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734Chapter 10Topics in Analytic Geometry54. Conveyor Design A moving conveyor is built so that itrises 1 meter for each 3 meters of horizontal travel.(a) Draw a diagram that gives a visual representation ofthe problem.(b) Find the inclination of the conveyor.(c) The conveyor runs between two floors in a factory. Thedistance between the floors is 5 meters. Find the lengthof the conveyor.55. Truss Find the angles and shown in the drawing ofthe roof truss.α6 ft6 ftβ9 ft36 ftModel It56. Inclined Plane The Johnstown Inclined Plane inJohnstown, Pennsylvania is an inclined railway that wasdesigned to carry people to the hilltop community ofWestmont. It also proved useful in carrying people andvehicles to safety during severe floods. The railway is896.5 feet long with a 70.9% uphill grade (see figure).58. To find the angle between two lines whose angles ofinclination 1 and 2 are known, substitute 1 and 2 for m1and m2, respectively, in the formula for the angle betweentwo lines.59. Exploration 0, 4 .Consider a line with slope m and y-intercept(a) Write the distance d between the origin and the line asa function of m.(b) Graph the function in part (a).(c) Find the slope that yields the maximum distancebetween the origin and the line.(d) Find the asymptote of the graph in part (b) andinterpret its meaning in the context of the problem.60. Exploration 0, 4 .Consider a line with slope m and y-intercept(a) Write the distance d between the point 3, 1 and theline as a function of m.(b) Graph the function in part (a).(c) Find the slope that yields the maximum distancebetween the point and the line.(d) Is it possible for the distance to be 0? If so, what is theslope of the line that yields a distance of 0?(e) Find the asymptote of the graph in part (b) andinterpret its meaning in the context of the problem.Skills ReviewIn Exercises 61– 66, find all x -intercepts and y-intercepts ofthe graph of the quadratic function.896.5 ft61. f x x 7 262. f x x 9 263. f x x 5 2 5θNot drawn to scale(a) Find the inclination of the railway.(b) Find the change in elevation from the base to thetop of the railway.(c) Using the origin of a rectangular coordinate systemas the base of the inclined plane, find the equationof the line that models the railway track.(d) Sketch a graph of the equation you found in part (c).Synthesis64. f x x 11 2 1265. f x x 2 7x 166. f x x 2 9x 22In Exercises 67–72, write the quadratic function in standardform by completing the square. Identify the vertex of thefunction.67. f x 3x 2 2x 1668. f x 2x 2 x 2169. f x 5x 34x 770. f x x 2 8x 15271. f x 6x 2 x 1272. f x 8x 2 34x 21True or False? In Exercises 57 and 58, determine whetherthe statement is true or false. Justify your answer.In Exercises 73–76, graph the quadratic function.57. A line that has an inclination greater than 2 radians hasa negative slope.73. f x x 4 2 374. f x 6 x 1 275. g x 2x 3x 176. g x x2 6x 82

Section 10.2Introduction to Conics: Parabolas73510.2 Introduction to Conics: ParabolasWhat you should learn Recognize a conic as theintersection of a plane anda double-napped cone. Write equations of parabolasin standard form and graphparabolas. Use the reflective property ofparabolas to solve real-lifeproblems.Why you should learn itConicsConic sections were discovered during the classical Greek period, 600 to 300 B.C.The early Greeks were concerned largely with the geometric properties of conics.It was not until the 17th century that the broad applicability of conics becameapparent and played a prominent role in the early development of calculus.A conic section (or simply conic) is the intersection of a plane and a doublenapped cone. Notice in Figure 10.8 that in the formation of the four basicconics, the intersecting plane does not pass through the vertex of the cone. Whenthe plane does pass through the vertex, the resulting figure is a degenerate conic,as shown in Figure 10.9.Parabolas can be used tomodel and solve many types ofreal-life problems. For instance,in Exercise 62 on page 742, aparabola is used to model thecables of the Golden GateBridge.CircleFIGURE10.8EllipseBasic Conics10.9Degenerate ConicsParabolaHyperbolaCosmo Condina/Getty ImagesPointFIGURELineTwo IntersectingLinesThere are several ways to approach the study of conics. You could begin bydefining conics in terms of the intersections of planes and cones, as the Greeksdid, or you could define them algebraically, in terms of the general seconddegree equationAx 2 Bxy Cy 2 Dx Ey F 0.However, you will study a third approach, in which each of the conics is definedas a locus (collection) of points satisfying a geometric property. For example, inSection 1.2, you learned that a circle is defined as the collection of all points x, y that are equidistant from a fixed point h, k . This leads to the standard formof the equation of a circle x h 2 y k 2 r 2.Equation of circle

736Chapter 10Topics in Analytic GeometryParabolasIn Section 2.1, you learned that the graph of the quadratic functionf x ax2 bx cis a parabola that opens upward or downward. The following definition of aparabola is more general in the sense that it is independent of the orientation ofthe parabola.Definition of ParabolaA parabola is the set of all points x, y in a plane that are equidistant froma fixed line (directrix) and a fixed point (focus) not on the line.yThe midpoint between the focus and the directrix is called the vertex, and theline passing through the focus and the vertex is called the axis of the parabola.Note in Figure 10.10 that a parabola is symmetric with respect to its axis. Usingthe definition of a parabola, you can derive the following standard form of theequation of a parabola whose directrix is parallel to the x-axis or to the y-axis.d2Focusd1d1Vertexd2Standard Equation of a ParabolaDirectrixxFIGURE10.10ParabolaThe standard form of the equation of a parabola with vertex at h, k is asfollows. x h 2 4p y k , p 0Vertical axis, directrix: y k p y k 4p x h , p 0Horizontal axis, directrix: x h p2The focus lies on the axis p units (directed distance) from the vertex. If thevertex is at the origin 0, 0 , the equation takes one of the following forms.x 2 4pyVertical axisy 2 4pxHorizontal axisSee Figure 10.11.For a proof of the standard form of the equation of a parabola, see Proofs inMathematics on page 807.Axis:x hFocus:(h , k p )Axis: x hDirectrix: y k pVertex: (h, k)p 0p 0Vertex:(h , k)Directrix:y k p(a) x h 2 4p y k Vertical axis: p 0FIGURE10.11Directrix: x h pp 0Directrix:x h pp 0Focus:(h, k p)(b) x h 2 4p y k Vertical axis: p 0Focus:(h p , k)Axis:y kVertex: (h, k)(c) y k 2 4p x h Horizontal axis: p 0Focus:(h p, k)Axis:y kVertex:(h, k)(d) y k 2 4p x h Horizontal axis: p 0

Section 10.2Example 1Te c h n o l o g yUse a graphing utility to confirmthe equation found in Example 1.In order to graph the equation,you may have to use two separateequations:y1 8xIntroduction to Conics: Parabolas737Vertex at the OriginFind the standard equation of the parabola with vertex at the origin and focus 2, 0 .SolutionThe axis of the parabola is horizontal, passing through 0, 0 and 2, 0 , as shownin Figure 10.12.Upper partyandy2 8x.Lower part2y 2 8x1Vertex1 1Focus(2, 0)23x4(0, 0) 2You may want to review thetechnique of completing thesquare found in Appendix A.5,which will be used to rewriteeach of the conics in standardform.FIGURE10.12So, the standard form is y 2 4px, where h 0, k 0, and p 2. So, theequation is y 2 8x.Now try Exercise 33.Example 2Finding the Focus of a ParabolaFind the focus of the parabola given by y 12 x 2 x 12.SolutionTo find the focus, convert to standard form by completing the square.y 12 x 2 x 12y 2y x 2 2x 121 2y x 2 2xVertex ( 1, 1)Focus 1, 121( 3) 2x 11 1y 12 x2 x 12 2FIGURE10.131 1 2y x 2 2x 12 2y x 2 2x 1 2 y 1 x 1 2Write original equation.Multiply each side by –2.Add 1 to each side.Complete the square.Combine like terms.Standard formComparing this equation with x h 2 4p y k you can conclude that h 1, k 1, and p 12. Because p is negative,the parabola opens downward, as shown in Figure 10.13. So, the focus of theparabola is h, k p 1, 12 .Now try Exercise 21.

738Chapter 10Topics in Analytic Geometryy8(x 2)2 12(y 1)6Focus(2, 4)Find the standard form of the equation of the parabola with vertex 2, 1 andfocus 2, 4 .Solution4Vertex(2, 1) 4x 22468 2Because the axis of the parabola is vertical, passing through 2, 1 and 2, 4 ,consider the equation x h 2 4p y k where h 2, k 1, and p 4 1 3. So, the standard form is 4FIGUREFinding the Standard Equation of a ParabolaExample 3 x 2 2 12 y 1 .10.14You can obtain the more common quadratic form as follows. x 2 2 12 y 1 x 2 4x 4 12y 12x2 4x 16 12y1 2 x 4x 16 y12Light sourceat focusWrite original equation.Multiply.Add 12 to each side.Divide each side by 12.The graph of this parabola is shown in Figure 10.14.Now try Exercise 45.AxisFocusApplicationParabolic reflector:Light is reflectedin parallel rays.FIGURE10.15AxisPαFocusαTangentlineA line segment that passes through the focus of a parabola and has endpoints onthe parabola is called a focal chord. The specific focal chord perpendicular to theaxis of the parabola is called the latus rectum.Parabolas occur in a wide variety of applications. For instance, a parabolicreflector can be formed by revolving a parabola around its axis. The resultingsurface has the property that all incoming rays parallel to the axis are reflectedthrough the focus of the parabola. This is the principle behind the construction ofthe parabolic mirrors used in reflecting telescopes. Conversely, the light raysemanating from the focus of a parabolic reflector used in a flashlight are allparallel to one another, as shown in Figure 10.15.A line is tangent to a parabola at a point on the parabola if the line intersects,but does not cross, the parabola at the point. Tangent lines to parabolas have special properties related to the use of parabolas in constructing reflective surfaces.Reflective Property of a ParabolaThe tangent line to a parabola at a point P makes equal angles with thefollowing two lines (see Figure 10.16).1. The line passing through P and the focus2. The axis of the parabolaFIGURE10.16

Section 10.2Example 4y x2d2(0, )14α(0, b)FIGUREFor this parabola, p 14 and the focus is 0, 14 , as shown in Figure 10.17. Youcan find the y-intercept 0, b of the tangent line by equating the lengths of thetwo sides of the isosceles triangle shown in Figure 10.17:1αd1Finding the Tangent Line at a Point on a ParabolaSolution(1, 1)x 1739Find the equation of the tangent line to the parabola given by y x 2 at the point 1, 1 .y1Introduction to Conics: Parabolas10.17d1 1 b4d2 1 0 1 14 and225 .4Note that d1 14 b rather than b 14. The order of subtraction for the distanceis important because the distance must be positive. Setting d1 d2 producesTe c h n o l o g yUse a graphing utility to confirmthe result of Example 4. Bygraphingy1 x 2andy2 2x 1in the same viewing window, youshould be able to see that the linetouches the parabola at the point 1, 1 .15 b 44b 1.So, the slope of the tangent line ism 1 1 21 0and the equation of the tangent line in slope-intercept form isy 2x 1.Now try Exercise 55.WRITING ABOUTMATHEMATICSTelevision Antenna Dishes Cross sections of television antenna dishes are parabolicin shape. Use the figure shown to write a paragraph explaining why these dishesare parabolic.AmplifierDish reflectorCable to radioor TV

740Chapter 10Topics in Analytic Geometry10.2 ExercisesVOCABULARY CHECK: Fill in the blanks.1. A is the intersection of a plane and a double-napped cone.2. A collection of points satisfying a geometric property can also be referred to as a of points.3. A is defined as the set of all points x, y in a plane that are equidistant from a fixed line,called the , and a fixed point, called the , not on the line.4. The line that passes through the focus and vertex of a parabola is called the of the parabola.5. The of a parabola is the midpoint between the focus and the directrix.6. A line segment that passes through the focus of a parabola and has endpoints on the parabola is calleda .7. A line is to a parabola at a point on the parabola if the line intersects, but does not cross, the parabola at the point.PREREQUISITE SKILLS REVIEW: Practice and review algebra skills needed for this section at www.Eduspace.com.In Exercises 1– 4, describe in words how a plane couldintersect with the double-napped cone shown to form theconic section.y(e)(f)y4 6 44x 2 4 2x2 45. y 2 4x7.x2 8y9. y 1 2 4 x 3 1. Circle2. Ellipse3. Parabola4. Hyperbolay11. y 12x 213.y2 6x15. x 2 6y 0y(b)8. y 2 12x10. x 3 2 2 y 1 In Exercises 11–24, find the vertex, focus, and directrix ofthe parabola and sketch its graph.In Exercises 5–10, match the equation with its graph. [Thegraphs are labeled (a), (b), (c), (d), (e), and (f).](a)6. x 2 2y12. y 2x 214. y 2 3x16. x y 2 017. x 1 8 y 2 024624319. x 2 4 y 2 2120. x 2 4 y 1 2121. y 4 x 2 2x 5 122. x 4 y 2 2y 33 x218. x 5 y 1 2 06 2 4y(c)x 224y(d)In Exercises 25–28, find the vertex, focus, and directrix ofthe parabola. Use a graphing utility to graph the parabola.2 4x 2 4 4 623. y 2 6y 8x 25 024. y 2 4y 4x 02 62x4 2 425. x 2 4x 6y 2 026. x 2 2x 8y 9 027. y 2 x y 028. y 2 4x 4 0

Section 10.2In Exercises 29–40, find the standard form of the equationof the parabola with the given characteristic(s) and vertexat the origin.y29.6y30.(3, 6) 8x2x 4453. 52, 0 35. Directrix: y 140. Vertical axis and passes through the point 3, 3 59. Revenue The revenue R (in dollars) generated by the saleof x units of a patio furniture set is given byIn Exercises 41–50, find the standard form of the equationof the parabola with the given characteristics.y(2, 0)(4, 0)(3, 1)Use a graphing utility to graph the function and approximate the number of sales that will maximize revenue.(4.5, 4)x4660. Revenue The revenue R (in dollars) generated by the saleof x units of a digital camera is given by(5, 3)2 4x2y4812( 4, 0)8(0, 4)x(0, 0)8 4x 45 x 135 2 R 25,515 .7Use a graphing utility to graph the function and approximate the number of sales that will maximize revenue.y44. 84 x 106 2 R 14,045 .5y42.4x y 3 058. y 2x 2, 2, 8 39. Horizontal axis and passes through the point 4, 6 43.x y 2 057. y 2x 2, 1, 2 38. Directrix: x 3 2 8x 0956. x 2 2y, 3, 2 37. Directrix: x 24Tangent Line55. x 2 2y, 4, 8 36. Directrix: y 32ParabolaIn Exercises 55–58, find an equation of the tangent line tothe parabola at the given point, and find the x -intercept ofthe line.34. Focus: 0, 2 2y254. x2 12y 033. Focus: 2, 0 41.In Exercises 53 and 54, the equations of a parabola and atangent line to the parabola are given. Use a graphingutility to graph both equations in the same viewing window. Determine the coordinates of the point of tangency. 8431. Focus: 0, 32 32. Focus:51. y 3 2 6 x 1 ; upper half of parabola52. y 1 2 2 x 4 ; lower half of parabola( 2, 6)2 2In Exercises 51 and 52, change the equation of the parabolaso that its graph matches the description.84 4741Introduction to Conics: Parabolas61. Satellite Antenna The receiver in a parabolic televisiondish antenna is 4.5 feet from the vertex and is located at thefocus (see figure). Write an equation for a cross section ofthe reflector. (Assume that the dish is directed upward andthe vertex is at the origin.)8y(3, 3)45. Vertex: 5, 2 ; focus: 3, 2 46. Vertex: 1, 2 ; focus: 1, 0 47. Vertex: 0, 4 ; directrix: y 248. Vertex: 2, 1 ; directrix: x 149. Focus: 2, 2 ; directrix: x 250. Focus: 0, 0 ; directrix: y 8Receiver4.5 ftx

742Chapter 10Topics in Analytic GeometryyModel It(1000, 800)400x400(a) Draw a sketch of the bridge. Locate the origin of arectangular coordinate system at the center of theroadway. Label the coordinates of the known points.0250400500100012001600(1000, 800) 800(c) Complete the table by finding the height y of thesuspension cables over the roadway at a distance ofx meters from the center of the bridge.Height, y800 400(b) Write an equation that models the cables.Distance, xInterstate80062. Suspension Bridge Each cable of the Golden GateBridge is suspended (in the shape of a parabola)between two towers that are 1280 meters apart. The topof each tower is 152 meters above the roadway. Thecables touch the roadway midway between the towers.StreetFIGURE FOR6465. Satellite Orbit A satellite in a 100-mile-high circularorbit around Earth has a velocity of approximately 17,500miles per hour. If this velocity is multiplied by 2, thesatellite will have the minimum velocity necessary toescape Earth’s gravity and it will follow a parabolic pathwith the center of Earth as the focus (see figure).Circularorbity4100milesParabolicpathx63. Road Design Roads are often designed with parabolicsurfaces to allow rain to drain off. A particular road that is32 feet wide is 0.4 foot higher in the center than it is on thesides (see figure).Not drawn to scale(a) Find the escape velocity of the satellite.(b) Find an equation of the parabolic path of the satellite(assume that the radius of Earth is 4000 miles).66. Path of a SoftballThe path of a softball is modeled by 12.5 y 7.125 x 6.25 2where the coordinates x and y are measured in feet, withx 0 corresponding to the position from which the ballwas thrown.32 ft0.4 ftNot drawn to scaleCross section of road surface(a) Find an equation of the parabola that models the roadsurface. (Assume that the origin is at the center of theroad.)(b) How far from the center of the road is the road surface0.1 foot lower than in the middle?64. Highway Design Highway engineers design a paraboliccurve for an entrance ramp from a straight street to aninterstate highway (see figure). Find an equation of theparabola.(a) Use a graphing utility to graph the trajectory of thesoftball.(b) Use the trace feature of the graphing utility to approximate the highest point and the range of the trajectory.Projectile Motion In Exercises 67 and 68, consider thepath of a projectile projected horizontally with a velocity ofv feet per second at a height of s feet, where the model forthe path isv2 y s .16In this model (in which air resistance is disregarded), y isthe height (in feet) of the projectile and x is the horizontaldistance (in feet) the projectile travels.x2

Section 10.267. A ball is thrown from the top of a 75-foot tower with avelocity of 32 feet per second.(a) Find the equation of the parabolic path.(b) How far does the ball travel horizontally beforestriking the ground?68. A cargo plane is flying at an altitude of 30,000 feet and aspeed of 540 miles per hour. A supply crate is dropped fromthe plane. How many feet will the crate travel horizontallybefore it hits the ground?Introduction to Conics: Parabolas743(a) Find the area when p 2 and b 4.(b) Give a geometric explanation of why the area

Dec 08, 2005 · 728 Chapter 10 Topics in Analytic Geometry What you should learn F ind the inclination of a line. F ind the angle between two lines. F ind the distance between a point and a line. Why you should learn it The inclination of a line can be used to measure heights indirectly.For instance