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- 1. D EPED C O PY 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.
- 2. D EPED C O PY Precalculus Learner’s Material First Edition 2016 Republic Act 8293. Section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties. Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this learning resource are owned by their respective copyright holders. DepEd is represented by the Filipinas Copyright Licensing Society (FILCOLS), Inc. in seeking permission to use these materials from their respective copyright owners. All means have been exhausted in seeking permission to use these materials. The publisher and authors do not represent nor claim ownership over them. Only institutions and companies which have entered an agreement with FILCOLS and only within the agreed framework may copy from this Manual. Those who have not entered in an agreement with FILCOLS must, if they wish to copy, contact the publishers and authors directly. Undersecretary: Dina S. Ocampo, PhD Printed in the Philippines by Sunshine Interlinks Publishing House, Inc. 3F Maine City Tower, 236 Tomas Morato Avenue, Brgy. South Triangle, Quezon City Department of Education-Bureau of Learning Resources (DepEd-BLR) Office Address: Ground Floor Bonifacio Building, DepEd Complex Meralco Avenue, Pasig City, Philippines 1600 Telefax: (02) 634-1054, 634-1072, 631-4985 E-mail Address: blr.lrqad@deped.gov.ph / blr.lrpd@deped.gov.ph Dr. Flordeliza F. Francisco Mark Anthony J. Vidallo Carly Mae Casteloy Angela Dianne Agustin (02) 435-5258, respectively. Published by the Department of Education Secretary: Br. Armin A. Luistro FSC Authors and publishers may email or contact FILCOLS at filcols@gmail.com or Cover Art Illustrator: Quincy D. Gonzales Team Leader: Ian June L. Garces, Ph.D. Management Team of the Precalculus Learner’s Material Bureau of Curriculum Development Bureau of Learning Resources Development Team of the Precalculus Learner’s Material Joy P. Ascano Jesus Lemuel L. Martin Jr. Jerico B. Bacani, Ph.D Richard B. Eden, Ph.D Arnel D. Olofernes Mark Anthony C. Tolentino, Ph.D Reviewers: 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.
- 3. D EPED C O PY 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 iii 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.
- 4. D EPED C O PY 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 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.
- 5. D EPED C O PY 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 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.
- 6. D EPED C O PY 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 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.
- 7. D EPED C O PY 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 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.
- 8. D EPED C O PY 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 2 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.
- 9. D EPED C O PY Kto12BASICEDUCATIONCURRICULUM SENIORHIGHSCHOOL–SCIENCE,TECHNOLOGY,ENGINEERINGANDMATHEMATICS(STEM)SPECIALIZEDSUBJECT Kto12SeniorHighSchoolSTEMSpecializedSubject–Pre-CalculusDecember2013Page2of4 CONTENT CONTENT STANDARDS PERFORMANCE STANDARDS LEARNINGCOMPETENCIESCODE 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 3 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.
- 10. D EPED C O PY Kto12BASICEDUCATIONCURRICULUM SENIORHIGHSCHOOL–SCIENCE,TECHNOLOGY,ENGINEERINGANDMATHEMATICS(STEM)SPECIALIZEDSUBJECT Kto12SeniorHighSchoolSTEMSpecializedSubject–Pre-CalculusDecember2013Page3of4 CONTENT CONTENT STANDARDS PERFORMANCE STANDARDS LEARNINGCOMPETENCIESCODE 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 accuratelysituational 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 accuratelysituational problemsinvolvingthe polarcoordinatesystem 20.locatepointsinpolarcoordinatesystemSTEM_PC11T-IIj-1 21.convertthecoordinatesofapointfromrectangulartopolar systemsandviceversa STEM_PC11T-IIj-2 22.solvesituationalproblemsinvolvingpolarcoordinatesystemSTEM_PC11T-IIj-3 ***SuggestionforICT-enhancedlessonwhenavailableandwhereappropriate 4 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.
- 11. D EPED C O PY Kto12BASICEDUCATIONCURRICULUM SENIORHIGHSCHOOL–SCIENCE,TECHNOLOGY,ENGINEERINGANDMATHEMATICS(STEM)SPECIALIZEDSUBJECT Kto12SeniorHighSchoolSTEMSpecializedSubject–Pre-CalculusDecember2013Page4of4 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 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.
- 12. D EPED C O PY 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 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.
- 13. D EPED C O PY 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 7 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.
- 14. D EPED C O PY • 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. 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). As we get to know more about a circle, we will also be able to distinguish more between these two conics. 8 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.
- 15. D EPED C O PY 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 . 9 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.
- 16. D EPED C O PY 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. 10 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.
- 17. D EPED C O PY 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 11 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.
- 18. D EPED C O PY 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 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. 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. 12 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.
- 19. D EPED C O PY 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. 13 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.
- 20. D EPED C O PY 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 14 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.
- 21. D EPED C O PY (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. 15 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.
- 22. D EPED C O PY Figure 1.17 Figure 1.18 Figure 1.19 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. Figure 1.21 Figure 1.22 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, 16 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.
- 23. D EPED C O PY 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) 17 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.
- 24. D EPED C O PY 18. has a diameter with endpoints 9 2 , 4 and − 3 2 , −2 19. concentric with x2 + 20x + y2 − 14y + 145 = 0, diameter 12 20. concentric with x2 − 2x + y2 − 2y − 23 = 0 and has 1/5 the area 21. concentric with x2 + 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 18 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.
- 25. D EPED C O PY Lesson 1.2. Parabolas 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 (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. 19 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.
- 26. D EPED C O PY 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 20 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.
- 27. D EPED C O PY 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. 21 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.
- 28. D EPED C O PY (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 22 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.
- 29. D EPED C O PY 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 23 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.
- 30. D EPED C O PY 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 24 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.
- 31. D EPED C O PY 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. 25 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.
- 32. D EPED C O PY (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 26 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.
- 33. D EPED C O PY 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? 27 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.
- 34. D EPED C O PY 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 More Solved Examples 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. Figure 1.28 Figure 1.29 28 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.
- 35. D EPED C O PY 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 29 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.
- 36. D EPED C O PY 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. 30 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.
- 37. D EPED C O PY 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. Supplementary Problems 1.2 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 31 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.
- 38. D EPED C O PY 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 32 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.
- 39. D EPED C O PY Lesson 1.3. Ellipses 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 (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. 33 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.
- 40. D EPED C O PY 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. 34 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.
- 41. D EPED C O PY (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. 35 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.
- 42. D EPED C O PY 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. 36 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.
- 43. D EPED C O PY 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. 37 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.
- 44. D EPED C O PY (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 38 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.
- 45. D EPED C O PY 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. 39 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.
- 46. D EPED C O PY 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 40 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.
- 47. D EPED C O PY 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 41 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.
- 48. D EPED C O PY More Solved Examples 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 42 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.
- 49. D EPED C O PY 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 43 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.
- 50. D EPED C O PY 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. 44 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.
- 51. D EPED C O PY Supplementary Problems 1.3 Give the coordinates of the center, foci, vertices, and covertices of the ellipse with the given equation. Sketch the graph, and include these points. 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? 45 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.
- 52. D EPED C O PY 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 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 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 46 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.
- 53. D EPED C O PY 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 47 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.
- 54. D EPED C O PY 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. 48 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.
- 55. D EPED C O PY Figure 1.44 Figure 1.45 (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 49 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.
- 56. D EPED C O PY 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, 50 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.
- 57. D EPED C O PY 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 51 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.
- 58. D EPED C O PY 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. 52 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.
- 59. D EPED C O PY (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. 53 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.
- 60. D EPED C O PY 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. 54 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.
- 61. D EPED C O PY 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 More Solved Examples 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 55 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.
- 62. D EPED C O PY 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 56 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.
- 63. D EPED C O PY 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 57 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.
- 64. D EPED C O PY 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 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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