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Precalculus
Learner’s Material
Department of Education Republic of the Philippines
This learning resource was collaboratively developed and reviewed by educators from public and private schools, colleges, and/or universities. We encourage teachers and other education stakeholders to email their feedback, comments and recommendations to the Department of Education at [email protected].
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PrecalculusLearner’s Material First Edition 2016
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Table of Contents
To the Precalculus Learners 1
DepEd Curriculum Guide for Precalculus 2
Unit 1: Analytic Geometry 6
Lesson 1.1: Introduction to Conic Sections and Circles . . . 7
1.1.1: An Overview of Conic Sections . . . 7
1.1.2: Definition and Equation of a Circle . . . 8
1.1.3: More Properties of Circles . . . 10
1.1.4: Situational Problems Involving Circles . . . 12
Lesson 1.2: Parabolas . . . 19
1.2.1: Definition and Equation of a Parabola . . . 19
1.2.2: More Properties of Parabolas . . . 23
1.2.3: Situational Problems Involving Parabolas . . . 26
Lesson 1.3: Ellipses . . . 33
1.3.1: Definition and Equation of an Ellipse . . . 33
1.3.2: More Properties of Ellipses . . . 36
1.3.3: Situational Problems Involving Ellipses . . . 40
Lesson 1.4: Hyperbolas . . . 46
1.4.1: Definition and Equation of a Hyperbola . . . 46
1.4.2: More Properties of Hyperbolas . . . 50
1.4.3: Situational Problems Involving Hyperbolas . . . 54
Lesson 1.5: More Problems on Conic Sections . . . 60
1.5.1: Identifying the Conic Section by Inspection . . . 60
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Lesson 1.6: Systems of Nonlinear Equations . . . 67
1.6.1: Review of Techniques in Solving Systems of Linear Equations . . . 68
1.6.2: Solving Systems of Equations Using Substitution . . . 69
1.6.3: Solving Systems of Equations Using Elimination . . . 70
1.6.4: Applications of Systems of Nonlinear Equations . . . 73
Unit 2: Mathematical Induction 80 Lesson 2.1: Review of Sequences and Series . . . 81
Lesson 2.2: Sigma Notation . . . 86
2.2.1: Writing and Evaluating Sums in Sigma Notation . . . 87
2.2.2: Properties of Sigma Notation . . . 89
Lesson 2.3: Principle of Mathematical Induction . . . 96
2.3.1: Proving Summation Identities . . . 97
2.3.2: Proving Divisibility Statements . . . 101
?2.3.3: Proving Inequalities . . . 102
Lesson 2.4: The Binomial Theorem . . . 108
2.4.1: Pascal’s Triangle and the Concept of Combination . . . 109
2.4.2: The Binomial Theorem . . . 111
2.4.3: Terms of a Binomial Expansion . . . 114
?2.4.4: Approximation and Combination Identities . . . 116
Unit 3: Trigonometry 123 Lesson 3.1: Angles in a Unit Circle . . . 124
3.1.1: Angle Measure . . . 124
3.1.2: Coterminal Angles . . . 128
3.1.3: Arc Length and Area of a Sector . . . 129
Lesson 3.2: Circular Functions . . . 135
3.2.1: Circular Functions on Real Numbers . . . 136
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Lesson 3.3: Graphs of Circular Functions and SituationalProblems . . . 144
3.3.1: Graphs of y = sin x and y = cos x . . . 145
3.3.2: Graphs of y = a sin bx and y = a cos bx . . . 147
3.3.3: Graphs of y = a sin b(x − c) + d and y = a cos b(x − c) + d . . . 151
3.3.4: Graphs of Cosecant and Secant Functions . . . 154
3.3.5: Graphs of Tangent and Cotangent Functions . . . 158
3.3.6: Simple Harmonic Motion . . . 160
Lesson 3.4: Fundamental Trigonometric Identities . . . 171
3.4.1: Domain of an Expression or Equation . . . 171
3.4.2: Identity and Conditional Equation . . . 173
3.4.3: The Fundamental Trigonometric Identities . . . 174
3.4.4: Proving Trigonometric Identities . . . 176
Lesson 3.5: Sum and Difference Identities . . . 181
3.5.1: The Cosine Difference and Sum Identities . . . 181
3.5.2: The Cofunction Identities and the Sine Sum and Difference Identities . . . 183
3.5.3: The Tangent Sum and Difference Identities . . . 186
Lesson 3.6: Double-Angle and Half-Angle Identities . . . 192
3.6.1: Double-Angle Identities . . . 192
3.6.2: Half-Angle Identities . . . 195
Lesson 3.7: Inverse Trigonometric Functions . . . 201
3.7.1: Inverse Sine Function . . . 202
3.7.2: Inverse Cosine Function . . . 205
3.7.3: Inverse Tangent Function and the Remaining Inverse Trigonometric Functions . . . 208
Lesson 3.8: Trigonometric Equations . . . 220
3.8.1: Solutions of a Trigonometric Equation . . . 221
3.8.2: Equations with One Term . . . 224
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Lesson 3.9: Polar Coordinate System . . . 236
3.9.1: Polar Coordinates of a Point . . . 237
3.9.2: From Polar to Rectangular, and Vice Versa . . . 241
3.9.3: Basic Polar Graphs and Applications . . . 244
Answers to Odd-Numbered Exercises in Supplementary Problems
and All Exercises in Topic Tests 255
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To the Precalculus LearnersThe Precalculus course bridges basic mathematics and calculus. This course completes your foundational knowledge on algebra, geometry, and trigonometry. It provides you with conceptual understanding and computational skills that are prerequisites for Basic Calculus and future STEM courses.
Based on the Curriculum Guide for Precalculus of the Department of Edu-cation (see pages 2-5), the primary aim of this Learning Manual is to give you an adequate stand-alone material that can be used for the Grade 11 Precalculus course.
The Manual is divided into three units: analytic geometry, summation no-tation and mathematical induction, and trigonometry. Each unit is composed of lessons that bring together related learning competencies in the unit. Each lesson is further divided into sub-lessons that focus on one or two competencies for effective learning.
At the end of each lesson, more examples are given in Solved Examples to reinforce the ideas and skills being developed in the lesson. You have the oppor-tunity to check your understanding of the lesson by solving the Supplementary Problems. Finally, two sets of Topic Test are included to prepare you for the exam.
Answers, solutions, or hints to odd-numbered items in the Supplementary Problems and all items in the Topic Tests are provided at the end of the Manual to guide you while solving them. We hope that you will use this feature of the Manual responsibly.
Some items are marked with a star. A starred sub-lesson means the discussion and accomplishment of the sub-lesson are optional. This will be decided by your teacher. On the other hand, a starred example or exercise means the use of calculator is required.
We hope that you will find this Learning Manual helpful and convenient to use. We encourage you to carefully study this Manual and solve the exercises yourselves with the guidance of your teacher. Although great effort has been put into this Manual for technical correctness and precision, any mistake found and reported to the Team is a gain for other students. Thank you for your cooperation.
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K t o 1 2 B A S IC EDU CA T ION CU R R IC U LU M S ENIOR H IGH S CH O O L – S CI ENCE, TECH NO LO GY, EN GINEER ING A ND MA THE M A TICS (S TEM) S P ECI A LIZED S U B JECT K to 12 Se nio r H ig h Sc ho ol STEM S pe cia liz ed S ub je ct – Pre -C alc ulus De ce m be r 2013 P ag e 1 o f 4 Gra de : 11 S eme st er : Fi rst Sem est er C or e S u b je ct Tit le : P re -C alc ulu s No. of H ou rs/ S eme st er: 8 0 hou rs/ se m est er P re -re q uisi te ( if n ee de d ): Sub je ct Desc ri pt io n : At t he e nd of t he c ou rse , th e st ude nt s m ust be a ble t o apply con ce pt s an d so lv e pr oble m s in vol vin g co nic se ct ion s, sy st em s of n on lin ea r equat ion s, se rie s a nd m at he m at ic al in duc tio n, c irc ula r a nd t rigon ome tric f un ct ion s, tr igon om et ric ide nt itie s, a nd p ola r co ordina te sy st em. C ONT E NT C ONT E NT S TA NDA R DS P ER FO R MAN C E S TA NDA R DS LEARNING C OMP ETE NC IES C OD E Ana lyt ic Geo m et ry Th e le arner s de m on st ra te an un de rst an di ng of ... ke y con ce pt s of con ic se ct io ns a nd sy st em s o f non lin ea r equat ion s Th e le arner s sh all be a ble to.. . m ode l sit ua tion s appr opria te ly a nd sol ve prob le m s ac cu ra te ly u sin g con ic se ct io ns a nd sy st em s of n on lin ea r e quat ion s Th e le arner s... 1. illu st ra te t he dif fe re nt t ype s of c on ic se ct io ns: pa ra bola , ellip se , circle, hy pe rbo la , an d de ge ne ra te case s.* ** S TEM_P C 11 AG -Ia -1 2. de fin e a ci rcle. S TEM_P C 11 AG -Ia -2 3. de te rm in e th e st an da rd f or m of e quat io n of a ci rcle S TEM_P C 11 AG -Ia -3 4. gr aph a ci rcle in a re ct an gu la r coo rdina te sy st em S TEM_P C 11 AG -Ia -4 5. de fin e a pa ra bola S TEM_P C 11 AG -Ia -5 6. de te rm in e th e st an da rd f or m of e quat io n of a pa ra bola S TEM_P C 11 AG -Ib -1 7. gra ph a pa ra bola in a re ct an gular c oordinat e s yst em S TEM_P C 11 AG -Ib -2 8. de fin e an e lli pse S TEM_P C 11 AG -Ic -1 9. de te rm in e th e st an da rd f or m of e quat io n of a n e llipse S TEM_P C 11 AG -Ic -2 10. gr aph a n e llipse in a re ct an gular co ordi na te sy st em S TEM_P C 11 AG -Ic -3 11. de fin e a h ype rbola S TEM_P C 11 AG -Id -1 12. de te rm in e th e st an da rd f or m of e quat io n of a h ype rbo la S TEM_P C 11 AG -Id -2DEPED COPY
K t o 1 2 B A S IC EDU CA T ION CU R R IC U LU M S ENIOR H IGH S CH O O L – S CI ENCE, TECH NO LO GY, EN GINEER ING A ND MA THE M A TICS (S TEM) S P ECI A LIZED S U B JECT K to 12 Se nio r H ig h Sc ho ol STEM S pe cia liz ed S ub je ct – Pre -C alc ulus De ce m be r 2013 P ag e 2 o f 4 C ONT E NT C ONT E NT S TA NDA R DS P ER FO R MAN C E S TA NDA R DS LEARNING C OMP ETE NC IES C OD E 13. gr aph a h ype rbola in a re ct an gular co ordina te sy st em S TEM_P C 11 AG -Id -3 14. re cogn ize t he e quat ion a nd im port an t ch ar ac te rist ic s of t he dif fe re nt t ype s of c on ic se ct ion s S TEM_P C 11 AG -Ie -1 15. solv es sit ua tio na l prob le m s in volv in g con ic se ct io ns S TEM_P C 11 AG -Ie -2 16. illu st ra te sy st em s o f n on lin ea r e quat ion s S TEM_P C 11 AG -If -1 17. de te rm in e th e solu tion s o f sy st em s o f n on lin ea r e quat ion s u sin g te ch niq ue s su ch a s su bst itu tion , elim in at io n, a nd gr aphin g* ** S TEM_P C 11 AG -If -g -1 18. solv e sit ua tio na l prob le m s i nv ol vin g sy st em s of n on lin ea r e quat ion s S TEM_P C 11 AG -Ig -2 S erie s a nd Mat h ema tica l Ind uc ti on ke y c on ce pt s of se rie s a nd m at he m at ic al in du ct ion a nd t he Bino m ia l Th eore m . ke en ly obse rve an d in ve st iga te pa tt erns, a nd form ula te a ppr opria te m at he m at ic al st at em en ts an d pr ov e th em u si ng m at he m at ic al in duc tion an d/ or Binomia l T he ore m . 1. illu st ra te a se rie s S TEM_P C 11 S M I-Ih -1 2. dif fe re nt ia te a se rie s f rom a se quenc e S TEM_P C 11 S M I-Ih -2 3. use t he sigm a n ota tion t o r epre se nt a se rie s S TEM_P C 11 S M I-Ih -3 4. illu st ra te t he P rinc iple of Ma th em at ic al In duc tio n S TEM_P C 11 S M I-Ih -4 5. appl y m at he m at ic al in du ct ion in provi ng ide nt itie s S TEM_P C 11 S M I-Ih -i -1 6. illu st ra te P asc al’ s T ria ngle in t he e xpa nsi on o f 𝑥 + 𝑦 𝑛for sm al l posit iv e i nt egr al v alu es of 𝑛 S TEM_P C 11 S M I-Ii -2 7. prov e th e Bi nomia l T he ore m S TEM_P C 11 S M I-Ii -3 8. de te rm in e an y t erm o f 𝑥 + 𝑦 𝑛, w he re 𝑛 is a posit iv e in te ge r, w ith ou t e xpa nding S TEM_P C 11 S M I-Ij -1 9. solv e prob le m s usi ng m at he m at ic al in duc tio n a nd t he Bino m ia l Th eore m S TEM_P C 11 S M I-Ij -2DEPED COPY
K t o 1 2 B A S IC EDU CA T ION CU R R IC U LU M S ENIOR H IGH S CH O O L – S CI ENCE, TECH NO LO GY, EN GINEER ING A ND MA THE M A TICS (S TEM) S P ECI A LIZED S U B JECT K to 12 Se nio r H ig h Sc ho ol STEM S pe cia liz ed S ub je ct – Pre -C alc ulus De ce m be r 2013 P ag e 3 o f 4 C ONT E NT C ONT E NT S TA NDA R DS P ER FO R MAN C E S TA NDA R DS LEARNING C OMP ETE NC IES C OD E Trigo no m et ry ke y c on ce pt s of circu la r f un ct io ns, trigonome tric ide nt itie s, in ve rse trigonome tric fu nc tio ns , an d th e pola r coordi na te sy st em 1. f orm ula te a nd sol ve ac cu ra te ly sit ua tion al prob le m s in vol vin g circ ula r f un ct ion s 1. illu st ra te t he u nit ci rcle a nd t he re la tion sh ip be tw ee n t he lin ea r an d a ngular m ea su re s o f a c en tr al a ngle in a u nit ci rcle S TEM_P C 11 T -I Ia -1 2. con ve rt de gre e m ea su re t o r adia n m ea su re a nd vic e ve rsa S TEM_P C 11 T -I Ia -2 3. illu st ra te a ngle s in st an da rd posit ion a nd cote rm in al a ngle s S TEM_P C 11 T -I Ia -3 4. illu st ra te t he di ffe re nt c ircu la r f un ct io ns S TEM_P C 11 T -I Ib -1 5. use s re fe re nc e a ng le s t o f in d e xa ct v al ue s of c irc ula r f un ct io ns S TEM_P C 11 T -I Ib -2 6. de te rm in e th e dom ain a nd ra nge of t he di ff ere nt ci rc ul ar f un ct io ns S TEM_P C 11 T -I Ic -1 7. gr aph t he six c irc ula r f un ct ion s (a ) a m plit ude , (b) pe rio d, an d (c) phase sh ift S TEM_P C 11 T -I Ic -d -1 8. solv e prob le m s i nv ol vin g cir cu la r fu nc tio ns S TEM_P C 11 T -I Id -2 2. apply a ppr opria te trigonome tric ide nt itie s i n solv in g sit ua tio na l prob le m s 9. de te rm in e w he th er a n e qua tion is a n ide nt ity or a c on dit ion al equat ion S TEM_P C 11 T -I Ie -1 10. de riv e th e fu nda m en ta l t rig on ome tric ide nt itie s S TEM_P C 11 T -I Ie -2 11. de riv e trigo nome tric ide nt iti es in vol vin g s um a nd di ff er en ce of an gle s S TEM_P C 11 T -I Ie -3 12. de riv e th e do uble a nd ha lf-an gle f orm ula s S TEM_P C 11 T -I If -1 13. sim pli fy t rigon ome tric expr ession s S TEM_P C 11 T -I If -2 14. prove other t rig on ome tri c i de nt itie s S TEM_P C 11 T -I If -g -1 15. solv e sit ua tio na l prob le m s i nv ol vin g trig on ome tri c ide nt itie s S TEM_P C 11 T -I Ig -2 3. form ula te a nd sol ve ac cu ra te ly sit ua tion al prob le m s in vol vin g appr opria te t rigonome tri c fu nc tio ns 16. illu st ra te t he dom ain a nd ra nge of t he in ve rse t rigon om et ric fu nc tio ns. S TEM_P C 11 T -I Ih -1 17. ev alu at e an in ve rse t rigono m et ric expr ession . S TEM_P C 11 T -I Ih -2 18. solv e trig on ome tri c e quat io ns. S TEM_P C 11 T -I Ih -i -1 19. solv e sit ua tion al prob le m s in volv in g i nv erse t rigon ome tric fu nc tio ns a nd t rig on ome tri c e quat ion s S TEM_P C 11 T -I Ii -2 4. form ula te a nd sol ve ac cu ra te ly sit ua tion al prob le m s in vol vin g t he pola r coo rdi na te sy st em 20. loc at e poin ts i n pola r c oordi na te sy st em S TEM_P C 11 T -I Ij -1 21. con ve rt t he c oordi na te s of a poin t f rom re ct an gular t o pola r sy st em s a nd vic e ve rsa S TEM_P C 11 T -I Ij -2 22. solv e sit ua tion al prob le m s i nv olv in g pola r coordinat e s yst em S TEM_P C 11 T -I Ij -3 ** *Su gge st ion f or IC T-en ha nc ed le sso n w he n a va ila ble a nd w he re a ppr opria teDEPED COPY
K t o 1 2 B A S IC EDU CA T ION CU R R IC U LU M S ENIOR H IGH S CH O O L – S CI ENCE, TECH NO LO GY, EN GINEER ING A ND MA THE M A TICS (S TEM) S P ECI A LIZED S U B JECT K to 12 Se nio r H ig h Sc ho ol STEM S pe cia liz ed S ub je ct – Pre -C alc ulus De ce m be r 2013 P ag e 4 o f 4 C o de Boo k Le ge nd S amp le : S TE M_P C 11 A G -Ia -1 DO MAIN/ C OMP ONE NT C OD E Ana ly tic G eome try AG Serie s a nd Ma th em at ic al In duc tio n SMI Tr ig on om et ry T LEGEND S AMP LE Fi rst En try Le arnin g A re a a nd Stra nd/ Su bj ec t or Specia liza tion Sc ie nc e, Te ch nolog y, En ginee ring a nd Ma th em at ic s Pre -Ca lc ulu s S TE M_P C 11 AG G ra de L ev el G ra de 1 1 U pp erca se Le tt er/s D oma in /Co nt en t/ Com pon en t/ T opi c Ana ly tic G eome try - R om a n N um era l *Z ero i f n o spe ci fic quar te r Qua rt er Fi rst Q ua rt er I Lowe rca se Le tt er /s *P ut a h yphen ( -) in be tw ee n le tt ers t o in dica te m ore t ha n a spe cif ic w ee k We ek We ek on e a - Ara b ic N um b er Com pe te nc y illu st ra te t he dif fe re nt t ype s of c on ic se ct io ns: pa ra bola , ellipse , c ircle , h ype rbola , an d de ge ne ra te case s 1DEPED COPY
Unit 1
Analytic Geometry
San Juanico Bridge, by Morten Nærbøe, 21 June 2009,
https://commons.wikimedia.org/wiki/File%3ASan Juanico Bridge 2.JPG. Public Domain.
Stretching from Samar to Leyte with a total length of more than two kilome-ters, the San Juanico Bridge has been serving as one of the main thoroughfares of economic and social development in the country since its completion in 1973. Adding picturesque effect on the whole architecture, geometric structures are subtly built to serve other purposes. The arch-shaped support on the main span of the bridge helps maximize its strength to withstand mechanical resonance and aeroelastic flutter brought about by heavy vehicles and passing winds.
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Lesson 1.1. Introduction to Conic Sections and Circles
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate the different types of conic sections: parabola, ellipse, circle, hyper-bola, and degenerate cases;
(2) define a circle;
(3) determine the standard form of equation of a circle; (4) graph a circle in a rectangular coordinate system; and (5) solve situational problems involving conic sections (circles). Lesson Outline
(1) Introduction of the four conic sections, along with the degenerate conics (2) Definition of a circle
(3) Derivation of the standard equation of a circle (4) Graphing circles
(5) Solving situational problems involving circles Introduction
We present the conic sections, a particular class of curves which sometimes appear in nature and which have applications in other fields. In this lesson, we first illustrate how each of these curves is obtained from the intersection of a plane and a cone, and then discuss the first of their kind, circles. The other conic sections will be covered in the next lessons.
1.1.1. An Overview of Conic Sections
We introduce the conic sections (or conics), a particular class of curves which oftentimes appear in nature and which have applications in other fields. One of the first shapes we learned, a circle, is a conic. When you throw a ball, the trajectory it takes is a parabola. The orbit taken by each planet around the sun is an ellipse. Properties of hyperbolas have been used in the design of certain telescopes and navigation systems. We will discuss circles in this lesson, leaving parabolas, ellipses, and hyperbolas for subsequent lessons.
• Circle (Figure 1.1) - when the plane is horizontal
• Ellipse (Figure 1.1) - when the (tilted) plane intersects only one cone to form a bounded curve
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• Parabola (Figure 1.2) - when the plane intersects only one cone to form anunbounded curve
• Hyperbola (Figure 1.3) - when the plane (not necessarily vertical) intersects both cones to form two unbounded curves (each called a branch of the hyper-bola)
Figure 1.1 Figure 1.2 Figure 1.3
We can draw these conic sections (also called conics) on a rectangular co-ordinate plane and find their equations. To be able to do this, we will present equivalent definitions of these conic sections in subsequent sections, and use these to find the equations.
There are other ways for a plane and the cones to intersect, to form what are referred to as degenerate conics: a point, one line, and two lines. See Figures 1.4, 1.5 and 1.6.
Figure 1.4 Figure 1.5 Figure 1.6
1.1.2. Definition and Equation of a Circle
A circle may also be considered a special kind of ellipse (for the special case when the tilted plane is horizontal). As we get to know more about a circle, we will also be able to distinguish more between these two conics.
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See Figure 1.7, with the point C(3, 1) shown. From the figure, the distance of A(−2, 1) from C is AC = 5. By the distance formula, the distance of B(6, 5) from C is BC = p(6 − 3)2+ (5 − 1)2 = 5. There are other points P such thatP C = 5. The collection of all such points which are 5 units away from C, forms a circle.
Figure 1.7 Figure 1.8
Let C be a given point. The set of all points P having the same distance from C is called a circle. The point C is called the center of the circle, and the common distance its radius.
The term radius is both used to refer to a segment from the center C to a point P on the circle, and the length of this segment.
See Figure 1.8, where a circle is drawn. It has center C(h, k) and radius r > 0. A point P (x, y) is on the circle if and only if P C = r. For any such point then, its coordinates should satisfy the following.
P C = r p
(x − h)2+ (y − k)2 = r
(x − h)2+ (y − k)2 = r2
This is the standard equation of the circle with center C(h, k) and radius r. If the center is the origin, then h = 0 and k = 0. The standard equation is then x2+ y2 = r2.
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Example 1.1.1. In each item, give thestandard equation of the circle satisfy-ing the given conditions.
(1) center at the origin, radius 4 (2) center (−4, 3), radius √7 (3) circle in Figure 1.7 (4) circle A in Figure 1.9 (5) circle B in Figure 1.9
(6) center (5, −6), tangent to the
y-axis Figure 1.9
(7) center (5, −6), tangent to the x-axis
(8) It has a diameter with endpoints A(−1, 4) and B(4, 2). Solution. (1) x2+ y2 = 16
(2) (x + 4)2+ (y − 3)2 = 7
(3) The center is (3, 1) and the radius is 5, so the equation is (x − 3)2+ (y − 1)2 = 25.
(4) By inspection, the center is (−2, −1) and the radius is 4. The equation is (x + 2)2+ (y + 1)2 = 16.
(5) Similarly by inspection, we have (x − 3)2+ (y − 2)2 = 9.
(6) The center is 5 units away from the y-axis, so the radius is r = 5 (you can make a sketch to see why). The equation is (x − 5)2 + (y + 6)2 = 25.
(7) Similarly, since the center is 6 units away from the x-axis, the equation is (x − 5)2+ (y + 6)2 = 36.
(8) The center C is the midpoint of A and B: C = −1+42 ,4+22 = 32, 3. The radius is then r = AC = q −1 −3 2 2 + (4 − 3)2 = q29
4. The circle has
equation x − 322+ (y − 3)2 = 29
4 . 2
1.1.3. More Properties of Circles After expanding, the standard equation
x − 3 2 2 + (y − 3)2 = 29 4 can be rewritten as x2+ y2 − 3x − 6y + 4 = 0, an equation of the circle in general form.
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If the equation of a circle is given in the general formAx2+ Ay2+ Cx + Dy + E = 0, A 6= 0, or
x2+ y2+ Cx + Dy + E = 0,
we can determine the standard form by completing the square in both variables. Completing the square in an expression like x2 + 14x means determining
the term to be added that will produce a perfect polynomial square. Since the coefficient of x2 is already 1, we take half the coefficient of x and square it, and
we get 49. Indeed, x2+ 14x + 49 = (x + 7)2 is a perfect square. To complete
the square in, say, 3x2+ 18x, we factor the coefficient of x2 from the expression: 3(x2 + 6x), then add 9 inside. When completing a square in an equation, any
extra term introduced on one side should also be added to the other side.
Example 1.1.2. Identify the center and radius of the circle with the given equa-tion in each item. Sketch its graph, and indicate the center.
(1) x2+ y2− 6x = 7
(2) x2+ y2− 14x + 2y = −14 (3) 16x2 + 16y2+ 96x − 40y = 315
Solution. The first step is to rewrite each equation in standard form by complet-ing the square in x and in y. From the standard equation, we can determine the center and radius.
(1) x2− 6x + y2 = 7 x2− 6x + 9 + y2 = 7 + 9 (x − 3)2+ y2 = 16 Center (3, 0), r = 4, Figure 1.10 (2) x2− 14x + y2+ 2y = −14 x2− 14x + 49 + y2+ 2y + 1 = −14 + 49 + 1 (x − 7)2+ (y + 1)2 = 36 Center (7, −1), r = 6, Figure 1.11 (3) 16x2+ 96x + 16y2− 40y = 315
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16(x2 + 6x) + 16 y2− 5 2y = 315 16(x2+ 6x + 9) + 16 y2−5 2y + 25 16 = 315 + 16(9) + 16 25 16 16(x + 3)2+ 16 y − 5 4 2 = 484 (x + 3)2 + y − 5 4 2 = 484 16 = 121 4 = 11 2 2 Center −3,54, r = 5.5, Figure 1.12. 2Figure 1.10 Figure 1.11 Figure 1.12
In the standard equation (x − h)2 + (y − k)2 = r2, both the two squared
terms on the left side have coefficient 1. This is the reason why in the preceding example, we divided by 16 at the last equation.
1.1.4. Situational Problems Involving Circles
Let us now take a look at some situational problems involving circles.
?Example 1.1.3. A street with two lanes, each 10 ft wide, goes through a semicircular tunnel with radius 12 ft. How high is the tunnel at the edge of each lane? Round off to 2 decimal places.
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Solution. We draw a coordinate system with origin at the middle of the highway, as shown. Because of the given radius, the tunnel’s boundary is on the circle x2+ y2 = 122. Point P is the point on the arc just above the edge of a lane, soits x-coordinate is 10. We need its y-coordinate. We then solve 102 + y2 = 122
for y > 0, giving us y = 2√11 ≈ 6.63 ft. 2
Example 1.1.4. A piece of a broken plate was dug up in an archaeological site. It was put on top of a grid, as shown in Figure 1.13, with the arc of the plate passing through A(−7, 0), B(1, 4) and C(7, 2). Find its center, and the standard equation of the circle describing the boundary of the plate.
Figure 1.13
Figure 1.14
Solution. We first determine the center. It is the intersection of the perpendicular bisectors of AB and BC (see Figure 1.14). Recall that, in a circle, the perpen-dicular bisector of any chord passes through the center. Since the midpoint M of AB is −7+12 ,0+42 = (−3, 2), and mAB = 4−01+7 = 12, the perpendicular bisector
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Since the midpoint N of BC is 1+72 , 4+2 2 = (4, 3), and mBC = 2−4 7−1 = − 1 3,
the perpendicular bisector of BC has equation y − 3 = 3(x − 4), or equivalently, y = 3x − 9.
The intersection of the two lines y = 2x − 4 and y = 3x − 9 is (1, −6) (by solving a system of linear equations). We can take the radius as the distance of this point from any of A, B or C (it’s most convenient to use B in this case). We then get r = 10. The standard equation is thus (x − 1)2+ (y + 6)2 = 100. 2
More Solved Examples
1. In each item, give the standard equation of the circle satisying the given con-ditions.
(a) center at the origin, contains (0, 3) (b) center (1, 5), diameter 8
(c) circle A in Figure 1.15 (d) circle B in Figure 1.15 (e) circle C in Figure 1.15
(f) center (−2, −3), tangent to the y-axis
(g) center (−2, −3), tangent to the x-axis
(h) contains the points (−2, 0) and (8, 0), radius 5
Figure 1.15
Solution:
(a) The radius is 3, so the equation is x2+ y2 = 9.
(b) The radius is 8/2 = 4, so the equation is (x − 1)2+ (y − 5)2 = 16.
(c) The center is (−2, 2) and the radius is 2, so the equation is (x + 2)2+ (y − 2)2 = 4.
(d) The center is (2, 3) and the radius is 1, so the equation is (x − 2)2+ (y − 3)2 = 1. (e) The center is (1, −1) and by the Pythagorean Theorem, the radius (see Figure 1.16) is √22+ 22 = √8, so the
equation is (x − 1)2+ (x + 1)2 = 8.
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(f) The radius is 3, so the equation is (x + 2)2+ (y + 3)2 = 9.(g) The radius is 2, so the equation is (x + 2)2+ (y + 3)2 = 4.
(h) The distance between (−2, 0) and (8, 0) is 10; since the radius is 5, these two points are endpoints of a diameter. Then the circle has center at (3, 0) and radius 5, so its equation is (x − 3)2+ y2 = 25.
2. Identify the center and radius of the circle with the given equation in each item. Sketch its graph, and indicate the center.
(a) x2+ y2+ 8y = 33 (b) 4x2+ 4y2− 16x + 40y + 67 = 0 (c) 4x2+ 12x + 4y2+ 16y − 11 = 0 Solution: (a) x2+ y2+ 8y = 33 x2+ y2+ 8y + 16 = 33 + 16 x2+ (y + 4)2 = 49 Center (0, −4), radius 7, see Figure 1.17. (b) 4x2+ 4y2− 16x + 40y + 67 = 0 x2− 4x + y2+ 10y = −67 4 x2− 4x + 4 + y2+ 10y + 25 = −67 4 + 4 + 25 (x − 2)2+ (y + 5)2 = 49 4 = 7 2 2
Center (2, −5), radius 3.5, see Figure 1.18. (c) 4x2+ 12x + 4y2+ 16y − 11 = 0 x2+ 3x + y2+ 4y = 11 4 x2+ 3x + 9 4 + y 2 + 4y + 4 = 11 4 + 9 4+ 4 x + 3 2 2 + (y + 2)2 = 9 Center −3 2, −2
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Figure 1.17 Figure 1.18 Figure 1.19
3. A circular play area with radius 3 m is to be partitioned into two sections using a straight fence as shown in Figure 1.20. How long should the fence be?
Solution: To determine the length of the fence, we need to determine the coordi-nates of its endpoints. From Figure 1.20, the endpoints have x coordinate −1 and are on the circle x2 + y2 = 9. Then
1 + y2 = 9, or y = ±2√2. Therefore,
the length of the fence is 4√2 ≈ 5.66 m. Figure 1.20
4. A Cartesian coordinate system was used to identify locations on a circu-lar track. As shown in Figure 1.21, the circucircu-lar track contains the points A(−2, −4), B(−2, 3), C(5, 2). Find the total length of the track.
Figure 1.21 Figure 1.22
Solution: The segment AB is vertical and has midpoint (−2, −0.5), so its perpendicular bisector has equation y = −0.5. On the other hand, the segment BC has slope −1/7 and midpoint (1.5, 2.5), so its perpendicular bisector has equation y − 2.5 = 7(x − 1.5), or 7x − y − 8 = 0.
The center of the circle is the intersection of y = −0.5 and 7x − y − 8 = 0; that is, the center is at 1514, −12.
The radius of the circle is the distance from the center to any of the points A, B, or C; by the distance formula, the radius is
r 2125 98 = 5 14 √ 170. Therefore,
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the total length of the track (its circumference), is2 × π × 5 14
√
170 ≈ 29.26 units.
Supplementary Problems 1.1
Identify the center and radius of the circle with the given equation in each item. Sketch its graph, and indicate the center.
1. x2+ y2 = 1 4 2. 5x2+ 5y2 = 125 3. (x + 4)2 + y − 3 4 2 = 1 4. x2− 4x + y2− 4y − 8 = 0 5. x2+ y2− 14x + 12y = 36 6. x2+ 10x + y2− 16y − 11 = 0 7. 9x2+ 36x + 9y2+ 72y + 155 = 0 8. 9x2+ 9y2− 6x + 24y = 19 9. 16x2+ 80x + 16y2− 112y + 247 = 0
Find the standard equation of the circle which satisfies the given conditions. 10. center at the origin, radius 5√3
11. center at (17, 5), radius 12
12. center at (−8, 4), contains (−4, 2) 13. center at (15, −7), tangent to the x-axis 14. center at (15, −7), tangent to the y-axis
15. center at (15, −7), tangent to the line y = −10 16. center at (15, −7), tangent to the line x = 8 17. has a diameter with endpoints (3, 1) and (−7, 6)
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18. has a diameter with endpoints 92, 4 and −3 2, −2
19. concentric with x2+ 20x + y2− 14y + 145 = 0, diameter 12
20. concentric with x2− 2x + y2− 2y − 23 = 0 and has 1/5 the area
21. concentric with x2+ 4x + y2− 6y + 9 = 0 and has the same circumference as
x2+ 14x + y2+ 10y + 62 = 0
22. contains the points (3, 3), (7, 1), (0, 2) 23. contains the points (1, 4), (−1, 2), (4, −3)
24. center at (−3, 2) and tangent to the line 2x − 3y = 1 25. center at (−5, −1) and tangent to the line x + y + 10 = 0
26. has center with x-coordinate 4 and tangent to the line −x + 3y = 9 at (3, 4) 27. A stadium is shaped as in Figure 1.23, where its left and right ends are circular
arcs both with center at C. What is the length of the stadium 50 m from one of the straight sides?
Figure 1.23 28. A waterway in a theme park has a
semicircular cross section with di-ameter 11 ft. The boats that are going to be used in this waterway have rectangular cross sections and are found to submerge 1 ft into the water. If the waterway is to be filled with water 4.5 ft deep, what is the maximum possible width of the boats?
Figure 1.24
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Lesson 1.2. Parabolas
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to: (1) define a parabola;
(2) determine the standard form of equation of a parabola; (3) graph a parabola in a rectangular coordinate system; and (4) solve situational problems involving conic sections (parabolas). Lesson Outline
(1) Definition of a parabola
(2) Derivation of the standard equation of a parabola (3) Graphing parabolas
(4) Solving situational problems involving parabolas Introduction
A parabola is one of the conic sections. We have already seen parabolas which open upward or downward, as graphs of quadratic functions. Here, we will see parabolas opening to the left or right. Applications of parabolas are presented at the end.
1.2.1. Definition and Equation of a Parabola
Consider the point F (0, 2) and the line ` having equation y = −2, as shown in Figure 1.25. What are the distances of A(4, 2) from F and from `? (The latter is taken as the distance of A from A`, the point on ` closest to A). How about
the distances of B(−8, 8) from F and from ` (from B`)?
AF = 4 and AA` = 4
BF =p(−8 − 0)2 + (8 − 2)2 = 10 and BB ` = 10
There are other points P such that P F = P P` (where P` is the closest point on
line `). The collection of all such points forms a shape called a parabola. Let F be a given point, and ` a given line not containing F . The set of all points P such that its distances from F and from ` are the same, is called a parabola. The point F is its focus and the line ` its directrix.
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Figure 1.25Figure 1.26
Consider a parabola with focus F (0, c) and directrix ` having equation y = −c. See Figure 1.26. The focus and directrix are c units above and below, respectively, the origin. Let P (x, y) be a point on the parabola so P F = P P`, where P` is the
point on ` closest to P . The point P has to be on the same side of the directrix as the focus (if P was below, it would be closer to ` than it is from F ).
P F = P P`
p
x2 + (y − c)2 = y − (−c) = y + c
x2+ y2− 2cy + c2 = y2 + 2cy + c2 x2 = 4cy
The vertex V is the point midway between the focus and the directrix. This equation, x2 = 4cy, is then the standard equation of a parabola opening upward
with vertex V (0, 0).
Suppose the focus is F (0, −c) and the directrix is y = c. In this case, a point P on the resulting parabola would be below the directrix (just like the focus). Instead of opening upward, it will open downward. Consequently, P F = px2+ (y + c)2 and P P
` = c − y (you may draw a version of Figure 1.26 for
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x2 = −4cy.We collect here the features of the graph of a parabola with standard equation x2 = 4cy or x2 = −4cy, where c > 0.
(1) vertex : origin V (0, 0)
• If the parabola opens upward, the vertex is the lowest point. If the parabola opens downward, the vertex is the highest point.
(2) directrix : the line y = −c or y = c
• The directrix is c units below or above the vertex. (3) focus: F (0, c) or F (0, −c)
• The focus is c units above or below the vertex.
• Any point on the parabola has the same distance from the focus as it has from the directrix.
(4) axis of symmetry: x = 0 (the y-axis)
• This line divides the parabola into two parts which are mirror images of each other.
Example 1.2.1. Determine the focus and directrix of the parabola with the given equation. Sketch the graph, and indicate the focus, directrix, vertex, and axis of symmetry.
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(1) x2 = 12y (2) x2 = −6y
Solution. (1) The vertex is V (0, 0) and the parabola opens upward. From 4c = 12, c = 3. The focus, c = 3 units above the vertex, is F (0, 3). The directrix, 3 units below the vertex, is y = −3. The axis of symmetry is x = 0.
(2) The vertex is V (0, 0) and the parabola opens downward. From 4c = 6, c = 32. The focus, c = 32 units below the vertex, is F 0, −32. The directrix, 32 units above the vertex, is y = 32. The axis of symmetry is x = 0.
Example 1.2.2. What is the standard equation of the parabola in Figure 1.25? Solution. From the figure, we deduce that c = 2. The equation is thus x2 =
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1.2.2. More Properties of ParabolasThe parabolas we considered so far are “vertical” and have their vertices at the origin. Some parabolas open instead horizontally (to the left or right), and some have vertices not at the origin. Their standard equations and properties are given in the box. The corresponding computations are more involved, but are similar to the one above, and so are not shown anymore.
In all four cases below, we assume that c > 0. The vertex is V (h, k), and it lies between the focus F and the directrix `. The focus F is c units away from the vertex V , and the directrix is c units away from the vertex. Recall that, for any point on the parabola, its distance from the focus is the same as its distance from the directrix.
(x − h)2 = 4c(y − k) (y − k)2 = 4c(x − h)
(x − h)2 = −4c(y − k) (y − k)2 = −4c(x − h)
directrix `: horizontal directrix `: vertical
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Note the following observations:• The equations are in terms of x − h and y − k: the vertex coordinates are subtracted from the corresponding variable. Thus, replacing both h and k with 0 would yield the case where the vertex is the origin. For instance, this replacement applied to (x − h)2 = 4c(y − k) (parabola opening upward) would yield x2 = 4cy, the first standard equation we encountered (parabola opening
upward, vertex at the origin).
• If the x-part is squared, the parabola is “vertical”; if the y-part is squared, the parabola is “horizontal.” In a horizontal parabola, the focus is on the left or right of the vertex, and the directrix is vertical.
• If the coefficient of the linear (non-squared) part is positive, the parabola opens upward or to the right; if negative, downward or to the left.
Example 1.2.3. Figure 1.27 shows the graph of parabola, with only its focus and vertex indicated. Find its standard equation. What is its directrix and its axis of symmetry?
Solution. The vertex is V (5, −4) and the focus is F (3, −4). From these, we deduce the following: h = 5, k = −4, c = 2 (the distance of the focus from the vertex). Since the parabola opens to the left, we use the template (y − k)2 =
−4c(x − h). Our equation is
(y + 4)2 = −8(x − 5).
Its directrix is c = 2 units to the right of V , which is x = 7. Its axis is the horizontal line through V : y = −4.
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The standard equation (y + 4)2 = −8(x − 5) from the preceding example canbe rewritten as y2 + 8x + 8y − 24 = 0, an equation of the parabola in general
form.
If the equation is given in the general form Ax2+ Cx + Dy + E = 0 (A and C
are nonzero) or By2+ Cx + Dy + E = 0 (B and C are nonzero), we can determine the standard form by completing the square in both variables.
Example 1.2.4. Determine the vertex, focus, directrix, and axis of symmetry of the parabola with the given equation. Sketch the parabola, and include these points and lines.
(1) y2− 5x + 12y = −16 (2) 5x2+ 30x + 24y = 51
Solution. (1) We complete the square on y, and move x to the other side. y2 + 12y = 5x − 16
y2+ 12y + 36 = 5x − 16 + 36 = 5x + 20 (y + 6)2 = 5(x + 4)
The parabola opens to the right. It has vertex V (−4, −6). From 4c = 5, we get c = 54 = 1.25. The focus is c = 1.25 units to the right of V : F (−2.75, −6). The (vertical) directrix is c = 1.25 units to the left of V : x = −5.25. The (horizontal) axis is through V : y = −6.
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(2) We complete the square on x, and move y to the other side.5x2 + 30x = −24y + 51
5(x2+ 6x + 9) = −24y + 51 + 5(9)
5(x + 3)2 = −24y + 96 = −24(y − 4) (x + 3)2 = −24
5 (y − 4)
In the last line, we divided by 5 for the squared part not to have any coeffi-cient. The parabola opens downward. It has vertex V (−3, 4).
From 4c = 245, we get c = 65 = 1.2. The focus is c = 1.2 units below V : F (−3, 2.8). The (horizontal) directrix is c = 1.2 units above V : y = 5.2. The (vertical) axis is through V : x = −3.
Example 1.2.5. A parabola has focus F (7, 9) and directrix y = 3. Find its standard equation.
Solution. The directrix is horizontal, and the focus is above it. The parabola then opens upward and its standard equation has the form (x − h)2 = 4c(y − k). Since the distance from the focus to the directrix is 2c = 9 − 3 = 6, then c = 3. Thus, the vertex is V (7, 6), the point 3 units below F . The standard equation is
then (x − 7)2 = 12(y − 6). 2
1.2.3. Situational Problems Involving Parabolas
Let us now solve some situational problems involving parabolas.
Example 1.2.6. A satellite dish has a shape called a paraboloid, where each cross-section is a parabola. Since radio signals (parallel to the axis) will bounce
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off the surface of the dish to the focus, the receiver should be placed at the focus. How far should the receiver be from the vertex, if the dish is 12 ft across, and 4.5 ft deep at the vertex?Solution. The second figure above shows a cross-section of the satellite dish drawn on a rectangular coordinate system, with the vertex at the origin. From the problem, we deduce that (6, 4.5) is a point on the parabola. We need the distance of the focus from the vertex, i.e., the value of c in x2 = 4cy.
x2 = 4cy 62 = 4c(4.5)
c = 6
2
4 · 4.5 = 2
Thus, the receiver should be 2 ft away from the vertex. 2 Example 1.2.7. The cable of a suspension bridge hangs in the shape of a parabola. The towers supporting the cable are 400 ft apart and 150 ft high. If the cable, at its lowest, is 30 ft above the bridge at its midpoint, how high is the cable 50 ft away (horizontally) from either tower?
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Solution. Refer to the figure above, where the parabolic cable is drawn with its vertex on the y-axis 30 ft above the origin. We may write its equation as (x − 0)2 = a(y − 30); since we don’t need the focal distance, we use the simpler variable a in place of 4c. Since the towers are 150 ft high and 400 ft apart, we deduce from the figure that (200, 150) is a point on the parabola.x2 = a(y − 30) 2002 = a(150 − 30) a = 200 2 120 = 1000 3
The parabola has equation x2 = 10003 (y − 30), or equivalently, y = 0.003x2 + 30. For the two points on the parabola 50 ft away from the
towers, x = 150 or x = −150. If x = 150, then
y = 0.003(1502) + 30 = 97.5.
Thus, the cable is 97.5 ft high 50 ft away from either tower. (As expected, we
get the same answer from x = −150.) 2
More Solved Examples
For Examples 1 and 2, determine the focus and directrix of the parabola with the given equation. Sketch the graph, and indicate the focus, directrix, and vertex. 1. y2 = 20x
Solution:
Vertex: V (0, 0), opens to the right 4c = 20 ⇒ c = 5
Focus: F (5, 0), Directrix: x = −5 See Figure 1.28.
2. 3x2 = −12y
Solution: 3x2 = −12y ⇔ x2 = −4y
Vertex: V (0, 0), opens downward 4c = 4 ⇒ c = 1
Focus: F (0, −1), Directrix: y = 1 See Figure 1.29.
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3. Determine the standard equation of theparabola in Figure 1.30 given only its vertex and focus. Then determine its di-rectix and axis of symmetry.
Solution: V −3 2, 4 , F (−4, 4) c = 5 2 ⇒ 4c = 10
Parabola opens to the left Equation: (y − 4)2 = −10
x +3
2
Directrix: x = 1, Axis: y = 4 Figure 1.30 4. Determine the standard equation of the
parabola in Figure 1.31 given only its vertex and diretrix. Then determine its focus and axis of symmetry.
Solution: V 5,13 2 , directrix: y = 15 2 c = 1 ⇒ 4c = 4
Parabola opens downward Equation: y − 13 2 2 = −4 (x − 5) Focus: 5,11 2 , Axis: x = 5 Figure 1.31
For Examples 5 and 6, determine the vertex, focus, directrix, and axis of sym-metry of the parabola with the given equation. Sketch the parabola, and include these points and lines.
5. x2− 6x − 2y + 9 = 0
Solution:
x2 − 6x = 2y − 9 x2− 6x + 9 = 2y
(x − 3)2 = 2y V (3, 0), parabola opens upward 4c = 2 ⇒ c = 1 2, F 3,1 2 , directrix: y = −1 2, axis: x = 3 See Figure 1.32. Figure 1.32
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6. 3y2+ 8x + 24y + 40 = 0 Solution: 3y2+ 24y = −8x − 40 3(y2 + 8y) = −8x − 40 3(y2+ 8y + 16) = −8x − 40 + 48 3(y + 4)2 = −8x + 8 (y + 4)2 = −8 3(x − 1) V (1, −4), parabola opens to the left 4c = 8 3 ⇒ c = 2 3, F 1 3, −4 , directrix: x = 5 3, axis: y = −4 See Figure 1.33. Figure 1.337. A parabola has focus F (−11, 8) and directrix x = −17. Find its standard equation.
Solution: Since the focus is 6 units to the right of the directrix, the parabola opens to the right with 2c = 6. Then c = 3 and V (−14, 8). Hence, the equation is (y − 8)2 = 12(x + 14).
8. A flashlight is shaped like a paraboloid and the light source is placed at the focus so that the light bounces off parallel to the axis of symmetry; this is done to maximize illumination. A particular flashlight has its light source located 1 cm from the base and is 6 cm deep; see Figure 1.34. What is the width
of the flashlight’s opening? Figure 1.34
Solution: Let the base (the vertex) of the flashlight be the point V (0, 0). Then the light source (the focus) is at F (0, 1); so c = 1. Hence, the parabola’s equation is x2 = 4y. To get the width of the opening, we need the x coordinates
of the points on the parabola with y coordinate 6. x2 = 4(6) ⇒ x = ±2√6
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9. An object thrown from a height of 2m above the ground follows a parabolic path until the object falls to the ground; see Figure 1.35. If the object reaches a maximum height (measured from the ground) of 7 m after travelling a hor-izontal distance of 4 m, determine the horizontal distance between the object’s initial and final positions.
Figure 1.35
Solution: Let V (0, 7) be the parabola’s vertex, which corresponds to the high-est point reached by the object. Then the parabola’s equation is of the form x2 = −4c(y − 7) and the object’s starting point is at (−4, 2). Then
(−4)2 = −4c(2 − 7) ⇒ c = 16 20 =
4 5. Hence, the equation of the parabola is x2 = −16
5 (y − 7). When the object hits the ground, the y coordinate is 0 and
x2 = −16 5 (0 − 7) = 112 5 ⇒ x = ±4 r 7 5.
Since this point is to the right of the vertex, we choose x = +4r 7
5. Therefore, the total distance travelled is 4r 7
5− (−4) ≈ 8.73 m.
Supplementary Problems 1.2
Determine the vertex, focus, directrix, and axis of symmetry of the parabola with the given equation. Sketch the graph, and include these points and lines.
1. y2 = −36x 2. 5x2 = 100y 3. y2+ 4x − 14y = −53 4. y2− 2x + 2y − 1 = 0 5. 2x2− 12x + 28y = 38 6. (3x − 2)2 = 84y − 112
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Find the standard equation of the parabola which satisfies the given conditions. 7. vertex (7, 11), focus (16, 11) 8. vertex (−10, −5), directrix y = −1 9. focus −10,23 2 , directrix y = −11 2 10. focus −3 2, 3 , directrix x = −37 211. axis of symmetry y = 9, directrix x = 24, vertex on the line 3y − 5x = 7 12. vertex (0, 7), vertical axis of symmetry, through the point P (4, 5)
13. vertex (−3, 8), horizontal axis of symmetry, through the point P (−5, 12) 14. A satellite dish shaped like a paraboloid has its receiver located at the focus.
How far is the receiver from the vertex if the dish is 10 ft across and 3 ft deep at the center?
15. A flashlight shaped like a paraboloid has its light source at the focus located 1.5 cm from the base and is 10 cm wide at its opening. How deep is the flashlight at its center?
16. The ends of a rope are held in place at the top of two posts, 9 m apart and each one 8 m high. If the rope assumes a parabolic shape and touches the ground midway between the two posts, how high is the rope 2 m from one of the posts?
17. Radiation is focused to an unhealthy area in a patient’s body using a parabolic reflector, positioned in such a way that the target area is at the focus. If the reflector is 30 cm wide and 15 cm deep at the center, how far should the base of the reflector be from the target area?
18. A rectangular object 25 m wide is to pass under a parabolic arch that has a width of 32 m at the base and a height of 24 m at the center. If the vertex of the parabola is at the top of the arch, what maximum height should the rectangular object have?
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Lesson 1.3. Ellipses
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to: (1) define an ellipse;
(2) determine the standard form of equation of an ellipse; (3) graph an ellipse in a rectangular coordinate system; and (4) solve situational problems involving conic sections (ellipses). Lesson Outline
(1) Definition of an ellipse
(2) Derivation of the standard equation of an ellipse (3) Graphing ellipses
(4) Solving situational problems involving ellipses Introduction
Unlike circle and parabola, an ellipse is one of the conic sections that most stu-dents have not encountered formally before. Its shape is a bounded curve which looks like a flattened circle. The orbits of the planets in our solar system around the sun happen to be elliptical in shape. Also, just like parabolas, ellipses have reflective properties that have been used in the construction of certain structures. These applications and more will be encountered in this lesson.
1.3.1. Definition and Equation of an Ellipse
Consider the points F1(−3, 0) and F2(3, 0), as shown in Figure 1.36. What is the
sum of the distances of A(4, 2.4) from F1 and from F2? How about the sum of
the distances of B (and C(0, −4)) from F1 and from F2?
AF1+ AF2 = 7.4 + 2.6 = 10
BF1+ BF2 = 3.8 + 6.2 = 10
CF1+ CF2 = 5 + 5 = 10
There are other points P such that P F1+ P F2 = 10. The collection of all such
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Figure 1.36Figure 1.37
Let F1 and F2 be two distinct points. The set of all points P , whose
distances from F1 and from F2 add up to a certain constant, is called
an ellipse. The points F1 and F2 are called the foci of the ellipse.
Given are two points on the x-axis, F1(−c, 0) and F2(c, 0), the foci, both c
units away from their center (0, 0). See Figure 1.37. Let P (x, y) be a point on the ellipse. Let the common sum of the distances be 2a (the coefficient 2 will make computations simpler). Thus, we have P F1+ P F2 = 2a.
P F1 = 2a − P F2 p (x + c)2+ y2 = 2a −p(x − c)2+ y2 x2+ 2cx + c2+ y2 = 4a2− 4ap(x − c)2+ y2+ x2 − 2cx + c2+ y2 ap(x − c)2+ y2 = a2− cx a2x2− 2cx + c2+ y2 = a4− 2a2cx + c2x2 (a2− c2)x2+ a2y2 = a4− a2c2 = a2(a2− c2) b2x2+ a2y2 = a2b2 by letting b =√a2− c2, so a > b x2 a2 + y2 b2 = 1
When we let b =√a2− c2, we assumed a > c. To see why this is true, look at
4P F1F2 in Figure 1.37. By the Triangle Inequality, P F1 + P F2 > F1F2, which
implies 2a > 2c, so a > c.
We collect here the features of the graph of an ellipse with standard equation x2 a2 + y2 b2 = 1, where a > b. Let c = √ a2− b2.
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(1) center : origin (0, 0)(2) foci : F1(−c, 0) and F2(c, 0)
• Each focus is c units away from the center.
• For any point on the ellipse, the sum of its distances from the foci is 2a. (3) vertices: V1(−a, 0) and V2(a, 0)
• The vertices are points on the ellipse, collinear with the center and foci. • If y = 0, then x = ±a. Each vertex is a units away from the center. • The segment V1V2 is called the major axis. Its length is 2a. It divides
the ellipse into two congruent parts. (4) covertices: W1(0, −b) and W2(0, b)
• The segment through the center, perpendicular to the major axis, is the minor axis. It meets the ellipse at the covertices. It divides the ellipse into two congruent parts.
• If x = 0, then y = ±b. Each covertex is b units away from the center. • The minor axis W1W2 is 2b units long. Since a > b, the major axis is
longer than the minor axis.
Example 1.3.1. Give the coordinates of the foci, vertices, and covertices of the ellipse with equation
x2 25+
y2 9 = 1. Sketch the graph, and include these points.
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Solution. With a2 = 25 and b2 = 9, we have a = 5, b = 3, and c =√a2− b2 = 4.foci: F1(−4, 0), F2(4, 0) vertices: V1(−5, 0), V2(5, 0)
covertices: W1(0, −3), W2(0, 3)
Example 1.3.2. Find the (standard) equation of the ellipse whose foci are F1(−3, 0) and F2(3, 0), such that for any point on it, the sum of its distances
from the foci is 10. See Figure 1.36.
Solution. We have 2a = 10 and c = 3, so a = 5 and b = √a2− c2 = 4. The
equation is
x2
25 + y2
16 = 1. 2
1.3.2. More Properties of Ellipses
The ellipses we have considered so far are “horizontal” and have the origin as their centers. Some ellipses have their foci aligned vertically, and some have centers not at the origin. Their standard equations and properties are given in the box. The derivations are more involved, but are similar to the one above, and so are not shown anymore.
In all four cases below, a > b and c = √a2− b2. The foci F
1 and F2 are c
units away from the center. The vertices V1 and V2 are a units away from the
center, the major axis has length 2a, the covertices W1 and W2 are b units away
from the center, and the minor axis has length 2b. Recall that, for any point on the ellipse, the sum of its distances from the foci is 2a.
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Center Corresponding Graphs
(0, 0) x2 a2 + y2 b2 = 1, a > b x2 b2 + y2 a2 = 1, b > a (h, k) (x − h)2 a2 + (y − k)2 b2 = 1 (x − h)2 b2 + (y − k)2 a2 = 1 a > b b > a
major axis: horizontal major axis: vertical minor axis: vertical minor axis: horizontal In the standard equation, if the x-part has the bigger denominator, the ellipse is horizontal. If the y-part has the bigger denominator, the ellipse is vertical. Example 1.3.3. Give the coordinates of the center, foci, vertices, and covertices of the ellipse with the given equation. Sketch the graph, and include these points.
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(1) (x + 3) 2 24 + (y − 5)2 49 = 1 (2) 9x2+ 16y2− 126x + 64y = 71Solution. (1) From a2 = 49 and b2 = 24, we have a = 7, b = 2√6 ≈ 4.9, and
c =√a2− b2 = 5. The ellipse is vertical.
center: (−3, 5) foci: F1(−3, 0), F2(−3, 10) vertices: V1(−3, −2), V2(−3, 12) covertices: W1(−3 − 2 √ 6, 5) ≈ (−7.9, 5) W2(−3 + 2 √ 6, 5) ≈ (1.9, 5)
(2) We first change the given equation to standard form. 9(x2− 14x) + 16(y2+ 4y) = 71 9(x2− 14x + 49) + 16(y2 + 4y + 4) = 71 + 9(49) + 16(4) 9(x − 7)2+ 16(y + 2)2 = 576 (x − 7)2 64 + (y + 2)2 36 = 1
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We have a = 8 and b = 6. Thus, c = √a2− b2 = 2√7 ≈ 5.3. The ellipse ishorizontal. center: (7, −2) foci: F1(7 − 2 √ 7, −2) ≈ (1.7, −2) F2(7 + 2 √ 7, −2) ≈ (12.3, −2) vertices: V1(−1, −2), V2(15, −2) covertices: W1(7, −8), W2(7, 4)
Example 1.3.4. The foci of an ellipse are (−3, −6) and (−3, 2). For any point on the ellipse, the sum of its distances from the foci is 14. Find the standard equation of the ellipse.
Solution. The midpoint (−3, −2) of the foci is the center of the ellipse. The ellipse is vertical (because the foci are vertically aligned) and c = 4. From the given sum, 2a = 14 so a = 7. Also, b = √a2 − c2 = √33. The equation is
(x + 3)2
33 +
(y + 2)2
49 = 1. 2
Example 1.3.5. An ellipse has vertices (2 −√61, −5) and (2 +√61, −5), and its minor axis is 12 units long. Find its standard equation and its foci.
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Solution. The midpoint (2, −5) of the vertices is the center of the ellipse, which is horizontal. Each vertex is a = √61 units away from the center. From the length of the minor axis, 2b = 12 so b = 6. The standard equation is (x − 2)2
61 +
(y + 5)2
36 =
1. Each focus is c = √a2− b2 = 5 units away from (2, −5), so their coordinates
are (−3, −5) and (7, −5). 2
1.3.3. Situational Problems Involving Ellipses
Let us now apply the concept of ellipse to some situational problems.
?Example 1.3.6. A tunnel has the shape of a semiellipse that is 15 ft high at the center, and 36 ft across at the base. At most how high should a passing truck be, if it is 12 ft wide, for it to be able to fit through the tunnel? Round off your answer to two decimal places.
Solution. Refer to the figure above. If we draw the semiellipse on a rectangular coordinate system, with its center at the origin, an equation of the ellipse which contains it, is
x2
182 +
y2
152 = 1.
To maximize its height, the corners of the truck, as shown in the figure, would have to just touch the ellipse. Since the truck is 12 ft wide, let the point (6, n) be the corner of the truck in the first quadrant, where n > 0, is the (maximum) height of the truck. Since this point is on the ellipse, it should fit the equation. Thus, we have 62 182 + n2 152 = 1 n = 10√2 ≈ 14.14 ft 2
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Example 1.3.7. The orbit of a planet has the shape of an ellipse, and on one of the foci is the star around which it revolves. The planet is closest to the star when it is at one vertex. It is farthest from the star when it is at the other vertex. Suppose the closest and farthest distances of the planet from this star, are 420 million kilometers and 580 million kilometers, respectively. Find the equation of the ellipse, in standard form, with center at the origin and the star at the x-axis. Assume all units are in millions of kilometers.Solution. In the figure above, the orbit is drawn as a horizontal ellipse with center at the origin. From the planet’s distances from the star, at its closest and farthest points, it follows that the major axis is 2a = 420 + 580 = 1000 (million kilometers), so a = 500. If we place the star at the positive x-axis, then it is c = 500 − 420 = 80 units away from the center. Therefore, we get b2 = a2− c2 = 5002− 802 = 243600. The equation then is
x2 250000+
y2
243600 = 1.
The star could have been placed on the negative x-axis, and the answer would
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More Solved Examples
1. Give the coordinates of the foci, vertices, and covertices of the ellipse with equa-tion x
2
169 + y2
144 = 1. Then sketch the graph and include these points.
Solution: The ellipse is horizontal. a2 = 169 ⇒ a = 13, b2 = 144 ⇒ b = 12,
c =√169 − 144 = 5 Foci: F1(−5, 0), F2(5, 0)
Vertices: V1(−13, 0), V2(13, 0)
Covertices: W1(0, −12), W2(0, 12)
See Figure 1.38. Figure 1.38
2. Find the standard equation of the ellipse whose foci are F1(0, −8) and F2(0, 8),
such that for any point on it, the sum of its distances from the foci is 34. Solution: The ellipse is vertical and has center at (0, 0).
2a = 34 ⇒ a = 17 c = 8 ⇒ b =√172− 82 = 15 The equation is x 2 225 + y2 289 = 1.
For Examples 3 and 4, give the coordinates of the center, foci, vertices, and covertices of the ellipse with the given equation. Sketch the graph, and include these points. 3. (x − 7) 2 64 + (y + 2)2 25 = 1
Solution: The ellipse is horizontal. a2 = 64 ⇒ a = 8, b2 = 25 ⇒ b = 5 c =√64 − 25 =√39 ≈ 6.24 center: (7, −2) foci: F1(7 − √ 39, −2) ≈ (0.76, −2) F2(7 + √ 39, −2) ≈ (13.24, −2) vertices: V1(−1, −2), V2(15, −2) covertices: W1(7, −7), W2(7, 3) See Figure 1.39. Figure 1.39
