Complete Solutions Manual. Technical Calculus with Analytic Geometry FIFTH EDITION. Peter Kuhfittig Milwaukee School of Engineering.

Complete Solutions Manual Technical Calculus with Analtic Geometr FIFTH EDITION Peter Kuhfittig Milwaukee School of Engineering Australia Brazil Meico Singapore United Kingdom United States

213 Cengage Learning ALL RIGHTS RESERVED No part of this work covered b the copright herein ma be reproduced, transmitted, stored, or used in an form or b an means graphic, electronic, or mechanical, including but not limited to photocoping, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval sstems, ecept as permitted under Section 17 or 18 of the 1976 United States Copright Act, without the prior written permission of the publisher ecept as ma be permitted b the license terms below ISBN-13: 978-1-285-5342-4 ISBN-1: 1-285-5342-7 Cengage Learning 2 First Stamford Place, 4th Floor Stamford, CT 692 USA Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Meico, Brazil, and Japan Locate our local office at: wwwcengagecom/global For product information and technolog assistance, contact us at Cengage Learning Customer & Sales Support, 1-8-354-976 For permission to use material from this tet or product, submit all requests online at wwwcengagecom/permissions Further permissions questions can be emailed to permissionrequest@cengagecom Cengage Learning products are represented in Canada b Nelson Education, Ltd To learn more about Cengage Learning Solutions, visit wwwcengagecom Purchase an of our products at our local college store or at our preferred online store wwwcengagebraincom NOTE: UNDER NO CIRCUMSTANCES MAY THIS MATERIAL OR ANY PORTION THEREOF BE SOLD, LICENSED, AUCTIONED, OR OTHERWISE REDISTRIBUTED EXCEPT AS MAY BE PERMITTED BY THE LICENSE TERMS HEREIN Dear Professor or Other Supplement Recipient: Cengage Learning has provided ou with this product (the Supplement ) for our review and, to the etent that ou adopt the associated tetbook for use in connection with our course (the Course ), ou and our students who purchase the tetbook ma use the Supplement as described below Cengage Learning has established these use limitations in response to concerns raised b authors, professors, and other users regarding the pedagogical problems stemming from unlimited distribution of Supplements Cengage Learning hereb grants ou a nontransferable license to use the Supplement in connection with the Course, subject to the following conditions The Supplement is for our personal, noncommercial use onl and ma not be reproduced, posted electronicall or distributed, ecept that portions of the Supplement ma be provided to our students IN PRINT FORM ONLY in connection with our instruction of the Course, so long as such students are advised that the ma not cop or distribute READ IMPORTANT LICENSE INFORMATION an portion of the Supplement to an third part You ma not sell, license, auction, or otherwise redistribute the Supplement in an form We ask that ou take reasonable steps to protect the Supplement from unauthorized use, reproduction, or distribution Your use of the Supplement indicates our acceptance of the conditions set forth in this Agreement If ou do not accept these conditions, ou must return the Supplement unused within 3 das of receipt All rights (including without limitation, coprights, patents, and trade secrets) in the Supplement are and will remain the sole and eclusive propert of Cengage Learning and/or its licensors The Supplement is furnished b Cengage Learning on an as is basis without an warranties, epress or implied This Agreement will be governed b and construed pursuant to the laws of the State of New York, without regard to such State s conflict of law rules Thank ou for our assistance in helping to safeguard the integrit of the content contained in this Supplement We trust ou find the Supplement a useful teaching tool Printed in the United States of America 1 2 3 4 5 6 7 17 16 15 14 13

Contents 1 Introduction to Analtic Geometr 1 11 The Cartesian Coordinate Sstem 1 12 The Slope 3 13 The Straight Line 5 14 Curve Sketching 1 15 Curves with Graphing Utilities 23 17 The Circle 27 18 The Parabola 34 19 The Ellipse 4 11 The Hperbola 5 111 Translation of Aes; Standard Equations of the Conics 56 Chapter 1 Review 72 2 Introduction to Calculus: The Derivative 81 21 Functions and Intervals 81 22 Limits 84 24 The Derivative b the Four-Step Process 89 25 Derivatives of Polnomials 95 26 Instantaneous Rates of Change 97 27 Differentiation Formulas 12 28 Implicit Differentiation 113 29 Higher Derivatives 119 Chapter 2 Review 121 3 Applications of the Derivative 127 31 The First-Derivative Test 127 32 The Second-Derivative Test 136 33 Eploring with Graphing Utilities 153 34 Applications of Minima and Maima 158 35 Related Rates 171 36 Differentials 18 Chapter 3 Review 182 Not For iii Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

iv CONTENTS 4 The Integral 193 41 Antiderivatives 193 42 The Area Problem 194 43 The Fundamental Theorem of Calculus 196 45 Basic Integration Formulas 197 46 Area Between Curves 22 47 Improper Integrals 215 48 The Constant of Integration 223 49 Numerical Integration 232 Chapter 4 Review 238 5 Applications of the Integral 245 51 Means and Root Mean Squares 245 52 Volumes of Revolution: Disk and Washer Methods 247 53 Volumes of Revolution: Shell Method 256 54 Centroids 27 55 Moments of Inertia 288 56 Work and Fluid Pressure 31 Chapter 5 Review 318 6 Derivatives of Transcendental Functions 327 61 Review of Trigonometr 327 62 Derivatives of Sine and Cosine Functions 331 63 Other Trigonometric Functions 336 64 Inverse Trigonometric Functions 341 65 Derivatives of Inverse Trigonometric Functions 346 66 Eponential and Logarithmic Functions 35 67 Derivative of the Logarithmic Function 354 68 Derivative of the Eponential Function 358 69 L Hospital s Rule 362 61 Applications 365 611 Newton s Method 373 Chapter 6 Review 376 7 Integration Techniques 383 71 The Power Formula Again 383 72 The Logarithmic and Eponential Forms 386 73 Trigonometric Forms 394 74 Further Trigonometric Forms 399 75 Inverse Trigonometric Forms 49 76 Integration b Trigonometric Substitution 414 77 Integration b Parts 425 78 Integration of Rational Functions 431 79 Integration b Use of Tables 439 Chapter 7 Review 443 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

CONTENTS v 8 Parametric Equations, Vectors, and Polar Coordinates 451 81 Vectors and Parametric Equations 451 82 Arc Length 459 83 Polar Coordinates 463 84 Curves in Polar Coordinates 469 85 Areas in Polar Coordinates; Tangents 476 Chapter 8 Review 488 9 Three Dim Space; Partial Derivatives; Multiple Integrals 493 91 Surfaces in Three Dimensions 493 92 Partial Derivatives 56 93 Applications of Partial Derivatives 514 94 Iterated Integrals 525 95 Volumes b Double Integration 535 96 Mass, Centroids, and Moments of Inertia 545 97 Volumes in Clindrical Coordinates 558 Chapter 9 Review 564 1 Infinite Series 575 11 Introduction to Infinite Series 575 12 Tests for Convergence 579 13 Maclaurin Series 584 14 Operations with Series 588 15 Computations with Series; Applications 591 16 Fourier Series 61 Chapter 1 Review 615 11 First-Order Differential Equations 621 111 What is a Differential Equation? 621 112 Separation of Variables 624 113 First-Order Linear Differential Equations 634 114 Applications of First-Order Differential Equations 647 115 Numerical Solutions 662 Chapter 11 Review 666 12 Higher-Order Linear Differential Equations 675 121 Higher-Order Homogeneous Differential Equations 675 122 Auiliar Equations with Repeating or Comple Roots 68 123 Nonhomogeneous Equations 686 124 Applications of Second-Order Equations 74 Chapter 12 Review 72 13 The Laplace Transform 727 Sections 131-133 727 134 Solution of Linear Equations b Laplace Transforms 74 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part Chapter 13 Review 759

Chapter 1 Introduction to Analtic Geometr 11 The Cartesian Coordinate Sstem 1 Let ( 2, 2 ) = (2, 4) and ( 1, 1 ) = (5, 2) From the distance formula we get d = ( 2 1 ) 2 + ( 2 1 ) 2 d = (2 5) 2 + (4 2) 2 = ( 3) 2 + 2 2 = 9 + 4 = 13 2 Let ( 2, 2 ) = ( 3, 2) and ( 1, 1 ) = (5, 4) From the distance formula d = ( 2 1 ) 2 +( 2 1 ) 2 we get d = ( 3 5) 2 +[2 ( 4)] 2 = ( 8) 2 +(6) 2 = 64+36 = 1 3 Let ( 2, 2 ) = ( 3, 6) and ( 1, 1 ) = (5, 2) Then d = ( 3 5) 2 + [ 6 ( 2)] 2 = ( 8) 2 + ( 4) 2 = 64 + 16 = 16 5 = 4 5 4 Let ( 2, 2 ) = ( 5, 2) and ( 1, 1 ) = (, ) Then d = ( 5 ) 2 +(2 ) 2 = 5+4 = 3 5 Let ( 2, 2 ) = ( 3, 4) and ( 1, 1 ) = (, 2) Then d = ( 3 ) 2 + (4 2) 2 = 3 + 4 = 7 6 d = ( 2 2) 2 +( 5 ) 2 = +5 = 5 7 d = [1 ( 1)] 2 + ( 2 ) 2 = 4 + 2 = 6 8 d = [2 ( 2)] 2 + (7 3) 2 = 16 + 16 = 2 16 = 4 2 9 d = [ 9 ( 11)] 2 + ( 1 1) 2 = 2 2 + ( 2) 2 = 8 = 2 4 = 2 2 1 Distance from (, ) to (4, 3): (4 ) 2 +(3 ) 2 = 16 + 9 = 5 Distance from (, ) to (6, )= 6 Distance from (4, 3) to (6, ): (6 4) 2 +( 3) 2 = 4 + 9 = 13 Perimeter = 5 + 6 + 13 = 11 + 13 Not For 1 Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

2 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 11b / is negative whenever and have opposite signs: quadrants II and IV 12 If 2 > 1, then 2 1 = P 1 P 3 = 2 1 If 2 < 1, then 2 1 = P 1 P 3 and P 1 P 3 = P 1 P 3 = 2 1 13a An point on the -ais has coordinates of the form (, ) 14 Distance from (11, 2) to origin: (11 ) 2 + (2 ) 2 = 125 Distance from ( 5, 1) to origin: ( 5 ) 2 + (1 ) 2 = 125 Distance from ( 1, 11) to origin: ( 1 ) 2 + ( 11 ) 2 = 122 Answer: ( 1, 11) 15 Let A = ( 2, 5), B = ( 4, 1) and C = (5, 4); then AB = [ 2 ( 4)] 2 + ( 5 1) 2 = 4, AC = ( 2 5)2 + ( 5 4) 2 = 13, and BC = ( 4 5) 2 + (1 4) 2 = 9 Since (AB) 2 + (BC) 2 = (AC) 2, the triangle must be a right triangle 16 Let A = ( 1, 1), B = ( 2, 3), and C = (6, 5) After calculating AB = 17, BC = 68, and AC = 85, we observe that (AB) 2 + (BC) 2 = (AC) 2 17 The points (12, ), ( 4, 8) and ( 1, 13) are all 5 5 units from (1, 2) 18 Distance from ( 2, 1) to (3, 2): 13 Distance from (15, 3) to (3, 2): 13 19 Distance from ( 1, 1) to (2, 8): Distance from (2, 8) to (5, 17): ( 1 2)2 + ( 1 8) 2 = 9 + 81 = 9 = 9 1 = 3 1 Distance from ( 1, 1) to (5, 17): (5 2)2 + (17 8) 2 = 9 = 3 1 62 + 18 2 = 36 = 6 1 Total distance 6 1 = 3 1 + 3 1, the sum of the other two distances 2 Distance from (, ) to ( 1, 2): d = ( + 1) 2 + ( 2) 2 = 3 ( + 1) 2 + ( 2) 2 = (3) 2 squaring both sides 2 + 2 + 1 + 2 4 + 4 = 9 2 + 2 + 2 4 = 4 21 Distance from (, ) to -ais: units Distance from (, ) to (2, ): ( 2) 2 + ( ) 2 = ( 2) 2 + 2 B assumption, ( 2)2 + 2 = ( 2) 2 + 2 = 2 squaring both sides 2 4 + 4 + 2 = 2 2 4 + 4 = 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

12 THE SLOPE 3 22 Let ( ( 1, 1 ) = ( 3, 5) and ( 2, 2 ) = ( 1, 7) Then from the midpoint formula 1 + 2, ) ( 1 + 2 3 + ( 1) =, 5 + 7 ) = ( 2, 1) 2 2 2 2 23 Let ( 1, 1 ) = ( 2, 6) and ( 2, 2 ) = (2, 4) Then from the midpoint formula ( 1 + 2, ) 1 + 2 2 2 24 we get ( 2 + 2 2 ( 1 + 2 2, 6 + ( 4) ) = (, 1) 2, ) ( 1 + 2 3 + ( 2) =, 5 + 9 ) = ( 52 ) 2 2 2,7 25 Let ( 1, 1 ) = (5, ) and ( 2, 2 ) = (9, 4) Then from the midpoint formula ( 1 + 2, ) 1 + 2 2 2 we get ( 5 + 9 2, + 4 ) = (7, 2) 2 ( 1 + 2 26, ) ( 1 + 2 4 + ( 1) =, 3 + ( 7) ) = ( 52 ) 2 2 2 2, 2 ( ) 2 + 6 1 + 11 27 The center is the midpoint:, = (2, 5) 2 2 28 Midpoint of given line segment: (2, 6) Midpoint of line segment from (2, 6) to ( 2, 4): (, 5) 12 The Slope 1 Let ( 2, 2 ) = (1, 7) and ( 1, 1 ) = (2, 6) Then, b formula (14), we get m = 2 1 2 1 m = 7 6 1 2 = 1 1 = 1 2 Let ( 2, 2 ) = ( 3, 1) and ( 1, 1 ) = ( 5, 2) Then b formula (14) m = 2 1 = 1 2 2 1 3 ( 5) = 12 = 6 2 3 Let ( 1, 1 ) = (, 2) and ( 2, 2 ) = ( 4, 4) Then m = 4 2 4 = 6 4 = 3 2 4 Let ( 2, 2 ) = (6, 3) and ( 1, 1 ) = (4, ) Then m = 2 1 2 1 = 3 6 4 = 3 2 = 3 2 5 Let ( 2, 2 ) = (7, 8) and ( 1, 1 ) = ( 3, 4) Then m = 8 ( 4) 7 ( 3) = 8 + 4 7 + 3 = 12 1 = 6 5 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

4 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 6 m = 4 8 = 1 2 7 m = 43 ( 1) 1 1 8 m = 31 1 ( 1) = 3 = 44 2 = 22 9 m = 5 4 3 3 = 9 (undefined) 1 m = 6 6 5 ( 3) = 8 = 11 m = 3 ( 3) 9 5 12 m 3 ( 2) 4 4 = 4 = 13 m = 3 2 12 ( 2) = 1 14 = 5 (undefined) 14 (a) tan = ; (b) tan 3 = 3 3 ; (c) tan 15 = 3 3 ; (d) tan 9 is undefined; (e) tan 45 = 1; (f) tan 135 = 1 15 See answer section of book 16 Slope of AB = 2 2 ( 1) = 2 1 = 2; of BC = 5 3 ; of AC = 3 4 17 Slope of given line is 1 ( 5) 7 6 = 6 13 = 6 Slope of perpendicular is given b the negative 13 1 reciprocal and is therefore ( 6/13) = 13 6 3 6 18 Slope of line through ( 4, 6) and ( 1, 3): 1 ( 4) = 3 9 6 Slope of line through ( 4, 6) and (1, 9): 1 ( 4) = 3 Since the lines have the same slope and pass through ( 4, 6), the must coincide 6 1 19 Slope of line through ( 4, 6) and (6, 1): 4 6 = 4 1 = 2 5 1 Slope of line through (6, 1) and (1, ): 6 1 = 1 4 = 5 2 Since the slopes are negative reciprocals, the lines are perpendicular 2 Slope of line through ( 4, 2) and ( 1, 8): 2; Slope of line through (9, 4) and (6, 2): 2; Slope of line through ( 1, 8) and (9, 4): 2 / 5 ; Slope of line through ( 4, 2) and (6, 2): 2 / 5 ; 3 3 21 Slope of line through (, 3) and ( 2, 3): ( 2) = 6 2 = 3 6 Slope of line through (7, 6) and (9, ): 7 9 = 6 2 = 3 3 6 Slope of line through ( 2, 3) and (7, 6): 2 7 = 3 9 = 1 3 3 Slope of line through (, 3) and (9, ): 9 = 1 3 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

13 THE STRAIGHT LINE 5 Since 3 and 1 3 are negative reciprocals, adjacent sides are perpendicular and opposite sides are parallel ( 2 + 8 22 Midpoint of line segment:, 4 + 8 ) = (3, 2) Point: (6, 4), m = 4 2 2 2 6 3 = 2 ( 3 + 9 23 Midpoint:, 2 + ) = (3, 1) 2 2 6 ( 1) Slope of line through (5, 6) and (3, 1): = 7 5 3 2 24 Let (, ) be the other end of the diameter Since the center is the midpoint, we get + 4 2 and 3 = 2; solving, = 2, = 7 2 25 tan θ = rise run θ = 36 = 1 ft 16 ft = 625 26 tan θ = rise run = 25 m 1316; θ = 75 19 m 1 ( 5) 27 Slope of line through ( 1, 1) and (3, 5): = 4 1 3 4 = 1 2 + 6 Slope of line through (, 2) and (4, 6): 4 = 8 4 Since the two slopes must be equal, we have: 8 = 1 4 8 = + 4 multipling both sides b 4 = 4 28 Slope of line segment (2, 1) to ( 3, 2): 3 5 ; Slope of other line segment: 2 + 7 4 ; so 5 4 = 5 (negative reciprocals); solving, = 7 3 13 The Straight Line 1 Since ( 1, 1 ) = ( 7, 2) and m = 1/2, we get 2 = 1 2 ( + 7) 1 = m( 1) 2 4 = + 7 clearing fractions 2 + 11 = = 2 + 7 + 4 = 1 2 Since ( 1, 1 ) = (, 3) and m = 4, we get 3 = 4( ) 1 = m( 1) 4 + 3 = 3 1 = m( 1 ) + 4 = 3( 3) ( 1, 1) = (3, 4); m = 3 + 4 = 3 9 3 13 = 4 B Not (18), = 2 For Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

6 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 5 1 = m( 1 ) = 1 3 ( ) (1, 1) = (, ); m = 1/3 3 = + 3 = 6 B (19), = 3 7 The line = 1 = + 1 has slope 1 = m( 1 ) = ( + 4) ( 1, 1) = ( 4, ); m = = -ais 8 First determine the slope: m = 2 1 2 1 = 2 6 3 7 = 2 5 Let ( 1, 1 ) = ( 3, 2), then 2 = 2 5 ( + 3) 5 1 = 2 + 6 multipling b 5 2 5 + 16 = 9 First determine the slope using m = 2 1 to get m = 4 ( 6) 2 1 3 3 = 1 6 = 5 3 Then let ( 1, 1 ) = ( 3, 4) to get 4 = 5 3 ( + 3) 1 = m( 1) 3 12 = 5 15 multipling b 3 5 + 3 + 3 = 1 m = 6 3 4 2 = 1 2 3 = 1 2 ( 2) ( 1, 1 ) = (2, 3) 2 6 = + 2 multipling b 2 + 2 8 = 11 m = 4 9 5 = 1 = 1( 5) choosing ( 1, 1) = (5, ) + 5 = 12 m = 5 7 3 1 = 1 2 5 = 1 2 ( + 3) ( 1, 1 ) = ( 3, 5) 2 1 = + 3 multipling b 2 2 + 13 = 13 6 + 2 = 5 2 = 6 + 5 = 3 + 5 2 = m + b m = 3, -intercept = 5 2 ; see graph in answer section of book 14 Solving for, we get = 1 Slope:1, -intercept : 1 15 Since 2 = 3, = 2 3 From the form = m + b, m = 2 3 and b = The line passes through the origin and has slope 2 3 See graph in answer section of book 16 From = 4 + 12, we get m = 4 and b = 12 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

13 THE STRAIGHT LINE 7 17 2 7 = = + 7 2 = m + b m =, -intercept = 7 2 ; see graph in answer section of book 18 Solving for : = 1 4 3 2 slope: 1 4 ; -intercept : 3 2 19 2 3 = 1 4 6 + 3 = 3 = 2 + 1 6 = 4 3 = 2 3 1 3 = 4 6 + 3 6 = 2 3 + 1 2 From the form = m + b, m = 2 3 in both cases, so that the lines are parallel 2 2 + 4 + 3 = 2 = 2 = 1 2 3 4 = 2 + 2 slope = 1 2 slope = 2 Answer: The lines are perpendicular 21 3 4 = 1 3 4 = 3 4 = 3 + 1 3 = 4 + 3 = 3 4 1 4 = 4 3 + 1 The lines are neither parallel nor perpendicular 22 7 1 = 6 4 = = 7 1 3 5 = 4 slope = 7 1 slope = Answer: neither 23 + 3 = 5 3 2 = 3 = + 5 = 3 + 2 = 1 3 + 5 3 The slopes are 1 3 and 3, respectivel Since the slopes are negative reciprocals, the lines are perpendicular 24 2 + 5 = 2 6 + 15 = 1 Slope is 2 5 = 2 5 + 2 5 = 2 5 + 1 15 in each case; the lines are parallel 25 3 5 = 6 9 15 = 4 5 = 3 + 6 15 = 9 + 4 = 3 5 6 5 = 9 15 + 4 15 = 3 5 4 15 From the form = m + b, the slope m is 3 5 in both cases; so the lines are parallel 26 4 3 + 6 = 6 8 + 1 = 4 = 3 6 8 = 6 1 = 3 4 3 2 = 6 8 + 1 8 = 3 4 + 1 8 slope = 3 4 slope = 3 4 Answer: the lines are parallel 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

8 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 27 2 3 = 6 6 + 4 = 5 2 = 3 + 6 6 = 4 + 5 = 3 2 + 3 = 4 6 + 5 6 = 2 3 5 6 The respective slopes are 3 2 and 2 3 Since the slopes are negative reciprocals, the lines are perpendicular 28 4 2 = 2 + 8 7 = 4 = + 2 2 = 8 + 7 = 1 4 1 = 4 + 7 2 2 The respective slopes are 1 4 and 4 Since the slopes are negative reciprocals, the lines are perpendicular 29 3 2 12 = 2 + 3 4 = 3 = 2 + 12 3 = 2 + 4 = 2 3 + 4 = 2 3 + 4 3 The lines are neither parallel nor pependicular 3 3 + 4 4 = 6 8 3 = 4 = 3 + 4 8 = 6 + 3 = 3 4 + 1 = 3 4 3 8 The lines are neither parallel nor perpendicular 31 3 + 4 = 5 (given line) = 3 4 + 5 4 slope = 3 4 1 = m( 1 ) point-slope form 1 = 3 4 ( + 2) point: ( 2, 1) 4 4 = 3 6 3 + 4 + 2 = 32 The given line can be written = 3 4 + 5 4 So the slope is 3 4 Slope of perpendicular: 4 3 1 = 4 3 ( + 2) 1 = m( 1 ) 3 3 = 4 + 8 4 3 + 11 = 33 To find the coordinates of the point of intersection, solve the equations simultaneousl: 2 4 = 1 3 + 4 = 4 5 = 5 adding = 1 From the second equation, 3(1) + 4 = 4, and = 1 4 So the point of intersection is (1, 1 4 ) From the equation 5 + 7 + 3 =, we get 7 = 5 3 = 5 7 3 7 slope= 5/7 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

13 THE STRAIGHT LINE 9 Thus ( 1, 1 ) = (1, 1 4 ) and m = 5 7 The desired line is 1 4 = 5 7 ( 1) To clear fractions, we multipl both sides b 28: 28 7 = 2( 1) 28 7 = 2 + 2 2 + 28 27 = 34 The first two lines are perpendicular: = 1 3 + 1 and = 3 25 4 35 See graph in answer section of book 36 F = 3, slope = 3, passing through the origin 37 From F = k, we get 3 = k 1 2 Thus k = 6 and F = 6 38 -intercept : initial value; t-intercept : the ear the value becomes zero 39 F = mc + b 212 = m(1) + b F = 212, C = 1 32 = m() + b F = 32, C = b = 32 second equation 212 = m(1) + 32 substituting into first equation m = 18 1 = 9 5 Solution: F = 9 5 C + 32 4 C = 18 + 5t 41 R = at + b 51 = a 1 + b R = 51, T = 1 54 = a 4 + b R = 54, T = 4 3 = 3a subtracting 3 a = 3 = 1 From the first equation, 51 = a 1 + b, we get 51 = (1)(1) + b (a = 1) b = 5 So the formula R = at + b becomes R = 1T + 5 42 P = k; let P = 1872 lb and = 3 ft Then So the relationship is P = 624 1872 = k(3) and k = 1872 3 = 624 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

1 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 14 Curve Sketching In the answers below, the intercepts are given first, followed b smmetr, asmptotes, and etent 2 = 2, = 1 2 ; none; none; all 3 Intercepts If =, then = 9 If =, then = 2 9 2 = 9 solving for = ±3 = 3 and = 3 Smmetr If is replaced b, we get = ( ) 2 9, which reduces to the given equation = 2 9 The graph is therefore smmetric with respect to the -ais There is no other tpe of smmetr Asmptotes Since the equation is not in the form of a quotient with a variable in the denominator, there are no asmptotes Etent is defined for all Graph 4 = 1; -ais; none; all 5 Intercepts If =, then = 1 If =, then = 1 2 2 = 1 solving for = ±1 = 1 and = 1 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

14 CURVE SKETCHING 11 Smmetr If is replaced b, we get = 1 ( ) 2, which reduces to the given equation = 1 2 The graph is therefore smmetric with respect to the -ais There is no other tpe of smmetr Asmptotes Since the equation is not in the form of a quotient with a variable in the denominator, there are no asmptotes Etent is defined for all Graph 6 = 5, = ± 5; -ais; none; all 7 Intercepts If =, then =, and if =, then = So the onl intercept is the origin Smmetr If we replace b, we get 2 =, which does not reduce to the given equation So there is no smmetr with respect to the -ais If is replaced b, we get ( ) 2 =, which reduces to 2 =, the given equation It follows that the graph is smmetric with respect to the -ais To check for smmetr with respect to the origin, we replace b and b : ( ) 2 = The resulting equation, 2 =, does not reduce to the given equation So there is no smmetr with respect to the origin Asmptotes Since the equation is not in the form of a fraction with a variable in the denominator, there are no asmptotes Etent Solving the equation for in terms of, we get = ± 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part Note that to avoid imaginar values, cannot be negative It follows that the etent is

12 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY Graph 8 origin; -ais; none; 9 Intercepts If =, then = ±1 If =, then = 1 Smmetr If we replace b we get 2 = + 1, which does not reduce to the given equation So there is no smmetr with respect to the -ais If is replaced b, we get ( ) 2 = + 1, which reduces to 2 = + 1, the given equation It follows that the graph is smmetric with respect to the -ais The graph is not smmetric with respect to the origin Asmptotes Since the equation is not in the form of a fraction with a variable in the denominator, there are no asmptotes Etent Solving the equation for, we get = ± + 1 To avoid imaginar values, we must have + 1 or 1 Therefore the etent is 1 Graph 1 = ± 2; = 2; -ais; none; 2 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

14 CURVE SKETCHING 13 11 Intercepts If =, then = ( 3)( + 5) = 15 If =, then = ( 3)( + 5) 3 = + 5 = = 3 = 5 Smmetr If is replaced b, we get = ( 3)( + 5), which does not reduce to the given equation So there is no smmetr with respect to the -ais Similarl, there is no other tpe of smmetr Asmptotes Since the equation is not in the form of a quotient with a variable in the denominator, there are no asmptotes Etent is defined for all Graph (,-15) 12 = 24; = 6, 4; none; none; all 13 Intercepts If =, then = If =, then = ( + 3)( 2) =, 3, 2 Smmetr If is replaced b, we get = ( + 3)( 2), which does not reduce to the given equation So the graph is not smmetric with respect to the -ais There is no other tpe of smmetr Asmptotes Since the equation is not in the form of a quotient with a variable in the denominator, there are no asmptotes Etent Not is defined for all For Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

14 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY Graph 14 = ; =,1, 4; none; none; all 15 Intercepts If =, = ; if =, then ( 1)( 2) 2 = =, 1, 2 Smmetr If is replaced b, we get = ( 1)( 2) 2, which does not reduce to the given equation So there is no smmetr with respect to the -ais Similarl, there is no other tpe of smmetr Asmptotes None (the equation does not have the form of a fraction) Etent is defined for all Graph 16 = ; = 2,, 3; none; none; all 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

14 CURVE SKETCHING 15 17 Intercepts If =, then = If =, then = ( 1) 2 ( 2) =, 1, 2 Smmetr If is replaced with, we get = ( 1) 2 ( 2), which does not reduce to the given equation Therefore there is no smmetr with respect to the -ais There is no other tpe of smmetr Asmptotes None (the equation does not have the form of a fraction) Etent is defined for all Graph 18 = ; = 2,, 3; none; none; all 19 Intercepts If =, = 1; if =, we have This equation has no solution Smmetr Replacing b, we get = 2 + 2 = 2 + 2 which does not reduce to the given equation So there is no smmetr with respect to the -ais Similarl, there is no other tpe of smmetr Asmptotes Setting the denominator equal to, we get + 2 = or = 2 It follows that = 2 is a vertical asmptote Also, as gets large, approaches So the -ais is a horizontal asmptote 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part Etent To avoid division b, cannot be equal to 2 So the etent is all ecept = 2

16 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY Graph 2 = 1; none; = 3; = ; 3 21 Intercepts If =, then = 2 If =, then This equation has no solution Smmetr Replacing b, we get 2 ( 1) 2 = = 2 ( 1) 2 which does not reduce to the given equation There are no other tpes of smmetr Asmptotes Setting the denominator equal to gives ( 1) 2 = or = 1 It follows that = 1 is a vertical asmptote Also, as gets large, approaches So the -ais is a horizontal asmptote Etent To avoid division b, cannot be equal to 1 So the etent is the set of all ecept = 1 Graph 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

14 CURVE SKETCHING 17 22 = 1 4 ; none; none; = 2; = ; 2 6 4 2 2 23 Intercepts If =, then = If =, then The onl solution is = Smmetr Replacing b ields = 2 1 = ( )2 1 = 2 1 which is not the same as the given equation So the graph is not smmetric with respect to the -ais Replacing b, we have = 2 1 which does not reduce to the given equation So the graph is not smmetric with respect to the -ais Similarl, there is no smmetr with respect to the origin Asmptotes Setting the denominator equal to, we get 1 =, or = 1 So = 1 is a vertical asmptote There are no horizontal asmptotes (Observation: for ver large the 1 in the denominator becomes insignificant So the graph gets ever closer to = 2 = ; the line = is a slant asmptote) Etent To avoid division b, cannot be equal to 1 So the etent is all ecept = 1 Graph 24 = ; = ; none; = 2; = 1; 2 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

18 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 25 Intercepts If =, then = 1/2 If =, then The onl solution is = 1 Smmetr Replacing b ields = + 1 ( 1)( + 2) = + 1 ( 1)( + 2) which is not the same as the given equation There are no tpes of smmetr Asmptotes Setting the denominator equal to, we get ( 1)( + 2) = So = 1 and = 2 are the vertical asmptotes As gets large, approaches, so the -ais is a horizontal asmptote Etent To avoid division b, the etent is all ecept = 1 and = 2 Graph 26 =, = 1, ; none; = 1, 2, = 1; 1, 2 27 Intercepts If =, then = 4 If =, then 2 4 2 1 = 2 4 = multipling b 2 1 = ±2 solution Smmetr Replacing b reduces to the given equation So there is smmetr with respect to the -ais There is no other tpe of smmetr Asmptotes Vertical: setting the denominator equal to, we have 2 1 = or = ±1 Horizontal: dividing numerator and denominator b 2, the equation becomes = 1 4 2 1 1 2 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

14 CURVE SKETCHING 19 As gets large, approaches 1 So = 1 is a horizontal asmptote Etent All ecept = ±1 (to avoid division b ) Graph 28 = 1 4, = ±1; -ais; = ±2, = 1; ±2 3 2 1 3 2 1 1 1 2 3 2 3 29 Intercepts If =, then 2 = 4 1 = 4, or = ±2 If =, then = 2 4 2 1 which is possible onl if 2 4 =, or = ±2 Smmetr The even powers on and tell us that if is replaced b and is replaced b, the resulting equation will reduce to the given equation The graph is therefore smmetric with respect to both aes and the origin Asmptotes Vertical: setting the denominator equal to, we get 2 1 = or = ±1 Horizontal: dividing numerator and denominator b 2, we get 2 = 1 4 2 1 1 2 The right side approaches 1 as gets large Thus 2 approaches 1, so that = ±1 are the horizontal Not asmptotes For Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

2 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY Etent From 2 4 = ± 2 1 we conclude that 2 4 ( 2)( + 2) 2 = 1 ( 1)( + 1) Since the signs change onl at = 2, 2, 1, and 1, we need to use arbitrar test values between these points The results are summarized in the following chart test values 2 1 + 1 + 2 ( 2)( + 2) ( 1)( + 1) > 2 3 + + + + + 1 < < 2 3/2 + + + 1 < < 1 + + + 2 < < 1 3/2 + < 2 3 + Note that the fraction is positive onl when > 2, 1 < < 1 and < 2 Since = when = ±2, the etent is 2, 1 < < 1, 2 Graph 3 = ± 1 2, = ±1; both aes; = ±2, = ±1; < 2, 1 1, > 2 31 Intercepts If =, 2 = ( 3)(5) = 15, or = ± 15 j, which is a pure imaginar number If =, ( 3)( + 5) = = 3, 5 Smmetr Replacing b, we get ( ) 2 = ( 3)( + 5), which reduces to the given equation Hence the graph is smmetric with respect to the -ais Asmptotes None (no fractions) Etent From = ± ( 3)( + 5), we conclude that ( 3)( + 5) If 3, ( 3)( + 5) If 5, ( 3)( + 5), since both factors are negative (or zero) If 5 < < 3, ( 3)( + 5) < [For eample, if =, we get ( 3)(5) = 15] These observations are summarized in the following chart 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

14 CURVE SKETCHING 21 test values 3 + 5 ( 3)( + 5) > 3 4 + + + 5 < < 3 + < 5 6 + Etent: 5, 3 Graph (-5,) (3,) 32 =, = ; -ais; = 2, = ±1; < 2, 33 Intercepts If =, = ; if =, = Smmetr Replacing b leaves the equation unchanged So there is smmetr with respect to the -ais There is no other tpe of smmetr Asmptotes Vertical: setting the denominator equal to, we get ( 3)( 2) = or = 3, 2 Horizontal: as gets large, approaches (-ais) Etent From we conclude that = ± ( 3)( 2) ( 3)( 3) Since signs change onl at =, 2 and 3, we need to use test values between these points The results are summarized in the following chart test values 2 3 ( 3)( 2) < 1 < < 2 1 + + 2 < < 3 5/2 + + > 3 4 + + + + 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part So the inequalit is satisfied for < < 2 and > 3 In addition, = when =

22 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY So the etent is < 2 and > 3 Graph 34 = 1; -ais; = 2, 1, = ; 2 < 1, > 1 35 C = 1 2 C 1 C 1 + 1 2, C 1 The onl intercept is the origin Dividing numerator and denominator b C 1, the equation becomes C = 1 2 1 + 1 2 /C 1 As C 1 gets large, C approaches 1 2 ; so C = 1 2 is a horizontal asmptote See graph in answer section of book 37 Intercepts If t =, S = ; if S =, we get 38 = 6t 5t 2 = 5t(12 t) or t =, 12 Smmetr None Asmptotes None Etent t b assumption Graph See graph in answer section of book 39 Etent L See graph in answer section of book 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

15 CURVES WITH GRAPHING UTILITIES 23 4 2 units ( = 2) 15 Curves with Graphing Utilities Graphs are from the answer section of the book 1 If =, then 2 ( 1)( 2) = Setting each factor equal to, we get =, 1, 2 [ 1, 3] b [ 2, 2] 2 2,, 1 [ 4, Not 2] b [ 4, 2] For Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

24 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 4 2, 1,, 1 [ 3, 3] b [ 5, 5] 5 4 2 3 = 3 ( 2) = =, 2 [ 1, 3] b [ 2, 2] 6 6 ± 2 [ 2, 2] b [ 15, 15] 8, 1 2 [ 1, 1] b [ 2, 2] 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

15 CURVES WITH GRAPHING UTILITIES 25 9 Domain: (to avoid imaginar values) Vertical asmptotes: None (The denominator is alwas positive, that is, 1 + ) [, 1] b [, 1] 1 = 1, > 1 [ 2, 4] b [ 4, 4] 12 = 1, [ 5, 2] b [ 4, 4] 13 To find the vertical asmptotes, we set the denominator equal to : 2 2 3 = 2 2 = 3 2 = 3 2 2 2 = 6 4 6 = ± 2 Domain: Not is defined for all For ecept = ± Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part 6 2

26 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY [ 3, 3] b [ 2, 2] 14 = ± 1 1 2, ± 2 [ 3, 3] b [ 2, 2] 16 =, 1 [ 2, 2] b [ 1, 1] 17 See graph in answer section of book 18 ( 65, 436) [ 5, 1] b [ 1, 1] 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

17 THE CIRCLE 27 2 45125 m [, 2] b [, 5] 21 See graph in answer section of book 22 773 [, 2] b [, 2] 17 The Circle 1 Since (h, k) = (, ) and r = 4, we get from the form the equation ( h) 2 + ( k) 2 = r 2 2 + 2 = 16 2 Since (h, k) = (, ) and r = 8, we get from the form ( h) 2 + ( k) 2 = r 2 the equation 2 + 2 = 64 3 The radius of the circle is the distance from the origin to ( 6, 8) Hence r 2 = ( + 6) 2 + ( 8) 2 = 1 From the standard form of the equation of the circle we get ( ) 2 + ( ) 2 = 1 center: (, ) 2 + 2 = 1 4 The radius of the circle is the distance from the origin to the point (1, 4) Hence r 2 = ( 1) 2 + ( + 4) 2 = 17 Equation: 2 + 2 = 17, (h, k) = (, ) 5 ( h) 2 + ( k) 2 = r 2 ( + 2) 2 + ( 5) 2 = 1 2 Not 2 + 2 + 4 1 + 28 = For Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

28 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 6 ( h) 2 + ( k) 2 = r 2 ( 2) 2 + ( + 3) 2 = ( 2 ) 2 2 4 + 4 + 2 + 6 + 9 = 2 2 + 2 4 + 6 + 11 = 7 The radius is the distance from ( 1, 4) to the origin: r 2 = ( 1 ) 2 + ( 4 ) 2 = 1 + 16 = 17 Hence, ( + 1) 2 + ( + 4) 2 = 17 ( h) 2 + ( k) 2 = r 2 2 + 2 + 1 + 2 + 8 + 16 = 17 2 + 2 + 2 + 8 = 8 The radius is the distance from the center to a point on the circle Thus r 2 = (5 3) 2 + (1 4) 2 = 4 ( 3) 2 + ( 4) 2 = 4 2 6 + 9 + 2 8 + 16 = 4 2 + 2 6 8 15 = 9 Diameter: distance from ( 2, 6) to (1, 5) Hence r = 1 2 ( 2 1)2 + ( 6 5) 2 = 1 1 9 + 121 = 13 2 2 and thus, r 2 = 1 65 (13) = 4 2 Center: midpoint of the line segment, whose coordinates are ( 2 + 1 Thus ( + 1 2 )2 + ( + 1 2 )2 = 65 2 2 + + 1 4 + 2 + + 1 4 = 65 2 2 + 2 + + 32 = 1 Distance from ( 1, 2) to the -ais: 1 2, 6 + 5 2 ) = ( 1 2, 1 2 ( + 1) 2 + ( 2) 2 = 1 ) 11 Since r = 5, we get ( 4) 2 + ( + 5) 2 = 25 or 2 + 2 8 + 1 + 16 = radius = 5 (4,-5) 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

17 THE CIRCLE 29 12 Distance to the line = 1: 3 ( 3) 2 + ( 4) 2 = 9 13 From the diagram, r 2 = 1 2 + 1 2 = 2; so 2 + 2 = 2 r 1 + = 2 1 14 Since the circle is tangent to the ais with radius 2, h = 2 or h = 2 The equation = 3 2 ields two possibilities for the center: (2, 3) and ( 2, 3) Thus or 15 2 + 2 2 2 2 = ( 2) 2 + ( 3) 2 = 2 2 and ( + 2) 2 + ( + 3) 2 = 2 2 2 + 2 4 6 + 9 = and 2 + 2 + 4 + 6 + 9 = 2 2 + 2 2 = 2 We now add to each side the square of one-half the coefficient of : [ 1 2 ( 2)] 2 = 1 2 2 + 1 + 2 2 = 2 + 1 Similarl, we add 1 (the square of one-half the coefficient of ): ( 2 2 + 1) + ( 2 2 + 1) = 2 + 1 + 1 ( 1) 2 + ( 1) 2 = 4 Center: (h, k) = (1, 1); radius: 4 = 2 16 2 + 2 2 4 + 4 = 2 2 + 2 4 = 4 We now add to each side the square of one-half the coefficient of : [ 1 2 ( 2)] 2 = 1: 2 2 + 1 + 2 4 = 4 + 1 Similarl, we add to each side the square of one-half the coefficient of : [ 1 2 ( 4)] 2 = 4 2 2 + 1 + 2 4 + 4 = 4 + 1 + 4 ( 1) 2 + ( 2) 2 = 1; Center: (h, k) = (1, 2); radius: 1 = 1 17 2 + 2 + 4 8 + 4 = 2 + 4 + 2 8 = 4 Since ( ) 2 1 2 4 = 4 and we get ( 2 + 4 + 4) + ( 2 8 + 16) = 4 + 4 + 16 [ ] 2 1 2 ( 8) = 16 Not ( + 2) 2 + ( For 4) 2 = 16 Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

3 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY The equation can be written [ ( 2)] 2 + ( 4) 2 = 4 2 It follows that (h, k) = ( 2, 4) and r = 4 18 2 + 2 + 2 + 6 + 3 = 2 + 2 + 2 + 6 = 3 Observe that [ 1 2 (2)] 2 = 1 and [ 1 2 (6)] 2 = 9 Adding 1 and 9 to each side, we get 2 + 2 + 1 + 2 + 6 + 9 = 3 + 1 + 9 or ( + 1) 2 + ( + 3) 2 = 7 [ ( 1)] 2 + [ ( 3)] 2 = 7; Center: ( 1, 3); radius: 7 19 2 + 2 4 + + 9 4 = 2 4 + 2 + = 9 4 We add to each side the square of one-half the coefficient of : 2 4 + 4 + 2 + = 9 4 + 4 [ ] 2 1 2 ( 4) = 4 This gives [ ] 2 1 Similarl, we add the square of one-half the coefficient of : 2 1 = 1 This gives 4 2 4 + 4 + 2 + + 1 4 = 9 4 + 4 + 1 4 ( 2) 2 + ( + 1 2 )2 = 8 4 + 4 ( 2) 2 + ( + 1 2 )2 = 2 Center: (2, 1 2 ); radius: 2 2 2 + 2 + 4 + 5 4 = 2 + + 2 4 = 5 4 Add to each side: [ 1 2 (1)] 2 = 1 4 and [ 1 2 ( 4)] 2 = 4 2 + + 1 4 + 2 4 + 4 = 5 4 + 1 4 + 4 ( + 1 2 + ( 2) 2) 2 = 3; Center: ( 12 ), 2 ; radius: 3 21 4 2 + 4 2 8 12 + 9 = 2 + 2 2 3 + 9 4 = dividing b 4 2 2 + 2 3 = 9 4 [ ] 2 1 Add to each side: 2 ( 2) = 1 and [ ] 2 1 2 ( 3) = 9 4 ( 2 2 + 1) + ( 2 3 + 9 4 ) = 9 4 + 1 + 9 4 ( 1) 2 + ( 3 2 )2 = 1 Center: (1, 3 2 ); radius: 1 to get 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

17 THE CIRCLE 31 22 4 2 + 4 2 2 8 + 25 = First divide both sides b 4: 2 + 2 5 2 + 25 = 4 2 5 + 2 2 = 25 4 Add to each side: [ 1 2 ( 5)] 2 = 25 4 and [ 1 2 ( 2)] 2 = 1 2 5 + 25 4 + 2 2 + 1 = 25 4 + 25 4 + 1 ( ) ( 5 2 5 2) + ( 1) 2 = 1; Center: 2, 1 ; radius: 1 23 2 + 2 + 4 2 4 = 2 + 4 + 2 2 = 4 Note that ( ) 2 1 2 4 = 4 and Adding 4 and 1, respectivel, we get ( 2 + 4 + 4) + ( 2 2 + 1) = 4 + 4 + 1 ( + 2) 2 + ( 1) 2 = 9 The equation can be written So the center is ( 2, 1) and the radius is 3 24 2 + 2 + 2 + 8 + 1 = 2 + 2 + 2 + 8 = 1 [ ] 2 1 2 ( 2) = 1 [ ( 2)] 2 + ( 1) 2 = 3 2 Add to each side: [ 1 2 (2)] 2 = 1 and [ 1 2 (8)] 2 = 16 2 + 2 + 1 + 2 + 8 + 16 = 1 + 1 + 16 ( + 1) 2 + ( + 4) 2 = 16; Center: ( 1, 4); radius: 4 25 2 + 2 2 + 1 4 = 2 + 2 2 = 1 4 [ ] 2 1 Add to each side: 2 ( 1) = 1 4 and ( 2 + 1 4 ) + (2 2 + 1) = 1 4 + 1 4 + 1 ( 1 2 )2 + ( 1) 2 = 1 Center: ( 1 2, 1); radius: 1 [ ] 2 1 2 ( 2) = 1 to get 26 2 + 2 6 + 8 + 19 = 2 6 + 2 8 = 19 Add to each side: [ 1 2 ( 6)] 2 = 9 and [ 1 2 ( 8)] 2 = 16 2 6 + 9 + 2 8 + 16 = 19 + 9 + 16 ( Not 3) 2 + ( 4) 2 = 6; Center: For (3, 4); radius: 6 Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

32 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 27 2 + 2 4 + + 9 4 = 2 4 + 2 + = 9 4 Note that [ ] 2 1 2 ( 4) = 4 and Adding 4 and 1 4, respectivel, we get ( ) 2 1 2 1 = 1 4 ( 2 4 + 4) + ( 2 + + 1 4 ) = 9 4 + 4 + 1 4 ( 2) 2 + ( + 1 2 )2 = 2 The equation can be written [ ( ( 2) 2 + 1 2 = ( 2)] 2) 2 Center: (2, 1 2 ); radius: 2 28 2 + 2 + 1 2 = 2 + + 2 = 1 2 Add to each side: [ 1 2 (1)] 2 = 1 4 and [ 1 2 ( 1)] 2 = 1 4 2 + + 1 4 + 2 + 1 4 = 1 2 + 1 4 + 1 4 ( ( ) + 1 2 ( ) 2 + 1 2 2 = 1; Center: 1 2, 1 ) ; radius: 1 2 29 4 2 + 4 2 + 12 + 16 + 5 = 2 + 2 + 3 + 4 + 5 4 = 2 + 3 + 2 + 4 = 5 4 [ ] 2 1 Add to each side: 2 3 = 9 ( ) 2 1 4 and 2 4 = 4 to get 2 + 3 + 9 4 + 2 + 4 + 4 = 5 4 + 9 4 + 4 ( + 3 2 )2 + ( + 2) 2 = 5 Center: ( 3 2, 2); radius: 5 3 36 2 + 36 2 144 12 + 219 = First divide both sides b 36: 2 + 2 4 1 3 + 219 = 36 2 4 + 2 1 3 = 219 36 Add to each side: [ 1 2 ( 4)] 2 [ = 4 and 1 ( )] 2 1 2 3 = 25 9 2 4 + 4 + 2 1 3 + 25 9 = 219 36 + 4 + 25 9 = 219 36 + 144 36 + 1 36 ( ( 2) 2 + 5 ) 2 = 25 ( 3 36 ; Center 2, 5 ) 5 ; radius: 3 6 31 4 2 + 4 2 2 4 + 26 = 2 + 2 5 + 26 4 = dividing b 4 2 5 + 2 = 26 4 2 5 + 25 4 + 2 + 1 4 = 26 4 + 25 4 + 1 4 ( 5 2 )2 + ( 1 2 )2 = Locus is the single point ( 5 2, 1 2 ) 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

17 THE CIRCLE 33 32 2 + 2 + 4 2 + 7 = 2 + 4 + 2 2 = 7 2 + 4 + 4 + 2 2 + 1 = 7 + 4 + 1 ( + 2) 2 + ( 1) 2 = 2 Imaginar circle 33 2 + 2 6 + 8 + 25 = 2 6 + 2 + 8 = 25 ( 2 6 + 9) + ( 2 + 8 + 16) = 25 + 9 + 16 ( 3) 2 + ( + 4) 2 = Locus is the single point (3, 4) 34 2 + 2 + 2 + 4 + 5 = 2 + 2 + 2 + 4 = 5 2 + 2 + 1 + 2 + 4 + 4 = 5 + 1 + 4 ( + 1) 2 + ( + 2) 2 = Point circle: ( 1, 2) 35 2 + 2 6 8 + 3 = 2 6 + 2 8 = 3 2 6 + 9 + 2 8 + 16 = 3 + 9 + 16 ( 3) 2 + ( 4) 2 = 5 (imaginar circle) 36 2 + 2 6 + 4 + 13 = 2 6 + 2 + 4 = 13 2 6 + 9 + 2 + 4 + 4 = 13 + 9 + 4 ( 3) 2 + ( + 2) 2 = Point circle: (3, 2) 37 2 + 2 + 4 + 17 4 = 2 + 2 + 4 = 17 4 [ ] 2 1 We add to each side 2 ( 1) and [ ] 2 1 2 (4) : 2 + 1 4 + 2 + 4 + 4 = 17 4 + 1 4 + 4 ( 1 2) 2 + ( + 2) 2 = (point circle) 38 From the given circle, we have 2 6 + 2 4 = 12 2 6 + 9 + 2 4 + 4 = 12 + 9 + 4 ( 3) 2 + ( 2) 2 = 25 Center: (3, 2); Desired circle: ( 3) 2 + ( 2) 2 = 9 39 2 + 2 = (2) 2 = 4; 2 + 2 = (34) 2 = 116 4 Center: (5,5); radius: 21 ( 5) 2 + ( 5) 2 = (21) 2 2 1 + 25 + 2 1 + 25 = 441 2 + 2 1 1 + 456 = 41 The Not radius is 22, 3 + 4 For =26,3 mi Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

34 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 43 (h, k) = (, ) and r = 3 2 ft: 2 + 2 = 9 4 and = 9 4 2 18 The Parabola 1 Since the focus is on the -ais, the form is 2 = 4p Since the focus is at (3, ), p = 3 (positive) Thus 2 = 4(3), or 2 = 12 2 Since the focus is on the ais, the form is 2 = 4p Since the focus is at ( 3,), p = 3 (negative) Thus 2 = 4( 3), or 2 = 12 3 Since the focus is on the -ais, the form is 2 = 4p Since the focus is at (, 5), p = 5 (negative) Thus 2 = 4( 5), or 2 = 2 4 Since the focus is on the ais, the form is 2 = 4p Since the focus is at (,4), p = 4 (positive) Thus 2 = 4(4), or 2 = 16 5 Since the focus is on the -ais, the form is 2 = 4p The focus is on the left side of the origin, at ( 4, ) So p = 4 (negative) It follows that 2 = 4( 4), or 2 = 16 6 Since the focus is on the ais, the form is 2 = 4p Since the focus is at (, 6), p = 6 (negative) Thus 2 = 4( 6), or 2 = 24 7 Since the directri is = 1, the focus is at (1, ) So the form is 2 = 4p with p = 1, and the equation is 2 = 4 8 Since the directri is = 2, the focus is at (, 2), so that p = 2 From the form 2 = 4p, we get 2 = 8 9 Since the directri is = 2, the focus is at ( 2, ) So the form is 2 = 4p with p = 2 Thus 2 = 8 1 Directri: = 2; focus: (2,); form: 2 = 4p; equation: 2 = 8 11 Form: 2 = 4p Substituting the coordinates of the point ( 2, 4), we get ( 4) 2 = 4p( 2) and 4p = 8 Thus 2 = 8 12 Form: 2 = 4p; the coordinates of the point ( 1,1) satisf the equation: ( 1) 2 = 4p(1), so 4p = 1, and we get 2 = 13 The form is either 2 = 4p or 2 = 4p Substituting the coordinates of the point (1, 1), we get 1 2 = 4p 1 or 1 2 = 4p 1 In either case, p = 1 4 So the equations are 2 = and 2 = 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

18 THE PARABOLA 35 14 Both forms are possible; substituting the coordinates of the point (2, 1), we get 2 = 4p 2 = 4p 2 2 = 4p( 1) ( 1) 2 = 4p(2) 4p = 4 4p = 1 2 2 = 4 2 = 1 2 15 The form is either 2 = 4p or 2 = 4p Substituting the coordinates of the point ( 2, 4), we get 4 2 = 4p( 2) or ( 2) 2 = 4p(4) The respective values of p are 2 and 1 4 ; so the equations are 2 = 8 and 2 = 16 Form: 2 = 4p or 2 = 4p Substituting (3, 5): 25 = 4p(3) or 9 = 4p( 5) p = 25 12 or p = 9 2 The equations are 2 = 25 3 and 2 = 9 5 17 From 2 = 12, we have 2 = 4(3) Thus p = 3 and the focus is at (, 3) 18 2 = 2 (,3) 2 = 4(5) p = 5 focus: (, 5) (,5) = -5 19 From 2 = 8, we have 2 = 4( 2) So p = 2 and the focus is at (, 2) (,-2) 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

36 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 2 2 = 24 2 = 4( 6) =6 p = 6 (,-6) 21 2 = 24 = 4(6); p = 6 and the focus is at (6, ) 22 2 = 12 2 = 4(3) p = 3 (6,) = -3 (3,) 23 From 2 = 4, 2 = 4( 1) So p = 1 and the focus is at ( 1, ) (-1,) 24 2 = 12 2 = 4( 3) p = 3 (-3,) =3 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

18 THE PARABOLA 37 25 2 = 4 = 4(1); p = 1 and the focus is at (, 1) (,1) 26 2 = 12 2 = 4( 3) focus: (, 3), directri: 3 = 27 From 2 = 9, 2 = 4( 9 4 ) (inserting 4) So p = 9 4 and the focus is at ( 9 4, ) 28 2 = 1 ( ) 5 2 = 4 2 p = 5 2 focus: 29 2 = = 4( 1 4 ); p = 1 4 and the focus is at ( 1 4, ) 3 2 = 3 2( ) 3 2 = 4 8 p = 3 8 31 3 2 + 2 = 2 = 2 3 2 = 4( 2 3 1 4 ) 2 = 4( 1 6 ) So the focus is at ( 1 6, ) 32 2 = 2a ( a ) 2 = 4 2 p = a 2 focus: focus: ( ) 5 2,, directri: + 5 2 = (, 3 ), directri: + 3 8 8 = ( a 2, ), directri: + a 2 = 33 2 = 4(3); p = 3 Focus: (, 3); directri: = 3 (-6,3) (6,3) 6 6 = -3 Observe that the points (6, 3) and ( 6, 3) lie on the curve because the distance to the focus must be equal to the distance to the directri 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

38 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY Circle: ( h) 2 + ( k) 2 = r 2 ( ) 2 + ( 3) 2 = 6 2 2 + 2 6 + 9 = 36 2 + 2 6 27 = 34 Let (,) be a point on the parabola in Figure 146 Distance to (,p): ( ) 2 + ( p) 2 Distance to the line = p: ( p) = + p B definition, the distances are equal So ( )2 + ( p) 2 = + p 2 + ( p) 2 = ( + p) 2 (squaring both sides) 2 + 2 2p + p 2 = 2 + 2p + p 2 2 = 4p after collecting terms 35 We need to find the locus of points (, ) equidistant from (4, 1) and the -ais Since the distance from (, ) to the -ais is units, we get ( 4)2 + ( 1) 2 = ( 4) 2 + ( 1) 2 = 2 2 8 + 16 + 2 2 + 1 = 2 2 2 8 + 17 = 36 From the figure, d 1 = d 2 : ( 4)2 + ( 7) 2 = + 1 ( 4) 2 + ( 7) 2 = ( + 1) 2 (squaring both sides) 2 8 + 16 + 2 14 + 49 = 2 + 2 + 1 2 8 16 + 64 = 37 If the origin is the lowest point on the cable, then the top of the right supporting tower is at (1, 7) (1,7) From the equation 2 = 4p, we get (1) 2 = 4p(7) 1, 4p = = 1 7 7 The equation is therefore 2 = 1 7 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

18 THE PARABOLA 39 To find the length of the cable 3 m from the center, we let = 3: 3 2 = 1 63 and = 7 1 = 63 So the length of the cable is 2 + 63 = 263 m 39 4 38 Place the parabola with verte at the origin and ais along the ais From the given information the point (12,1) lies on the curve Substituting coordinates, 2 = 4p (1) 2 = 4p(12) 4p = 1 and p = 28 cm 12 So the light bulb is placed at the focus, 28 cm from the verte -25 2 (2,-13) The required minimum clearance of 12 ft ields the point (2, 13) in the figure 2 = 4p 2 2 = 4p( 13) or 4p = 22 13 Equation: 2 = 22 13 When = 25, 2 = 22 13 ( 25) 22 25 = = 2 5 = 1 13 13 13 2 = 2 555 ft 13 The problem is to find in the figure; observe that + 1 corresponds to the maimum clearance There are two equations from the form 2 = 4p: (3) 2 = 4p (2) 2 = 4p( + 1) (3) 2 (2) 2 = 4p(1) (subtracting) Not 4p = 5 For Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

4 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY Equation: 2 = 5; to find, let = 3: (3) 2 = 5 = 32 5 = 18 So the height of the arch is 18 m 41 We place the verte of the parabola at the origin, so that one point on the parabola is (3, 3) (from the given dimensions) Substituting in the equation 2 = 4p, we get 3 2 = 4p( 3) 4p = 3 The equation is therefore seen to be 2 = 3 (,-1) (3,-3) The right end of the beam 2 m above the base is at (, 1) To find, let = 1: 2 = 3( 1) = 3 = ± 3 Hence the length of the beam is 2 = 2 3 m 42 = 15 1 2 = when = 15 ft 43 2 = 4p From Figure 155, we see that the point (4, 1) lies on the curve: 4 2 = 4p(1) So p = 4 ft 44 Place the parabola with verte at the origin and ais along the ais From the given information the point (17,) lies on the curve The problem is to find, given that p = 119 2 = 4p = 4 119 17 19 The Ellipse Diameter= 2 = 2 4 119 17 = 18 in 1 The equation is 2 25 + 2 16 = 1 So b (116), a 2 = 25 and b 2 = 16; thus a = 5 and b = 4 Since the major ais is horizontal, the vertices are at (±5, ) From b 2 = a 2 c 2, 16 = 25 c 2 c 2 = 9 c = ±3 The foci are therefore at (±3, ), on the major ais Finall, the length of the semi-minor ais is equal to b = 4 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

19 THE ELLIPSE 41 4 5 2 2 16 + 2 9 = 1 B (116), a 2 = 16 and b 2 = 9; so a = 4 and b = 3 Since the major ais is horizontal, the vertices are at (±4,) From b 2 = a 2 c 2, 9 = 16 c 2 c 2 = 7 c = ± 7 So the foci are at ( ± 7, ), on the major ais Finall, the length of the semiminor ais is b = 3 3 The equation is 2 9 + 2 4 = 1 So b (116), a 2 = 9 and b 2 = 4; thus a = 3 and b = 2 Since the major ais is horizontal, the vertices are at (±3, ) From b 2 = a 2 c 2, 4 = 9 c 2 c 2 = 5 c = ± 5 The foci are therefore at ( ± 5, ), on the major ais Finall, the length of the semi-minor ais is equal to b = 2 2 3 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

42 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 2 4 4 + 2 9 = 1 B (117), a = 3 and b = 2 From b 2 = a 2 c 2, 4 = 9 c 2 c 2 = 5 c = ± 5 Since the major ais is vertical, the vertices are at (, ± 3) and the foci are at (, ± 5 ) The length of the semiminor ais is b = 2 5 The equation is 6 2 16 + 2 = 1 B (116), a = 4 and b = 1 From b 2 = a 2 c 2, 1 = 16 c 2 c 2 = 15 c = ± 15 Since the major ais is horizontal, the vertices and foci lie on the -ais The vertices are therefore at (±4, ) and the foci are at (± 15, ) The length of the semi-minor ais is b = 1 2 2 + 2 4 = 1 1 4 B (117), a = 2 and b = 2, major ais vertical b 2 = a 2 c 2 2 = 4 c 2 c = ± 2 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

19 THE ELLIPSE 43 Vertices:(, ± 2), foci: (, ± 2 ), semiminor ais: 2 7 16 2 + 9 2 = 144 2 9 + 2 = 1 16 So b (117), a 2 = 16 and b 2 = 9; so a = 4 and b = 3 Since the major ais is vertical, the vertices are at (, ±4) From b 2 = a 2 c 2 9 = 16 c 2 c = ± 7 The foci are therefore at (, ± 7) Semi-minor ais: b = 3 4 8 2 + 2 2 = 4 2 4 + 2 2 = 1 3 B (116), a = 2 and b = 2, major ais horizontal b 2 = a 2 c 2 2 = 4 c 2 c = ± 2 Vertices:(±2,), foci: ( ± 2, ), semiminor ais: 2 9 5 2 + 2 2 = 2 5 2 2 + 22 = 1 2 2 4 + 2 = 1 1 B (117), a = 1 and b = 2 Since the major ais is vertical, the vertices are at (, ± 1) From b 2 = a 2 c 2 4 = 1 c 2 c = ± 6 The foci, also on the major ais, are therefore at (, ± 6), while the length of the semi-minor ais Not is b = 2 (See sketch in For answer section of book) Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

44 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 1 5 2 + 9 2 = 45 2 9 + 2 5 = 1 B (116), a = 3 and b = 5, major ais horizontal b 2 = a 2 c 2 5 = 9 c 2 c = ±2 Vertices: (±3,), foci: (±2,), semiminor ais: 5 11 5 2 + 2 = 5 2 1 + 2 = 1 major ais vertical 5 Vertices: (, ± 5), foci: (, ±2) Length of semi-minor ais: b = 1 (See sketch in answer section of book) 12 2 + 4 2 = 4 2 4 + 2 1 b 2 = a 2 c 2 1 = 4 c 2 c = ± 3 = 1 a = 2 and b = 1, major ais horizontal Vertices: (±2, ), foci: ( ± 3, ), semiminor ais: 1 13 2 + 2 2 = 6 2 6 + 22 = 1 6 2 6 + 2 = 1 major ais horizontal 3 Thus a = 6 and b = 3 From b 2 = a 2 c 2, 3 = 6 c 2 c = ± 3 Vertices: (± 6, ); foci: (± 3, ) Length of semi-minor ais: b = 3 (See sketch in answer section of book) 14 9 2 + 2 2 = 18 2 2 + 2 = 1 a = 3 and b = 2, major ais vertical 9 b 2 = a 2 c 2 2 = 9 c 2 c = ± 7 Vertices: (, ±3), foci: (, ± 7 ), semiminor ais: 2 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

19 THE ELLIPSE 45 15 15 2 + 7 2 = 15 2 7 + 2 = 1 major ais vertical 15 Thus a = 15 and b = 7 From b 2 = a 2 c 2 7 = 15 c 2 c = ± 8 = ±2 2 Vertices: (, ± 15), foci: (, ±2 2) Length of semi-minor ais: b = 7 16 9 2 + 2 = 27 2 3 + 2 = 1 a = 27 = 3 3 and b = 3, major ais vertical 27 b 2 = a 2 c 2 3 = 27 c 2 c = ± 24 = ±2 6 Vertices: (, ±3 3), foci: (, ±2 6 ), semiminor ais: 3 17 3 2 + 4 2 = 12 18 2 4 + 2 = 1 dividing b 12 3 Since a 2 = 4, a = ±2, so the vertices are at (±2, ) From b 2 = 3, we get b = 3 (semiminor ais) b 2 = a 2 c 2 ields c = ±1 2 5 + 2 12 = 1 ields a = ± 5, b = ±2 3 From b 2 = a 2 c 2, we get c = ± 7 2 Graph the two equations = 22 Graph the two equations = ± 1 1 4 2 1 2 (5 62 1 ) and = 2 (5 62 ) 23 Since the foci (±2, ) lie along the major ais, the major ais is horizontal So the form of the equation is 2 a 2 + 2 b 2 = 1 Since the vertices are at (±3, ), a = 3 From b 2 = a 2 c 2 (with c = 2), we get b 2 = 9 4 = 5 So the equation is 2 9 + 2 5 = 1 24 Since the foci are at (, ±3), the major ais is vertical 2 b 2 + 2 a 2 = 1 (form) Since the length of the major ais is 8, a = 4, while c = 3 Hence b 2 = a 2 c 2 = 16 9 = 7 and Not 2 7 + 2 16 = 1 For Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

46 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 25 Since the foci (, ±2) lie on the major ais, the major ais is vertical So b (117) the form of the equation is 2 b 2 + 2 a 2 = 1 Since the length of the major ais is 8, a = 4 From b 2 = a 2 c 2 with c = 2, b 2 = 16 4 = 12 Hence 2 12 + 2 16 = 1 or 42 + 3 2 = 48 26 Since the length of the entire major ais is 6, we get a = 3 Foci at (, ± 2) implies that c = 2 and that the major ais is vertical From b 2 = a 2 c 2 = 9 4 = 5 and from the form 2 b 2 + 2 2 = 1 we get a2 5 + 2 9 = 1 27 Since the foci are at (, ±3), c = 3, and the major ais is vertical B (117) 2 b 2 + 2 a 2 = 1 Since the length of the minor ais is 6, b = 3 From b 2 = a 2 c 2, 9 = a 2 9 and a 2 = 18 Equation: 2 9 + 2 18 = 1 or 22 + 2 = 18 28 Since the length of the entire minor ais is 4, b = 2 Foci at (, ± 2) implies that c = 2 and that the major ais is vertical From b 2 = a 2 c 2, 4 = a 2 4 and a 2 = 8 B (117) 2 4 + 2 8 = 1 29 Since the vertices and foci are on the -ais, the form of the equation is, b (117), 2 b 2 + 2 a 2 = 1 From b 2 = a 2 c 2 with a = 8 and c = 5, b 2 = 64 25 = 39 Hence 2 39 + 2 64 = 1 3 Foci at (±3, ) implies that c = 3 and that the major ais is horizontal; b = 4 (semiminor ais) From b 2 = a 2 c 2, we have 16 = a 2 9, or a 2 = 25 B (116) 31 Form: 2 25 + 2 16 = 1 or 162 + 25 2 = 4 2 a 2 + 2 b 2 = 1; c = 2 3, b = 2 From b 2 = a 2 c 2, 4 = a 2 (2 3) 2 and a 2 = 16 Equation: 2 16 + 2 4 = 1 or 2 + 4 2 = 16 32 Since the foci and vertices lie on the major ais, the major ais is horizontal Moreover, c = 5 and a = 7 So b 2 = 7 5 = 2 and b (116), 2 7 + 2 2 = 1 or 22 + 7 2 = 14 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

19 THE ELLIPSE 47 33 From the form we get (since b = 2) 2 a 2 + 2 b 2 = 1 a 2 + 2 4 = 1 Substituting the coordinates of the point (3, 1) ields So Equation: 2 9 a 2 + 1 4 = 1 and 9 a 2 = 3 4 a 2 9 = 4 3 and a2 = 12 2 12 + 2 4 = 1 or 2 + 3 2 = 12 34 Since b = 3, we get from (117), 2 9 + 2 = 1 Since (1,4) lies on the curve, the values = 1 a2 and = 4 satisfies the equation: 1 2 9 + 42 a 2 = 1 whence a2 = 18 and the final result is 2 9 + 2 18 = 1 or 22 + 2 = 18 35 From the original derivation of the ellipse, 2a = 16 and a = 8 Since the foci are at (±6, ), c = 6 Thus b 2 = a 2 c 2 = 64 36 = 28 B (116) the equation is 2 64 + 2 28 = 1 36 Vertices at (±4,) tell us that a = 4 and that the major ais is horizontal The definition of eccentricit gives the following equation: e = 1 2 = c a or 1 2 = c 4 which ields c = 2 Finall, b 2 = a 2 c 2 = 16 4 = 12 B (116), 2 16 + 2 12 = 1 or 32 + 4 2 = 48 37 9 2 + 5 2 = 45 or 2 5 + 2 9 = 1; a = 3; b = 5 From b 2 = a 2 c 2, 5 = 9 c 2 and c = 2 Thus e = c a = 2 3 38 Distance from (, ) to (, ): ( ) 2 + ( ) 2 = 2 + 2 Distance from (, ) to (3, ): ( 3) 2 + ( ) 2 = ( 3) 2 + 2 From the given condition: 2 + 2 = 2 ( 3) 2 + 2 2 + 2 = 4[( 3) 2 + 2 ] squaring both sides 2 + 2 = 4( 2 6 + 9 + 2 ) 2 + 2 = 4 2 24 + 36 + 4 2 = 3 2 24 + 36 + 3 2 3 2 + 3 2 24 + 36 = Not 2 + 2 8 + 12 = For Sale The locus is a circle 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

48 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 39 We want the center of the ellipse to be at the origin with the center of the earth at one of the foci Stud the following diagram: b 2 = 41 2 2 2 = 16, 89, 6 4 Let A = the maimum distance and P = the minimum distance, as shown A is also the distance from the left focus to the right verte So A P is the distance between foci Therefore 1 2 (A P ) is the distance from the center to the sun (the focus), or c = 1 2 (A P ) Now a = c + P = 1 2 (A P ) + P = 1 2 (A + P ) so e = c 1 a = 2 (A P ) 1 2 (A + P ) = A P A + P In our problem e = 946 17 914 1 7 947 1 7 + 914 1 7 = 172 1 6 41 Let A = the maimum distance and P = the minimum distance as shown A is also the distance from the left focus to the right verte So A P is the distance between the foci Therefore 1 2 (A P ) is the distance from the center to the sun (the focus), or c = 1 2 (A P ) Now a = c + P = 1 2 (A P ) + P = 1 2 (A + P ) So e = c 1 a = 2 (A P ) 1 2 (A + P ) = A P A + P 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

19 THE ELLIPSE 49 In our problem 42 A = π a b = π 15 12 565 ft 2 43 Since a = 2 and b = 3 2, we get 2 e = 3285 19 548 1 7 3285 1 9 + 548 1 7 = 967 and 9 2 + 16 2 = 36 4 + 2 9 4 = 1 or 2 4 + 42 9 = 1, 44 From the figure, a = 15 and b = 5 From (117) 2 5 2 + 2 2 = 1 The net step is to find when = 1: 15 2 25 = 225 225 1 225 = 125 225 and 5 25 25 2 = 15 2 5 25 5 So = and the length of the beam is 2 = 1 15 3 75ft 45 Placing the center at the origin, the vertices are at (±6, ) The road etends from ( 4, ) to (4, ) Since the clearance is 4 m, the point (4, 4) lies on the ellipse, as shown B (116), 2 a 2 + 2 b 2 = 1 2 4 (4,4) 6 36 + 2 b 2 = 1 To find b, we substitute the coordinates of (4, 4) in the equation: 16 36 + 16 b 2 = 1 16 b 2 = 36 36 16 36 = 2 36 = 5 9 b 2 16 = 9 5 b 2 = (9)(16) 5 b = (3)(4) 5 = 12 Not For 5 Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part 5 = 12 5

5 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY So the height of the arch is 12 5 = 54 m to two significant digits 5 46 Since the sun is at one of the foci, a c = 12 (in millions of miles) when the comet is closest to the sun From e = c, we get a 99992 = c a or c = 99992a So a c = a 99992a = 12; solving, a = 15 and c = 149988 Farthest point: (15 + 149988)(1) = 3 1 11 mi 11 The Hperbola 1 Comparing the given equation, 2 16 2 9 = 1 to form (122), we see that the transverse ais is horizontal, with a 2 = 16 and b 2 = 9 So a = 4 and b = 3 From b 2 = c 2 a 2, we get 9 = c 2 16 c = ±5 So the vertices are at (±4, ) and the foci are at (±5, ) Using a = 4 and b = 3, we draw the auiliar rectangle and sketch the curve: 2 Comparing the given equation 2 9 2 = 1 to the form (122), we see that the transverse ais 4 is horizontal, with a 2 = 9 and b 2 = 4 So a = 3 and b = 2 From b 2 = c 2 a 2, we get 4 = c 2 9 and c 2 = 13 It follows that the vertices are at (±3,) and the foci at ( ± 13, ) Using a = 3 and b = 2, we draw the auiliar rectangle and sketch the curve 3 2 9 2 16 = 1; b Equation (122), the transverse ais is horizontal with a2 = 9 and b 2 = 16 So a = 3 and b = 4 From b 2 = c 2 a 2, we have 16 = c 2 9 or c = ±5 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

11 THE HYPERBOLA 51 It follows that the vertices are at (±3, ) and the foci are at (±5, ) Using a = 3 and b = 4, we draw the auiliar rectangle and the asmptotes, and then sketch the curve 4 Equation: 2 16 2 4 = 1 B (122), a = 4 and b = 2, transverse ais horizontal From b 2 = c 2 a 2, 4 = c 2 16 and c = ± 2 = ±2 5 So the vertices are at (±4,) and the foci at ( ±2 5, ) Using a = 4 and b = 2, we draw the auiliar rectangle and sketch the curve 5 B (123), a = 2 and b = 2, transverse ais vertical along the -ais From b 2 = c 2 a 2, 4 = c 2 4 and c = ± 8 = ±2 2 So the vertices are at (, ±2) and the foci are at (, ±2 2) Using a = 2 and b = 2, we draw the auiliar rectangle and sketch the curve: 6 Equation: 2 4 2 8 = 1 B (123), a = 2 and b = 8 = 2 2, transverse ais vertical From b 2 = c 2 a 2, 8 = c 2 4 and c = ± 12 = ±2 3 So the vertices are at (, ± 2) and the foci at (, ± 2 3 ) Using a = Not 2 and b = 2 2, we draw For the auiliar rectangle Sale and sketch the curve 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

52 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 7 2 2 = 1 transverse ais horizontal 5 a 2 = 1 and b 2 = 5; so a = 1 and b = 5 From b 2 = c 2 a 2, 5 = c 2 1 and c = ± 6 Vertices: (±1, ); foci: (± 6, ) Using a = 1 and b = 5, we draw the auiliar rectangle and sketch the curve 8 Equation: 9 2 2 2 = 18 or 2 2 2 9 = 1 B (123), a = 2 and b = 3, transverse ais vertical From b 2 = c 2 a 2, 9 = c 2 2 and c = ± 11 So the vertices are at (, ± 2 ) and the foci at (, ± 11 ) Using a = 2 and b = 3, we draw the auiliar rectangle and sketch the curve 9 2 2 3 2 = 24 2 2 24 32 24 = 1 2 12 2 = 1 8 B (123), a = 12 = 2 3 and b = 8 = 2 2 From b 2 = c 2 a 2, 8 = c 2 12, so that c = ± 2 = ±2 5 Since the transverse ais lies along the -ais, the vertices are at (, ±2 3) and the foci at (, ±2 5) Using a = 2 3 and b = 2 2, we draw the auiliar rectangle and sketch the curve: 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

11 THE HYPERBOLA 53 2 1 Equation: 6 2 6 = 1 B (122), a = 6 and b = 6, transverse ais horizontal From b 2 = c 2 a 2, 6 = c 2 6 and c = ± 12 = ±2 3 So the vertices are at ( ± 6, ) and the foci at ( ±2 3, ) Using a = 6 and b = 6, we draw the auiliar rectangle and sketch the curve 11 3 2 2 2 = 6 or 2 2 2 3 = 1 B (123) the transverse ais is vertical with a = 2 and b = 3 From b 2 = c 2 a 2, 3 = c 2 2 or c = ± 5 So the vertices are at (, ± 2) and the foci at (, ± 5) Using a = 2 and b = 3, we draw the auiliar rectangle and sketch the curve 12 Equation:11 2 7 2 = 77 or 2 7 2 11 = 1 B (122), a = 7 and b = 11, transverse ais horizontal From b 2 = c 2 a 2, 11 = c 2 7 and c = ± 18 = ±3 2 So the vertices are at ( ± 7, ) and the foci at ( ±3 2, ) Using a = 7 and b = 11, we draw the auiliar rectangle and sketch the curve 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

54 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 13 Since the foci (and hence the vertices) lie on the -ais, the transverse ais is horizontal B (122), 2 a 2 2 b 2 = 1 Since the length of the transverse ais is 4, a = 2, and since the length of the conjugate ais is 2, b = 1 It follows that 2 4 2 1 = 1 and 2 4 2 = 4 14 Since the foci (and hence the vertices) lie on the -ais, the transverse ais is vertical B (123), 2 a 2 2 = 1 Since the length of the transverse ais is 4, a = 2, and since the length b2 of the conjugate ais is 8, b = 4 It follows that 2 4 2 16 = 1 or 42 2 = 16 15 Since the foci (and hence the vertices) lie on the -ais, the form is 2 a 2 2 b 2 = 1 Since the length of the transverse ais is 8, a = 4, while c = 6 From b 2 = c 2 a 2, we get b 2 = 36 16 = 2 So the equation is 2 16 2 2 = 1 16 Since the foci (and hence the vertices) lie on the -ais, the transverse ais is horizontal and the form is 2 a 2 2 b 2 = 1 Since the length of the transverse ais is 6, a = 3, while c = 4 From b 2 = c 2 a 2, we get b 2 = 16 9 = 7 So the equation is 17 Since the vertices lie on the -ais, the form is 2 9 2 7 = 1 or 72 9 2 = 63 2 a 2 2 b 2 = 1 The conjugate ais of length 8 implies that b = 4 and the position of the vertices impl that a = 4 Equation: 2 16 2 16 = 1 or 2 2 = 16 18 Since the vertices are on the -ais, the form is 2 a 2 2 b 2 = 1 Since a = 5 and b = 6, the equation is 2 25 2 36 = 1 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

11 THE HYPERBOLA 55 19 Since the vertices are on the -ais, the form is 2 a 2 2 b 2 = 1 Since a = 5 and c = 7, we get b 2 = c 2 a 2 = 49 25 = 24 Thus 2 25 2 24 = 1 2 Since the foci lie on the -ais, the transverse ais is vertical B (123) 2 a 2 2 b 2 = 1 Since the length of the conjugate ais is 8, b = 4 The foci are at (, ±5), so that c = 5 From b 2 = c 2 a 2, we have 16 = 25 a 2 and a 2 = 9 Equation: 21 Since the foci are on the -ais, the form is 2 9 2 16 = 1 or 162 9 2 = 144 2 a 2 2 b 2 = 1 Since the length of the conjugate ais is 1, b = 5, while c = 6 From b 2 = c 2 a 2, we get a 2 = c 2 b 2 = 36 25 = 11 So the equation is 22 Start with Eq (119): ( c)2 + 2 ( + c) 2 + 2 = ±2a ( c)2 + 2 = ±2a + ( + c) 2 + 2 2 11 2 25 = 1 isolating the radical ( c) 2 + 2 = 4a 2 ± 4a ( + c) 2 + 2 + ( + c) 2 + 2 squaring both sides 2 2c + c 2 + 2 = 4a 2 ± 4a ( + c) 2 + 2 + 2 + 2c + c 2 + 2 4c 4a 2 = ±4a ( + c) 2 + 2 c + a 2 = ±a ( + c) 2 + 2 c 2 2 + 2ca 2 + a 4 = a 2 ( 2 + 2c + c 2 + 2) collecting terms squaring both sides c 2 2 + 2ca 2 + a 4 = a 2 2 + 2ca 2 + a 2 c 2 + a 2 2 c 2 2 a 2 2 a 2 2 = a 2 c 2 a 4 collecting terms ( c 2 a 2) 2 a 2 2 = a 2 ( c 2 a 2) factoring 2 a 2 2 c 2 a 2 = 1 dividing b ( a2 c 2 a 2) 23 B the original derivation of the equation of the hperbola, 2a = 6 and a = 3 Since (, ±5) are the foci, c = 5 Thus b 2 = 25 9 = 16 B (123) 2 a 2 2 b 2 = 1 2 9 2 16 Not = 1 or 16 2 For 9 2 = 144 Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

56 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 24 Since the foci are at (±3,), we know that c = 3 and the transverse ais is horizontal From Eq (121), = ± b a = ± 4 3 and b a = 4 3 So we get b = 4 3 a; from b2 = c 2 a 2, we get 16 9 a2 = 9 a 2 25 9 a2 = 9 a 2 = 81 25 and b2 = 16 9 81 25 = 9 16 = 144 25 25 2 B (122), 81/25 2 252 = 1 or 144/25 81 252 144 = 1 25 B (123), the equation has the form 2 a 2 2 = 1 Since a = 12, we have b2 26 2 144 2 b 2 = 1 To find b, we substitute the coordinates of ( 1, 13) in the last equation: 169 144 1 b 2 = 1 1 b 2 = 144 144 169 144 = 25 144 Thus b 2 = 144 The equation is 25 27 pv = k 2 144 2 144/25 = 1 or 2 144 252 144 = 1 (12)(3) = k V = 3 m 3, p = 12 Pa So pv = 36 (See graph in answer section of book) 28 b 2 = c 2 a 2 33 = c 2 16 c = ±7 So the reflected ra crosses the -ais at = 7 111 Translation of Aes; Standard Equations of the Conics 1 Circle, center at (1, 2), r = 3 2 The equation ma be viewed as the ellipse 2 9 + 2 5 with center at the origin to the center at (1,2) = 1 rigidl translated from its position 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

111 TRANSLATION OF AXES; STANDARD EQUATIONS OF THE CONICS 57 3 ( + 3) 2 = 8( 2) ( + 3) 2 = 4(2)( 2) p = +2 Verte at (2, 3), focus at (2 + 2, 3) = (4, 3) 4 The equation ma be viewed as the hperbola 2 4 2 = 1 rigidl translated from its position 9 with center at the origin to center at (3,) 5 2 2 3 2 + 8 12 + 14 = 2 2 + 8 3 2 12 + 14 = 2( 2 + 4 ) 3( 2 + 4 ) = 14 factoring 2 and 3 ( ) 2 1 Note that the square of one-half the coefficient of and is 2 4 = 4 Inserting these values inside the parentheses and balancing the equation, we get 2( 2 + 4 + 4) 3( 2 + 4 + 4) = 14 + 2 4 3 4 2( + 2) 2 3( + 2) 2 = 18 3( + 2) 2 2( + 2)2 = 1 18 18 ( + 2) 2 ( + 2)2 = 1 6 9 The equation represents a hperbola with transverse ais vertical Center: ( 2, 2), a = Not 6, b = 3 For Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

58 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 6 4 2 4 48 + 193 = 4 2 4 = 48 193 2 = 12 193 [ ] 4 2 1 Adding 2 ( 1) = 1 to each side, 4 dividing b 4 2 + 1 = 12 193 4 4 + 1 = 12 48 4 ( 1 2 = 12( 4) 2) ( 1 2) 2 = 4(3)( 4) p = 3 This is the equation of a parabola with verte at ( ) ( ) 1 1 2, 4 + 3 = 2, 7 ( ) 1 2, 4 Since p = 3, the focus is at 7 16 2 + 4 2 + 64 12 + 57 = 16 2 + 64 + 4 2 12 + 57 = 16( 2 + 4 ) + 4( 2 3 ) = 57 factoring 16 and 4 Note that ( ) 2 [ ] 2 1 1 2 4 = 4 and 2 ( 3) = 9 4 Inserting these values inside the parentheses and balancing the equation, we get 16( 2 + 4 + 4) + 4( 2 3 + 9 4 ) = 57 + 16 4 + 4( 9 4 ) 16( + 2) 2 + 4( 3 2 )2 = 16 ( + 2) 2 ( 3/2)2 + 1 4 = 1 The equation represents an ellipse with major ais vertical Center: ( 2, 3 2 ), a = 2, b = 1 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

111 TRANSLATION OF AXES; STANDARD EQUATIONS OF THE CONICS 59 8 2 12 5 + 41 = 2 12 = 5 41 [ ] 2 1 Adding 2 ( 12) = 36 to each side, 2 12 + 36 = 5 41 + 36 ( 6) 2 = 5 5 = 5( 1) factoring ( 6) 2 = 4 5 4 ( 1) p = 5 4 This is a parabola with verte at (1, 6) and focus at (1 + 54 ) ( ) 9, 6 = 4, 6 9 2 + 2 + 2 2 + 2 = 2 + 2 + 2 2 = 2 ( 2 + 2 + 1) + ( 2 2 + 1) = 2 + 1 + 1 ( + 1) 2 + ( 1) 2 = Point: ( 1, 1) 1 2 + 2 2 6 + 4 + 1 = 2 6 + 2 2 + 4 = 1 2 6 + 2( 2 + 2 ) = 1 factoring [ ] 2 [ ] 2 1 1 Add 2 ( 6) = 9 and 2 (2) = 1 to each side (inside the parentheses): 2 6 + 9 + 2( 2 + 2 + 1) = 1 + 9 + 2(1) ( 3) 2 + 2( 1) 2 = 1 ( 3) 2 ( + 1)2 5 Not For + Sale = 1 1 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

6 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY Ellipse, center at (3, 1) with a = 1 and b = 5 11 2 2 12 2 + 6 63 = 2 2 12( 2 5 ) = 63 The square of one-half the coefficient of is parentheses, we get 2 2 12( 2 5 + 25 4 2 6( 2 5 + 25 4 ) = 6 2 6( 5 2 )2 = 6 ) = 63 12( 25 4 ) = 12 [ ] 2 1 2 ( 5) = 25 Inserting this number inside the 4 ( 5 2 )2 2 = 1 6 Hperbola, center at (, 5 2 ), transverse ais vertical with a = 1 and b = 6 12 2 2 + 3 2 8 18 + 35 = 2 2 8 + 3 2 18 = 35 2( 2 4 ) + 3( 2 6 ) = 35 [ ] 2 [ ] 2 1 1 Observe that 2 ( 4) = 4 and 2 ( 6) = 9 Inserting these values inside the parentheses and balancing the equation, we get 2( 2 4 + 4) + 3( 2 6 + 9) = 35 + 2(4) + 3(9) 2( 2) 2 + 3( 3) 2 = Single point (2, 3) 13 64 2 + 64 2 16 96 27 = 64 2 16 + 64 2 96 27 = 64( 2 4 ) + 64( 2 3 2 ) = 27 64( 2 4 + 1 64 ) + 64(2 3 2 + 9 16 ) = 27 + 1 + 36 64( 1 8 )2 + 64( 3 4 )2 = 64 ( 1 8 )2 + ( 3 4 )2 = 1 ( 1 Circle of radius 1 centered at 8, 3 ) 4 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

111 TRANSLATION OF AXES; STANDARD EQUATIONS OF THE CONICS 61 14 4 2 4 16 + 5 = 4 2 4 = 16 5 2 = 4 5 [ ] 4 2 1 Add 2 ( 1) = 1 to each side: 4 dividing b 4 Parabola with verte at 2 + 1 4 ( 1 2 = 4 2) 15 3 2 + 2 18 + 2 + 29 = = 4 5 4 + 1 4 = 4 1 ( 1 ) factoring 4 ( 1 2 ( = 4(1) 2) 1 ) 4 p = 1 ( 1 2, 1 ) ( 1 and focus at 4 2, 1 ) ( 1 4 + 1 = 2, 5 ) 4 3 2 18 + 2 + 2 = 29 3( 2 6 ) + ( 2 + 2 ) = 29 [ ] 2 ( ) 2 1 1 Observe that 2 ( 6) = 9 and 2 2 = 1 Adding these values inside the parentheses and balancing the equation, we get 3( 2 6 + 9) + ( 2 + 2 + 1) = 29 + 3 9 + 1 3( 3) 2 + ( + 1) 2 = 1 which is an imaginar locus 16 1 2 18 1 + 81 = Add 1 2 18 = 1 81 2 9 5 = 81 [ ( 1 1 9 )] 2 = 81 to each side: 2 5 1 2 9 5 + 81 = 1 ( 9 ) 2 = 4 1 1 4 p = 1 4 ( ) ( 9 9 1,, focus at 1, 1 ) 4 Parabola, Not verte at For Sale 213 Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

62 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 17 2 + 2 12 + 25 = 2 + 2 = 12 25 We add to each side of the equation the square of one-half the coefficient of, 2 + 2 + 1 = 12 25 + 1 ( + 1) 2 = 12 24 ( + 1) 2 = 12( 2) ( + 1) 2 = 4 3( 2) p = +3 Verte at ( 1, 2), focus at ( 1, 5) 18 2 2 2 8 2 + 3 = 2 2 8 2 2 = 3 [ ] 2 1 2 2 = 1: 2( 2 4 ) ( 2 + 2 ) = 3 factoring [ ] 2 [ ] 2 1 1 Observe that 2 ( 4) = 4 and 2 (2) = 1 Inserting these values inside the parentheses and balancing the equation, we get 2( 2 4 + 4) ( 2 + 2 + 1) = 3 + 2(4) 1 2( 2) 2 ( + 1) 2 = 4 ( 2) 2 2 ( + 1)2 4 = 1 dividing b 4 Hperbola, center at (2, 1) 213 Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part