Pre-Calculus_Consolidated_Lecture Notes.pdf

ATENEO DE DAVAO UNIVERSITY SENIOR HIGH SCHOOL GRADE 11 – PRE-CALCULUS LECTURE NOTES Prepared by: Orville J. Evardo Jr. Jeffrey T. Marabolas Kharelle Mae M. Naduma Reil Benedict S. Obinque Aries James L. Toledo 2...

ATENEO DE DAVAO UNIVERSITY SENIOR HIGH SCHOOL GRADE 11 – PRE-CALCULUS LECTURE NOTES Prepared by: Orville J. Evardo Jr. Jeffrey T. Marabolas Kharelle Mae M. Naduma Reil Benedict S. Obinque Aries James L. Toledo 2022 Table of Contents Lecture Notes No. 1 Introduction to Conic Sections....................................................................... 2 Lecture Notes No. 2 The Circle....................................................................................................... 9 Lecture Notes No. 3 The Parabola................................................................................................. 15 Lecture Notes No. 4 The Ellipse.................................................................................................... 23 Lecture Notes No. 5 The Hyperbola.............................................................................................. 28 Lecture Notes No. 6 Systems of Nonlinear Equations................................................................... 33 Lecture Notes No. 7 The Unit Circle............................................................................................. 41 Lecture Notes No. 8 The Six Circular Functions........................................................................... 47 Lecture Notes No. 9 Graph of Circular Functions......................................................................... 56 AdDU -Senior High School | Pre-Calculus | Page 1 of 73  ATENEO DE DAVAO UNIVERSITY Km 7 Central Park Blvd, Talomo, 8016 Davao City, Philippines Tel No. +63 (82) 221.2411 local 8608 E-Mail: [email protected] * www.addu.edu.ph In Consortium with Ateneo de Zamboanga University and Xavier University SENIOR HIGH SCHOOL – PRE-CALCULUS Topic: Introduction to the Conic Sections CONIC SECTIONS Definition: Conic Section or Conic - The section obtained if a plane is made to cut a double right circular cone. - If the plane does not pass through the vertex of the cone, then the conic sections formed are: a circle, an ellipse, a parabola and a hyperbola. Degenerate Conic - If the plane passes through the vertex of the cone. - A degenerate conic may be a point, line, or two intersecting lines. Circle Ellipse Parabola Hyperbola plane perpendicular plane oblique to plane parallel to plane cuts both to cone axis cone axis side of cone halves of cone Figure 1: Conic Sections Point Single line Pair of intersecting plane through cone plane tangent to lines vertex only cone (degenerate (degenerate ellipse) (degenerate hyperbola) parabola) Figure 2: Degenerate Conics AdDU -Senior High School | Pre-Calculus | Page 2 of 73 CLASSIFYING CONICS General Equation of a Conic 𝑨𝒙𝟐 + 𝑩𝒙𝒚 + 𝑪𝒚𝟐 + 𝑫𝒙 + 𝑬𝒚 + 𝑭 = 𝟎 To classify a Conic Section based from its general equation where 𝐵 = 0, we look at the coefficients of the square terms. Conic Coefficients Circle 𝐴=𝐶 Parabola 𝐴 = 0 or 𝐶 = 0 Ellipse 𝐴 and 𝐶 have the same signs Hyperbola 𝐴 and 𝐶 have opposite signs Examples Classify the conic based on the give general equation. 1. 𝑥 2 + 3𝑦 2 − 4𝑥 + 6𝑦 − 13 = 0 𝐴 = 1 and 𝐶 = 3, since both 𝐴 and 𝐶 have the same sign, therefore it is an Ellipse. 2. 3𝑥 2 + 3𝑦 2 + 6𝑥 + 6𝑦 − 4 = 0 𝐴 = 3 and 𝐶 = 3, since 𝐴 = 𝐶, therefore it is a Circle. 3. 4𝑦 2 − 2𝑥 + 8𝑦 + 10 = 0 𝐴 = 0 and 𝐶 = 4, since 𝐴 = 0, therefore it is a Parabola. 4. 3𝑥 2 + 5𝑥 − 𝑦 + 2 = 0 𝐴 = 3 and 𝐶 = 0, since 𝐶 = 0, therefore it is a Parabola. 5. 5𝑥 2 − 4𝑦 2 − 10𝑥 + 16𝑦 − 25 = 0 𝐴 = 5 and 𝐶 = −4, since 𝐴 and 𝐶 have different signs, therefore it is a Hyperbola. 6. −𝑥 2 − 𝑦 2 + 3𝑥 − 5𝑦 + 20 = 0 𝐴 = −1 and 𝐶 = −1, since 𝐴 = 𝐶, therefore it is a Circle. AdDU -Senior High School | Pre-Calculus | Page 3 of 73 STANDARD EQUATION Each conic has a unique standard equation. Below is the type of conic, its graph, and its standard equation. Conic Graph Standard Equation Circle (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 Parabola (𝑥 − ℎ)2 = ±4𝑝(𝑦 − 𝑘) Ellipse (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Hyperbola (𝑥 − ℎ)2 (𝑦 − 𝑘)2 − =1 𝑎2 𝑏2 The general equation of a conic can be transformed into its standard form and vice versa. To transform a general equation into its standard equation, it is done by using completing the square. AdDU -Senior High School | Pre-Calculus | Page 4 of 73 COMPLETING THE SQUARE Say we have an expression like 𝑥 2 + 𝑏𝑥, we can visualize this geometrically as This can be rearranged to form a figure that is almost a square but having the same area as above, The resulting figure is almost a square but is not yet a square, to complete the square, we add 𝑏 2 (2) , so we have 𝑏 2 The resulting figure is now a square. Algebraically, we express it as 𝑥 2 + 𝑏𝑥 + (2). The 𝑏 2 resulting expression is a perfect square trinomial which then can be factored out as (𝑥 + 2). Keeping the Balance 𝑏 2 We can’t just add (2) on one side, to maintain the balance of the equation, whatever is added to the left must also be added to the right. Examples Complete the Square a. 𝑥 2 + 8𝑥 = 0 a. 𝑦 2 − 6𝑦 = 0 Solution: Solution: 8 2 8 2 6 2 6 2 𝑥 2 + 8𝑥 + ( ) = 0 + ( ) 𝑦 2 − 6𝑦 + ( ) = 0 + ( ) 2 2 2 2 𝑥 2 + 8𝑥 + (4)2 = (4)2 𝑦 2 − 6𝑦 + (3)2 = (3)2 𝑥 2 + 8𝑥 + 16 = 16 𝑦 2 − 6𝑦 + 9 = 9 (𝒙 + 𝟒)𝟐 = 𝟏𝟔 (𝒚 − 𝟑)𝟐 = 𝟗 AdDU -Senior High School | Pre-Calculus | Page 5 of 73 Transforming General Equation to Standard Equation To transform the general equation of a conic into standard equation, we use completing the square. Examples Express the following general equations into its standard equation. 1. 𝑥 2 + 𝑦 2 − 2𝑥 − 4𝑦 + 1 = 0 Solution: 𝐴 = 1 and 𝐶 = 1, since 𝐴 = 𝐶, the equation is a circle and the standard equation of a circle is of the form (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2. (𝑥 2 − 2𝑥) + (𝑦 2 − 4𝑦) = −1 2 2 2 2 4 2 2 2 4 2 (𝑥 − 2𝑥 + ( ) ) + (𝑦 − 4𝑦 + ( ) ) = −1 + ( ) + ( ) 2 2 2 2 (𝑥 2 − 2𝑥 + 1) + (𝑦 2 − 4𝑦 + 4) = −1 + 1 + 4 (𝒙 − 𝟏)𝟐 + (𝒚 − 𝟐)𝟐 = 𝟒 2. 𝑥 2 + 𝑦 2 + 4𝑥 + 2𝑦 − 11 = 0 Solution: 𝐴 = 1 and 𝐶 = 1, since 𝐴 = 𝐶, the equation is a circle and the standard equation of a circle is of the form (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2. (𝑥 2 + 4𝑥) + (𝑦 2 + 2𝑦) = 11 2 4 2 2 2 2 4 2 2 2 (𝑥 + 4𝑥 + ( ) ) + (𝑦 + 2𝑦 + ( ) ) = 11 + ( ) + ( ) 2 2 2 2 (𝑥 2 + 4𝑥 + 4) + (𝑦 2 + 2𝑦 + 1) = 11 + 4 + 1 (𝒙 + 𝟐)𝟐 + (𝒚 + 𝟏)𝟐 = 𝟏𝟔 AdDU -Senior High School | Pre-Calculus | Page 6 of 73 3. 9𝑥 2 + 4𝑦 2 − 36𝑥 − 108 = 0 Solution: 𝐴 = 9 and 𝐶 = 4, since both 𝐴 and 𝐶 have the same sign, the equation is an ellipse (𝑥−ℎ)2 (𝑦−𝑘)2 and the standard equation of an ellipse is of the form + = 1. 𝑎2 𝑏2 (9𝑥 2 − 36𝑥) + 4𝑦 2 = 108 9(𝑥 2 − 4𝑥) + 4𝑦 2 = 108 2 4 2 2 4 2 9 (𝑥 − 4𝑥 + ( ) ) + 4𝑦 = 108 + 9 ( ) 2 2 9(𝑥 2 − 4𝑥 + 4) + 4𝑦 2 = 108 + 36 9(𝑥 − 2)2 + 4𝑦 2 = 144 9(𝑥 − 2)2 4𝑦 2 144 + = 144 144 144 (𝒙 − 𝟐)𝟐 𝒚𝟐 + =𝟏 𝟏𝟔 𝟑𝟔 4. 4𝑥 2 − 𝑦 2 − 16𝑥 − 2𝑦 + 11 = 0 Solution: 𝐴 = 4 and 𝐶 = −1, since 𝐴 and 𝐶 have different signs, the equation is a hyperbola (𝑥−ℎ)2 (𝑦−𝑘)2 and the standard equation of a hyperbola is of the form − = 1. 𝑎2 𝑏2 (4𝑥 2 − 16𝑥) + (−𝑦 2 − 2𝑦) = −11 4(𝑥 2 − 4𝑥) − (𝑦 2 + 2𝑦) = −11 4 2 2 2 4 2 2 2 4 (𝑥 2 − 4𝑥 + ( ) ) − (𝑦 2 + 2𝑦 + ( ) ) = −11 + 4 ( ) − ( ) 2 2 2 2 4(𝑥 2 − 4𝑥 + 4) − (𝑦 2 + 2𝑦 + 1) = −11 + 16 − 1 4(𝑥 − 2)2 − (𝑦 + 1)2 = 4 4(𝑥 − 2)2 (𝑦 + 1)2 4 − = 4 4 4 (𝒚 + 𝟏)𝟐 (𝒙 − 𝟐)𝟐 − =𝟏 𝟒 AdDU -Senior High School | Pre-Calculus | Page 7 of 73 5. 16𝑥 2 + 25𝑦 2 − 32𝑥 + 150𝑦 − 159 = 0 Solution: 𝐴 = 16 and 𝐶 = 25, since both 𝐴 and 𝐶 have the same sign, the equation is an ellipse (𝑥−ℎ)2 (𝑦−𝑘)2 and the standard equation of an ellipse is of the form + = 1. 𝑎2 𝑏2 (16𝑥 2 − 32𝑥) + (25𝑦 2 + 150𝑦) = 159 16(𝑥 2 − 2𝑥) + 25(𝑦 2 + 6𝑦) = 159 2 2 2 2 6 2 2 2 6 2 16 (𝑥 − 2𝑥 + ( ) ) + 25 (𝑦 + 6𝑦 + ( ) ) = 159 + 16 ( ) + 25 ( ) 2 2 2 2 16(𝑥 2 − 2𝑥 + 1) + 25(𝑦 2 + 6𝑦 + 9) = 159 + 16 + 225 16(𝑥 − 1)2 + 25(𝑦 + 3)2 = 400 16(𝑥 − 1)2 25(𝑦 + 3)2 400 + = 400 400 400 (𝑥 − 1)2 (𝑦 + 3)2 + =1 25 16 6. 𝑥 2 + 2𝑥 + 𝑦 − 1 = 0 Solution: 𝐴 = 1 and 𝐶 = 0, since 𝐶 = 0, the equation is a parabola and the standard equation of the parabola is (𝑥 − ℎ)2 = ±4𝑝(𝑦 − 𝑘). 𝑥 2 + 2𝑥 = −𝑦 + 1 2 2 2 2 2 (𝑥 + 2𝑥 + ( ) ) = −𝑦 + 1 + ( ) 2 2 (𝑥 2 + 2𝑥 + 1) = −𝑦 + 2 (𝒙 + 𝟏)𝟐 = −(𝒚 − 𝟐) AdDU -Senior High School | Pre-Calculus | Page 8 of 73  ATENEO DE DAVAO UNIVERSITY Km 7 Central Park Blvd, Talomo, 8016 Davao City, Philippines Tel No. +63 (82) 221.2411 local 8608 E-Mail: [email protected] * www.addu.edu.ph In Consortium with Ateneo de Zamboanga University and Xavier University SENIOR HIGH SCHOOL – PRE-CALCULUS Topic: The Circle THE CIRCLE Definition: Circle - the set of all points P(x,y) in a plane which moves at a constant distance from a fixed point called its center C(h,k) and the constant distance of any point from the center is called the radius, r. Figure 1: Circle with center at C(h, k) Figure 2: Circle with center at C(0, 0) Equation of a Circle An equation of the second degree in which the xy-term is missing and the coefficients of the squared terms are always equal. Standard Equation of the circle: (𝒙 − 𝒉)𝟐 + (𝒚 − 𝒌)𝟐 = 𝒓𝟐 where: center at C(h, k) General Equation of the circle: 𝑥 2 + 𝑦 2 − 2ℎ𝑥 − 2𝑘𝑦 + 𝑐 = 0 or 𝟐 𝟐 𝒙 + 𝒚 + 𝑫𝒙 + 𝑬𝒚 + 𝑭 = 𝟎 AdDU -Senior High School | Pre-Calculus | Page 9 of 73 The general form of equation of a circle can be used to find for the center and the radius: 𝐷 𝐸 Center: ℎ =−2 𝑘 = −2 Radius: 𝑟 = √ℎ2 + 𝑘 2 − 𝐹 Where the radicand h2 + k 2 − F should be greater than zero. Otherwise, the following cases (degenerate circles) is observed: a) if ℎ2 + 𝑘 2 − 𝐹 = 0, then we have a point (or a point-circle). b) If ℎ2 + 𝑘 2 − 𝐹 < 0, then we have an empty (or null) set. If the equation of the circle is given in general form, completing the square can be used to transform it to center-radius form. EXAMPLES Example 1: Find the equations of the circle with center (𝟐, −𝟑) and radius 5 units. Sketch its graph. From the standard form, (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 ℎ = 2; 𝑘 = −3; and 𝑟 = 5 Substituting these values to the standard form, we have: 2 (𝑥 − 2)2 + (𝑦 − (−3)) = 52 (𝒙 − 𝟐)𝟐 + (𝒚 + 𝟑)𝟐 = 𝟐𝟓 To get the general equation of the circle, expand the standard equation. From (𝑥 − 2)2 + (𝑦 + 3)2 = 25 (𝑥 − 2)2 + (𝑦 + 3)2 = 25 (𝑥 2 − 4𝑥 + 4) + (𝑦 2 + 6𝑦 + 9) − 25 = 0 𝑥 2 + 𝑦 2 − 4𝑥 + 6𝑦 + 4 + 9 − 25 = 0 𝒙𝟐 + 𝒚𝟐 − 𝟒𝒙 + 𝟔𝒚 − 𝟏𝟐 = 𝟎 AdDU -Senior High School | Pre-Calculus | Page 10 of 73 Example 2: Find the equation of the circle with center (𝟑, 𝟐) and radius 2 units. Sketch its graph. From the standard form, (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 , ℎ = 3; 𝑘 = 2; and 𝑟 = 2 Substituting these values to the standard form, we have: (𝑥 − 3)2 + (𝑦 − 2)2 = 22 (𝒙 − 𝟑)𝟐 + (𝒚 − 𝟐)𝟐 = 𝟒 To get the general equation of the circle, expand the standard equation. (𝑥 − 3)2 + (𝑦 − 2)2 = 4 (𝑥 2 − 6𝑥 + 9) + (𝑦 2 − 4𝑦 + 4) − 4 = 0 𝑥 2 + 𝑦 2 − 6𝑥 + 4𝑦 + 9 + 4 − 4 = 0 𝒙𝟐 + 𝒚𝟐 − 𝟔𝒙 + 𝟒𝒚 + 𝟗 = 𝟎 Example 3: Find the equation of the circle with center 𝟓 𝟑 (𝟐 , − 𝟐) and radius 4 units. Graph. From the standard form, (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 , 5 3 ℎ = 2;𝑘 = − 2; and 𝑟 = 4 the equation is 2 5 2 3 (𝑥 − ) + (𝑦 − (− )) = 42 2 2 𝟐 𝟓 𝟑 𝟐 (𝒙 − ) + (𝒚 + ) = 𝟏𝟔 𝟐 𝟐 To get the general equation of the circle, expand the standard equation. 5 2 3 2 (𝑥 − ) + (𝑦 + ) = 16 2 2 25 9 (𝑥 2 − 5𝑥 + ) + (𝑦 2 + 3𝑦 + ) − 16 = 0 4 4 25 9 𝑥 2 + 𝑦 2 − 5𝑥 + 3𝑦 + + − 16 = 0 4 4 15 𝑥 2 + 𝑦 2 − 5𝑥 + 3𝑦 − =0 2 Multiply both sides by 2 to eliminate fraction. 𝟐𝒙𝟐 + 𝟐𝒚𝟐 − 𝟏𝟎𝒙 + 𝟔𝒚 − 𝟏𝟓 = 𝟎 AdDU -Senior High School | Pre-Calculus | Page 11 of 73 Example 4: Find the center and the radius of the circle given the equation below. Sketch its graph. 𝒙𝟐 + 𝒚𝟐 + 𝟒𝒙 − 𝟔𝒚 − 𝟐𝟑 = 𝟎 By completing the square, 𝑥 2 + 𝑦 2 + 4𝑥 − 6𝑦 − 23 = 0 (𝑥 2 + 4𝑥) + (𝑦 2 − 6𝑦) = 23 (𝑥 2 + 4𝑥 + 4) + (𝑦 2 − 6𝑦 + 9) = 23 + 4 + 9 (𝒙 + 𝟐)𝟐 + (𝒚 − 𝟑)𝟐 = 𝟑𝟔 From the standard equation, ℎ = −2, 𝑘 = 3, and 𝑟 = √36 = 6. Therefore, Center: (−𝟐, 𝟑) Radius: 𝟔 units Example 5: Find the center and the radius of the circle given the equation below. Sketch its graph. 𝒙𝟐 + 𝒚𝟐 + 𝟔𝒙 + 𝟏𝟎𝒚 − 𝟐 = 𝟎 By completing the square, 𝑥 2 + 𝑦 2 + 6𝑥 + 10𝑦 − 2 = 0 (𝑥 2 + 6𝑥) + (𝑦 2 + 10𝑦) = 2 (𝑥 2 + 6𝑥 + 9) + (𝑦 2 + 10𝑦 + 25) = 2 + 9 + 25 (𝒙 + 𝟑)𝟐 + (𝒚 + 𝟓)𝟐 = 𝟑𝟔 From the standard equation, ℎ = −3, 𝑘 = −5, and 𝑟 = √36 = 6. Therefore, Center: (−𝟑, −𝟓) Radius: 𝟔 units AdDU -Senior High School | Pre-Calculus | Page 12 of 73 Example 6: Find the center and the radius of the circle given the equation below. Sketch its graph. 𝒙𝟐 + 𝒚𝟐 − 𝟔𝒙 + 𝟐𝒚 − 𝟔 = 𝟎 By completing the square, 𝑥 2 + 𝑦 2 − 6𝑥 + 2𝑦 − 6 = 0 (𝑥 2 − 6𝑥) + (𝑦 2 + 2𝑦) = 6 (𝑥 2 − 6𝑥 + 9) + (𝑦 2 + 2𝑦 + 1 ) = 6 + 9 + 1 (𝒙 − 𝟑)𝟐 + (𝒚 + 𝟏)𝟐 = 𝟏𝟔 From the standard equation, ℎ = 3 , 𝑘 = −1 , 𝑎𝑛𝑑 𝑟 = √16 = 4 Therefore, Center: (𝟑, −𝟏) Radius: 𝟒 units Example 7: Find the center and the radius of the circle given the equation below. Sketch its graph 𝟑𝒙𝟐 + 𝟑𝒚𝟐 − 𝟔𝒙 + 𝟏𝟐𝒚 + 𝟏𝟐 = 𝟎 Simplify the equation by dividing both sides by 3 𝑥 2 + 𝑦 2 − 2𝑥 + 4𝑦 + 4 = 0 By completing the square, (𝑥 2 − 2𝑥) + (𝑦 2 + 4𝑦) = −4 (𝑥 2 − 2𝑥 + 1) + (𝑦 2 + 4𝑦 + 4) = −4 + 1 + 4 (𝒙 − 𝟏)𝟐 + (𝒚 + 𝟐)𝟐 = 𝟏 From the standard equation, ℎ = 1 , 𝑘 = −2 , 𝑎𝑛𝑑 𝑟 = √1 = 1. Therefore, Center: (𝟏, −𝟐) Radius: 𝟏 unit AdDU -Senior High School | Pre-Calculus | Page 13 of 73 Example 8: Find the center and the radius of the circle given the equation below. Sketch its graph 𝟏 𝟐 𝟏 𝟐 𝒙 + 𝒚 + 𝟒𝒙 + 𝟐𝒚 + 𝟖 = 𝟎 𝟐 𝟐 Simplify the equation by multiplying both sides by 2 𝑥 2 + 𝑦 2 + 8𝑥 + 4𝑦 + 16 = 0 By completing the square, (𝑥 2 + 8𝑥) + (𝑦 2 + 4𝑦) = −16 (𝑥 2 + 8𝑥 + 16) + (𝑦 2 + 4𝑦 + 4) = −16 + 16 + 4 (𝒙 + 𝟒)𝟐 + (𝒚 + 𝟐)𝟐 = 𝟒 From the standard equation, ℎ = −4 , 𝑘 = −2 , 𝑎𝑛𝑑 𝑟 = √4 = 2. Therefore, Center: (−𝟒, −𝟐) Radius: 𝟐 unit AdDU -Senior High School | Pre-Calculus | Page 14 of 73  ATENEO DE DAVAO UNIVERSITY Km 7 Central Park Blvd, Talomo, 8016 Davao City, Philippines Tel No. +63 (82) 221.2411 local 8608 E-Mail: [email protected] * www.addu.edu.ph In Consortium with Ateneo de Zamboanga University and Xavier University SENIOR HIGH SCHOOL – PRE-CALCULUS Topic: The Parabola THE PARABOLA Definition: Parabola - is the set of all points Q(x, y) in the plane at equal distance from a fixed point called the focus F, and a fixed line called the directrix D. - The line that passes through the focus and is perpendicular to the directrix D is called the axis of symmetry. - The axis of symmetry intersects the parabola at a point called the vertex V. - The line segment joining two points on the parabola passing through the focus and perpendicular to the axis is called the focal width, also known as the latus rectum. Figure 1: Characteristics of a Parabola Properties of the parabola: a. The parabola is shaped like the letter U. b. The vertex V is the midpoint of the line segment between the focus F and the directrix D. c. |p| is the focal distance from the vertex to the focus. d. |4p| is the length of the focal width or latus rectum. e. The focus F is not a point on the parabola nor a point on the directrix. F is a point on the axis of symmetry. f. We will only consider parabolas where D is either vertical or horizontal. The axis of symmetry will also be either vertical or horizontal. AdDU -Senior High School | Pre-Calculus | Page 15 of 73 Different cases for the parabola with vertex V(h, k) Standard Form Properties Graph 𝑉(ℎ, 𝑘) 𝐹(ℎ, 𝑘 + 𝑝) (𝑥 − ℎ)2 = 4𝑝(𝑦 − 𝑘) 𝐷: 𝑦 = 𝑘 − 𝑝 Axis of symmetry: 𝑥 = ℎ 𝑉(ℎ, 𝑘) 𝐹(ℎ, 𝑘 − 𝑝) (𝑥 − ℎ)2 = −4𝑝(𝑦 − 𝑘) 𝐷: 𝑦 = 𝑘 + 𝑝 Axis of symmetry: 𝑥 = ℎ 𝑉(ℎ, 𝑘) 𝐹(ℎ + 𝑝, 𝑘) (𝑦 − 𝑘)2 = 4𝑝(𝑥 − ℎ) 𝐷: 𝑥 = ℎ − 𝑝 Axis of symmetry: 𝑦 = 𝑘 𝑉(ℎ, 𝑘) 𝐹(ℎ − 𝑝, 𝑘) (𝑦 − 𝑘)2 = −4𝑝(𝑥 − ℎ) 𝐷: 𝑥 = ℎ + 𝑝 Axis of symmetry: 𝑦 = 𝑘 General Equation of the parabola: 𝐴𝑥 2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 or 𝐶𝑦 2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 An equation of the second degree in which the xy-term is missing and only one square term is present represents a parabola with its axis parallel to a coordinate axis. To transform the equation of the parabola in general form to standard form, use completing the squares. AdDU -Senior High School | Pre-Calculus | Page 16 of 73 EXAMPLES Example 1. Given the equation of the parabola: 𝑦 2 = 8𝑥, find the (a) coordinate of the vertex, (b) coordinate of the focus, (c) coordinates of the latus rectum, (d) equation of the directrix, (e) equation of the axis of symmetry and (f) sketch the graph. Solution: If we rewrite it: (𝑦 − 0)2 = 8(𝑥 − 0), it follows the form (𝑦 − 𝑘)2 = 4𝑝(𝑥 − ℎ), which means that the parabola opens to the right. This also implies that ℎ = 0; 𝑘 = 0; and 4𝑝 = 8 which will then give us 2𝑝 = 4 and 𝑝 = 2. (a) Vertex From the standard equation: ℎ = 0 and 𝑘 = 0 𝑉(ℎ, 𝑘) 𝑽(𝟎, 𝟎) (b) Focus The focus is 𝑝 units away from the vertex. Since 𝑝 = 2 and the opening of the parabola is to the right, the focus must be located 2 units to the right of 𝑉(0,0). Thus, 𝐹( 0 + 2, 0 ) 𝑭( 𝟐, 𝟎 ) (c) Ends of the Latus Rectum The ends of the latus rectum are 2𝑝 units away from the focus. Since 2𝑝 = 4 and the opening of the parabola is to the right, the ends of the latus rectum must be located 4 units up and down of the focus 𝐹(2,0). Thus, 𝐿𝑅1 ( 2 , 0 + 4); 𝐿𝑅2 (2 , 0 − 4) 𝑳𝑹𝟏 (𝟐, 𝟒); 𝑳𝑹𝟐 (𝟐, −𝟒) (d) Equation of the Directrix The directrix is a line located 𝑝 units away from the vertex but opposite direction of the focus. Since 𝑝 = 2, then the equation of the directrix must be 𝒙 = −𝟐 (e) Equation of the Axis of Symmetry The axis of symmetry is the equation of the line that divides the parabola into two symmetrical parts, thus the axis of symmetry is 𝒚=𝟎 (f) Graph Refer to the figure at the right. AdDU -Senior High School | Pre-Calculus | Page 17 of 73 Example 2. Given the standard equation of the parabola: (𝑥 + 4)2 = −8(𝑦 − 3), find the (a) coordinate of the vertex, (b) coordinate of the focus, (c) coordinates of the latus rectum, (d) equation of the directrix, (e) equation of the axis of symmetry, (f) general equation, and (g) sketch the graph. Solution: Rewriting the given:[𝑥 − (−4)]2 = −8(𝑦 − 3), it follows the form (𝑥 − ℎ)2 = −4𝑝(𝑦 − 𝑘), which means that the parabola opens downward. This also implies that ℎ = −4; 𝑘 = 3; and 4𝑝 = 8 which will then give us 2𝑝 = 4 and 𝑝 = 2. (a) Vertex From the standard equation: ℎ = −4 and 𝑘 = 3 𝑉(ℎ, 𝑘) 𝑽(−𝟒, 𝟑) (b) Focus The focus is 𝑝 units away from the vertex. Since 𝑝 = 2 and the opening of the parabola is downward, the focus must be located 2 units below of 𝑉(−4,3). Thus, 𝐹(−4, 3 − 2 ) 𝑭(−𝟒, 𝟏) (c) Ends of the Latus Rectum The ends of the latus rectum are 2𝑝 units away from the focus. Since 2𝑝 = 4 and the opening of the parabola is downward, the ends of the latus rectum must be located 4 units to the right and left from the focus 𝐹(−4, 1 ). Thus, 𝐿𝑅1 (−4 + 4, 1 ); 𝐿𝑅2 (−4 − 4, 1 ) 𝑳𝑹𝟏 ( 𝟎, 𝟏 ); 𝑳𝑹𝟐 (−𝟖, 𝟏 ) (d) Equation of the Directrix The directrix is a line located 𝑝 units away from the vertex but opposite direction of the focus. Since 𝑝 = 2, then the equation of the directrix must be 𝒚=𝟓 (e) Equation of the Axis of Symmetry The axis of symmetry is the equation of the line that divides the parabola into two symmetrical parts, thus the axis of symmetry is 𝒙 = −𝟒 (f) General Equation (𝑥 + 4)2 = −8(𝑦 − 3) 2 𝑥 + 8𝑥 + 16 = −8𝑦 + 24 𝑥 2 + 8𝑥 + 8𝑦 + 16 − 24 = 0 𝒙𝟐 + 𝟖𝒙 + 𝟖𝒚 − 𝟖 = 𝟎 (g) Graph Refer to the figure at the right. AdDU -Senior High School | Pre-Calculus | Page 18 of 73 Example 3. Given the general equation of the parabola: 𝑥 2 − 12𝑥 − 4𝑦 + 16 = 0, solve for the (a) standard equation, (b) coordinate of the vertex, (c) coordinate of the focus, (d) coordinates of the latus rectum, (e) equation of the directrix, (f) equation of the axis of symmetry and (g) sketch the graph. Solution: (a) Standard Equation To solve the standard equation given the general equation, we use completing the square: 𝑥 2 − 12𝑥 − 4𝑦 + 16 = 0 𝑥 2 − 12𝑥 = 4𝑦 − 16 𝑥 2 − 12𝑥 + 36 = 4𝑦 − 16 + 36 (𝑥 − 6)2 = 4𝑦 + 20 (𝒙 − 𝟔)𝟐 = 𝟒(𝒚 + 𝟓) If we rewrite it this form: (𝑥 − 6)2 = 4[𝑦 − (−5)], it follows the form (𝑥 − ℎ)2 = 4𝑝(𝑦 − 𝑘), which means that the parabola opens upward. This also implies that ℎ = 6; 𝑘 = −5; and 4𝑝 = 4 which will then give us 2𝑝 = 2 and 𝑝 = 1. (b) Vertex From the standard equation: ℎ = 6 and 𝑘 = −5 𝑉(ℎ, 𝑘) 𝑽(𝟔, −𝟓) (c) Focus The focus is 𝑝 units away from the vertex. Since 𝑝 = 1 and the opening of the parabola is upward, the focus must be located 1 unit above of 𝑉(6, −5). Thus, 𝐹(6, −5 + 1) 𝑭(𝟔, −𝟒) (d) Ends of the Latus Rectum The ends of the latus rectum are 2𝑝 units away from the focus. Since 2𝑝 = 2 and the opening of the parabola is upward, the ends of the latus rectum must be located 2 units to the right and left from the focus 𝐹(6, −4). Thus, 𝐿𝑅1 (6 + 2, −4); 𝐿𝑅2 (6 − 2, −4) 𝑳𝑹𝟏 (𝟖, −𝟒); 𝑳𝑹𝟐 (𝟒, −𝟒) (e) Equation of the Directrix The directrix is a line located 𝑝 units away from the vertex but opposite direction of the focus. Since 𝑝 = 1, then the equation of the directrix must be 𝒚 = −𝟔 (f) Equation of the Axis of Symmetry The axis of symmetry is the equation of the line that divides the parabola into two symmetrical parts, thus the axis of symmetry is 𝒙=𝟔 (g) Graph Refer to the figure at the right. AdDU -Senior High School | Pre-Calculus | Page 19 of 73 Example 4. Given the general equation of the parabola 𝑦 2 + 4𝑦 + 6𝑥 + 10 = 0, solve for the (a) standard equation, (b) coordinate of the vertex, (c) coordinate of the focus, (d) coordinates of the latus rectum, (e) equation of the directrix, (f) equation of the axis of symmetry and (g) sketch the graph. Solution: (a) Standard Equation To solve the standard equation given the general equation, we use completing the square: 𝑦 2 + 4𝑦 + 6𝑥 + 10 = 0 𝑦 2 + 4𝑦 = −6𝑥 − 10 𝑦 2 + 4𝑦 + 4 = −6𝑥 − 10 + 4 (𝒚 + 𝟐)𝟐 = −𝟔(𝒙 + 𝟏) Rewriting, [𝑦 − (−2)]2 = −6[𝑥 − (−1)], it follows the form (𝑦 − 𝑘)2 = −4𝑝(𝑥 − ℎ), which means that the parabola opens to the left. This also implies that ℎ = −1; 𝑘 = −2; and 4𝑝 = 6 3 which will then give us 2𝑝 = 3 and 𝑝 = 2. (b) Vertex From the standard equation: ℎ = −1 and 𝑘 = −2 𝑉(ℎ, 𝑘) 𝑽(−𝟏, −𝟐) (c) Focus 3 The focus is 𝑝 units away from the vertex. Since 𝑝 = 2 and the opening of the parabola is 3 to the left, the focus must be located 2 units to the left of 𝑉(−1, −2). Thus, 3 𝐹 (−1 − , −2) 2 𝟓 𝑭 (− , −𝟐) 𝟐 (d) Ends of the Latus Rectum The ends of the latus rectum are 2𝑝 units away from the focus. Since 2𝑝 = 3 and the opening of the parabola is to the left, the ends of the latus rectum must be located 3 units up and 5 down from the focus 𝐹 (− 2 , −2). Thus, 5 5 𝐿𝑅1 (− , −2 + 3) ; 𝐿𝑅2 (− , −2 − 3) 2 2 𝟓 𝟓 𝑳𝑹𝟏 (− , 𝟏) ; 𝑳𝑹𝟐 (− , −𝟓) 𝟐 𝟐 (e) Equation of the Directrix The directrix is a line located 𝑝 units away from the vertex but opposite direction of 3 the focus. Since 𝑝 = 2, then the equation of the directrix must be 𝟏 𝒙= 𝟐 (f) Equation of the Axis of Symmetry The axis of symmetry is the equation of the line that divides the parabola into two symmetrical parts, thus the axis of symmetry is 𝒚 = −𝟐 (g) Graph Refer to the figure at the right. AdDU -Senior High School | Pre-Calculus | Page 20 of 73  1 Example 5. Given the general equation of the parabola 4 𝑥 2 − 2𝑥 + 𝑦 + 6 = 0, solve for the (a) standard equation, (b) coordinate of the vertex, (c) coordinate of the focus, (d) coordinates of the latus rectum, (e) equation of the directrix, (f) equation of the axis of symmetry and (g) sketch the graph. Solution: (a) Standard Equation To solve the standard equation given the general equation, we use completing the square: 1 2 𝑥 − 2𝑥 + 𝑦 + 6 = 0 Multiply both sides by 4 4 𝑥 2 − 8𝑥 + 4𝑦 + 24 = 0 𝑥 2 − 8𝑥 = −4𝑦 − 24 𝑥 2 − 8𝑥 + 16 = −4𝑦 − 24 + 16 (𝑥 − 4)2 = −4𝑦 − 8 (𝒙 − 𝟒)𝟐 = −𝟒(𝒚 + 𝟐) Rewriting, (𝑥 − 4)2 = −4[𝑦 − (−2)], it follows the form (𝑥 − ℎ)2 = −4𝑝(𝑦 − 𝑘), which means that the parabola downward. This also implies that ℎ = 4; 𝑘 = −2; and 4𝑝 = 4 which will then give us 2𝑝 = 2 and 𝑝 = 1. (b) Vertex From the standard equation: ℎ = 4 and 𝑘 = −2 𝑉(ℎ, 𝑘) 𝑽(𝟒, −𝟐) (c) Focus The focus is 𝑝 units away from the vertex. Since 𝑝 = 1 and the opening of the parabola is downward, the focus must be located 1 unit down of 𝑉(4, −2). Thus, 𝐹(4 , −2 − 1) 𝑭(𝟒, −𝟑) (d) Ends of the Latus Rectum The ends of the latus rectum are 2𝑝 units away from the focus. Since 2𝑝 = 2 and the opening of the parabola is downward, the ends of the latus rectum must be located 2 units to the left and right from the focus 𝐹(4, −3). Thus, 𝐿𝑅1 (4 + 2 , −3); 𝐿𝑅2 (4 − 2 , −3) 𝑳𝑹𝟏 ( 𝟔 , −𝟑); 𝑳𝑹𝟐 ( 𝟐 , −𝟑) (e) Equation of the Directrix The directrix is a line located 𝑝 units away from the vertex but opposite direction of the focus. Since 𝑝 = 1, then the equation of the directrix must be 𝒚 = −𝟏 (f) Equation of the Axis of Symmetry The axis of symmetry is the equation of the line that divides the parabola into two symmetrical parts, thus the axis of symmetry is 𝒙=𝟒 (g) Graph Refer to the figure at the right. AdDU -Senior High School | Pre-Calculus | Page 21 of 73  1 Example 6. Find the equations of the parabola that opens downward, has directrix 𝑦 = − 2, if the vertex is on line 2𝑥 − 𝑦 + 1 = 0 and the focus is on line 2𝑦 − 5𝑥 + 1 = 0. Let the vertex be at (h,k). With the parabola opening downward, the directrix must have an equation 𝑦 = 𝑘 + 𝑝 and the focus must have the coordinates (ℎ, 𝑘 − 𝑝). We can then generate these equations: 1 𝑘+𝑝 =− 2 2ℎ − 𝑘 + 1 = 0 2(𝑘 − 𝑝) − 5ℎ + 1 = 0 5 Solving the system, we obtain 𝑝 = 2 , ℎ = −2, 𝑎𝑛𝑑 𝑘 = −3. Therefore, the equation is (𝒙 + 𝟐)𝟐 = −𝟏𝟎(𝒚 + 𝟑). Example 7. A satellite dish has a cross section that is a parabola. The signals from a satellite strike the surface of the dish and are reflected to the focus of the parabola where the receive is positioned. If the dish is 6 ft across at its opening and 2 ft deep at its center, where should the receiver be placed? The situation may be illustrated in a Cartesian plane. Setting the vertex at (0,0) and letting the parabola open upward, we get this graph: Since the parabola opens upward, its equation must be of the form 𝑥 2 = 4𝑝𝑦. We see that (2,3) is 9 on the parabola, so we have (3)2 = 4𝑝(2), making 𝑝 = 8. This means that the receiver should be 9 9 at the focus (0, 8). Therefore, the receiver should be placed 8 𝑓𝑡 above the lowest point of the dish. AdDU -Senior High School | Pre-Calculus | Page 22 of 73  ATENEO DE DAVAO UNIVERSITY Km 7 Central Park Blvd, Talomo, 8016 Davao City, Philippines Tel No. +63 (82) 221.2411 local 8608 E-Mail: [email protected] * www.addu.edu.ph In Consortium with Ateneo de Zamboanga University and Xavier University SENIOR HIGH SCHOOL – PRE-CALCULUS Topic: The Ellipse THE ELLIPSE Definition: Ellipse - is the set of all points (𝑥, 𝑦) in the plane, the sum of their distances from two distinct fixed points, called foci (𝐹1 and 𝐹2 ), is a constant. - If 𝑄 and 𝑅 are arbitrary points on the ellipse, then from the definition: ̅̅̅̅̅1 | + |𝑄𝐹 |𝑄𝐹 ̅̅̅̅̅2 | = |𝑅𝐹 ̅̅̅̅̅1 | + |𝑅𝐹 ̅̅̅̅̅2 | Properties of the ellipse: (refer to Figure 1) a. The ellipse is a closed curve and is symmetrical with respect to both its axes. b. Two points on the ellipse, which are points of intersection of the ellipse and the line passing through the two foci 𝐹1 and 𝐹2 , are called the vertices 𝑉1 and 𝑉2. c. The line that passes through the foci 𝐹1 and 𝐹2 , and the vertices 𝑉1 and 𝑉2, is called the principal axis. Ellipse with either horizontal or vertical principal axis will only be considered. d. The line segment connecting 𝑉1 and 𝑉2 passing through the foci is called the major axis of the ellipse. The midpoint of the major axis is the center 𝐶(ℎ, 𝑘) of the ellipse. Let the distance from a vertex to the center be 𝑎. Then, the length of the major axis is 𝟐𝒂, where 𝑎 is a constant. The coordinates of the vertices are 𝑉1 (ℎ − 𝑎, 𝑘) and 𝑉2 (ℎ + 𝑎, 𝑘). e. The line segment perpendicular to the major axis which passes through the center 𝐶 is called the minor axis. The two points on the ellipse which intersect the minor axis are called co-vertices. Let the distance from a co-vertex to the center be 𝑏. Then, the length of the minor axis is 𝟐𝒃, where 𝑏 is a constant. f. We denote the distance from the center 𝐶(ℎ, 𝑘) to one focus as 𝑐. Thus, the coordinates of the foci are 𝐹1 (ℎ − 𝑐, 𝑘) and 𝐹2 (ℎ + 𝑐, 𝑘). ̅̅̅̅̅1 | + |𝑃𝐹 g. For any point P on the ellipse: |𝑃𝐹 ̅̅̅̅̅2 | = 2𝑎 AdDU -Senior High School | Pre-Calculus | Page 23 of 73  Figure 1: Characteristics of an Ellipse with a horizontal principal axis Standard equation for the ellipse with center C(h, k) Standard Form Properties Graph 𝐶(ℎ, 𝑘) Vertices: 𝑉1 (ℎ − 𝑎, 𝑘) 𝑉2 (ℎ + 𝑎, 𝑘) Co-vertices: (ℎ, 𝑘 ± 𝑏) (𝑥 − ℎ)2 (𝑦 − 𝑘)2 Foci: 𝐹1 (ℎ − 𝑐, 𝑘) + 𝐹1 (ℎ + 𝑐, 𝑘) 𝑎2 𝑏2 =1 Length of major axis: 2𝑎 Length of minor axis: 2𝑏 Principal axis: horizontal 0 0) Transverse Axis: Horizontal Length = 2a Asymptotes: 𝑏 𝑦 − 𝑘 = ± (𝑥 − ℎ) 𝑎 Center: 𝐶(ℎ, 𝑘) Vertices: 𝑉1 (ℎ, 𝑘 − 𝑎) 𝑉2 (ℎ, 𝑘 + 𝑎) Endpoints of the (𝑦 − 𝑘)2 (𝑥 − ℎ)2 conjugate axis: (ℎ ± − 𝑏, 𝑘) 𝑎2 𝑏2 Foci: 𝐹1 (ℎ, 𝑘 − 𝑐) =1 𝐹1 (ℎ, 𝑘 + 𝑐) (a > 0, b > 0) 𝑐 2 = 𝑎2 + 𝑏 2 Transverse Axis: Vertical Length = 2a Asymptotes: 𝑎 𝑦 − 𝑘 = ± (𝑥 − ℎ) 𝑏 AdDU -Senior High School | Pre-Calculus | Page 29 of 73  General equation of the hyperbola: 𝐴𝑥 2 + 𝐶𝑦 2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 An equation of the second degree in which the xy-term is missing and coefficients of x2 and y2 are not numerically equal and of opposite sign represents a hyperbola with axes parallel to the coordinate axes. Steps in sketching the graph of a hyperbola: i. Draw a rectangle with sides parallel to the axes, and of lengths 2a and 2b. The midpoint of the two diagonals of the rectangle should be the center C(h, k) of the hyperbola. This rectangle is called the auxiliary rectangle (or central box). ii. Extend the two diagonals of the auxiliary rectangle to get the graphs of the two asymptotes. iii. Plot the vertices 𝑉1 (ℎ − 𝑎, 𝑘) and 𝑉2 (ℎ + 𝑎, 𝑘) iv. From one vertex, draw a smooth curve, which approaches the asymptotes. This is the one branch of the hyperbola. Apply this to the other vertex. Example 1: Graph the equation 16𝑥 2 − 9𝑦 2 − 96𝑥 − 18𝑦 − 9 = 0. Determine the coordinates of the center, vertices, covertices, foci, and equations of the asymptotes. By completing the square, we get 16𝑥 2 − 9𝑦 2 − 96𝑥 − 18𝑦 − 9 = 0 16(𝑥 2 − 6𝑥) − 9(𝑦 2 + 2𝑦) = 9 16(𝑥 − 6𝑥 + 9) − 9(𝑦 2 + 2𝑦 + 1) = 9 + 144 − 9 2 16(𝑥 − 3)2 − 9(𝑦 + 1)2 = 144 (𝒙 − 𝟑)𝟐 (𝒚 + 𝟏)𝟐 − =𝟏 𝟗 𝟏𝟔 We can infer from the equation that the hyperbola has a horizontal principal axis, with 𝑎 = 3, 𝑏 = 4, and 𝑐 = √16 + 9 = √25 = 5. It has the center (3, −1), vertices (0, −1) and (6, −1), covertices (3,3) and (3, −5), and foci (−2, −1) & (8, −1). The equations of the asymptotes are 4 (𝑦 + 1) = ± (𝑥 − 3) 3 4 ⟹ 𝑦 = 3 𝑥 − 5 and 𝑦 = 4 − 3 𝑥 + 3. AdDU -Senior High School | Pre-Calculus | Page 30 of 73  𝑥2 𝑦2 Example 2: Draw a sketch of the graph of the equation − = 1. 9 4 We can infer from the equation that the center of the hyperbola is at (0,0) with a horizontal principal axis. Given that 𝑎 = 3 and 𝑏 = 2, the vertices of the hyperbola are at (±3,0) and 2 2 the covertices are at (0, ±2). The hyperbola is asymptotic at 𝑦 = − 3 𝑥 and 𝑦 = 3 𝑥. Now, 𝑐 = √𝑎2 + 𝑏 2 = √9 + 4 = √13. Then, the foci are at (±√13, 0). Example 3: Find an equation of the hyperbola with vertices at (1,6) and (1, −2), and foci at (1,7) and (1, −3). To find the center, we obtain the midpoint of the transverse axis (whose endpoints are the vertices). Thus, 1+1 6−2 the center is ( 2 , 2 ) = (1,2). The transeve axis has the length 2𝑎 = 6 + 2 = 8 while the distance between the two foci is 2𝑐 = 7 + 3 = 10. Then, 𝑎 = 4 and 𝑐 = 5. Since 𝑐 = √𝑎2 + 𝑏 2 , then, 𝑏 = √𝑐 2 − 𝑎2 = √25 − 16 = 3. Since the principal axis is vertical, then the equation of the (𝑦−2)2 (𝑥−1)2 parabola is − = 1. 16 9 AdDU -Senior High School | Pre-Calculus | Page 31 of 73 Example 4: Find the equation of a hyperbola with foci (2,2) and (2, −3) if the conjugate axis has length 4. The given foci imply that the principal axis is vertical. To locate the center, we obtain the 1 midpoint of the line segment joining the foci. Thus, 𝐶 = (2, − 2). The distance between the two 5 foci is 2𝑐 = 5, so 𝑐 = 2. The conjugate axis has the length 2𝑏 = 4, so 𝑏 = 2. Then, 𝑎 = 1 2 25 9 3 4(𝑦+ ) (𝑥−2)2 √𝑐 2 − 𝑏 2 = √ 4 − 4 = √4 = 2. Thus, the equation of the hyperbola is 2 − = 1. 9 4 Example 5: Find the equation of a hyperbola that has covertex at (2√2 − 2,3) and asympototes 𝑦 + 3𝑥 + 3 = 0 and 𝑦 − 3𝑥 − 9 = 0. Since the asymptotes intersect at the center of the hyperbola, we solve the system: 𝑦 + 3𝑥 + 3 = 0 𝑦 − 3𝑥 − 9 = 0 We obtain 𝑦 = 3 and 𝑥 = −2, so the center is (−2,3). Since the y-coordinate of the center and the given covertex is the same, it follows that the hyperbola has a vertical principal axis. The distance between the covertex and the center is 𝑏 = 2√2. To get the value of 𝑎, we use the slope 𝑎 of the asymptotes ± 𝑏 = ±3 ⟹ 𝑎 = 3𝑏 = 6√2. Therefore, the equation of the hyperbola is (𝑦−3)2 (𝑥+2)3 (𝑦−3)2 (𝑥+2)2 2 − 2 =1⟹ − = 1. (6√2) (2√2) 72 8 Example 6: Given the graph, find the standard equation of the hyperbola. Since it is a vertical hyperbola, we will have this equation: (𝑦−𝑘)2 (𝑥−ℎ)2 − =1 𝑎2 𝑏2 Identify the coordinates of the center (h, k): (−1, −2) Identify the values of 𝑎 (transverse axis) and 𝑏 (conjugate axis): 𝑎 = 5 𝑎𝑛𝑑 𝑏 = 3 Substitute the identified values to the standard form (𝑦−𝑘)2 (𝑥−ℎ)2 of the vertical hyperbola: − = 1. 𝑎2 𝑏2 Thus, the standard equation of the hyperbola is: (𝑦+2)2 (𝑥+1)2 − =1 25 9 AdDU -Senior High School | Pre-Calculus | Page 32 of 73  ATENEO DE DAVAO UNIVERSITY Km 7 Central Park Blvd, Talomo, 8016 Davao City, Philippines Tel No. +63 (82) 221.2411 local 8608 E-Mail: [email protected] * www.addu.edu.ph In Consortium with Ateneo de Zamboanga University and Xavier University SENIOR HIGH SCHOOL – PRE-CALCULUS Topic: Systems of Non-Linear Equations SYSTEMS OF NON-LINEAR EQUATIONS Definition: - A system of equations is nonlinear if it involves at least one equation that is not linear. - A solution to a system of equations of two variables is an ordered pair that makes all the equations true. Graphically, this/these is/are the point/s of intersection between graphs of conics (Figure 1). The number of solutions correspond to the number of intersections. - The collection of all solutions enclosed in braces and separated by commas is what we call a solution set (SS). - Substitution and elimination methods are used to find solutions to a given system and verify the solutions using graphical method. Figure 1. Solutions to a system of equations are graphically expressed as intersections among graphs. STEPS IN SOLVING SYSTEMS OF NON-LINEAR EQUATIONS By substitution method: 1. Solve for one variable in terms of the other variable of the equations. 2. Substitute the expression obtained in step 1 to the other equation. 3. Solve the equation obtained in step 2. 4. Substitute back the value obtained in step 3 to the equation in step 1 to get the value of the remaining variable. By elimination method: 1. Multiply one or both equations by appropriate constant/s so that one of the variables in each equation will have the same coefficient with opposite signs. 2. Add the corresponding sides of the two equations. 3. Solve the equation of one variable obtained in step 2. 4. Substitute the value obtained in step 3 to one of the original equations. Generally, we use elimination method when there is one variable we can readily eliminate. Otherwise, we use substitution method. AdDU -Senior High School | Pre-Calculus | Page 33 of 73 Examples 1. Solve the system of equations: 9𝑥 2 − 25𝑦 2 = 225 (1) { 𝑥+𝑦+4=0 (2) We examine the variables in (1) and (2) and find no variable we can readily eliminate. We use substitution method. Step 1: Solve for one variable in terms of the other variable of the equations. Solve for 𝑥 in (2) 𝑥+𝑦+4=0 𝑥 = −𝑦 − 4 Step 2: Substitute the expression obtained in step 1 to the other equation. Substitute −𝑦 − 4 to 𝑥 in (1) 9𝑥 2 − 25𝑦 2 = 225 9(−𝑦 − 4)2 − 25𝑦 2 = 225 Step 3: Solve the equation obtained in step 2. Solve for 𝑦 9(−𝑦 − 4)2 − 25𝑦 2 = 225 9(𝑦 + 8𝑦 + 16) − 25𝑦 2 = 225 2 16𝑦 2 − 72𝑦 + 81 = 0 (4𝑦 − 9)2 = 0 𝟗 𝒚= 𝟒 Step 4: Substitute back the value obtained in step 3 to the equation in step 1 to get the value of the remaining variable. 9 Substitute 4 to 𝑦 in 𝑥 = −𝑦 − 4 𝑥 = −𝑦 − 4 9 𝑥 = −( )−4 4 𝟐𝟓 𝒙=− 𝟒 25 9 Therefore, the solution set is {(− , 4)}. This is where the two graphs intersect (Figure 2). 4 AdDU -Senior High School | Pre-Calculus | Page 34 of 73  25 9 Figure 2. The hyperbola and the line intersect at 𝐴 (− , ) 𝑜𝑟 𝐴(6.25, 2.25) 4 4 2. Solve the system of equations: 𝑦 − 𝑥 2 = −1 (1) { 𝑥+𝑦 =1 (2) We see that in the given system, we can readily eliminate 𝑦. We use elimination method. Step 1: Multiply one or both equations by appropriate constant/s so that one of the variables in each equation will have the same coefficient with opposite signs. Multiply (1) by −1 (−1)(𝑦 − 𝑥 2 ) = (−1)(−1) −𝑦 + 𝑥 2 = 1 Step 2: Add the corresponding sides of the two equations. Add the equation obtained above to (2), thereby eliminating y −𝑦 + 𝑥 2 = 1 𝑥+𝑦 =1 ⇓ 𝑥 + 𝑥2 = 2 Step 3: Solve the equation of one variable obtained in step 2. Solve for 𝑥 𝑥 + 𝑥2 = 2 𝑥2 + 𝑥 − 2 = 0 (𝑥 + 2)(𝑥 − 1) = 0 𝑥 = −2 , 𝑥 = 1 Step 4: Substitute the value obtained in step 3 to one of the original equations. Substitute −2 and 1 to x in (2) 𝑥+𝑦 =1 𝑥+𝑦 =1 −2 + 𝑦 = 1 1+𝑦 =1 𝒚=𝟑 𝒚=𝟎 Therefore, the solution set is {(−2,3), (1,0)}. These are where the graphs intersect (Figure 3). AdDU -Senior High School | Pre-Calculus | Page 35 of 73  Figure 3. The parabola and the line intersect at 𝐴(−2,3) and 𝐵(1,0). 3. Solve the system of equations: (𝑥 + 2)2 − 4(𝑦 − 1)2 = 1 (1) { 4(𝑥 + 2)2 + (𝑦 − 1)2 = 4 (2) We can use substitution to make the equations look simpler. Let 𝑎 = (𝑥 + 2)2 and 𝑏 = (𝑦 − 1)2. We will have: 𝑎 − 4𝑏 = 1 (1) { 4𝑎 + 𝑏 = 4 (2) We see that we can readily eliminate either 𝑎2 or 𝑏 2. We use elimination method. Step 1: Multiply one or both equations by appropriate constant/s so that one of the variables in each equation will have the same coefficient with opposite signs. Multiply (1) by −4 (−4)(𝑎 − 4𝑏) = (1)(−4) −4𝑎 + 16𝑏 = −4 Step 2: Add the corresponding sides of the two equations. Add the equation obtained above to (2), thereby eliminating a −4𝑎 + 16𝑏 = −4 4𝑎 + 𝑏 = 4 ⇓ 17𝑏 = 0 Step 3: Solve the equation of one variable obtained in step 2. Solve for 𝑏 17𝑏 = 0 𝒃=𝟎 Step 4: Substitute the value obtained in step 3 to one of the original equations. Substitute 0 to b in (1) AdDU -Senior High School | Pre-Calculus | Page 36 of 73  𝑎 − 4𝑏 = 1 𝑎 − 4(0) = 1 𝒂=𝟏 Since 𝑎 = (𝑥 + 2)2 and 𝑏 = (𝑦 − 1)2 , then 𝑎 = (𝑥 + 2)2 𝑏 = (𝑦 − 1)2 1 = (𝑥 + 2)2 0 = (𝑦 − 1)2 ±1 = 𝑥 + 2 0=𝑦−1 𝒙 = −𝟑 , 𝒙 = −𝟏 𝒚=𝟏 Therefore, the solution set is {(−3,1), (−1,1)}. These are where the graphs intersect (Figure 4) Figure 4. The hyperbola and the ellipse intersect at 𝐴(−3,1) and 𝐵(−1,1) 4. Solve the system of equations: 𝑥2 − 𝑦2 = 5 (1) { 2 𝑥 + 𝑦 2 = 13 (2) We see that we can readily eliminate 𝑦 2. We use elimination method. Step 1: Multiply one or both equations by appropriate constant/s so that one of the variables in each equation will have the same coefficient with opposite signs. We skip this step, since we already have −𝑦 2 and 𝑦 2 in the first and second equation, respectively. Step 2: Add the corresponding sides of the two equations. Add both (1) and (2) 𝑥2 − 𝑦2 = 5 𝑥 2 + 𝑦 2 = 13 ⇓ 2 2𝑥 = 18 Step 3: Solve the equation of one variable obtained in step 2. AdDU -Senior High School | Pre-Calculus | Page 37 of 73  Solve for 𝑥 2𝑥 2 = 18 𝑥2 = 9 𝒙 = ±𝟑 Step 4: Substitute the value obtained in step 3 to one of the original equations. Substitute 3 and −3 to 𝑥 in (2) 𝑥 2 + 𝑦 2 = 13 𝑥 2 + 𝑦 2 = 13 32 + 𝑦 2 = 13 (−3)2 + 𝑦 2 = 13 9 + 𝑦 2 = 13 9 + 𝑦 2 = 13 𝑦2 = 4 𝑦2 = 4 𝒚 = ±𝟐 𝒚 = ±𝟐 Therefore, the solution set is {(−3, −2), (−3,2), (3, −2), (3,2)}. These are where the graphs intersect (Figure 5). Figure 5. The hyperbola and the circle intersect at A(-3,-2), B(-3,2), C(3,2) and D(3,-2). 5. Solve the system of equations: 2 2 𝑦 − 6𝑦 − 4𝑥 − 8𝑥 + 21 = 0 (1) { 2𝑥 − 𝑦 2 + 6𝑦 − 3 = 0 (2) We see that we can readily eliminate 𝑦 2 and 6𝑦from both equations. We use elimination method. Step 1: Multiply one or both equations by appropriate constant/s so that one of the variables in each equation will have the same coefficient with opposite signs. We skip this step, since we already have −𝑦 2 and 𝑦 2 , and −6𝑦 and 6y in the first and second equation, respectively. Step 2: Add the corresponding sides of the two equations. Add both (1) and (2) 𝑦 2 − 6𝑦 − 4𝑥 2 − 8𝑥 + 21 = 0 2𝑥 − 𝑦 2 + 6𝑦 − 3 = 0 ⇓ −4𝑥 2 − 6𝑥 + 18 = 0 Step 3: Solve the equation of one variable obtained in step 2. Solve for x AdDU -Senior High School | Pre-Calculus | Page 38 of 73  −4𝑥 2 − 6𝑥 + 18 = 0 −4𝑥 2 − 6𝑥 + 18 0 = −2 2 2𝑥 2 + 3𝑥 − 9 = 0 (2𝑥 − 3)(𝑥 + 3) = 0 2𝑥 − 3 = 0; 𝑥 + 3 = 0 3 𝑥 = ; 𝑥 = −3 2 Step 4: Substitute the value obtained in step 3 to one of the original equations. 3 Substitute 2 and −3 to 𝑥 in (2) 2𝑥 − 𝑦 2 + 6𝑦 − 3 = 0 2𝑥 − 𝑦 2 + 6𝑦 − 3 = 0 3 2(−3) − 𝑦 2 + 6𝑦 − 3 = 0 2 ( ) − 𝑦 2 + 6𝑦 − 3 = 0 2 −6 − 𝑦 2 + 6𝑦 − 3 = 0 3 − 𝑦 2 + 6𝑦 − 3 = 0 −𝑦 2 + 6𝑦 − 9 = 0 −𝑦 2 + 6𝑦 = 0 𝑦 2 − 6𝑦 + 9 = 0 −𝑦(𝑦 − 6) = 0 (𝑦 − 3)2 = 0 𝒚=𝟎, 𝒚=𝟔 𝑦−3=0 𝒚=𝟑 3 3 Therefore, the solution set is {(−3,3), (2 , 0) , (2 , 6)}. These are where the graphs intersect (Figure 6). Figure 6. The hyperbola and the parabola 3 3 intersect at A(-3,3), 𝐵 ( , 6), and 𝐶 ( , 0). 2 2 AdDU -Senior High School | Pre-Calculus | Page 39 of 73 6. Solve the system of equations: 4𝑥 2 + 3𝑦 2 = 48 (1) { 2 𝑥 + 𝑦2 = 4 (2) We can manipulate the equations to eliminate either 𝑦 2 or 𝑥 2. We use elimination method. Step 1: Multiply one or both equations by appropriate constant/s so that one of the variables in each equation will have the same coefficient with opposite signs. Multiply −3 to (2), thereby eliminating 𝑦 2. (−3)(𝑥 2 + 𝑦 2 ) = (4)(−3) −3𝑥 2 − 3𝑦 2 = −12 Step 2: Add the corresponding sides of the two equations. Add both (1) and the equation obtained above. 4𝑥 2 + 3𝑦 2 = 48 −3𝑥 2 − 3𝑦 2 = −12 ⇓ 𝑥 2 = 36 Step 3: Solve the equation of one variable obtained in step 2. Solve for x 𝑥 2 = 36 𝑥 = ±6 Step 4: Substitute the value obtained in step 3 to one of the original equations. Substitute 6 and −6 to x in (2) 𝑥2 + 𝑦2 = 4 𝑥2 + 𝑦2 = 4 (6)2 + 𝑦 2 = 4 (−6)2 + 𝑦 2 = 4 𝑦 2 = 4 − 36 𝑦 2 = 4 − 36 𝑦 = ±√−32 𝑦 = ±√−32 We get imaginary numbers as values of 𝑦. Therefore, the solution set is an empty set or {∅}. This means that the graphs do not intersect at all (Figure 7). Figure 7. The ellipse and the circle do not intersect at all. AdDU -Senior High School | Pre-Calculus | Page 40 of 73  ATENEO DE DAVAO UNIVERSITY Km 7 Central Park Blvd, Talomo, 8016 Davao City, Philippines Tel No. +63 (82) 221.2411 local 8608 E-Mail: [email protected] * www.addu.edu.ph In Consortium with Ateneo de Zamboanga University and Xavier University SENIOR HIGH SCHOOL – PRE-CALCULUS Topic: The Unit Circle Unit Circle is a circle with center at the origin Central Angle is an angle whose vertex is at and with radius equal to 1 unit (see Figure 1). the center of the circle (see Figure 2). Parts: Equation of a unit circle: 𝒙𝟐 + 𝒚𝟐 = 𝟏 Initial Side – stationary ray of the angle. Terminal Side – rotating ray of the angle. An angle is in Standard Position if it has a vertex on the origin and the initial side lies on the positive side of x-axis (see Figure 3). Figure 3: Angles in Standard Position Figure 4 shows angles that are NOT in Standard Position. Figure 4: Angles NOT in standard position AdDU -Senior High School | Pre-Calculus | Page 41 of 73 When angle is obtained from the initial side to a certain direction it corresponds to an angle measure. See Figure 5 below. Figure 5: Positive and Negative Angles As the terminal side of an angle rotates in a counterclockwise direction it has a positive angle measure while if it rotates in a clockwise direction then it has a negative angle measure. (See table below) Direction Angle Measure Counter-clockwise Positive Clockwise Negative There are two ways of measuring an angle in a unit circle: 1 1 a. Degree measure – 1 revolution = 360°, 2 revolution = 180°, 4 revolution = 90° b. Radian measure – 2π radian = 360°, π radian = 180° Converting Degree to Radians, and vice versa; 𝜋 rad 1. To convert a degree measure to radian, multiply it by. 180° 180° 2. To convert a radian measure to degree, multiply it by 𝜋 rad. Examples A. Convert the following degree measure to radian measure. 𝜋 rad Example 1: 15° Solution: By multiplying with the given degree measure, we have: 180° 𝜋 rad 𝝅 15° = 15° × = 𝟏𝟐 rad 180° 𝝅 Therefore, 15° in radian measure is 𝐫𝐚𝐝. 𝟏𝟐 𝜋 rad 5𝜋 Example 2: - 255° Solution: −255° × =− rad 180° 4 𝟓𝝅 Therefore, -255° in radian measure is − 𝐫𝐚𝐝. 𝟒 𝜋 rad 2𝜋 Example 3: 120° Solution: 120° × = rad 180° 3 𝟓𝝅 Therefore, -255° in radian measure is − 𝐫𝐚𝐝. 𝟒 AdDU -Senior High School | Pre-Calculus | Page 42 of 73  𝜋 rad 𝟕𝝅 Example 4: - 70° Solution: −70° × = − 𝟏𝟖 rad 180° 𝟕𝝅 Therefore, -70° in radian measure is − 𝟏𝟖 𝐫𝐚𝐝. B. Convert the following radian measure to degree measure. 5𝜋 180° Example 5: rad Solution: by multiplying 𝜋 rad with the given radian measure, we have: 9 5𝜋 5𝜋 180° rad = rad × 𝜋 rad = 𝟏𝟎𝟎° 9 9 𝟓𝝅 Therefore, 𝐫𝐚𝐝 in degree measure is 100°. 𝟗 6𝜋 6𝜋 180° Example 6: − rad Solution: − rad × 𝜋 rad = −𝟐𝟏𝟔° 5 5 𝟔𝝅 Therefore, − 𝐫𝐚𝐝 in degree measure is - 216°. 𝟓 180° 𝟗𝟎𝟎° Example 7: 5 rad Solution: 5 rad × 𝜋 rad = 𝝅 𝟗𝟎𝟎° Therefore, 𝟓 𝐫𝐚𝐝 in degree measure is. 𝝅 3𝜋 3𝜋 180° Example 8: − rad Solution: − rad × 𝜋 rad = −𝟏𝟑𝟓° 4 4 𝟑𝝅 Therefore, − 𝐫𝐚𝐝 in degree measure is - 135°. 𝟒 Figure 6 shows some special angles in standard position with their respective terminal sides and degree and radian measures. Figure 6: Special Angles in Standard Position AdDU -Senior High School | Pre-Calculus | Page 43 of 73 COTERMINAL ANGLES Two angles are said to be coterminal if they have the same terminal side (see Figures 7 and 8). Figure 7 Figure 8 Observe that the degree measures of coterminal angles differ by multiples of 360°. Angles coterminal with a given angle θ may be derived using the formula: Degrees Radians 𝜽 + 𝟑𝟔𝟎°(𝒏) 𝜽 + 𝟐𝝅(𝒏) where 𝑛 ∈ ℤ where 𝑛 ∈ ℤ Examples Example 9: Find the least positive coterminal angle of 410°. Solution: Use 𝜃 + 360°(𝑛) Let n = -1, 410° + 360° (-1) = ? Figure 9 410° - 360° = 50° Therefore, the least positive coterminal angle of 410° is 50° (see Figure 9). Example 10: Find two angles, one positive and one negative, which are coterminal with 75o. Solution: Use 𝜃 + 360°(𝑛). For the positive angle, let n = 1, we have: 75o + 360°(1) = ? 75o + 360° = 435° For the negative angle, let n = -1, we have: 75o + 360°(-1) = ? 75o + 360° = -285° Therefore, the positive and negative coterminal angle of 75o is 435° and -285°. AdDU -Senior High School | Pre-Calculus | Page 44 of 73  7𝜋 Example 11: Find the coterminal angle of rad that measures between 0 & 2𝜋 rad. 3 Solution: Use 𝜃 + 2𝜋(𝑛). If n = -1, then we have: 7𝜋 7𝜋 7𝜋 − 6𝜋 𝝅 𝜋 + 2𝜋(−1) = − 2𝜋 = = 𝐫𝐚𝐝 (0 < < 2𝜋 rad) 3 3 3 𝟑 3 𝟕𝝅 𝝅 Therefore, the coterminal angle of 𝐫𝐚𝐝 that measures between 𝟎 & 𝟐𝝅 𝐫𝐚𝐝 is 𝟑 𝐫𝐚𝐝. 𝟑 ARC LENGTH (Linear Measure of a Central Angle) In a circle of radius r, the length of s of an arc intercepted by the central angle with measure θ radians is given by 𝒔 = 𝒓𝜽 Example Example 12: Find the length of an arc of a circle with radius 10 m that subtends a central angle 30°. Given: 𝑟 = 10 m, 𝜃 = 30° Required: 𝑠 =? Solution: First, convert the given central angle in radians, 𝜋 𝜋 30° × = 180° 6 Then, solve for 𝑠 = 𝑟𝜃, 𝜋 𝟓𝝅 𝑠 = (10m) ( ) = 𝐦 6 𝟑 Therefore, the arc length of a circle that subtends a 30° central angle and 10m radius 𝟓𝝅 is 𝟑 𝐦. AdDU -Senior High School | Pre-Calculus | Page 45 of 73 Example 13: A central angle 𝜃 in a circle of radius 4 m is subtended by an arc of length 6 m. Find the measure of 𝜃 in radians. Given: 𝑟 = 4 m, 𝑠 = 6 m Required: 𝜃 =? Solution: Deriving the formula 𝒔 = 𝒓𝜽, by dividing both sides with r we have: 𝑠 𝑟𝜃 𝒔 = 𝜽=𝒓 𝑟 𝑟 𝑠 Using 𝜃 = 𝑟, substitute the given and solve: 6m 𝟑 𝜃= = 𝐫𝐚𝐝 4m 𝟐 𝟑 Therefore, the measurement of the angle in radian is 𝟐 𝐫𝐚𝐝. AdDU -Senior High School | Pre-Calculus | Page 46 of 73  ATENEO DE DAVAO UNIVERSITY Km 7 Central Park Blvd, Talomo, 8016 Davao City, Philippines Tel No. +63 (82) 221.2411 local 8608 E-Mail: [email protected] * www.addu.edu.ph In Consortium with Ateneo de Zamboanga University and Xavier University SENIOR HIGH SCHOOL – PRE-CALCULUS Topic: The 6 Circular Functions Any Angle θ in standard position, 𝑷 (𝜽) = (𝒙, 𝒚) is the unique point on the unit circle that lies on the terminal side of the angle. Suppose the angle θ has a radian measure 𝒂 𝑟𝑎𝑑. Then we define the six circular/trigonometric functions on the angle 𝜽, and on the number 𝒂 as follows; 1 𝑠𝑖𝑛 (𝜃) = 𝑠𝑖𝑛 (𝑎) = 𝑦 𝑐𝑠𝑐 (𝜃) = 𝑐𝑠𝑐 (𝑎) = 𝑦 , 𝑖𝑓 𝑦 ≠ 0 1 𝑐𝑜𝑠 (𝜃) = 𝑐𝑜𝑠 (𝑎) = 𝑥 𝑠𝑒𝑐 (𝜃) = 𝑠𝑒𝑐 (𝑎) = 𝑥 , 𝑖𝑓 𝑥 ≠ 0 𝑦 𝑥 𝑡𝑎𝑛 (𝜃) = 𝑡𝑎𝑛 (𝑎) = 𝑥 , 𝑖𝑓 𝑥 ≠ 0 𝑐𝑜𝑡 (𝜃) = 𝑐𝑜𝑡 (𝑎) = 𝑦 , 𝑖𝑓 𝑦 ≠ 0 Where 𝑐𝑠𝑐, 𝑠𝑒𝑐, and 𝑐𝑜𝑡 are reciprocal functions of 𝑠𝑖𝑛, 𝑐𝑜𝑠 and 𝑡𝑎𝑛, respectively. Moreover, 𝒕𝒂𝒏 is the quotient of 𝒔𝒊𝒏 over 𝒄𝒐𝒔 and 𝒄𝒐𝒕 is the quotient of 𝒄𝒐𝒔 over 𝒔𝒊𝒏. Examples A. Using unit circle, identify the coordinates and the 6 circular functions of the following angle measures. 1. θ = 60° (See figure 1) Solution: Recall the property of 30° − 60° − 90° triangle. 1 The side opposite the 30°- angle has length 2 of the hypotenuse, while the one opposite the 60°- angle √3 has length times the hypotenuse 1. So, P (60°) has 2 𝟏 √𝟑 coordinates; P (60°) = (𝟐 , ) 𝟐 The 6 circular functions are: Figure 1: 60° Angle √𝟑 sin(θ) = sin(60°) = y = 𝟐 𝟏 cos (θ) = cos (60°) = x = 𝟐 1 1 2 2 √3 𝟐√𝟑 csc(θ) = csc(60°) = = = 1( ) = ∙ = y √3 √3 √3 √3 𝟑 2 1 1 sec (θ) = sec (60°) = = = 1(2) = 𝟐 x 1 2 AdDU -Senior High School | Pre-Calculus | Page 47 of 73  √3 𝑦 √3 2 tan (θ) = tan (60°) = 𝑥 = 2 1 = ( ) 2 1 = √𝟑 2 1 x 1 2 1 √3 √𝟑 cot(θ) = cot(60°) = = 2 = ( ) = ∙ = y √3 2 √3 √3 √3 𝟑 2 𝜋 2. θ = 4 rad (See figure 2) Solution: 𝜋 rad – has degree measure 45°. Recall the 45°- 4 45°-90° triangle, both legs have the same length √2 equal to times the hypotenuse 1. Therefore, P 2 𝜋 𝜋 √2 √2 (4 rad) has coordinates; P (4 rad) = ( 2 , ) 2 The 6 circular functions are: π √𝟐 sin(θ) = sin ( ) = y = 4 𝟐 Figure 2: 45° Angle π √𝟐 cos (θ) = cos (4) = x = 𝟐 π 1 1 2 2 √2 2√2 csc(θ) = csc ( ) = = = 1( ) = ∙ = = √𝟐 4 y √2 √2 √2 √2 2 2 π 1 1 2 2 √2 2√2 sec(θ) = sec ( ) = = = 1( ) = ∙ = = √𝟐 4 x √2 √2 √2 √2 2 2 √2 π y √2 2 tan(θ) = tan ( ) = = 2 = ( )=𝟏 4 x √2 2 √2 2 √2 π x √2 2 cot (θ) = cot ( ) = = 2 = ( )=𝟏 4 y √2 2 √2 2 Here are some special angles with their terminal points: Angle in Degree Angle in Radian Terminal Points 0° 0 (1 , 0 ) 30° 𝜋 √3 1 6 ( , ) 2 2 45° 𝜋 √2 √2 4 ( , ) 2 2 60° 𝜋 1 √3 3 ( , ) 2 2 90° 𝜋 (0 , 1 ) 2 AdDU -Senior High School | Pre-Calculus | Page 48 of 73 Reference Angle - for a nonquadrantal angle greater than 90° , the smallest nonnegative angle between the terminal side and the x- axis when the angle is in standard position. - It is drawn from the x-axis to the terminal side of the angle and sometimes from the terminal side to the x-axis but always in positive direction. (See figure 3) - All angles in standard position less than 90° are reference angles. Figure 3: Reference Angle To find the reference angle, subtract 360° or 180°, whichever is closer, from the angle if θ is in degrees; or subtract π or 2π, whichever is closer, from the angle if θ is in radians. (See figure 4) Figure 4: Finding the Reference Angle Examples Find the reference angle (𝜃𝑅 ), given a value for 𝜃. 1. 405° Solution: 405° is in Quadrant I, and thus greater than 360°, so we have: 𝜃𝑅 = 𝜃 − 360° 𝜃𝑅 = 405° − 360° 𝜃𝑅 = 𝟒𝟓° 2. −150° Solution: −150° is in Quadrant III, we have: 𝜃𝑅 = 180° − θ 𝜃𝑅 = 180° − 150° 𝜃𝑅 = 𝟑𝟎° AdDU -Senior High School | Pre-Calculus | Page 49 of 73  2𝜋 3. 𝑟𝑎𝑑 3 Solution: 2𝜋 𝑟𝑎𝑑 is in Quadrant II, we have: 3 𝜃𝑅 = 𝜋 − θ 2𝜋 3𝜋−2𝜋 𝜋 𝜃𝑅 = 𝜋 − = = 3 3 3 𝝅 𝜃𝑅 = 𝟑 11𝜋 4. 𝑟𝑎𝑑 6 Solution: 11𝜋 𝑟𝑎𝑑 is in Quadrant IV, we have: 6 𝜃𝑅 = 2𝜋 − θ 11𝜋 12𝜋 − 11𝜋 𝜋 𝜃𝑅 = 2𝜋 − = =