088_0_p676-7 /7/0 :5 PM Page 676 (FPG International / Telegraph Colour Librar) Conic Sections CHAPTER OUTLINE. The Circle. Ellipses and Hperbolas.3 Second-Degree Inequalities and Nonlinear Sstems O ne of the curves we will stud in this chapter has interesting reflective properties. Figure (a) shows how ou can draw one of these curves (an ellipse) using thumbtacks, string, pencil, and paper. Elliptical surfaces will reflect sound waves that originate at one focus through the other focus. This propert of ellipses allows doctors to treat patients with kidne stones using a procedure called lithotrips. A lithotripter is an elliptical device that creates sound waves that crush the kidne stone into small pieces, without surger. The sound wave originates at one focus of the lithotripter. The energ from it reflects off the surface of the lithotripter and converges at the other focus, where the kidne stone is positioned. Figure (b) shows a cross-section of a lithotripter, with a patient positioned so the kidne stone is at the other focus. Pencil tracing out an ellipse from a string anchored b two tacks (a) (b) FIGURE B studing the conic sections in this chapter, ou will be better equipped to understand some of the more technical equipment that eists in the world outside of class. 676
. The Circle Conic sections include ellipses, circles, hperbolas, and parabolas. The are called conic sections because each can be found b slicing a cone with a plane as shown in Figure. We begin our work with conic sections b studing circles. Before we find the general equation of a circle, we must first derive what is known as the distance formula. Parabola Circle Ellipse Hperbola FIGURE Suppose (, ) and (, ) are an two points in the first quadrant. (Actuall, we could choose the two points to be anwhere on the coordinate plane. It is just more convenient to have them in the first quadrant.) We can name the points P and P, respectivel, and draw the diagram shown in Figure. d P (, ) P (, ) Q(, ) FIGURE 677
678 C HAPTER Conic Sections Notice the coordinates of point Q. The -coordinate is because Q is directl below point P. The -coordinate of Q is because Q is directl across from point P. It is evident from the diagram that the length of P Q is and the length of P Q is. Using the Pthagorean theorem, we have (P P ) (P Q) (P Q) or d ( ) ( ) Taking the square root of both sides, we have d ( ) ( ) We know this is the positive square root, because d is the distance from P to P and must therefore be positive. This formula is called the distance formula. Eample Find the distance between (3, 5) and (, ). Solution If we let (3, 5) be (, ) and (, ) be (, ) and appl the distance formula, we have d ( 3) ( 5) (3, 5) () (6) 36 d 37 5 3-5--3-- - 3 5 - (, ) -3 - -5 FIGURE 3 Eample Find if the distance from (, 5) to (3, ) is. Solution Using the distance formula, we have ( 3) (5 ) ( 3) 6 0 6 8 0 ( )( ) Distance formula Square each side. Epand ( 3). Simplif. Factor. or Set factors equal to 0.
Section. The Circle 67 The two solutions are and, which indicates that two points, (, 5) and (, 5), are units from (3, ). We can use the distance formula to derive the equation of a circle. Theorem. The equation of the circle with center at (a, b) and radius r is given b ( a) ( b) r Proof B definition, all points on the circle are a distance r from the center (a, b). If we let (, ) represent an point on the circle, then (, ) is r units from (a, b). Appling the distance formula, we have r ( a) ( b) Squaring both sides of this equation gives the equation of the circle: ( a) ( b) r We can use Theorem. to find the equation of a circle, given its center and radius, or to find its center and radius, given the equation. Eample 3 Find the equation of the circle with center at ( 3, ) having a radius of 5. Solution We have (a, b) ( 3, ) and r 5. Appling Theorem. ields [ (3)] ( ) 5 ( 3) ( ) 5 Eample at the origin. Give the equation of the circle with radius 3 whose center is Solution The coordinates of the center are (0, 0), and the radius is 3. The equation must be ( 0) ( 0) 3 We can see from Eample that the equation of an circle with its center at the origin and radius r will be r
680 C HAPTER Conic Sections Eample 5 Find the center and radius, and sketch the graph of the circle whose equation is ( ) ( 3) Solution Writing the equation in the form ( a) ( b) r we have ( ) [ (3)] The center is at (,3), and the radius is. The graph is shown in Figure. 5 3 ( ) + ( + 3) = -5--3-- - 3 5 - -3 - -5 FIGURE Eample 6 Sketch the graph of. Solution Because the equation can be written in the form ( 0) ( 0) 3 it must have its center at (0, 0) and a radius of 3. The graph is shown in Figure 5. 5 3 + = -5--3-- - 3 5 - -3 - -5 FIGURE 5
Section. The Circle 68 Eample 7 Sketch the graph of 6 0. Solution To sketch the graph, we must find the center and radius of our circle. We can do so easil if the equation is in standard form. That is, if it has the form ( a) ( b) r To put our equation in standard form, we start b using the addition propert of equalit to group all the constant terms together on the right side of the equation. In this case, we add to each side of the equation. We do this because we are going to add our own constants later to complete the square. 6 Net, we group all the terms containing together and all terms containing together, and we leave some space at the end of each group for the numbers we will add when we complete the square on each group. 6 To complete the square on, we add to each side of the equation. To complete the square on, we add to each side of the equation. 6 The first three terms on the left side can be written as ( 3). Likewise, the last three terms on the left side simplif to ( ). The right side simplifies to 5. ( 3) ( ) 5 Writing 5 as 5, we have our equation in standard form. ( 3) ( ) 5 From this last line, it is apparent that the center is at ( 3, ) and the radius is 5. Using this information, we create the graph shown in Figure 6. 0 8 6-0 -6 - - 6 0 - -6-8 -0 ( + 3) + ( ) = 5 FIGURE 6
68 C HAPTER Conic Sections Getting Read for Class After reading through the preceding section, respond in our own words and in complete sentences. A. Describe the distance formula in words, as if ou were eplaining to someone how the should go about finding the distance between two points. B. What is the mathematical definition of a circle? C. How are the distance formula and the equation of a circle related? D. When graphing a circle from its equation, wh is completing the square sometimes useful? PROBLEM SET. Find the distance between the following points.. (3, 7) and (6, 3). (, 7) and (8, ) 3. (0, ) and (5, 0). (3, 0) and (0, ) 5. (3, 5) and (, ) 6. (8, ) and (3, ) 7. (, ) and (0, 5) 8. (3, 8) and (, 6). Find so the distance between (, ) and (, 5) is 3. 0. Find so the distance between (, 3) and (, ) is 3.. Find so the distance between (, 5) and (3, ) is 5.. Find so the distance between (, ) and (, ) is 8. 3. Find so the distance between (, ) and (, 6) is 6.. Find so the distance between (3, ) and (7, 3 ) is 6. Write the equation of the circle with the given center and radius. 5. Center (3, ); r 3 6. Center (, ); r 7. Center (5, ); r 5 8. Center (7, 6); r 3. Center (0, 5); r 0. Center (0, ); r 7. Center (0, 0); r. Center (0, 0); r 5 Give the center and radius, and sketch the graph of each of the following circles. 3.. 6 5. ( ) ( 3) 5 6. ( ) ( ) 36 7. ( ) ( ) 8 8. ( 3) ( ). ( ) ( ) 7 30. ( ) 3. 3. 33. 6 7 3. 5 35. 36. 0 0 37. 6 38.
Section. Problem Set 683 3. 0. 6. 8 3. 36 36 3 Each of the following circles passes through the origin. In each case, find the equation. 3. A B C - - 3 - - (3, ) 6. Each of the following circles passes through the origin. The centers are as shown. Find the equation of each circle. (0, 0). C B A (, 3) (0, 0) 5. Find the equations of circles A, B, and C in the following diagram. The three points are the centers of the three circles. 7. Find the equation of the circle with center at the origin that contains the point (3, ). 8. Find the equation of the circle with center at the origin that contains the point ( 5, ).. Find the equation of the circle with center at the origin and -intercepts 3 and 3. 50. Find the equation of the circle with -intercepts and and center at the origin.
088_0_p676-7 /7/0 :5 PM Page 68 68 CHAPTER Conic Sections 5. A circle with center at (, 3) passes through the point (, 3). Find the equation. 5. A circle with center at (, 5) passes through the point (, ). Find the equation. foot 53. Find the equation of the circle with center at (, 5), which passes through the point (, 3). feet 5. Find the equation of the circle with center at (, ), which passes through the point (6, 5). 8 feet 55. Find the equation of the circle with center on the ais and -intercepts at and 6. 56. Find the equation of the circle with center on the ais and -intercepts at 8 and. 57. Find the circumference and area of the circle ( 3) 8. Leave our answer in terms of. 58. Find the circumference and area of the circle ( ) ( 6). Leave our answer in terms of. 5. Find the circumference and area of the circle 0. Leave our answer in terms of. 60. Find the circumference and area of the circle 6 6. Leave our answer in terms of. 63. Placing a Bubble Fountain A circular garden pond with a diameter of feet is to have a bubble fountain. The water from the bubble fountain falls in a circular pattern with a radius of.5 feet. If the center of the bubble fountain is placed feet West and 3 feet North of the center of the pond, will all the water from the fountain fall inside the pond? What is the farthest distance from the center of the pond that water from the fountain will fall? 6. Biccle Path A circular biccle path is to be built in a corner of the cit park. If the circumference of Appling the Concepts 6. Search Area A 3-ear-old child has wandered awa from home. The police have decided to search a circular area with a radius of 6 blocks. The child turns up at his grandmother s house, 5 blocks East and 3 blocks North of home. Was he found within the search area? 6. Sizing a Tunnel The highwa department is making plans for a semicircular tunnel. If the want the tunnel to be large enough that a truck feet tall and 8 feet wide can pass through the tunnel remaining in his own lane and having an etra foot clearance, what is the minimum radius of the tunnel? 005 ft.
088_0_p676-7 /7/0 :5 PM Page 685 685 Section. Problem Set 6m 8m 0m m m 6m 66. Swing Across the Circles Another piece of plaground equipment will be 6 feet tall and 0 feet across. Circles with a diameter of foot are connected to each other and the top. If the origin of the coordinate sstem is at the bottom left, write the equations for the first and last circles. 0 feet ft 6 feet the path is to be,005 feet and the origin of the coordinate sstem is at the lower left corner of the park, write an equation for the path. (Hint: Find the radius using the formula for circumference, C r.) 65. Climb the Circles Plaground equipment is being designed using circles. A ladder is to be made of si circles with diameters of 6, 8, 0,,, and 6 inches. The bottom center of the largest circle is placed at the origin of the coordinate sstem. The equation for this circle is ( 8) 6. What is the equation for the smallest circle at the top of the ladder? 67. Range of Two-Wa Radios Heath, Curt, John, and Eric received two-wa radios for Christmas. The radios have a circular range of 5 blocks. Imagine the town having an -coordinate sstem with each block being one unit. Suppose Heath is at the point (0, 0), Curt is at the point (8, ), John is at the point (6, 7), and Eric is at the point (, 8). Write an equation for the circular range of each person. 68. Range of Two-Wa Radios Heath, Curt, John, and Eric received two-wa radios for Christmas. The radios have a circular range of 5 blocks. Imagine the town having an -coordinate sstem with each block being one unit. Suppose Heath is at the point (0, 0), Curt is at the point (8, ), John is at the point (6, 7), and Eric is at the point (, 8). Who is close enough to talk to John? (Hint: Who is at a distance of less than 5 blocks from John?) 6. Tire Swing A tire swing is being built. Three chains are to be attached to the side of the tire equidistant from each other. Suppose the center of the -coordinate sstem is at the center of a tire. If the chains are attached at the points A (7, 7 3), B (7, 7 3), and C (, 0), how far is it between chains? (Hint: Find the distances AB, BC, and CA.) 70. Range of Two-Wa Radios Lee and Otto want to bu two-wa radios to communicate between their campsites. If Otto s campsite is 50 feet West and 75 feet South of Lee s campsite, how man feet range do the radios need to have?
686 C HAPTER Conic Sections Review Problems The following problems are a review of material we covered in Sections 8. and 8.. Find the general term of each sequence. [8.] 7. 5,,3,7,... 7. 3, 8, 5,,... Epand and simplif each series. [8.] 73. 5 7. 6 (i 5) i i i 3 Write using summation notation. [8.] 75. 3 5 7 76. 3 3 5 5 6 Etending the Concepts A circle is tangent to a line if it touches, but does not cross, the line. 77. Find the equation of the circle with center at (, 3) if the circle is tangent to the -ais. 78. Find the equation of the circle with center at (3, ) if the circle is tangent to the -ais. 7. Find the equation of the circle with center at (, 3) if the circle is tangent to the vertical line. 80. Find the equation of the circle with center at (3, ) if the circle is tangent to the horizontal line 6. Find the distance from the origin to the center of each of the following circles. 8. 6 8 8. 8 6 83. 6 8 8. 8 6 85. Find the equation of the circle that has the points P(, ) and Q(5, ) as the endpoints of a diameter. 86. Find the equation of the circle that has the points P(3, 8) and Q(7, ) as the endpoints of a diameter. 87. Find the area of the region that lies inside the circle and outside the circle. Leave our answer in terms of. 88. Find the area of the region that lies inside the circle 5 and outside the circle. Leave our answer in terms of. 8. A square with sides of length is centered at the origin. Find the equation of the circle that circumscribes the square. Note: Circumscribes means the smallest circle to surround the square. 0. A square with sides of length is centered at the origin. Find the equation of the circle that is inscribed in the square. Note: Inscribed means the largest circle to lie within the square.. Ellipses and Hperbolas This section is concerned with the graphs of ellipses and hperbolas. To begin, we will consider onl those graphs that are centered about the origin. Suppose we want to graph the equation We can find the -intercepts b letting 0, and the -intercepts b letting 0: When 0 0 5 5 3 When 5 0 0 5 5
Section. Ellipses and Hperbolas 687 The graph crosses the -ais at (0, 3) and (0, 3) and the -ais at (5, 0) and ( 5, 0). Graphing these points and then connecting them with a smooth curve gives the graph shown in Figure. 5 3 + = 5-5--3-- - 3 5 - -3 - -5 FIGURE We can find other ordered pairs on the graph b substituting in values for (or ) and then solving for (or ). For eample, if we let 3, then 3 5 5 0.36 0.6 Add 0.36 to each side. 5.76 Multipl each side b.. Square root of each side. This would give us the two ordered pairs (3,.) and (3,.). A graph of the tpe shown in Figure is called an ellipse. If we were to find some other ordered pairs that satisf our original equation, we would find that their graphs lie on the ellipse. Also, the coordinates of an point on the ellipse will satisf the equation. We can generalize these results as follows. The Ellipse The graph of an equation of the form Standard form a b (continued)
688 C HAPTER Conic Sections will be an ellipse centered at the origin. The ellipse will cross the -ais at (a, 0) and ( a, 0). It will cross the -ais at (0, b) and (0, b). When a and b are equal, the ellipse will be a circle. Each of the points (a, 0), ( a, 0), (0, b), and (0, b) is a verte (intercept) of the graph. The most convenient wa to graph an ellipse is to locate the intercepts (vertices). Eample Solution Sketch the graph of 36. To write the equation in the form we must divide both sides b 36: 36 a b 36 The graph crosses the -ais at (3, 0), ( 3, 0) and the -ais at (0, ), (0, ). (See Figure.) 36 36 5 3 + = 36-5--3-- - 3 5 - -3 - -5 FIGURE Hperbolas Consider the equation
Section. Ellipses and Hperbolas 68 If we were to find a number of ordered pairs that are solutions to the equation and connect their graphs with a smooth curve, we would have Figure 3. 5 3 = -5--3-- - 3 5 - -3 - -5 FIGURE 3 This graph is an eample of a hperbola. Notice that the graph has -intercepts at (3, 0) and ( 3, 0). The graph has no -intercepts and hence does not cross the -ais. We can show this b substituting 0 into the equation Substitute 0 for. Simplif left side. Multipl each side b. for which there is no real solution. We want to produce reasonable sketches of hperbolas without having to build etensive tables. We can produce the graphs we are after b using what are called asmptotes for our graphs. The discussion that follows is intended to give ou some insight as to wh these asmptotes eist. However, even if ou don t understand this discussion completel, ou will still be able to graph hperbolas. Asmptotes for Hperbolas Let s solve the equation we graphed above for : 0 Original equation Add to each side. Multipl each side b. Square root propert of equalit
60 C HAPTER Conic Sections To understand what comes net, ou need to see that for ver large values of, the following epressions are almost the same: and 3 This is because the becomes insignificant compared with / for ver large values of. In fact, the larger becomes, the closer these two epressions are to being equal. The table below is intended to help ou see this fact. 3 undefined 0.67 0 6.355 6.66667 00 66.63666 66.66667 000 666.66367 666.66667 0000 6666.66637 6666.66667 Etending the idea presented above, we can sa that, for ver large values of, the graphs of the equations will be close to each other. Further, the larger becomes, the closer the graphs are to one another. (Using a similar line of reasoning, we can draw the same conclusion for values of on the other side of the origin, 0, 00,,000, and 0,000.) Believe it or not, this helps us find the shape of our hperbola. We simpl note that the graph of crosses the -ais at 3 and 3, and that as gets further and further from the origin, the graph looks more like the graph of and. 3 3 and 3 5 Asmptotes 3-5--3-- - 3 5 - -3 - = -5 FIGURE
Section. Ellipses and Hperbolas 6 The lines and are asmptotes for the graph of the hperbola. The further we get from the origin, the closer the hperbola is to these 3 3 lines. Asmptotes from a Rectangle In Figure, note the rectangle that has its sides parallel to the - and -aes and that passes through the -intercepts and the points on the -ais corresponding to the square roots of the number below, and. The lines that connect opposite corners of the rectangle are the asmptotes for graph of the hperbola Eample Graph the equation Solution In this case the -intercepts are 3 and 3, and the -intercepts do not eist. We can use the square roots of the number below,however,to find the asmptotes associated with the graph. The sides of the rectangle used to draw the asmptotes must pass through 3 and 3 on the -ais, and and on the -ais. Figure 5 shows the rectangle, the asmptotes, and the hperbola. 6. 5 3 = 6 Asmptotes -5--3-- - 3 5 - -3 - -5 FIGURE 5 Here is a summar of what we have for hperbolas.
6 C HAPTER Conic Sections Hperbolas Centered at the Origin The graph of the equation a will be a hperbola centered at the origin. The graph will have -intercepts (vertices) at a and a. The graph of the equation As an aid in sketching either of these equations, the asmptotes can be found b graphing the lines b and b, or b drawing lines through opposite a a corners of the rectangle whose sides pass through a, a, b, and b on the aes. b b a will be a hperbola centered at the origin. The graph will have -intercepts (vertices) at b and b. Ellipses and Hperbolas not Centered at the Origin The following equation is that of an ellipse with its center at the point (, ): To see wh the center is at (, ) we substitute (read prime ) for and for in the equation. That is, If and the equation becomes ( ) ( ) ( ) This is the equation of an ellipse in a coordinate sstem with an -ais and a ais. We call this new coordinate sstem the -coordinate sstem. The center of our ellipse is at the origin in the -coordinate sstem. The question is this: What are the coordinates of the center of this ellipse in the original -coordinate sstem? To answer this question, we go back to our original substitutions: In the -coordinate sstem, the center of our ellipse is at 0, 0(the origin of the sstem). Substituting these numbers for and, we have 0 0 () Solving these equations for and will give us the coordinates of the center of our ( ) ()
Section. Ellipses and Hperbolas 63 ellipse in the -coordinate sstem. As ou can see, the solutions are and. Therefore, in the -coordinate sstem, the center of our ellipse is at the point (, ). Figure 6 shows the graph. 5 3 (, ) -5--3-- 3 5 - (, ) - -3 - -5 ( ) ( ) + = (, 3) (7, ) FIGURE 6 The coordinates of all points labeled in Figure 6 are given with respect to the -coordinate sstem. The - and -aes are shown simpl for reference in our discussion. Note that the horizontal distance from the center to the vertices is 3 the square root of the denominator of the ( ) term. Likewise, the vertical distance from the center to the other vertices is the square root of the denominator of the ( ) term. We summarize the information above with the following: An Ellipse With Center at (h, k) The graph of the equation ( h) a ( k) b will be an ellipse with center at (h, k). The vertices of the ellipse will be at the points (h a, k), (h a, k), (h, k b), and (h, k b). Eample 3 Graph the ellipse: 5 76 0 Solution To identif the coordinates of the center, we must complete the square on and also on. To begin, we rearrange the terms so that those containing are together, those containing are together, and the constant term is on the other side of the equal sign. Doing so gives us the following equation: 5 76
6 C HAPTER Conic Sections Before we can complete the square on, we must factor from each term containing : ( 6 ) 76 To complete the square on, we add to each side of the equation. To complete the square on, we add inside the parentheses. This increases the left side of the equation b 8 since each term within the parentheses is multiplied b. Therefore, we must add 8 to the right side of the equation also. ( 6 ) 76 8 ( ) ( 3) To identif the distances to the vertices, we divide each term on both sides b : ( ) ( ) The graph is an ellipse with center at (, 3), as shown in Figure 7. ( 3) ( 3) 5 ( + ) ( 3) (, ) + = ( 5, 3) 3 (, 3) (, ) -5--3-- 3 5 - - -3 - -5 FIGURE 7 The ideas associated with graphing hperbolas whose centers are not at the origin parallel the ideas just presented about graphing ellipses whose centers have been moved off the origin. Without showing the justification for doing so, we state the following guidelines for graphing hperbolas: Hperbolas With Centers at (h, k) The graphs of the equations ( h) a ( k) b and ( k) b ( h) a
Section. Ellipses and Hperbolas 65 will be hperbolas with their centers at (h, k). The vertices of the graph of the first equation will be at the points (h a, k) and (h a, k), and the vertices for the graph of the second equation will be at (h, k b) and (h, k b). In either case, the asmptotes can be found b connecting opposite corners of the rectangle that contains the four points (h a, k), (h a, k), (h, k b), and (h, k b). Eample Graph the hperbola: 0 0 Solution To identif the coordinates of the center of the hperbola, we need to complete the square on. (Because there is no linear term in, we do not need to complete the square on. The -coordinate of the center will be 0.) 0 0 0 Add 0 to each side. ( ) 0 Factor from each term containing. To complete the square on, we add to the terms inside the parentheses. Doing so adds to the left side of the equation because everthing inside the parentheses is multiplied b. To keep from changing the equation we must add to the right side also. ( ) 0 Add to each side. ( ) 6 ( ) 6 ( ) 6 ( ) 6 6 6 Divide each side b 6. Simplif each term. This is the equation of a hperbola with center at (0, ). The graph opens to the right and left as shown in Figure 8. 0 8 6 ( ) = 6 (, ) (, ) -0-6 - 6 0 - - -6-8 -0 FIGURE 8