4 (x − 5)
Focus: 5,
11
2
, Axis: x = 5
Figure 1.31
For Examples 5 and 6, determine the vertex, focus, directrix, and axis of sym-
metry of the parabola with the given equation. Sketch the parabola, and include
these points and lines.
5. x2
− 6x − 2y + 9 = 0
Solution:
x2
− 6x = 2y − 9
x2
− 6x + 9 = 2y
(x − 3)2
= 2y
V (3, 0), parabola opens upward
4c = 2 ⇒ c =
1
2
, F 3,
1
2
,
directrix: y = −
1
2
, axis: x = 3
See Figure 1.32.
Figure 1.32
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6. 3y2
+ 8x + 24y + 40 = 0
Solution:
3y2
+ 24y = −8x − 40
3(y2
+ 8y) = −8x − 40
3(y2
+ 8y + 16) = −8x − 40 + 48
3(y + 4)2
= −8x + 8
(y + 4)2
= −
8
3
(x − 1)
V (1, −4), parabola opens to the left
4c =
8
3
⇒ c =
2
3
, F
1
3
, −4 ,
directrix: x =
5
3
, axis: y = −4
See Figure 1.33.
Figure 1.33
7. A parabola has focus F(−11, 8) and directrix x = −17. Find its standard
equation.
Solution: Since the focus is 6 units to the right of the directrix, the parabola
opens to the right with 2c = 6. Then c = 3 and V (−14, 8). Hence, the
equation is (y − 8)2
= 12(x + 14).
8. A flashlight is shaped like a
paraboloid and the light source
is placed at the focus so that
the light bounces off parallel to
the axis of symmetry; this is
done to maximize illumination.
A particular flashlight has its
light source located 1 cm from
the base and is 6 cm deep; see
Figure 1.34. What is the width
of the flashlight’s opening?
Figure 1.34
Solution: Let the base (the vertex) of the flashlight be the point V (0, 0).
Then the light source (the focus) is at F(0, 1); so c = 1. Hence, the parabola’s
equation is x2
= 4y. To get the width of the opening, we need the x coordinates
of the points on the parabola with y coordinate 6.
x2
= 4(6) ⇒ x = ±2
√
6
Therefore, the width of the opening is 2 × 2
√
6 = 4
√
6 ≈ 9.8 cm.
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9. An object thrown from a height of 2
m above the ground follows a parabolic
path until the object falls to the ground;
see Figure 1.35. If the object reaches
a maximum height (measured from the
ground) of 7 m after travelling a hor-
izontal distance of 4 m, determine the
horizontal distance between the object’s
initial and ﬁnal positions.
Figure 1.35
Solution: Let V (0, 7) be the parabola’s vertex, which corresponds to the high-
est point reached by the object. Then the parabola’s equation is of the form
x2
= −4c(y − 7) and the object’s starting point is at (−4, 2). Then
(−4)2
= −4c(2 − 7) ⇒ c =
16
20
=
4
5
.
Hence, the equation of the parabola is x2
= −
16
5
(y −7). When the object hits
the ground, the y coordinate is 0 and
x2
= −
16
5
(0 − 7) =
112
5
⇒ x = ±4
7
5
.
Since this point is to the right of the vertex, we choose x = +4
7
5
. Therefore,
the total distance travelled is 4
7
5
− (−4) ≈ 8.73 m.
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.
1. y2
= −36x
2. 5x2
= 100y
3. y2
+ 4x − 14y = −53
4. y2
− 2x + 2y − 1 = 0
5. 2x2
− 12x + 28y = 38
6. (3x − 2)2
= 84y − 112
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Find the standard equation of the parabola which satisfies the given conditions.
7. vertex (7, 11), focus (16, 11)
8. vertex (−10, −5), directrix y = −1
9. focus −10,
23
2
, directrix y = −
11
2
10. focus −
3
2
, 3 , directrix x = −
37
2
11. axis of symmetry y = 9, directrix x = 24, vertex on the line 3y − 5x = 7
12. vertex (0, 7), vertical axis of symmetry, through the point P(4, 5)
13. vertex (−3, 8), horizontal axis of symmetry, through the point P(−5, 12)
14. A satellite dish shaped like a paraboloid has its receiver located at the focus.
How far is the receiver from the vertex if the dish is 10 ft across and 3 ft deep
at the center?
15. A flashlight shaped like a paraboloid has its light source at the focus located
1.5 cm from the base and is 10 cm wide at its opening. How deep is the
flashlight at its center?
16. The ends of a rope are held in place at the top of two posts, 9 m apart and
each one 8 m high. If the rope assumes a parabolic shape and touches the
ground midway between the two posts, how high is the rope 2 m from one of
the posts?
17. Radiation is focused to an unhealthy area in a patient’s body using a parabolic
reflector, positioned in such a way that the target area is at the focus. If the
reflector is 30 cm wide and 15 cm deep at the center, how far should the base
of the reflector be from the target area?
18. A rectangular object 25 m wide is to pass under a parabolic arch that has a
width of 32 m at the base and a height of 24 m at the center. If the vertex
of the parabola is at the top of the arch, what maximum height should the
rectangular object have?
4
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Lesson 1.3. Ellipses
(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
(3) Graphing ellipses
(4) Solving situational problems involving ellipses
Introduction
Unlike circle and parabola, an ellipse is one of the conic sections that most stu-
dents have not encountered formally before. 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.
These applications and more will be encountered in this lesson.
1.3.1. Definition and Equation of an Ellipse
Consider the points F1(−3, 0) and F2(3, 0), as shown in Figure 1.36. 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 PF1 + PF2 = 10. The collection of all such
points forms a shape called an ellipse.
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Figure 1.36
Figure 1.37
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.
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 1.37. 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 PF1 + PF2 = 2a.
PF1 = 2a − PF2
(x + c)2 + y2 = 2a − (x − c)2 + y2
x2
+ 2cx + c2
+ y2
= 4a2
− 4a (x − c)2 + y2 + x2
− 2cx + c2
+ y2
a (x − c)2 + y2 = a2
− cx
a2
x2
− 2cx + c2
+ y2
= a4
− 2a2
cx + c2
x2
(a2
− c2
)x2
+ a2
y2
= a4
− a2
c2
= a2
(a2
− c2
)
b2
x2
+ a2
y2
= a2
b2
by letting b =
√
a2 − c2, so a > b
x2
a2
+
y2
b2
= 1
When we let b =
√
a2 − c2, we assumed a > c. To see why this is true, look at
PF1F2 in Figure 1.37. By the Triangle Inequality, PF1 + PF2 > F1F2, which
implies 2a > 2c, so a > c.
We collect here the features of the graph of an ellipse with standard equation
x2
a2
+
y2
b2
= 1, where a > b. Let c =
√
a2 − b2.
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(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.
• If y = 0, then x = ±a. Each vertex is a units away from the center.
• The segment V1V2 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 W1W2 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
25
+
y2
9
= 1.
Sketch the graph, and include these points.
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Solution. With a2
= 25 and b2
= 9, we have a = 5, b = 3, and c =
√
a2 − b2 = 4.
foci: F1(−4, 0), F2(4, 0) 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 1.36.
Solution. We have 2a = 10 and c = 3, so a = 5 and b =
√
a2 − c2 = 4. The
equation is
x2
25
+
y2
16
= 1. 2
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.
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.
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Center Corresponding Graphs
(0, 0)
x2
a2
+
y2
b2
= 1, a > b
x2
b2
+
y2
a2
= 1, b > a
(h, k)
(x − h)2
a2
+
(y − k)2
b2
= 1
(x − h)2
b2
+
(y − k)2
a2
= 1
a > b b > a
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.
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(1)
(x + 3)2
24
+
(y − 5)2
49
= 1
(2) 9x2
+ 16y2
− 126x + 64y = 71
Solution. (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: (−3, 5)
foci: F1(−3, 0), F2(−3, 10)
vertices: V1(−3, −2), V2(−3, 12)
covertices: W1(−3 − 2
√
6, 5) ≈ (−7.9, 5)
W2(−3 + 2
√
6, 5) ≈ (1.9, 5)
(2) We ﬁrst change the given equation to standard form.
9(x2
− 14x) + 16(y2
+ 4y) = 71
9(x2
− 14x + 49) + 16(y2
+ 4y + 4) = 71 + 9(49) + 16(4)
9(x − 7)2
+ 16(y + 2)2
= 576
(x − 7)2
64
+
(y + 2)2
36
= 1
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We have a = 8 and b = 6. Thus, c =
√
a2 − b2 = 2
√
7 ≈ 5.3. The ellipse is
horizontal.
center: (7, −2)
foci: F1(7 − 2
√
7, −2) ≈ (1.7, −2)
F2(7 + 2
√
7, −2) ≈ (12.3, −2)
vertices: V1(−1, −2), V2(15, −2)
covertices: 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 aligned) and c = 4. From the
given sum, 2a = 14 so a = 7. Also, b =
√
a2 − c2 =
√
33. The equation is
(x + 3)2
33
+
(y + 2)2
49
= 1. 2
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.
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Solution. The midpoint (2, −5) of 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
the minor axis, 2b = 12 so b = 6. The standard equation is
(x − 2)2
61
+
(y + 5)2
36
=
1. Each focus is c =
√
a2 − b2 = 5 units away from (2, −5), so their coordinates
are (−3, −5) and (7, −5). 2
1.3.3. Situational Problems Involving Ellipses
Let us 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 off 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
182
+
y2
152
= 1.
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
182
+
n2
152
= 1
n = 10
√
2 ≈ 14.14 ft 2
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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.
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
250000
+
y2
243600
= 1.
The star could have been placed on the negative x-axis, and the answer would
still be the same. 2
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1. Give the coordinates of the foci, vertices,
and covertices of the ellipse with equa-
tion
x2
169
+
y2
144
= 1. Then sketch the
graph and include these points.
Solution: The ellipse is horizontal.
a2
= 169 ⇒ a = 13, b2
= 144 ⇒ b = 12,
c =
√
169 − 144 = 5
Foci: F1(−5, 0), F2(5, 0)
Vertices: V1(−13, 0), V2(13, 0)
Covertices: W1(0, −12), W2(0, 12)
See Figure 1.38. Figure 1.38
2. Find the standard equation of the ellipse whose foci are F1(0, −8) and F2(0, 8),
such that for any point on it, the sum of its distances from the foci is 34.
Solution: The ellipse is vertical and has center at (0, 0).
2a = 34 ⇒ a = 17
c = 8 ⇒ b =
√
172 − 82 = 15
The equation is
x2
225
+
y2
289
= 1.
For Examples 3 and 4, give the coordinates of the center, foci, vertices, and
covertices of the ellipse with the given equation. Sketch the graph, and include
these points.
3.
(x − 7)2
64
+
(y + 2)2
25
= 1
Solution: The ellipse is horizontal.
a2
= 64 ⇒ a = 8, b2
= 25 ⇒ b = 5
c =
√
64 − 25 =
√
39 ≈ 6.24
center: (7, −2)
foci: F1(7 −
√
39, −2) ≈ (0.76, −2)
F2(7 +
√
39, −2) ≈ (13.24, −2)
vertices: V1(−1, −2), V2(15, −2)
covertices: W1(7, −7), W2(7, 3)
See Figure 1.39.
Figure 1.39
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4. 16x2
+ 96x + 7y2
+ 14y + 39 = 0
Solution:
16x2
+ 96x + 7y2
+ 14y = −39
16(x2
+ 6x + 9) + 7(y2
+ 2y + 1) = −39 + 151
16(x + 3)2
+ 7(y + 1)2
= 112
(x + 3)2
7
+
(y + 1)2
16
= 1
The ellipse is vertical.
a2
= 16 ⇒ a = 4, b2
= 7 ⇒ b =
√
7 ≈
2.65
c =
√
16 − 7 = 3
center: (−3, −1)
foci: F1(−3, −4), F2(−3, 2)
vertices: V1(−3, −5), V2(−3, 3)
covertices: W1(−3 −
√
7, −1) ≈ (−5.65, −1)
W2(−3 +
√
7, −1) ≈ (−0.35, −1)
See Figure 1.40.
Figure 1.40
5. The covertices of an ellipse are (5, 6) and (5, 8). For any point on the ellipse,
the sum of its distances from the foci is 12. Find the standard equation of the
ellipse.
Solution: The ellipse is horizontal with center at the midpoint (5, 7) of the
covertices. Also, 2a = 12 so a = 6 while b = 1. The equation is
(x − 5)2
36
+
(y − 7)2
1
= 1.
6. An ellipse has foci (−4 −
√
15, 3) and (−4 +
√
15, 3), and its major axis is 10
units long. Find its standard equation and its vertices.
Solution: The ellipse is horizontal with center at the midpoint (−4, 3) of the
foci; also c =
√
15. Since the length of the major axis is 10, 2a = 10 and
a = 5. Thus b =
√
52 − 15 =
√
10. Therefore, the equation of the ellipse is
(x + 4)2
25
+
(y − 3)2
10
= 1 and its vertices are (−9, 3) and (1, 3).
7. A whispering gallery is an enclosure or room where whispers can be clearly
heard in some parts of the gallery. Such a gallery can be constructed by
making its ceiling in the shape of a semi-ellipse; in this case, a whisper from
one focus can be clearly heard at the other focus. If an elliptical whispering
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gallery is 90 feet long and the foci are 50 feet apart, how high is the gallery at
its center?
Solution: We set up a Cartesian coordinate system by assigning the center of
the semiellipse as the origin. The point on the ceiling right above the center is a
covertex of the ellipse. Since 2a = 90 and 2c = 50; then b2
= 452
−252
= 1400.
The height is given by b =
√
1400 ≈ 37.4 ft.
8. A spheroid (or oblate spheroid) is the surface obtained by rotating an ellipse
around its minor axis. The bowl in Figure 1.41 is in the shape of the lower half
of a spheroid; that is, its horizontal cross sections are circles while its vertical
cross sections that pass through the center are semiellipses. If this bowl is 10
in wide at the opening and
√
10 in deep at the center, how deep does a circular
cover with diameter 9 in go into the bowl?
Figure 1.41
Solution: We set up a Cartesian coordinate system by assigning the center
of the semiellipse as the origin. Then a = 5, b =
√
10, and the equation of
the ellipse is x2
25
+ y2
10
= 1. We want the y-coordinate of the points on the
ellipse that has x = ±4.5. This coordinate is y = − 10 1 − x2
25
≈ −1.38.
Therefore, the cover will go 1.38 inches into the bowl.
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Give the coordinates of the center, foci, vertices, and covertices of the ellipse with
the given equation. Sketch the graph, and include these points.
1.
x2
8
+
y2
4
= 1
2.
x2
16
+
(y − 2)2
25
= 1
3. (x − 1)2
+ (2y − 2)2
= 4
4.
(x + 5)2
49
+
(y − 2)2
121
= 1
5. 16x2
− 224x + 25y2
+ 250y − 191 = 0
6. 25x2
− 200x + 16y2
− 160y = 800
Find the standard equation of the ellipse which satisﬁes the given conditions.
7. foci (2 −
√
33, 8) and (2 +
√
33, 8), the sum of the distances of any point from
the foci is 14
8. center (−3, −7), vertical major axis of length 20, minor axis of length 12
9. foci (−21, 10) and (3, 10), contains the point (−9, 15)
10. a vertex at (−3, −18) and a covertex at (−12, −7), major axis is either hori-
zontal or vertical
11. a focus at (−9, 15) and a covertex at (1, 10), with vertical major axis
12. A 40-ft wide tunnel has the shape of a semiellipse that is 5 ft high a distance
of 2 ft from either end. How high is the tunnel at its center?
13. The moon’s orbit is an ellipse with Earth as one focus. If the maximum
distance from the moon to Earth is 405 500 km and the minimum distance is
363 300 km, ﬁnd the equation of the ellipse in a Cartesian coordinate system
where Earth is at the origin. Assume that the ellipse has horizontal major
axis and that the minimum distance is achieved when the moon is to the right
of Earth. Use 100 km as one unit.
14. Two friends visit a whispering gallery (in the shape of a semiellipsoid) where
they stand 100 m apart to be at the foci. If one of them is 6 m from the
nearest wall, how high is the gallery at its center?
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15. A jogging path is in the shape of an ellipse. If it is 120 ft long and 40 ft wide,
what is the width of the track 15 ft from either vertex?
16. Radiation is focused to an unhealthy area in a patient’s body using a semiel-
liptic reflector, positioned in such a way that the target area is at one focus
while the source of radiation is at the other. If the reflector is 100 cm wide
and 30 cm high at the center, how far should the radiation source and the
target area be from the ends of the reflector?
4
Lesson 1.4. Hyperbolas
(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
Just like ellipse, a hyperbola is one of the conic sections that most students
have not encountered formally before. 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
different 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 1.42. What is the
absolute value of the difference of the distances of A(3.75, −3) from F1 and from
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F2? How about the absolute value of the diﬀerence of the distances of B −5, 16
3
from F1 and from F2?
|AF1 − AF2| = |9.25 − 3.25| = 6
|BF1 − BF2| =
16
3
−
34
3
= 6
There are other points P such that |PF1 − PF2| = 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, PF2 − PF1 = 6; for those on the right branch,
PF1 − PF2 = 6.
Figure 1.42
Figure 1.43
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Let F1 and F2 be two distinct points. The set of all points P, whose
distances from F1 and from F2 differ by a certain constant, is called a
hyperbola. The points F1 and F2 are called the foci of the hyperbola.
In Figure 1.43, 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 difference of the distances of P from F1 and F2, be 2a (the coefficient
2 will make computations simpler). Thus, |PF1 − PF2| = 2a, and so
(x + c)2 + y2 − (x − c)2 + y2 = 2a.
Algebraic manipulations allow us to rewrite this into the much simpler
x2
a2
−
y2
b2
= 1, where b =
√
c2 − a2.
When we let b =
√
c2 − a2, we assumed c > a. To see why this is true, suppose
that P is closer to F2, so PF1 − PF2 = 2a. Refer to Figure 1.43. Suppose also
that P is not on the x-axis, so PF1F2 is formed. From the triangle inequality,
F1F2 + PF2 > PF1. Thus, 2c > PF1 − PF2 = 2a, so c > a.
Now we present a derivation. For now, assume P is closer to F2 so PF1 > PF2,
and PF1 − PF2 = 2a.
PF1 = 2a + PF2
(x + c)2 + y2 = 2a + (x − c)2 + y2
(x + c)2 + y2
2
= 2a + (x − c)2 + y2
2
cx − a2
= a (x − c)2 + y2
(cx − a2
)2
= a (x − c)2 + y2
2
(c2
− a2
)x2
− a2
y2
= a2
(c2
− a2
)
b2
x2
− a2
y2
= a2
b2
by letting b =
√
c2 − a2 > 0
x2
a2
−
y2
b2
= 1
We collect here the features of the graph of a hyperbola with standard equa-
tion
x2
a2
−
y2
b2
= 1.
Let c =
√
a2 + b2.
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(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 hyperbola, the absolute value of the diﬀerence 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 V1V2 is called the transverse axis. Its length is 2a.
(4) asymptotes: y = b
a
x and y = −b
a
x, the lines 1 and 2 in Figure 1.45
• The asymptotes of the hyperbola are two lines passing through the cen-
ter 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
equation, replace 1 by 0, and in the resulting equation x2
a2 − y2
b2 = 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 1.45. 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 V1V2. The other two sides are congruent and parallel
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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
9
−
y2
7
= 1.
Sketch the graph, and include these points and lines, the transverse and conjugate
axes, and the auxiliary rectangle.
Solution. With a2
= 9 and b2
= 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)
asymptotes: y =
√
7
3
x and y = −
√
7
3
x
The graph is shown at the right. The conju-
gate axis drawn has its endpoints b =
√
7 ≈
2.7 units above and below the center. 2
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
diﬀerence of its distances from the foci is 6. See Figure 1.42.
Solution. We have 2a = 6 and c = 5, so a = 3 and b =
√
c2 − a2 = 4. The
hyperbola then has equation
x2
9
−
y2
16
= 1. 2
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
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.
In all four cases below, 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 V1V2 has length 2a. The conjugate axis has length 2b and is
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,
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for any point on the hyperbola, the absolute value of the difference of its distances
from the foci is 2a.
Center Corresponding Hyperbola
(0, 0)
x2
a2
−
y2
b2
= 1
y2
a2
−
x2
b2
= 1
(h, k)
(x − h)2
a2
−
(y − k)2
b2
= 1
(y − k)2
a2
−
(x − h)2
b2
= 1
transverse axis: horizontal transverse axis: vertical
conjugate axis: vertical conjugate axis: horizontal
In the standard equation, aside from being positive, there are no other re-
strictions 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
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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 asymp-
totes of the hyperbola with the given equation. Sketch the graph, and include
these points and lines, the transverse and conjugate axes, and the auxiliary rect-
angle.
(1)
(y + 2)2
25
−
(x − 7)2
9
= 1
(2) 4x2
− 5y2
+ 32x + 30y = 1
Solution. (1) From a2
= 25 and b2
= 9, we have a = 5, b = 3, and c =√
a2 + b2 =
√
34 ≈ 5.8. The hyperbola is vertical. To determine the asymp-
totes, we write (y+2)2
25
− (x−7)2
9
= 0, which is equivalent to y + 2 = ±5
3
(x − 7).
We can then solve this for y.
center: C(7, −2)
foci: F1(7, −2 −
√
34) ≈ (7, −7.8) and F2(7, −2 +
√
34) ≈ (7, 3.8)
vertices: V1(7, −7) and V2(7, 3)
asymptotes: y = 5
3
x − 41
3
and y = −5
3
x + 29
3
The conjugate axis drawn has its endpoints b = 3 units to the left and right
of the center.
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(2) We ﬁrst change the given equation to standard form.
4(x2
+ 8x) − 5(y2
− 6y) = 1
4(x2
+ 8x + 16) − 5(y2
− 6y + 9) = 1 + 4(16) − 5(9)
4(x + 4)2
− 5(y − 3)2
= 20
(x + 4)2
5
−
(y − 3)2
4
= 1
We have a =
√
5 ≈ 2.2 and b = 2. Thus, c =
√
a2 + b2 = 3. The hyperbola
is horizontal. To determine the asymptotes, we write (x+4)2
5
− (y−3)2
4
= 0
which is equivalent to y − 3 = ± 2√
5
(x + 4), and solve for y.
center: C(−4, 3)
foci: F1(−7, 3) and F2(−1, 3)
vertices: V1(−4 −
√
5, 3) ≈ (−6.2, 3) and V2(−4 +
√
5, 3) ≈ (−1.8, 3)
asymptotes: y = 2√
5
x + 8√
5
+ 3 and y = − 2√
5
x − 8√
5
+ 3
The conjugate axis drawn has its endpoints b = 2 units above and below
the center.
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 diﬀerence of its of its distances from
the foci is 10. Find the standard equation of the hyperbola.
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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 diﬀerence, 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
24
= 1. 2
Example 1.4.5. 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 (−4, 2) of the vertices is the center of the hyperbola,
which is vertical (because the vertices are vertically aligned). Each vertex is
a = 7 units away from the center. The given focus is c =
√
65 units away from
the center. Thus, b2
= c2
− a2
= 16, and the standard equation is
(y − 2)2
49
−
(x + 4)2
16
= 1. 2
1.4.3. Situational Problems Involving Hyperbolas
Let us now give an example on an application of hyperbolas.
Example 1.4.6. An explosion was 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 can deduce that the sound traveled
340(2) = 680 m farther in reaching F2 than in reaching F1. This is then the
diﬀerence 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.
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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
x2
115600
− y2
244400
= 1. 2
1. Determine the foci, vertices, and asymptotes of the hyperbola with equation
x2
16
−
y2
33
= 1. Sketch the graph, and include these points and lines, the
transverse and conjugate axes, and the auxiliary rectangle.
Solution: The hyperbola is horizontal.
a2
= 16 ⇒ a = 4,
b2
= 33 ⇒ b =
√
33,
c =
√
16 + 33 = 7
center: (0, 0)
foci: F1(−7, 0), F2(7, 0)
vertices: V1(−4, 0), V2(4, 0)
asymptotes: y =
√
33
4
x, y = −
√
33
4
x
The conjugate axis has endpoints
(0, −
√
33) and (0,
√
33). See Figure
1.46.
Figure 1.46
2. Find the standard equation of the hyperbola whose foci are F1(0, −10) and
F2(0, 10), such that for any point on it, the absolute value of the diﬀerence of
its distances from the foci is 12.
Solution: The hyperbola is vertical and has center at (0, 0). We have 2a = 12,
so a = 6; also, c = 10. Then b =
√
102 − 62 = 8. The equation is
y2
36
−
x2
64
= 1.
For Examples 3 and 4, give the coordinates of the center, foci, vertices, and
asymptotes of the hyperbola with the given equation. Sketch the graph, and in-
clude these points and lines, the transverse and conjugate axes, and the auxiliary
rectangle.
3.
(y + 6)2
25
−
(x − 4)2
39
= 1
Solution: The hyperbola is vertical.
a2
= 25 ⇒ a = 5, b2
= 39 ⇒ b =
√
39, c =
√
25 + 39 = 8
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center: (4, −6)
foci: F1(4, −14), F2(4, 2)
vertices: V1(4, −11), V2(4, −1)
asymptotes:
(y + 6)2
25
−
(x − 4)2
39
= 0
⇔ y + 6 = ±
5
√
39
(x − 4)
The conjugate axis has endpoints b =
√
39 units to the left and to the right of
the center. See Figure 1.47.
Figure 1.47
4. 9x2
+ 126x − 16y2
− 96y + 153 = 0
Solution:
9x2
+ 126x − 16y2
− 96y = −153
9(x2
+ 14x + 49) − 16(y2
+ 6y + 9) = −153 + 9(49) − 16(9)
9(x + 7)2
− 16(y + 3)2
= 144
(x + 7)2
16
−
(y + 3)2
9
= 1
The hyperbola is horizontal.
a2
= 16 ⇒ a = 4, b2
= 9 ⇒ b = 3, c =
√
16 + 9 = 5
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center: (−7, −3)
foci: F1(−12, −3), F2(−2, −3)
vertices: V1(−11, −3), V2(−3, −3)
asymptotes:
(x + 7)2
16
−
(y + 3)2
9
= 0 ⇔ y + 3 = ±
3
4
(x + 7)
The conjugate axis have endpoints (−7, −6) and (−7, 0). See Figure 1.48.
Figure 1.48
5. The foci of a hyperbola are (−17, −3) and (3, −3). For any point on the
hyperbola, the absolute value of the diﬀerence of its distances from the foci is
14. Find the standard equation of the hyperbola.
Solution: The hyperbola is horizontal with center at the midpoint (−7, −3) of
the foci. Also, 2a = 14 so a = 7 while c = 10. Then b2
= 102
− 72
= 51. The
equation is
(x + 7)2
49
−
(y + 3)2
51
= 1.
6. The auxiliary rectangle of a hyperbola has vertices (−24, −15), (−24, 9), (10, 9),
and (10, −15). Find the equation of the hyperbola if its conjugate axis is hor-
izontal.
Solution: The hyperbola is vertical. Using the auxiliary rectangle’s dimen-
sions, we see that the length of the transverse axis is 2a = 24 while the
length of the conjugate axis is 2b = 34. Thus, a = 12 and b = 17. The
hyperbola’s vertices are the midpoints (−7, −15) and (−7, 9) of the bottom
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and top sides, respectively, of the auxiliary rectangle. Then the hyperbola’s
center is (−7, −3), which is the midpoint of the vertices. The equation is
(y + 3)2
144
−
(x + 7)2
289
= 1.
7. Two LORAN (long range navigation) stations A and B are situated along a
straight shore, where A is 200 miles west of B. These stations transmit radio
signals at a speed 186 miles per millisecond. The captain of a ship travelling
on the open sea intends to enter a harbor that is located 40 miles east of
station A.
Due to the its location, the harbor experiences a time difference in receiving
the signals from both stations. The captain navigates the ship into the harbor
by following a path where the ship experiences the same time difference as the
harbor.
(a) What time difference between station signals should the captain be look-
ing for in order the ship to make a successful entry into the harbor?
(b) If the desired time difference is achieved, determine the location of the
ship if it is 75 miles offshore.
Solution:
(a) Let H represent the harbor on the shoreline. Note that BH−AH = 160−
40 = 120. The time difference on the harbor is given by 120÷186 ≈ 0.645
milliseconds. This is the time difference needed to be maintained in order
to for the ship to enter the harbor.
(b) Situate the stations A and B on the Cartesian plane so that A (−100, 0)
and B (100, 0). Let P represent the ship on the sea, which has coordinates
(h, 75). Since PB − PA = 120, then it should follow that h < 0. More-
over, P should lie on the left branch of the hyperbola whose equation is
given by
x2
a2
−
y2
b2
= 1
where 2a = 120 ⇒ a = 60, and b =
√
c2 − a2 =
√
1002 − 602 = 80.
Therefore,
h2
602
−
752
802
= 1
h = − 1 +
752
802
602 ≈ −82.24
This means that the ship is around 17.76 miles to the east of station A.
58

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