Geometria Analítica respostas em Ingles 4 Atividades

NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 8 PRECALCULUS AND ADVANCED TOPICS Lesson 8 Curves from Geometry S48 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from PreCalM3TE130082015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Lesson 8 Curves from Geometry Classwork Exercises 1 Let 𝐹05 and 𝐺0 5 be the foci of a hyperbola Let the points 𝑃𝑥 𝑦 on the hyperbola satisfy either 𝑃𝐹 𝑃𝐺 6 or 𝑃𝐺 𝑃𝐹 6 Use the distance formula to derive an equation for this hyperbola writing your answer in the form 𝑥2 𝑎2 𝑦2 𝑏2 1 2 Where does the hyperbola described above intersect the 𝑦axis 3 Find an equation for the line that acts as a boundary for the portion of the curve that lies in the first quadrant 4 Sketch the graph of the hyperbola described above NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 8 PRECALCULUS AND ADVANCED TOPICS Lesson 8 Curves from Geometry S49 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from PreCalM3TE130082015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Problem Set 1 For each hyperbola described below 1 Derive an equation of the form 𝑥2 𝑎2 𝑦2 𝑏2 1 or 𝑦2 𝑏2 𝑥2 𝑎2 1 2 State any 𝑥 or 𝑦intercepts 3 Find the equations for the asymptotes of the hyperbola a Let the foci be 𝐴20 and 𝐵20 and let 𝑃 be a point for which either 𝑃𝐴 𝑃𝐵 2 or 𝑃𝐵 𝑃𝐴 2 b Let the foci be 𝐴50 and 𝐵50 and let 𝑃 be a point for which either 𝑃𝐴 𝑃𝐵 5 or 𝑃𝐵 𝑃𝐴 5 c Consider 𝐴0 3 and 𝐵03 and let 𝑃 be a point for which either 𝑃𝐴 𝑃𝐵 25 or 𝑃𝐵 𝑃𝐴 25 d Consider 𝐴0 2 and 𝐵0 2 and let 𝑃 be a point for which either 𝑃𝐴 𝑃𝐵 2 or 𝑃𝐵 𝑃𝐴 2 2 Graph the hyperbolas in parts ad in Problem 1 3 For each value of 𝑘 specified in parts ae plot the set of points in the plane that satisfy the equation 𝑥2 𝑦2 𝑘 a 𝑘 4 b 𝑘 1 c 𝑘 1 4 d 𝑘 0 e 𝑘 1 4 f 𝑘 1 g 𝑘 4 h Describe the hyperbolas 𝑥2 𝑦2 𝑘 for different values of 𝑘 Consider both positive and negative values of 𝑘 and consider values of 𝑘 close to zero and far from zero i Are there any values of 𝑘 so that the equation 𝑥2 𝑦2 𝑘 has no solution 4 For each value of 𝑘 specified in parts ae plot the set of points in the plane that satisfy the equation 𝑥2 𝑘 𝑦2 1 a 𝑘 1 b 𝑘 1 c 𝑘 2 d 𝑘 4 e 𝑘 10 f 𝑘 25 g Describe what happens to the graph of 𝑥2 𝑘 𝑦2 1 as 𝑘 NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 8 PRECALCULUS AND ADVANCED TOPICS Lesson 8 Curves from Geometry S50 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from PreCalM3TE130082015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License 5 For each value of 𝑘 specified in parts ae plot the set of points in the plane that satisfy the equation 𝑥2 𝑦2 𝑘 1 a 𝑘 1 b 𝑘 1 c 𝑘 2 d 𝑘 4 e 𝑘 10 f Describe what happens to the graph 𝑥2 𝑦2 𝑘 1 as 𝑘 6 An equation of the form 𝑎𝑥2 𝑏𝑥 𝑐𝑦2 𝑑𝑦 𝑒 0 where 𝑎 and 𝑐 have opposite signs might represent a hyperbola a Apply the process of completing the square in both 𝑥 and 𝑦 to convert the equation 9𝑥2 36𝑥 4𝑦2 8𝑦 4 0 to one of the standard forms for a hyperbola 𝑥ℎ2 𝑎2 𝑦𝑘2 𝑏2 1 or 𝑦𝑘2 𝑏2 𝑥ℎ2 𝑎2 1 b Find the center of this hyperbola c Find the asymptotes of this hyperbola d Graph the hyperbola 7 For each equation below identify the graph as either an ellipse a hyperbola two lines or a single point If possible write the equation in the standard form for either an ellipse or a hyperbola a 4𝑥2 8𝑥 25𝑦2 100𝑦 4 0 b 4𝑥2 16𝑥 9𝑦2 54𝑦 65 0 c 4𝑥2 8𝑥 𝑦2 2𝑦 5 0 d 49𝑥2 98𝑥 4𝑦2 245 0 e What can you tell about a graph of an equation of the form 𝑎𝑥2 𝑏𝑥 𝑐𝑦2 𝑑𝑦 𝑒 0 by looking at the coefficients NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 6 PRECALCULUS AND ADVANCED TOPICS Lesson 6 Curves in the Complex Plane S37 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from PreCalM3TE130082015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Lesson 6 Curves in the Complex Plane Classwork Opening Exercise a Consider the complex number 𝑧 𝑎 𝑏𝑖 i Write 𝑧 in polar form What do the variables represent ii If 𝑟 3 and 𝜃 90 where would 𝑧 be plotted in the complex plane iii Use the conditions in part ii to write 𝑧 in rectangular form Explain how this representation corresponds to the location of 𝑧 that you found in part ii b Recall the set of points defined by 𝑧 3cos𝜃 𝑖 sin 𝜃 for 0 𝜃 360 where 𝜃 is measured in degrees i What does 𝑧 represent graphically Why ii What does 𝑧 represent geometrically NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 6 PRECALCULUS AND ADVANCED TOPICS Lesson 6 Curves in the Complex Plane S38 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from PreCalM3TE130082015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License c Consider the set of points defined by 𝑧 5 cos𝜃 3𝑖 sin𝜃 i Plot 𝑧 for 𝜃 0 90 180 270 360 Based on your plot form a conjecture about the graph of the set of complex numbers ii Compare this graph to the graph of 𝑧 3cos𝜃 𝑖 sin𝜃 Form a conjecture about what accounts for the differences between the graphs NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 6 PRECALCULUS AND ADVANCED TOPICS Lesson 6 Curves in the Complex Plane S39 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from PreCalM3TE130082015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Example 1 Consider again the set of complex numbers represented by 𝑧 3cos𝜃 𝑖 sin𝜃 for 0 𝜃 360 𝜽 𝟑 𝐜𝐨𝐬𝜽 𝟑 𝐬𝐢𝐧𝜽 𝟑 𝐜𝐨𝐬𝜽 𝟑𝒊 𝐬𝐢𝐧𝜽 0 𝜋 4 𝜋 2 3𝜋 4 𝜋 5𝜋 4 3𝜋 2 7𝜋 4 2𝜋 NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 6 PRECALCULUS AND ADVANCED TOPICS Lesson 6 Curves in the Complex Plane S40 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from PreCalM3TE130082015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License a Use an ordered pair to write a representation for the points defined by 𝑧 as they would be represented in the coordinate plane b Write an equation that is true for all the points represented by the ordered pair you wrote in part a c What does the graph of this equation look like in the coordinate plane Exercises 12 1 Recall the set of points defined by 𝑧 5 cos𝜃 3𝑖 sin𝜃 a Use an ordered pair to write a representation for the points defined by 𝑧 as they would be represented in the coordinate plane b Write an equation in the coordinate plane that is true for all the points represented by the ordered pair you wrote in part a NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 6 PRECALCULUS AND ADVANCED TOPICS Lesson 6 Curves in the Complex Plane S41 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from PreCalM3TE130082015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License 2 Find an algebraic equation for all the points in the coordinate plane traced by the complex numbers 𝑧 2 cos𝜃 𝑖 sin𝜃 Example 2 The equation of an ellipse is given by 𝑥2 16 𝑦2 4 1 a Sketch the graph of the ellipse b Rewrite the equation in complex form NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 6 PRECALCULUS AND ADVANCED TOPICS Lesson 6 Curves in the Complex Plane S42 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from PreCalM3TE130082015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Exercise 3 3 The equation of an ellipse is given by 𝑥2 9 𝑦2 26 1 a Sketch the graph of the ellipse b Rewrite the equation of the ellipse in complex form NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 6 PRECALCULUS AND ADVANCED TOPICS Lesson 6 Curves in the Complex Plane S43 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from PreCalM3TE130082015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Example 3 A set of points in the complex plane can be represented in the complex plane as 𝑧 2 𝑖 7 cos𝜃 𝑖 sin𝜃 as 𝜃 varies a Find an algebraic equation for the points described b Sketch the graph of the ellipse NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 6 PRECALCULUS AND ADVANCED TOPICS Lesson 6 Curves in the Complex Plane S44 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from PreCalM3TE130082015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Problem Set 1 Write the real form of each complex equation a 𝑧 4 cos𝜃 9𝑖 sin𝜃 b 𝑧 6 cos𝜃 𝑖 sin𝜃 c 𝑧 5 cos𝜃 10𝑖 sin𝜃 d 𝑧 5 2𝑖 4 cos𝜃 7𝑖 sin𝜃 2 Sketch the graphs of each equation a 𝑧 3 cos𝜃 𝑖 sin𝜃 b 𝑧 2 3𝑖 4 cos𝜃 𝑖 sin𝜃 c 𝑥12 9 𝑦2 25 1 d 𝑥22 3 𝑦2 15 1 3 Write the complex form of each equation a 𝑥2 16 𝑦2 36 1 b 𝑥2 400 𝑦2 169 1 c 𝑥2 19 𝑦2 2 1 d 𝑥32 100 𝑦52 16 1 4 Carrie converted the equation 𝑧 7 cos𝜃 4𝑖 sin𝜃 to the real form 𝑥2 7 𝑦2 4 1 Her partner Ginger said that the ellipse must pass through the point 7 cos0 4 sin0 70 and this point does not satisfy Carries equation so the equation must be wrong Who made the mistake and what was the error Explain how you know 5 Cody says that the center of the ellipse with complex equation 𝑧 4 5𝑖 2 cos𝜃 3𝑖 sin𝜃 is 4 5 while his partner Jarrett says that the center of this ellipse is 45 Which student is correct Explain how you know Extension 6 Any equation of the form 𝑎𝑥2 𝑏𝑥 𝑐𝑦2 𝑑𝑦 𝑒 0 with 𝑎 0 and 𝑐 0 might represent an ellipse The equation 4𝑥2 8𝑥 3𝑦2 12𝑦 4 0 is such an equation of an ellipse a Rewrite the equation 𝑥ℎ2 𝑎2 𝑦𝑘2 𝑏2 1 in standard form to locate the center of the ellipse ℎ 𝑘 b Describe the graph of the ellipse and then sketch the graph c Write the complex form of the equation for this ellipse Assignment 74 1 What effect does the xy term have on the graph of a conic section 2 If the equation of a conic section is written in the form Ax2By2CxDyE0 and AB0 what can we conclude 3 If the equation of a conic section is written in the form Ax2BxyCy2DxEyF0 and B24AC0 what can we conclude 4 Given the equation ax24x3y2120 what can we conclude if a0 5 For the equation Ax2BxyCy2DxEyF0 the value of θ that satisfies cot2θ ACB gives us what information 6 For the following exercises determine which conic section is represented based on the given equation 7 For the following exercises find a new representation of the given equation after rotating through the given angle 8 For the following exercises rotate through the given angle based on the given equation Give the new equation and graph the original and rotated equation y x2 θ 45 x y2 θ 45 x24 y21 1 θ 45 y216 x29 1 θ 45 y2 x2 1 θ 45 y x22 θ 30 x y 12 θ 30 x29 y24 1 θ 30 NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 33 Algebra II Lesson 33 The Definition of a Parabola S174 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IIM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Lesson 33 The Definition of a Parabola Classwork Opening Exercise Suppose you are viewing the crosssection of a mirror Where would the incoming light be reflected in each type of design Sketch your ideas below Mirror 1 Incoming Light Mirror 3 Incoming Light Mirror 2 Incoming Light NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 33 Algebra II Lesson 33 The Definition of a Parabola S175 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IIM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Discussion Telescope Design When Newton designed his reflector telescope he understood two important ideas Figure 1 shows a diagram of this type of telescope The curved mirror needs to focus all the light to a single point that we will call the focus An angled flat mirror is placed near this point and reflects the light to the eyepiece of the telescope The reflected light needs to arrive at the focus at the same time Otherwise the image is distorted Definition A parabola with directrix 𝐿 and focus point 𝐹 is the set of all points in the plane that are equidistant from the point 𝐹 and line 𝐿 Figure 2 to the right illustrates this definition of a parabola In this diagram 𝐹𝑃1 𝑃1𝑄1 𝐹𝑃2 𝑃2𝑄2 𝐹𝑃3 𝑃3𝑄3 showing that for any point 𝑃 on the parabola the distance between 𝑃 and 𝐹 is equal to the distance between 𝑃 and the line 𝐿 All parabolas have the reflective property illustrated in Figure 3 Rays parallel to the axis reflect off the parabola and through the focus point 𝐹 Thus a mirror shaped like a rotated parabola would satisfy Newtons requirements for his telescope design Szőcs Tamás Figure 1 Figure 3 Figure 2 NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 33 Algebra II Lesson 33 The Definition of a Parabola S176 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IIM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License 𝑭𝟎 𝟐 𝑨𝒙 𝒚 𝑭𝒙 𝟎 Figure 4 below shows several different line segments representing the reflected light with one endpoint on the curved mirror that is a parabola and the other endpoint at the focus Anywhere the light hits this type of parabolic surface it always reflects to the focus 𝐹 at exactly the same time Figure 5 shows the same image with a directrix Imagine for a minute that the mirror was not there Then the light would arrive at the directrix all at the same time Since the distance from each point on the parabolic mirror to the directrix is the same as the distance from the point on the mirror to the focus and the speed of light is constant it takes the light the same amount of time to travel to the focus as it would have taken it to travel to the directrix In the diagram this means that 𝐴𝐹 𝐴𝐹𝐴 𝐵𝐹 𝐵𝐹𝐵 and so on Thus the light rays arrive at the focus at the same time and the image is not distorted Example Finding an Analytic Equation for a Parabola Given a focus and a directrix create an equation for a parabola Focus 𝐹02 Directrix 𝑥axis Parabola 𝑃 𝑥 𝑦 𝑥 𝑦 is equidistant from 𝐹 and the 𝑥axis Let 𝐴 be any point 𝑥 𝑦 on the parabola 𝑃 Let 𝐹 be a point on the directrix with the same 𝑥coordinate as point 𝐴 What is the length 𝐴𝐹 Use the distance formula to create an expression that represents the length 𝐴𝐹 Figure 4 Figure 5 NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 33 Algebra II Lesson 33 The Definition of a Parabola S177 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IIM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Create an equation that relates the two lengths and solve it for 𝑦 Verify that this equation appears to match the graph shown Exercises 1 Demonstrate your understanding of the definition of a parabola by drawing several pairs of congruent segments given the parabola its focus and directrix Measure the segments that you drew to confirm the accuracy of your sketches in either centimeters or inches NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 33 Algebra II Lesson 33 The Definition of a Parabola S178 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IIM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License 2 Derive the analytic equation of a parabola given the focus of 04 and the directrix 𝑦 2 Use the diagram to help you work this problem a Label a point 𝑥 𝑦 anywhere on the parabola b Write an expression for the distance from the point 𝑥 𝑦 to the directrix c Write an expression for the distance from the point 𝑥 𝑦 to the focus d Apply the definition of a parabola to create an equation in terms of 𝑥 and 𝑦 Solve this equation for 𝑦 e What is the translation that takes the graph of this parabola to the graph of the equation derived in Example 1 NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 33 Algebra II Lesson 33 The Definition of a Parabola S179 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IIM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Problem Set 1 Demonstrate your understanding of the definition of a parabola by drawing several pairs of congruent segments given each parabola its focus and directrix Measure the segments that you drew in either inches or centimeters to confirm the accuracy of your sketches a b Lesson Summary PARABOLA A parabola with directrix line 𝐿 and focus point 𝐹 is the set of all points in the plane that are equidistant from the point 𝐹 and line 𝐿 AXIS OF SYMMETRY The axis of symmetry of a parabola given by a focus point and a directrix is the perpendicular line to the directrix that passes through the focus VERTEX OF A PARABOLA The vertex of a parabola is the point where the axis of symmetry intersects the parabola In the Cartesian plane the distance formula can help in deriving an analytic equation for a parabola NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 33 Algebra II Lesson 33 The Definition of a Parabola S180 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IIM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License c d 2 Find the distance from the point 42 to the point 01 3 Find the distance from the point 42 to the line 𝑦 2 4 Find the distance from the point 13 to the point 3 4 5 Find the distance from the point 13 to the line 𝑦 5 6 Find the distance from the point 𝑥 4 to the line 𝑦 1 7 Find the distance from the point 𝑥 3 to the line 𝑦 2 8 Find the values of 𝑥 for which the point 𝑥 4 is equidistant from 01 and the line 𝑦 1 9 Find the values of 𝑥 for which the point 𝑥 3 is equidistant from 1 2 and the line 𝑦 2 10 Consider the equation 𝑦 𝑥2 a Find the coordinates of the three points on the graph of 𝑦 𝑥2 whose 𝑥values are 1 2 and 3 b Show that each of the three points in part a is equidistant from the point 0 1 4 and the line 𝑦 1 4 c Show that if the point with coordinates 𝑥 𝑦 is equidistant from the point 0 1 4 and the line 𝑦 1 4 then 𝑦 𝑥2 NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 33 Algebra II Lesson 33 The Definition of a Parabola S181 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IIM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License 11 Consider the equation 𝑦 1 2 𝑥2 2𝑥 a Find the coordinates of the three points on the graph of 𝑦 1 2 𝑥2 2𝑥 whose 𝑥values are 2 0 and 4 b Show that each of the three points in part a is equidistant from the point 2 3 2 and the line 𝑦 5 2 c Show that if the point with coordinates 𝑥 𝑦 is equidistant from the point 2 3 2 and the line 𝑦 5 2 then 𝑦 1 2 𝑥2 2𝑥 12 Derive the analytic equation of a parabola with focus 13 and directrix 𝑦 1 Use the diagram to help you work this problem a Label a point 𝑥 𝑦 anywhere on the parabola b Write an expression for the distance from the point 𝑥 𝑦 to the directrix c Write an expression for the distance from the point 𝑥 𝑦 to the focus 13 d Apply the definition of a parabola to create an equation in terms of 𝑥 and 𝑦 Solve this equation for 𝑦 e Describe a sequence of transformations that would take this parabola to the parabola with equation 𝑦 1 4 𝑥2 1 derived in Example 1 13 Consider a parabola with focus 0 2 and directrix on the 𝑥axis a Derive the analytic equation for this parabola b Describe a sequence of transformations that would take the parabola with equation 𝑦 1 4 𝑥2 1 derived in Example 1 to the graph of the parabola in part a 14 Derive the analytic equation of a parabola with focus 010 and directrix on the 𝑥axis ai The complex number z a bi can be expressed in polar form as z r cos θ i sin θ Where r z is the modulus or magnitude of the complex number and it represents the distance from the origin to the point z in the complex plane θ is the argument or angle of the complex number and it represents the angle that the line connecting the origin to the point z makes with the positive real axis α r cos θ is the real part of z b r sin θ is the imaginary part of z aii Given r 3 and θ 90 we can interpret this as follows The magnitude r 3 means that the point is 3 units away from the origin The angle θ 90 means that the point lies along the positive imaginary axis because an angle of 90 from the positive real axis is straight up along the imaginary axis Thus the point z would be plotted at a distance of 3 units above the origin on the imaginary axis or at 0 3 in Cartesian coordinates aiii To convert from polar to rectangular form we use the following formulas a r cos θ b r sin θ Substituting r 3 and θ 90 or θ π2 radians into these formulas a 3 cos 90 3 0 0 b 3 sin 90 3 1 3 So the rectangular form of z is z a bi 0 3i 3i This confirms that z lies on the imaginary axis at 0 3 which matches the location we found in part ii It is 3 units above the origin as expected when r 3 and θ 90 bi i What does z represent graphically Why The given complex number is in polar form z 3 cos θ i sin θ This form represents a complex number where the modulus or magnitude is r 3 and the argument or angle θ varies between 0 and 360 Graphically this set of points represents a circle in the complex plane The center of the circle is at the origin since theres no constant term shifting the position of z The radius of the circle is 3 because r 3 As θ takes on values from 0 to 360 the complex number z traces out every point on the circle with radius 3 centered at the origin This is because the pair cos θ sin θ defines points on a unit circle and multiplying by 3 scales the unit circle to a circle with radius 3 bii ii What does z represent geometrically Geometrically z represents a circle in the complex plane with A radius of 3 since r 3 this is the distance from the origin to any point on the circle The center at the origin because there are no real or imaginary parts added to shift the circle away from the origin This is the locus or set of all points that are exactly 3 units away from the origin in the complex plane As θ varies from 0 to 360 the points on this circle sweep around the origin forming the complete circle ci i Plot z 5 cosθ 3i sinθ for θ 0 90 180 270 360 This form of z looks like an ellipse equation in the complex plane because the coefficients of cosθ and sinθ are different The complex number can be interpreted as z x yi Where x 5 cosθ this represents the real part y 3 sinθ this represents the imaginary part Lets calculate z for the given values of θ Lets calculate z for the given values of θ 1 For θ 0 z 5 cos0 3i sin0 5 0i 5 The point is 5 0 2 For θ 90 z 5 cos90 3i sin90 0 3i 3i The point is 0 3 3 For θ 180 z 5 cos180 3i sin180 5 0i 5 The point is 5 0 4 For θ 270 z 5 cos270 3i sin270 0 3i 3i The point is 0 3 5 For θ 360 z 5 cos360 3i sin360 5 0i 5 The point is 5 0 which is the same as for θ 0 Conjecture based on the plot If we plot the points 5 0 0 3 5 0 0 3 it suggests that the graph is an ellipse The major axis the longer axis is along the real axis with length 10 since it stretches from 5 to 5 and the minor axis the shorter axis is along the imaginary axis with length 6 stretching from 3i to 3i Thus my conjecture is that the set of complex numbers z 5 cosθ 3i sinθ describes an ellipse centered at the origin with A major axis of length 10 along the real axis A minor axis of length 6 along the imaginary axis cii ii Compare this graph to the graph of z 3cosθ i sinθ The graph of z 3cosθ i sinθ represents a circle with a radius of 3 centered at the origin This circle is symmetric in both the real and imaginary directions with the same distance 3 units from the origin in all directions The key difference between the two graphs is the shape The graph of z 5 cosθ 3i sinθ is an ellipse where the real and imaginary parts are stretched differently 5 units along the real axis and 3 units along the imaginary axis The graph of z 3cosθ i sinθ is a circle where the real and imaginary parts are scaled equally 3 units in both directions Conjecture about the differences The difference in the shape between the ellipse and the circle is due to the coefficients of cosθ and sinθ In the ellipse 5 cosθ and 3 sinθ scale the real and imaginary components unequally leading to different distances along the real and imaginary axes thus forming an ellipse In the circle both the real and imaginary components are scaled equally by 3 so the distance from the origin is the same in all directions forming a circle The presence of unequal coefficients 5 for cosθ and 3 for sinθ stretches the shape along the real and imaginary axes differently resulting in an ellipse Graphs of z 5 cosθ 3i sinθ Ellipse and z 3cosθ i sinθ Circle Here are the plots of the two complex functions The blue curve represents the ellipse described by z 5 cosθ 3i sinθ with a major axis of 10 units along the real axis and a minor axis of 6 units along the imaginary axis The red dashed curve represents the circle described by z 3cosθ i sinθ which has a radius of 3 units 1 a In this context the complex number z x yi can be represented as an ordered pair x y where x is the real part of z which is given by 5 cosθ y is the imaginary part of z which is given by 3 sinθ Thus the ordered pair representation of z in the coordinate plane is x y 5 cosθ 3 sinθ This pair describes the xcoordinate as 5 cosθ and the ycoordinate as 3 sinθ corresponding to different values of θ between 0 and 360 b The ordered pair x y 5 costheta 3 sintheta represents parametric equations for an ellipse To write a Cartesian equation for this ellipse we need to eliminate the parameter theta Heres how we do it 1 From the parametric equations x 5 costheta y 3 sintheta 2 Solve each equation for costheta and sintheta costheta x5 sintheta y3 3 Use the Pythagorean identity cos2theta sin2theta 1 x52 y32 1 This gives us the equation of the ellipse x225 y29 1 Step 1 Interpret z in terms of ordered pairs The complex number z x yi can be expressed as an ordered pair x y where x sqrt2costheta is the real part y sintheta is the imaginary part Thus the ordered pair representation of the points is x y sqrt2costheta sintheta Step 2 Eliminate theta using trigonometric identities We need to find a relationship between x and y that eliminates the parameter Theta 1 From the parametric equations x sqrt2costheta y sintheta 2 Solve for costheta and sintheta costheta xsqrt2 sintheta y 3 Use the Pythagorean identity cos2theta sin2theta 1 xsqrt22 y2 1 Step 3 Simplify the equation x22 y2 1 This is the equation of an ellipse in standard form where The semimajor axis along the real axis is sqrt2 The semiminor axis along the imaginary axis is 1 Graph of the Ellipse x29 y226 1 Imaginary part y Real part x a The graph above represents the ellipse described by the equation x29 y226 1 This ellipse has A semimajor axis along the imaginary axis with a length of sqrt26 approximately 51 A semiminor axis along the real axis with a length of 3 b Rewriting the equation of the ellipse in complex form The general form of an ellipse in complex terms is given by z a costheta ib sintheta Where a and b are the semimajor and semiminor axes respectively For the given ellipse The semiminor axis along the real axis is a 3 The semimajor axis along the imaginary axis is b sqrt26 Thus the complex form of the equation is z 3 costheta i sqrt26 sintheta This represents the same ellipse in the complex plane where z x yi is a complex number Problem set 1 a z 4 cosθ 9i sinθ Real part x 4 cosθ Imaginary part y 9 sinθ So the real form is x y 4 cosθ 9 sinθ b z 6 cosθ i sinθ Real part x 6 cosθ Imaginary part y sinθ So the real form is x y 6 cosθ sinθ c z 5 cosθ 10 i sinθ Real part x 5 cosθ Imaginary part y 10 sinθ So the real form is x y 5 cosθ 10 sinθ d z 5 2i 4 cosθ 7i sinθ This equation contains constant terms along with the trigonometric terms The real and imaginary parts are Real part x 5 4 cosθ Imaginary part y 2 7 sinθ So the real form is x y 5 4 cosθ 2 7 sinθ 2 a z 3 cosθ i sinθ This represents an ellipse in the complex plane with Real part x 3 cosθ Imaginary part y sinθ This is an ellipse centered at the origin with a semimajor axis of 3 along the real axis and a semiminor axis of 1 along the imaginary axis b z 2 3i 4 cosθ i sinθ This represents an ellipse that is shifted Real part x 2 4 cosθ Imaginary part y 3 sinθ This ellipse is shifted by 2 3 and has a semimajor axis of 4 along the real axis and a semiminor axis of 1 along the imaginary axis c x129 y225 1 This is the equation of an ellipse centered at 10 with Semimajor axis 5 along the imaginary axis Semiminor axis 3 along the real axis d x223 y215 1 This is an ellipse centered at 20 with Semimajor axis 15 along the imaginary axis Semiminor axis 3 along the real axis 3 a x216 y236 1 This is an ellipse in the complex plane Semimajor axis 6 along the imaginary axis Semiminor axis 4 along the real axis The complex form is z 4 cosθ 6i sinθ b x2100 y2169 1 This is an ellipse Semimajor axis 13 along the imaginary axis Semiminor axis 20 along the real axis The complex form is z 20 cosθ 13i sinθ c x219 y22 1 This is an ellipse Semimajor axis 19 along the real axis Semiminor axis 2 along the imaginary axis The complex form is z 19 cosθ 2 i sinθ d x32100 y5216 1 This ellipse is centered at 3 5 Semimajor axis 10 along the real axis Semiminor axis 4 along the imaginary axis The complex form is z 3 5i 10 cosθ 4i sinθ 4 The presence of the xy term in the equation of a conic section indicates that the axes of the conic section are rotated relative to the standard coordinate axes The xy term introduces a crossproduct that causes the orientation of the conic section to change To elaborate Ellipse If the equation of an ellipse has an xy term then the major and minor axes of the ellipse are rotated with respect to the x and yaxes Hyperbola If the equation of a hyperbola has an xy term then its asymptotes are rotated with respect to the x and yaxes Parabola If the equation of a parabola has an xy term then the axis of symmetry of the parabola is tilted with respect to the x and yaxes In general the effect of the xy term is to rotate the conic section By applying a rotation of axes you can eliminate the xy term and simplify the equation of the conic section This makes it easier to analyze and graph The rotation transformation typically involves substituting x X cos θ Y sin θ y X sin θ Y cos θ where θ is the angle of rotation necessary to eliminate the xy term The appropriate angle of rotation θ can be found using the formula θ 12 arctan B A C Here A B and C come from the general form of the conic section equation Ax2 Bxy Cy2 Dx Ey F 0 33 If the equation of a conic section is written in the form Ax2 By2 Cx Dy E 0 and AB 0 it means that either A 0 or B 0 or possibly both Lets explore each case Case 1 A 0 B 0 The equation becomes By2 Cx Dy E 0 This equation represents a parabola When A 0 there is no x2 term indicating that the conic section opens either upwards or downwards depending on the sign and value of B This shape is a vertical parabola Case 2 B 0 A 0 The equation becomes Ax2 Cx Dy E 0 This equation represents another form of a parabola Here there is no y2 term indicating that the conic section opens either to the left or right depending on the sign and value of A This shape is a horizontal parabola Case 3 A 0 and B 0 The equation becomes Cx Dy E 0 This equation is a linear equation which represents a line Thus for any case where AB 0 the conic section is a parabola either vertical or horizontal If both A and B are zero simultaneously the equation represents a line not a conic section in the traditional sense 34 Carrie made a mistake when she converted z 7 cosθ 4i sinθ to x27 y24 1 This equation is incorrect because The semimajor axis should be squared The correct equation should be x249 y216 1 This matches the form of the complex equation Ginger was correct because the point 7 0 satisfies the correct equation 7249 1 5 The complex equation z 4 5i 2 cosθ 3i sinθ has A real part x 4 An imaginary part y 5 Thus the center of the ellipse is 4 5 so Cody is correct 6 a Rewrite in standard form to locate the center We begin by completing the square 1 For the xterms 4x2 8x 4x2 2x 4x 12 1 4x 12 4 2 For the yterms 3y2 12y 3y2 4y 3y 22 4 3y 22 12 Substitute back into the equation 4x 12 4 3y 22 12 4 0 4x 12 3y 22 12 Divide by 12 x 12 3 y 22 4 1 So the center is 1 2 b Describe the graph This is an ellipse with Center at 1 2 Semimajor axis 2 along the imaginary axis Semiminor axis 3 along the real axis c Complex form of the equation z 1 2i 3 cosθ 2i sinθ If the equation of a conic section is written in the form Ax2 Bxy Cy2 Dx Ey F 0 and the discriminant B2 4AC 0 we can conclude that the conic section is a hyperbola The value of the discriminant B2 4AC determines the type of conic section represented by the equation Specifically If B2 4AC 0 the conic section is a hyperbola If B2 4AC 0 the conic section is a parabola If B2 4AC 0 the conic section is an ellipse and if A C and B 0 it is a circle Thus in this case where B2 4AC 0 it definitively indicates that the conic section described by the given equation is a hyperbola Given the equation ax2 4x 3y2 12 0 we want to determine the type of conic section it represents when a 0 First lets rewrite the equation in the form Ax2 By2 Cx Dy E 0 ax2 3y2 4x 12 0 where A a B 3 C 4 D 0 and E 12 We can determine the type of conic section using the discriminant Δ given by B2 4AC Δ 02 4a3 12a Notice that since a 0 we have 12a 0 Since Δ B2 4AC 0 the equation represents an ellipse Therefore if a 0 the given equation ax2 4x 3y2 12 0 describes an ellipse For the equation Ax2 Bxy Cy2 Dx Ey F 0 the value of θ that satisfies cot2θ A C B provides the angle of rotation needed to eliminate the xy term from the equation This transformation aligns the coordinate axes with the principal axes of the conic section described by the equation By rotating the coordinate system by this angle θ the xy term is removed and the equation takes on a simpler form typically without the xy crossproduct term making it easier to identify and analyze the conic section ellipse hyperbola or parabola Specifically this angle θ is the angle between the original coordinate axes and the new axes aligned with the principal axes of the conic section Equation 3x2 xy 3y2 5 0 Angle θ 45 Transformation x x cos 45 y sin 45 x y 2 y x sin 45 y cos 45 x y 2 Substituting 3 x y 22 x y 2x y 2 3 x y 22 5 0 Simplifying 3 x2 2xy y2 2 x2 y2 2 3 x2 2xy y2 2 5 0 3 12 x2 3 12 y2 3 12 xy 12 x2 12 y2 3 12 x2 3 12 y2 3 12 xy 5 3x2 3y2 3xy x2 y2 3x2 3y2 3xy 10 2 0 7x2 5y2 10 0 Rotated Equation 7x2 5y2 10 0 Equation 4x2 xy 4y2 2 0 Angle θ 45 Transformation x x cos 45 y sin 45 x y 2 y x sin 45 y cos 45 x y 2 Substituting 4 x y 22 x y 2x y 2 4 x y 22 2 0 Simplifying 4 x2 2xy y2 2 x2 y2 2 4 x2 2xy y2 2 2 0 2x2 2xy y2 x2 y2 2x2 2xy y2 2 0 4x2 4xy 2y2 x2 y2 4x2 4xy 2y2 2 0 9x2 2y2 2 0 Equation 2x2 8xy 1 0 Angle θ 30 Transformation x x cos 30 y sin 30 3x y 2 y x sin 30 y cos 30 x 3y 2 Substituting 2 3x y 22 8 3x y 2x 3y 2 1 0 Simplifying 2 3x2 23xy y2 4 4 3x2 33xy y2 1 0 23x2 23xy y2 4 23x2 3xy y2 1 Equation 2x2 8xy 1 0 Angle θ 45 Transformation Same as problem 1 because using sin θ cos 90 θ Equation 4x2 2xy 4y2 y 2 0 Angle θ 45 Transformations Same as problem 1 above For each of these cases the more straightforward path is to use symbolic algebra software to substitute together with simplifying Robinson equation of the form x x45 1 y x2 θ 45 Convert 45 to radians θ π 4 Using the rotation formulas x x cos π 4 y sin π 4 2 2 x 2 2 y y x sin π 4 y cos π 4 2 2 x 2 2 y Substituting y x2 y 2 2 x 2 2 x2 y 2 2 x 2 2 x2 2 x y2 θ 45 Convert 45 to radians θ π4 Using the rotation formulas x x cosπ4 y sinπ4 22 x 22 y y x sinπ4 y cosπ4 22 x 22 y Substituting x y2 x 22 y2 22 y y 22 y2 22 y 3 x24 y21 1 θ 45 Using the rotation formulas x 22 x 22 y y 22 x 22 y Substituting into the ellipse equation 22 x 22 y24 22 x 22 y21 1 4 x216 y29 1 θ 45 Using the same rotation formulas x 22 x 22 y y 22 x 22 y Substituting into the ellipse equation 22 x 22 y29 22 x 22 y216 1 5 y2 x2 1 θ 45 Using the rotation formulas x 22 x 22 y y 22 x 22 y Substituting into the hyperbola equation 22 x 22 y2 22 x 22 y2 1 6 y x22 θ 30 Convert 30 to radians θ π6 Using the rotation formulas x x cosπ6 y sinπ6 32 x 12 y y x sinπ6 y cosπ6 12 x 32 y Substituting y x22 y 12 x 32 x22 7 x y 12 θ 30 Using the same rotation formulas x 32 x 12 y y 12 x 32 y Substituting x y 12 x 32 y 12 12 y 8 x29 y24 1 θ 30 Using the same rotation formulas x 32x 12y y 12x 32y Substituting into the ellipse equation 32x 12y29 12x 32y24 1 1 image of a parabola with focus F and directrix L 2 Step 1 Define the geometry of the problem We are given The focus of the parabola F0 4 The directrix y 2 A parabola is the set of points x y that are equidistant from the focus and the directrix Step 2 Label a point x y anywhere on the parabola Let Px y represent any point on the parabola Step 3 Write an expression for the distance from the point x y to the directrix The distance between a point Px y and the directrix y 2 is the vertical distance from Px y to the line y 2 This is given by ddirectrix y 2 This is the absolute difference between the ycoordinate of P and the directrix Step 4 Write an expression for the distance from the point x y to the focus The distance between a point Px y and the focus F0 4 is the Euclidean distance This can be calculated using the distance formula dfocus x 02 y 42 x2 y 42 Step 5 Apply the definition of a parabola By definition a parabola is the set of points equidistant from the focus and the directrix Therefore the distances from Px y to the focus and the directrix must be equal x2 y 42 y 2 Step 6 Square both sides to eliminate the square root To remove the square root square both sides of the equation x2 y 42 y 22 Step 7 Expand both sides Expand both sides of the equation Lefthand side x² y 4² x² y² 8y 16 Righthand side y 2² y² 4y 4 Thus the equation becomes x² y² 8y 16 y² 4y 4 Step 8 Simplify the equation Cancel out y² from both sides x² 8y 16 4y 4 Now move all terms involving y to one side x² 4y 12 Step 9 Solve for y Solve the equation for y 4y x² 12 y x² 124 Simplifying y 14x² 3 This is the equation of the parabola in standard form Final Answer The equation of the parabola is y 14x² 3 Problem set 2 We use the distance formula between two points x₁ y₁ and x₂ y₂ d x₂ x₁² y₂ y₁² Substituting the given values 4 2 and 0 1 d 0 4² 1 2² 4² 1² 16 1 17 Final Answer for Problem 2 The distance is 17 units 3 Find the distance from the point 4 2 to the line y 2 The formula for the distance from a point x₁ y₁ to a horizontal line y b is d y₁ b Here the point is 4 2 and the line is y 2 d 2 2 2 2 4 Final Answer for Problem 3 The distance is 4 units 4 Find the distance from the point 1 3 to the point 3 4 Using the distance formula between two points 1 3 and 3 4 d 3 1² 4 3² 3 1² 4 3² 4² 7² 16 49 65 Final Answer for Problem 4 The distance is 65 units 5 Find the distance from the point 1 3 to the line y 5 Using the distance formula from a point to a horizontal line d y₁ b Here the point is 1 3 and the line is y 5 d 3 5 2 2 Final Answer for Problem 5 The distance is 2 units 6 Find the distance from the point x 4 to the line y 1 Using the pointtoline formula d y₁ b The point is x 4 and the line is y 1 d 4 1 4 1 5 Final Answer for Problem 6 The distance is 5 units 7 Find the distance from the point x 3 to the line y 2 Using the pointtoline formula d y₁ b Here the point is x 3 and the line is y 2 d 3 2 5 5 Final Answer for Problem 7 The distance is 5 units 8 Find the values of x for which the point x 4 is equidistant from 0 1 and the line y 1 We need to set up an equation where the distance from x 4 to 0 1 equals the distance from x 4 to the line y 1 Step 1 Distance from x 4 to 01 using the distance formula dpoint x 0² 4 1² x² 3² x² 9 Step 2 Distance from x 4 to the line y 1 using the pointtoline formula dline 4 1 4 1 5 Now equate the two distances x² 9 5 Step 3 Square both sides x² 9 25 Step 4 Solve for x x² 16 x 4 Final Answer for Problem 8 The values of x are x 4 and x 4 9 Find the values of x for which the point x 3 is equidistant from 1 2 and the line y 2 Step 1 Distance from x 3 to 1 2 dpoint sqrtx 12 3 22 sqrtx 12 12 sqrtx 12 1 Step 2 Distance from x 3 to the line y 2 dline 3 2 5 5 Equating the two distances sqrtx 12 1 5 Step 3 Square both sides x 12 1 25 Step 4 Solve for x x 12 24 x 1 sqrt24 x 1 2 sqrt6 Final Answer for Problem 9 The values of x are x 1 2 sqrt6 and x 1 2 sqrt6 Problem 10 Consider the equation y x2 a Find the coordinates of the three points on the graph of y x2 whose xvalues are 1 2 and 3 We are given the equation y x2 Lets substitute x 1 x 2 and x 3 to find the corresponding yvalues When x 1 y 12 1 So the point is 1 1 When x 2 y 22 4 So the point is 2 4 When x 3 y 32 9 So the point is 3 9 Final Answer for 10a The points are 1 1 2 4 and 3 9 b Show that each of the three points in part a is equidistant from the point 0 14 and the line y 14 We need to show that the distance from each point x y to the point 0 14 equals the distance from that point x y to the line y 14 Step 1 Distance from x y to the point 0 14 The distance formula between two points is dpoint sqrtx 02 y 142 sqrtx2 y 142 Step 2 Distance from x y to the line y 14 The distance from a point x y to a horizontal line y b is dline y 14 y 14 Now for each of the three points we check the equality of distances For 1 1 Distance to point 0 14 dpoint sqrt12 1 142 sqrt1 342 sqrt1 916 sqrt2516 54 Distance to line y 14 dline 1 14 54 54 Both distances are equal For 2 4 Distance to point 0 14 dpoint sqrt22 4 142 sqrt4 1542 sqrt4 22516 sqrt28916 174 Distance to line y 14 dline 4 14 174 174 Both distances are equal For 3 9 Distance to point 0 14 dpoint sqrt32 9 142 sqrt9 3542 sqrt9 122516 sqrt136916 374 Distance to line y 14 dline 9 14 374 374 Both distances are equal Final Answer for 10b All three points are equidistant from the point 0 14 and the line y 14 c Show that if the point with coordinates x y is equidistant from the point 0 14 and the line y 14 then y x2 We set the distance from x y to 0 14 equal to the distance from x y to the line y 14 Step 1 Distance from x y to the point 0 14 dpoint sqrtx2 y 142 Step 2 Distance from x y to the line y 14 dline y 14 Now equate the two distances sqrtx2 y 142 y 14 Square both sides to eliminate the square root x2 y 142 y 142 Expand both sides x2 y2 2y14 116 y2 2y14 116 Simplify x2 y2 12y 116 y2 12 y 116 Cancel out y2 and 116 from both sides x2 12y 12y Solve for y x2 y Final Answer for 10c If the point x y is equidistant from 0 14 and y 14 then y x2 Problem 11 Consider the equation y 12 x2 2x a Find the coordinates of the three points on the graph of y 12 x2 2x whose xvalues are 2 0 and 4 When x 2 y 12 22 22 12 4 4 2 4 6 So the point is 2 6 When x 0 y 12 02 20 0 So the point is 0 0 When x 4 y 12 42 24 12 16 8 8 8 0 So the point is 4 0 Final Answer for 11a The points are 2 6 0 0 and 4 0 1 Deriving the equation of the hyperbola We are given two foci F0 5 and G0 5 and the condition PF PG 6 where Px y is a point on the hyperbola Using the distance formula the distances from Px y to the foci F0 5 and G0 5 are PF sqrtx2 y 52 PG sqrtx2 y 52 The hyperbola is defined by the equation PF PG 6 This gives us two cases 1 PF PG 6 2 PG PF 6 We solve for the first case PF PG 6 sqrtx2 y 52 sqrtx2 y 52 6 Square both sides sqrtx2 y 522 sqrtx2 y 52 62 x2 y 52 sqrtx2 y 522 12 sqrtx2 y 52 36 This simplifies to an equation for the hyperbola We can derive it more easily by remembering that the standard form of a vertical hyperbola is x2 a2 y2 b2 1 Here the distance between the foci is 10 so 2c 10 gives c 5 Since the difference in distances is 6 we have 2a 6 giving a 3 Using c2 a2 b2 we can solve for b2 52 32 b2 25 9 b2 b2 16 Thus the equation of the hyperbola is x2 9 y2 16 1 2 To find where the hyperbola intersects the yaxis set x 0 in the equation 02 9 y2 16 1 y2 16 1 y2 16 There is no real solution so the hyperbola does not intersect the yaxis 3 The asymptotes of a hyperbola are the boundary lines For a hyperbola of the form x2 a2 y2 b2 1 the asymptotes are given by y ba x For our hyperbola x2 9 y2 16 1 we have y 43 x Thus the boundary line in the first quadrant is y 43 x 4 To sketch the hyperbola plot The center at 0 0 The vertices at 0 3 The foci at 0 5 and 0 5 The asymptotes y 43 x Problem set 1 a Foci A2 0 and B2 0 PA PB 2 The distance between the foci is 4 so 2c 4 giving c 2 The difference in distances is 2 so 2a 2 giving a 1 Using c2 a2 b2 we solve for b2 22 12 b2 4 1 b2 b2 3 The equation of the hyperbola is x2 1 y2 3 1 Asymptotes y sqrt3 x b Foci A5 0 and B5 0 PA PB 5 The distance between the foci is 10 so 2c 10 giving c 5 The difference in distances is 5 so 2a 5 giving a 52 Using c2 a2 b2 we solve for b2 52 522 b2 25 254 b2 b2 754 The equation of the hyperbola is x2 522 y2 75 22 1 c Foci A0 3 and B0 3 PA PB 25 This is a vertical hyperbola with c 3 a 25 2 125 Using c2 a2 b2 we can solve for b2 d Foci A0 2 and B0 2 PA PB 2 This is also a vertical hyperbola The same process applies as in part c 2 Here are the graphs of the hyperbolas for the given equations 1 The hyperbola with equation x2 9 y2 16 1 in red 2 The hyperbola with equation x2 1 y2 3 1 in green 3 The hyperbola with equation x2 522 y2 75 22 1 in blue 4 The hyperbola with equation x2 2 y2 2 1 in magenta 3 This is the equation of a hyperbola where the shape and orientation depend on the value of k a k 4 The equation is x2 y2 4 This describes a standard hyperbola opening left and right centered at the origin with vertices at 2 0 b k 1 The equation is x2 y2 1 This describes a similar hyperbola to the previous one but with vertices at 1 0 c k 14 The equation is x2 y2 14 This hyperbola is narrower with vertices at 14 0 12 0 d k 0 The equation x2 y2 0 simplifies to x2 y2 which describes two intersecting lines x y and x y e k 14 The equation x2 y2 14 has no real solution because the lefthand side is always positive or zero while the righthand side is negative f k 1 The equation x2 y2 1 similarly has no real solution g k 4 Again x2 y2 4 has no real solution h Hyperbolas for different values of k For positive k the hyperbola opens horizontally with vertices at k 0 As k approaches zero the hyperbola becomes narrower until it becomes two intersecting lines at k 0 For negative k there is no real solution meaning the hyperbola doesnt exist i Any value of k that makes the equation x2 y2 k have no real solution When k 0 the equation has no real solution because x2 is always greater than or equal to zero while y2 is also nonnegative 4 This is another form of a hyperbola where the shape depends on the value of k a k 1 The equation becomes x2 1 y2 1 or x2 y2 1 which has no real solutions b k 1 The equation becomes x2 1 y2 1 or x2 y2 1 which is a standard hyperbola c k 2 The equation is x2 2 y2 1 describing a horizontally compressed hyperbola d k 4 The equation becomes x2 4 y2 1 which describes an even more compressed hyperbola e k 10 The equation becomes x2 10 y2 1 where the hyperbola is very narrow horizontally f k 25 The equation becomes x2 25 y2 1 describing a very narrow hyperbola horizontally g As k As k increases the hyperbola becomes more compressed horizontally and it approaches two vertical asymptotes essentially turning into vertical lines 5 This form also describes hyperbolas but with a vertical orientation a k 1 The equation becomes x2 y2 1 1 which describes an ellipse not a hyperbola b k 1 The equation becomes x2 y2 1 a standard hyperbola opening horizontally c k 2 The equation becomes x2 y2 2 1 which describes a vertically stretched hyperbola d k 4 The equation becomes x2 y2 4 1 a more stretched hyperbola vertically e k 10 The equation becomes x2 y2 10 1 a very vertically stretched hyperbola f As k As k increases the hyperbola becomes increasingly stretched vertically approaching two horizontal asymptotes 6 a Completing the square Start by rearranging the terms 9x2 36x 4y2 8y 4 Complete the square for x 9x2 4x 9x 22 4 9x 22 36 For y 4y2 2y 4y 12 1 4y 12 4 Substitute these back into the equation 9x 22 36 4y 12 4 4 Simplifying 9x 22 4y 12 36 Finally divide by 36 to get the standard form x 22 4 y 12 9 1 This is the standard form of a hyperbola b Center The center of the hyperbola is 2 1 c Asymptotes The asymptotes of the hyperbola are given by the equations x 22 y 13 These simplify to y 1 32 x 2 d Graph The hyperbola opens horizontally with center 2 1 vertices at 4 1 and 0 1 and asymptotes described above 7 a 4x2 8x 25y2 100y 4 0 This represents an ellipse Completing the square for both x and y confirms this b 4x2 16x 9y2 54y 65 0 This is a hyperbola as the coefficients of x2 and y2 have opposite signs c 4x2 8x y2 2y 5 0 This represents a single point as completing the square leads to a degenerate case where the ellipse collapses to a point d 49x2 98x 4y2 245 0 This is a hyperbola since the coefficients of x2 and y2 have opposite signs e Interpreting the coefficients in a x2 b x c y2 d y e 0 If a and c have the same sign the graph is an ellipse If a and c have opposite signs the graph is a hyperbola If either a or c is zero the equation may represent parabolas or lines