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

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

2 213 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: ISBN-1: 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, For permission to use material from this tet or product, submit all requests online at wwwcengagecom/permissions Further permissions questions can be ed 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

3 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 The Circle The Parabola The Ellipse 4 11 The Hperbola Translation of Aes; Standard Equations of the Conics 56 Chapter 1 Review 72 2 Introduction to Calculus: The Derivative Functions and Intervals Limits The Derivative b the Four-Step Process Derivatives of Polnomials Instantaneous Rates of Change Differentiation Formulas Implicit Differentiation Higher Derivatives 119 Chapter 2 Review Applications of the Derivative The First-Derivative Test The Second-Derivative Test Eploring with Graphing Utilities Applications of Minima and Maima Related Rates 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

4 iv CONTENTS 4 The Integral Antiderivatives The Area Problem The Fundamental Theorem of Calculus Basic Integration Formulas Area Between Curves Improper Integrals The Constant of Integration Numerical Integration 232 Chapter 4 Review Applications of the Integral Means and Root Mean Squares Volumes of Revolution: Disk and Washer Methods Volumes of Revolution: Shell Method Centroids Moments of Inertia Work and Fluid Pressure 31 Chapter 5 Review Derivatives of Transcendental Functions Review of Trigonometr Derivatives of Sine and Cosine Functions Other Trigonometric Functions Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Eponential and Logarithmic Functions Derivative of the Logarithmic Function Derivative of the Eponential Function L Hospital s Rule Applications Newton s Method 373 Chapter 6 Review Integration Techniques The Power Formula Again The Logarithmic and Eponential Forms Trigonometric Forms Further Trigonometric Forms Inverse Trigonometric Forms Integration b Trigonometric Substitution Integration b Parts Integration of Rational Functions Integration b Use of Tables 439 Chapter 7 Review Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

5 CONTENTS v 8 Parametric Equations, Vectors, and Polar Coordinates Vectors and Parametric Equations Arc Length Polar Coordinates Curves in Polar Coordinates Areas in Polar Coordinates; Tangents 476 Chapter 8 Review Three Dim Space; Partial Derivatives; Multiple Integrals Surfaces in Three Dimensions Partial Derivatives Applications of Partial Derivatives Iterated Integrals Volumes b Double Integration Mass, Centroids, and Moments of Inertia Volumes in Clindrical Coordinates 558 Chapter 9 Review Infinite Series Introduction to Infinite Series Tests for Convergence Maclaurin Series Operations with Series Computations with Series; Applications Fourier Series 61 Chapter 1 Review First-Order Differential Equations What is a Differential Equation? Separation of Variables First-Order Linear Differential Equations Applications of First-Order Differential Equations Numerical Solutions 662 Chapter 11 Review Higher-Order Linear Differential Equations Higher-Order Homogeneous Differential Equations Auiliar Equations with Repeating or Comple Roots Nonhomogeneous Equations Applications of Second-Order Equations 74 Chapter 12 Review The Laplace Transform 727 Sections Solution of Linear Equations b Laplace Transforms 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

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7 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) = = 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 = = 1 3 Let ( 2, 2 ) = ( 3, 6) and ( 1, 1 ) = (5, 2) Then d = ( 3 5) 2 + [ 6 ( 2)] 2 = ( 8) 2 + ( 4) 2 = = 16 5 = 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 = = 7 6 d = ( 2 2) 2 +( 5 ) 2 = +5 = 5 7 d = [1 ( 1)] 2 + ( 2 ) 2 = = 6 8 d = [2 ( 2)] 2 + (7 3) 2 = = 2 16 = d = [ 9 ( 11)] 2 + ( 1 1) 2 = ( 2) 2 = 8 = 2 4 = Distance from (, ) to (4, 3): (4 ) 2 +(3 ) 2 = = 5 Distance from (, ) to (6, )= 6 Distance from (4, 3) to (6, ): (6 4) 2 +( 3) 2 = = 13 Perimeter = = 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

8 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 = a 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): Distance from ( 1, 1) to (2, 8): Distance from (2, 8) to (5, 17): ( 1 2)2 + ( 1 8) 2 = = 9 = 9 1 = 3 1 Distance from ( 1, 1) to (5, 17): (5 2)2 + (17 8) 2 = 9 = = 36 = 6 1 Total distance 6 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 = = 4 21 Distance from (, ) to -ais: units Distance from (, ) to (2, ): ( 2) 2 + ( ) 2 = ( 2) B assumption, ( 2)2 + 2 = ( 2) = 2 squaring both sides = = 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

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

10 4 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 6 m = 4 8 = m = 43 ( 1) m = 31 1 ( 1) = 3 = 44 2 = 22 9 m = = 9 (undefined) 1 m = ( 3) = 8 = 11 m = 3 ( 3) m 3 ( 2) 4 4 = 4 = 13 m = ( 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 = 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) = Slope of line through ( 4, 6) and ( 1, 3): 1 ( 4) = 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 Slope of line through ( 4, 6) and (6, 1): 4 6 = 4 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 ; 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 = Slope of line through ( 2, 3) and (7, 6): 2 7 = 3 9 = Slope of line through (, 3) and (9, ): 9 = 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 13 THE STRAIGHT LINE 5 Since 3 and 1 3 are negative reciprocals, adjacent sides are perpendicular and opposite sides are parallel ( Midpoint of line segment:, ) = (3, 2) Point: (6, 4), m = = 2 ( Midpoint:, 2 + ) = (3, 1) ( 1) Slope of line through (5, 6) and (3, 1): = Let (, ) be the other end of the diameter Since the center is the midpoint, we get and 3 = 2; solving, = 2, = tan θ = rise run θ = 36 = 1 ft 16 ft = tan θ = rise run = 25 m 1316; θ = m 1 ( 5) 27 Slope of line through ( 1, 1) and (3, 5): = = Slope of line through (, 2) and (4, 6): 4 = 8 4 Since the two slopes must be equal, we have: 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: ; so 5 4 = 5 (negative reciprocals); solving, = 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 = = = 1 2 Since ( 1, 1 ) = (, 3) and m = 4, we get 3 = 4( ) 1 = m( 1) = 3 1 = m( 1 ) + 4 = 3( 3) ( 1, 1) = (3, 4); m = = = 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

12 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 5 Let ( 1, 1 ) = ( 3, 2), then 2 = 2 5 ( + 3) 5 1 = multipling b = 9 First determine the slope using m = 2 1 to get m = 4 ( 6) = 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 = 1 m = = = 1 2 ( 2) ( 1, 1 ) = (2, 3) 2 6 = + 2 multipling b = 11 m = = 1 = 1( 5) choosing ( 1, 1) = (5, ) + 5 = 12 m = = = 1 2 ( + 3) ( 1, 1 ) = ( 3, 5) 2 1 = + 3 multipling b = = 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 = , we get m = 4 and b = 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 13 THE STRAIGHT LINE = = = m + b m =, -intercept = 7 2 ; see graph in answer section of book 18 Solving for : = slope: 1 4 ; -intercept : = = 3 = = 4 3 = = = From the form = m + b, m = 2 3 in both cases, so that the lines are parallel = 2 = 2 = = slope = 1 2 slope = 2 Answer: The lines are perpendicular = = 3 4 = = = = The lines are neither parallel nor perpendicular = 6 4 = = = 4 slope = 7 1 slope = Answer: neither = = 3 = + 5 = = The slopes are 1 3 and 3, respectivel Since the slopes are negative reciprocals, the lines are perpendicular = = 1 Slope is 2 5 = = in each case; the lines are parallel = = 4 5 = = = = = From the form = m + b, the slope m is 3 5 in both cases; so the lines are parallel = = 4 = = 6 1 = = = 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

14 8 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY = = 5 2 = = = = = The respective slopes are 3 2 and 2 3 Since the slopes are negative reciprocals, the lines are perpendicular = = 4 = = = = The respective slopes are 1 4 and 4 Since the slopes are negative reciprocals, the lines are perpendicular = = 3 = = = = The lines are neither parallel nor pependicular = = 4 = = = = The lines are neither parallel nor perpendicular = 5 (given line) = slope = = m( 1 ) point-slope form 1 = 3 4 ( + 2) point: ( 2, 1) 4 4 = = 32 The given line can be written = So the slope is 3 4 Slope of perpendicular: = 4 3 ( + 2) 1 = m( 1 ) 3 3 = = 33 To find the coordinates of the point of intersection, solve the equations simultaneousl: 2 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 =, we get 7 = 5 3 = 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

15 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 = = 34 The first two lines are perpendicular: = and = 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 = 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 C = t 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 P = k; let P = 1872 lb and = 3 ft Then So the relationship is P = = k(3) and k = = Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

16 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 = = 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 solving for = ±1 = 1 and = 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 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

18 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; 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 14 CURVE SKETCHING 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

20 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

21 14 CURVE SKETCHING 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 = = 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

22 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

23 14 CURVE SKETCHING = 1 4 ; none; none; = 2; = ; 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; Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

24 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 = 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 = Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

25 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; ± Intercepts If =, then 2 = 4 1 = 4, or = ±2 If =, then = 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 = 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

26 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)( + 2) ( 1)( + 1) > < < 2 3/ < < < < 1 3/2 + < 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

27 14 CURVE SKETCHING 21 test values ( 3)( + 5) > < < 3 + < 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 < < < < 3 5/2 + + > 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 =

28 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 , C 1 The onl intercept is the origin Dividing numerator and denominator b C 1, the equation becomes C = /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

29 15 CURVES WITH GRAPHING UTILITIES 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

30 24 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 4 2, 1,, 1 [ 3, 3] b [ 5, 5] = 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

31 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 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

32 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

33 17 THE CIRCLE m [, 2] b [, 5] 21 See graph in answer section of book [, 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 = 16 2 Since (h, k) = (, ) and r = 8, we get from the form ( h) 2 + ( k) 2 = r 2 the equation = 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: (, ) = 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: = 17, (h, k) = (, ) 5 ( h) 2 + ( k) 2 = r 2 ( + 2) 2 + ( 5) 2 = 1 2 Not = 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

34 28 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 6 ( h) 2 + ( k) 2 = r 2 ( 2) 2 + ( + 3) 2 = ( 2 ) = = 7 The radius is the distance from ( 1, 4) to the origin: r 2 = ( 1 ) 2 + ( 4 ) 2 = = 17 Hence, ( + 1) 2 + ( + 4) 2 = 17 ( h) 2 + ( k) 2 = r = = 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 = = = 9 Diameter: distance from ( 2, 6) to (1, 5) Hence r = 1 2 ( 2 1)2 + ( 6 5) 2 = = and thus, r 2 = 1 65 (13) = 4 2 Center: midpoint of the line segment, whose coordinates are ( Thus ( )2 + ( )2 = = = 1 Distance from ( 1, 2) to the -ais: 1 2, ) = ( 1 2, 1 2 ( + 1) 2 + ( 2) 2 = 1 ) 11 Since r = 5, we get ( 4) 2 + ( + 5) 2 = 25 or = 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

35 17 THE CIRCLE Distance to the line = 1: 3 ( 3) 2 + ( 4) 2 = 9 13 From the diagram, r 2 = = 2; so = 2 r 1 + = 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 = ( 2) 2 + ( 3) 2 = 2 2 and ( + 2) 2 + ( + 3) 2 = = and = = 2 We now add to each side the square of one-half the coefficient of : [ 1 2 ( 2)] 2 = = Similarl, we add 1 (the square of one-half the coefficient of ): ( ) + ( ) = ( 1) 2 + ( 1) 2 = 4 Center: (h, k) = (1, 1); radius: 4 = = = 4 We now add to each side the square of one-half the coefficient of : [ 1 2 ( 2)] 2 = 1: = Similarl, we add to each side the square of one-half the coefficient of : [ 1 2 ( 4)] 2 = = ( 1) 2 + ( 2) 2 = 1; Center: (h, k) = (1, 2); radius: 1 = = = 4 Since ( ) = 4 and we get ( ) + ( ) = [ ] ( 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

36 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 = = = 3 Observe that [ 1 2 (2)] 2 = 1 and [ 1 2 (6)] 2 = 9 Adding 1 and 9 to each side, we get = or ( + 1) 2 + ( + 3) 2 = 7 [ ( 1)] 2 + [ ( 3)] 2 = 7; Center: ( 1, 3); radius: = = 9 4 We add to each side the square of one-half the coefficient of : = [ ] ( 4) = 4 This gives [ ] 2 1 Similarl, we add the square of one-half the coefficient of : 2 1 = 1 This gives = ( 2) 2 + ( )2 = ( 2) 2 + ( )2 = 2 Center: (2, 1 2 ); radius: = = 5 4 Add to each side: [ 1 2 (1)] 2 = 1 4 and [ 1 2 ( 4)] 2 = = ( ( 2) 2) 2 = 3; Center: ( 12 ), 2 ; radius: = = dividing b = 9 4 [ ] 2 1 Add to each side: 2 ( 2) = 1 and [ ] ( 3) = 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

37 17 THE CIRCLE = First divide both sides b 4: = = 25 4 Add to each side: [ 1 2 ( 5)] 2 = 25 4 and [ 1 2 ( 2)] 2 = = ( ) ( ) + ( 1) 2 = 1; Center: 2, 1 ; radius: = = 4 Note that ( ) = 4 and Adding 4 and 1, respectivel, we get ( ) + ( ) = ( + 2) 2 + ( 1) 2 = 9 The equation can be written So the center is ( 2, 1) and the radius is = = 1 [ ] ( 2) = 1 [ ( 2)] 2 + ( 1) 2 = 3 2 Add to each side: [ 1 2 (2)] 2 = 1 and [ 1 2 (8)] 2 = = ( + 1) 2 + ( + 4) 2 = 16; Center: ( 1, 4); radius: = = 1 4 [ ] 2 1 Add to each side: 2 ( 1) = 1 4 and ( ) + ( ) = ( 1 2 )2 + ( 1) 2 = 1 Center: ( 1 2, 1); radius: 1 [ ] ( 2) = 1 to get = = 19 Add to each side: [ 1 2 ( 6)] 2 = 9 and [ 1 2 ( 8)] 2 = = ( 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

38 32 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY = = 9 4 Note that [ ] ( 4) = 4 and Adding 4 and 1 4, respectivel, we get ( ) = 1 4 ( ) + ( ) = ( 2) 2 + ( )2 = 2 The equation can be written [ ( ( 2) = ( 2)] 2) 2 Center: (2, 1 2 ); radius: = = 1 2 Add to each side: [ 1 2 (1)] 2 = 1 4 and [ 1 2 ( 1)] 2 = = ( ( ) ( ) = 1; Center: 1 2, 1 ) ; radius: = = = 5 4 [ ] 2 1 Add to each side: 2 3 = 9 ( ) and 2 4 = 4 to get = ( )2 + ( + 2) 2 = 5 Center: ( 3 2, 2); radius: = First divide both sides b 36: = = Add to each side: [ 1 2 ( 4)] 2 [ = 4 and 1 ( )] = = = ( ( 2) ) 2 = 25 ( 3 36 ; Center 2, 5 ) 5 ; radius: = = dividing b = = ( 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

39 17 THE CIRCLE = = = ( + 2) 2 + ( 1) 2 = 2 Imaginar circle = = 25 ( ) + ( ) = ( 3) 2 + ( + 4) 2 = Locus is the single point (3, 4) = = = ( + 1) 2 + ( + 2) 2 = Point circle: ( 1, 2) = = = ( 3) 2 + ( 4) 2 = 5 (imaginar circle) = = = ( 3) 2 + ( + 2) 2 = Point circle: (3, 2) = = 17 4 [ ] 2 1 We add to each side 2 ( 1) and [ ] (4) : = ( 1 2) 2 + ( + 2) 2 = (point circle) 38 From the given circle, we have = = ( 3) 2 + ( 2) 2 = 25 Center: (3, 2); Desired circle: ( 3) 2 + ( 2) 2 = = (2) 2 = 4; = (34) 2 = Center: (5,5); radius: 21 ( 5) 2 + ( 5) 2 = (21) = = 41 The Not radius is 22, 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

40 34 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY 43 (h, k) = (, ) and r = 3 2 ft: = 9 4 and = 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

41 18 THE PARABOLA 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 = = 4 2 = 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 = or p = 9 2 The equations are 2 = 25 3 and 2 = 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) = 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

42 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

43 18 THE PARABOLA = 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 = = 2 = = 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: = (, 3 ), directri: = ( 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

44 38 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY Circle: ( h) 2 + ( k) 2 = r 2 ( ) 2 + ( 3) 2 = = = 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) p + 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 = = = 36 From the figure, d 1 = d 2 : ( 4)2 + ( 7) 2 = + 1 ( 4) 2 + ( 7) 2 = ( + 1) 2 (squaring both sides) = = 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 = = The equation is therefore 2 = Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

45 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 = 263 m 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 (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 = Equation: 2 = When = 25, 2 = ( 25) = = 2 5 = = 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

46 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 = = 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 = = 4p = The Ellipse Diameter= 2 = = 18 in 1 The equation is = 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 = Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

47 19 THE ELLIPSE = 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 = 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 = Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

48 42 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY = 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 = 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 = = B (117), a = 2 and b = 2, major ais vertical b 2 = a 2 c 2 2 = 4 c 2 c = ± Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

49 19 THE ELLIPSE 43 Vertices:(, ± 2), foci: (, ± 2 ), semiminor ais: = = 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 = = = 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: = = = 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

50 44 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY = = 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: = = 1 major ais vertical 5 Vertices: (, ± 5), foci: (, ±2) Length of semi-minor ais: b = 1 (See sketch in answer section of book) = 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 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) = = 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: Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

51 19 THE ELLIPSE = = 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 = = = 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: = = 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 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 = ± ( ) 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 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 = 1 24 Since the foci are at (, ±3), the major ais is vertical 2 b 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 = 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

52 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 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 = 1 or = 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 = 1 we get a = 1 27 Since the foci are at (, ±3), c = 3, and the major ais is vertical B (117) 2 b 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: = 1 or = 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) = 1 29 Since the vertices and foci are on the -ais, the form of the equation is, b (117), 2 b a 2 = 1 From b 2 = a 2 c 2 with a = 8 and c = 5, b 2 = = 39 Hence = 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: = 1 or = 4 2 a 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: = 1 or = 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), = 1 or = Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

53 19 THE ELLIPSE From the form we get (since b = 2) 2 a b 2 = 1 a = 1 Substituting the coordinates of the point (3, 1) ields So Equation: 2 9 a = 1 and 9 a 2 = 3 4 a 2 9 = 4 3 and a2 = = 1 or = Since b = 3, we get from (117), = 1 Since (1,4) lies on the curve, the values = 1 a2 and = 4 satisfies the equation: a 2 = 1 whence a2 = 18 and the final result is = 1 or = 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 = = 28 B (116) the equation is = 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), = 1 or = = 45 or = 1; a = 3; b = 5 From b 2 = a 2 c 2, 5 = 9 c 2 and c = 2 Thus e = c a = Distance from (, ) to (, ): ( ) 2 + ( ) 2 = Distance from (, ) to (3, ): ( 3) 2 + ( ) 2 = ( 3) From the given condition: = 2 ( 3) = 4[( 3) ] squaring both sides = 4( ) = = = Not = 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

54 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 = = 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 = = 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

55 19 THE ELLIPSE 49 In our problem 42 A = π a b = π ft 2 43 Since a = 2 and b = 3 2, we get 2 e = = 967 and = = 1 or = 1, 44 From the figure, a = 15 and b = 5 From (117) = 1 The net step is to find when = 1: = = and = So = and the length of the beam is 2 = ft 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 b 2 = (4,4) b 2 = 1 To find b, we substitute the coordinates of (4, 4) in the equation: b 2 = 1 16 b 2 = = 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

56 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 = c a or c = 99992a So a c = a 99992a = 12; solving, a = 15 and c = Farthest point: ( )(1) = mi 11 The Hperbola 1 Comparing the given equation, = 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 = 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 = 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

57 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: = 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: = 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

58 52 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY = 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: = 18 or = 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 = = = 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

59 11 THE HYPERBOLA Equation: = 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 = 6 or = 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: = 77 or = 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

60 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 = 1 and = 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 = 1 or 42 2 = 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 = = 2 So the equation is = 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 = 1 or = 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: = 1 or 2 2 = 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 = Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

61 11 THE HYPERBOLA 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 = = 24 Thus = 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 = 1 or = 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 = = 11 So the equation is 22 Start with Eq (119): ( c)2 + 2 ( + c) = ±2a ( c)2 + 2 = ±2a + ( + c) = 1 isolating the radical ( c) = 4a 2 ± 4a ( + c) ( + c) squaring both sides 2 2c + c = 4a 2 ± 4a ( + c) c + c c 4a 2 = ±4a ( + c) c + a 2 = ±a ( + c) c ca 2 + a 4 = a 2 ( 2 + 2c + c 2 + 2) collecting terms squaring both sides c ca 2 + a 4 = a ca 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 = 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

62 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 a2 = 9 a 2 = and b2 = = 9 16 = B (122), 81/ = 1 or 144/ = 1 25 B (123), the equation has the form 2 a 2 2 = 1 Since a = 12, we have b b 2 = 1 To find b, we substitute the coordinates of ( 1, 13) in the last equation: b 2 = 1 1 b 2 = = Thus b 2 = 144 The equation is pv = k /25 = 1 or = 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 = 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 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

63 111 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 = 1 rigidl translated from its position 9 with center at the origin to center at (3,) = = 2( ) 3( ) = 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( ) 3( ) = ( + 2) 2 3( + 2) 2 = 18 3( + 2) 2 2( + 2)2 = ( + 2) 2 ( + 2)2 = 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

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

65 111 TRANSLATION OF AXES; STANDARD EQUATIONS OF THE CONICS = 2 12 = 5 41 [ ] 2 1 Adding 2 ( 12) = 36 to each side, = ( 6) 2 = 5 5 = 5( 1) factoring ( 6) 2 = ( 1) p = 5 4 This is a parabola with verte at (1, 6) and focus at ( ) ( ) 9, 6 = 4, = = 2 ( ) + ( ) = ( + 1) 2 + ( 1) 2 = Point: ( 1, 1) = = ( ) = 1 factoring [ ] 2 [ ] Add 2 ( 6) = 9 and 2 (2) = 1 to each side (inside the parentheses): ( ) = (1) ( 3) 2 + 2( 1) 2 = 1 ( 3) 2 ( + 1)2 5 Not For + Sale = Cengage Learning All Rights Reserved Ma not be scanned, copied duplicated, or posted to a publicl accessible website, in whole or in part

66 6 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY Ellipse, center at (3, 1) with a = 1 and b = = ( 2 5 ) = 63 The square of one-half the coefficient of is parentheses, we get ( ( ) = 6 2 6( 5 2 )2 = 6 ) = 63 12( 25 4 ) = 12 [ ] ( 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 = = = 35 2( 2 4 ) + 3( 2 6 ) = 35 [ ] 2 [ ] Observe that 2 ( 4) = 4 and 2 ( 6) = 9 Inserting these values inside the parentheses and balancing the equation, we get 2( ) + 3( ) = (4) + 3(9) 2( 2) 2 + 3( 3) 2 = Single point (2, 3) = = 64( 2 4 ) + 64( ) = 27 64( ) + 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 ) Cengage Learning All Rights Reserved Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part

67 111 TRANSLATION OF AXES; STANDARD EQUATIONS OF THE CONICS = = = 4 5 [ ] Add 2 ( 1) = 1 to each side: 4 dividing b 4 Parabola with verte at ( 1 2 = 4 2) = = = 4 1 ( 1 ) factoring 4 ( 1 2 ( = 4(1) 2) 1 ) 4 p = 1 ( 1 2, 1 ) ( 1 and focus at 4 2, 1 ) ( = 2, 5 ) = 29 3( 2 6 ) + ( ) = 29 [ ] 2 ( ) Observe that 2 ( 6) = 9 and 2 2 = 1 Adding these values inside the parentheses and balancing the equation, we get 3( ) + ( ) = ( 3) 2 + ( + 1) 2 = 1 which is an imaginar locus = Add = = 81 [ ( )] 2 = 81 to each side: = 1 ( 9 ) 2 = 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

68 62 CHAPTER 1 INTRODUCTION TO ANALYTIC GEOMETRY = = We add to each side of the equation the square of one-half the coefficient of, = ( + 1) 2 = ( + 1) 2 = 12( 2) ( + 1) 2 = 4 3( 2) p = +3 Verte at ( 1, 2), focus at ( 1, 5) = = 3 [ ] = 1: 2( 2 4 ) ( ) = 3 factoring [ ] 2 [ ] Observe that 2 ( 4) = 4 and 2 (2) = 1 Inserting these values inside the parentheses and balancing the equation, we get 2( ) ( ) = 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