Chapter 7 Hooked on Conics 7. Introduction to Conics In this chapter, we stud the Conic Sections - literall sections of a cone. Imagine a doublenapped cone as seen below being sliced b a plane. If we slice the cone with a horizontal plane the resulting curve is a circle.
96 Hooked on Conics Tilting the plane ever so slightl produces an ellipse. If the plane cuts parallel to the cone, we get a parabola. If we slice the cone with a vertical plane, we get a hperbola. For a wonderful animation describing the conics as intersections of planes and cones, see Dr. Louis Talman s Mathematics Animated Website.
7. Introduction to Conics 97 If the slicing plane contains the verte of the cone, we get the so-called degenerate conics: a point, a line, or two intersecting lines. We will focus the discussion on the non-degenerate cases: circles, parabolas, ellipses, and hperbolas, in that order. To determine equations which describe these curves, we will make use of their definitions in terms of distances.
98 Hooked on Conics 7. Circles Recall from Geometr that a circle can be determined b fiing a point (called the center) and a positive number (called the radius) as follows. Definition 7.. A circle with center (h, k) and radius r>0 is the set of all points (, ) inthe plane whose distance to (h, k) isr. r (, ) (h, k) From the picture, we see that a point (, ) is on the circle if and onl if its distance to (h, k) isr. We epress this relationship algebraicall using the Distance Formula, Equation., as r = ( h) +( k) B squaring both sides of this equation, we get an equivalent equation (since r>0) which gives us the standard equation of a circle. Equation 7.. The Standard Equation of a Circle: The equation of a circle with center (h, k) and radius r>0is( h) +( k) = r. Eample 7... Write the standard equation of the circle with center (, ) and radius. Solution. Here, (h, k) =(, ) and r =,soweget ( ( )) +( ) = () ( +) +( ) = Eample 7... Graph ( +) +( ) =. Find the center and radius. Solution. From the standard form of a circle, Equation 7., wehavethat + is h, soh = and is k so k =. This tells us that our center is (, ). Furthermore, r =,sor =. Thus we have a circle centered at (, ) with a radius of. Graphing gives us
7. Circles 99 If we were to epand the equation in the previous eample and gather up like terms, instead of the easil recognizable ( +) +( ) =, we d be contending with + + +=0. If we re given such an equation, we can complete the square in each of the variables to see if it fits the form given in Equation 7. b following the steps given below. To Write the Equation of a Circle in Standard Form. Group the same variables together on one side of the equation and position the constant on the other side.. Complete the square on both variables as needed.. Divide both sides b the coefficient of the squares. (For circles, the will be the same.) Eample 7... Complete the square to find the center and radius of 6 + + =0. Solution. 6 + + = 0 6 + + = add to both sides ( ) + ( + ) = factor out leading coefficients ( + ) ( ) + + + ( ) = +()+ complete the square in, 9 9 ( ( ) + + ) = factor ( ( ) + + ) = divide both sides b 9 From Equation 7., weidentif as h, soh =,and + as k, sok =. Hence, the center is (h, k) = (, ). Furthermore, we see that r = 9 so the radius is r =.
00 Hooked on Conics It is possible to obtain equations like ( ) +( +) = 0 or ( ) +( +) =, neither of which describes a circle. (Do ou see wh not?) The reader is encouraged to think about what, if an, points lie on the graphs of these two equations. The net eample uses the Midpoint Formula, Equation., in conjunction with the ideas presented so far in this section. Eample 7... Write the standard equation of the circle which has (, ) and (, ) as the endpoints of a diameter. Solution. We recall that a diameter of a circle is a line segment containing the center and two points on the circle. Plotting the given data ields r (h, k) Since the given points are endpoints of a diameter, we know their midpoint (h, k) is the center of the circle. Equation. gives us ( + (h, k) =, ) + ( + =, + ) ( =, 7 ) The diameter of the circle is the distance between the given points, so we know that half of the distance is the radius. Thus, r = ( ) +( ) ( 0 Finall, since = ( ( )) +( ) = + 0 = ) = 0, our answer becomes ( ( + ) 7 ) = 0
7. Circles 0 We close this section with the most important circle in all of mathematics: the Unit Circle. Definition 7.. The Unit Circle is the circle centered at (0, 0) with a radius of. The standard equation of the Unit Circle is + =. Eample 7... Find the points on the unit circle with -coordinate Solution. We replace with in the equation + = to get + = ( ) + = + =. Our final answers are (, ) ( and, = = ± ) = ±. While this ma seem like an opinion, it is indeed a fact. See Chapters 0 and for details.
0 Hooked on Conics 7.. Eercises In Eercises - 6, find the standard equation of the circle and then graph it.. Center (, ), radius 0. Center (, ), radius. Center (, 7 ), radius. Center (, 9), radius ln(8). Center ( e, ), radius π 6. Center (π, e ), radius 9 In Eercises 7 -, complete the square in order to put the equation into standard form. Identif the center and the radius or eplain wh the equation does not represent a circle. 7. + +0 = 8. 6 = 0 9. + +8 0 = 0 0. + + =0. + + 6 = 0. + + 6 = In Eercises - 6, find the standard equation of the circle which satisfies the given criteria.. center (, ), passes through (, ). center (, 6), passes through (, ). endpoints of a diameter: (, 6) and (, ) 6. endpoints of a diameter: (, ), (, ) 7. The Giant Wheel at Cedar Point is a circle with diameter 8 feet which sits on an 8 foot tall platform making its overall height is 6 feet. Find an equation for the wheel assuming that its center lies on the -ais and that the ground is the -ais. 8. Verif that the following points lie on the Unit Circle: (±, 0), (0, ±), ( ) and ±, ± ( ) ( ±, ±, ±, ± 9. Discuss with our classmates how to obtain the standard equation of a circle, Equation 7., from the equation of the Unit Circle, + = using the transformations discussed in Section.7. (Thus ever circle is just a few transformations awa from the Unit Circle.) 0. Find an equation for the function represented graphicall b the top half of the Unit Circle. Eplain how the transformations is Section.7 can be used to produce a function whose graph is either the top or bottom of an arbitrar circle.. Find a one-to-one function whose graph is half of a circle. (Hint: Think piecewise.) ) Source: Cedar Point s webpage.
7. Circles 0 7.. Answers. ( +) +( +) = 00. ( ) +( +) =9 9 7. ( +) + ( 7 ) =. ( ) +( +9) = (ln(8)) ln(8) +ln(8) 7 6 7 7 6 9 +ln(8) 9 9 ln(8). ( + e) + ( ) = π +π 6. ( π) + ( e ) =9 e + 9 e π e e + π π e e 9 π 9 π π + 9
0 Hooked on Conics 7. ( ) +( +) = Center (, ), radius r = 9. ( +) +( ) = Center (, ), radius r =. +( ) =0 This is not a circle. 8. ( +9) + = Center ( 9, 0), radius r = 0. ( + ( ) ) + = 0 Center (, ), radius r = 0. ( + ( ) ) + = 6 00 Center (, ), radius r = 6 0. ( ) +( ) = 6. ( ) +( 6) =0. ( ) +( ) = 6. ( ) + ( ) = 7. +( 7) = 096
7. Parabolas 0 7. Parabolas We have alread learned that the graph of a quadratic function f() =a + b + c (a 0)is called a parabola. To our surprise and delight, we ma also define parabolas in terms of distance. Definition 7.. Let F be a point in the plane and D be a line not containing F.Aparabola is the set of all points equidistant from F and D. ThepointF is called the focus of the parabola and the line D is called the directri of the parabola. Schematicall, we have the following. F V Each dashed line from the point F to a point on the curve has the same length as the dashed line from the point on the curve to the line D. The point suggestivel labeled V is, as ou should epect, the verte. The verte is the point on the parabola closest to the focus. We want to use onl the distance definition of parabola to derive the equation of a parabola and, if all is right with the universe, we should get an epression much like those studied in Section.. Let p denote the directed distance from the verte to the focus, which b definition is the same as the distance from the verte to the directri. For simplicit, assume that the verte is (0, 0) and that the parabola opens upwards. Hence, the focus is (0,p) and the directri is the line = p. Our picture becomes D (, ) (0,p) (0, 0) = p (, p) From the definition of parabola, we know the distance from (0,p)to(, ) is the same as the distance from (, p) to(, ). Using the Distance Formula, Equation., weget We ll talk more about what directed means later.
06 Hooked on Conics ( 0) +( p) = ( ) +( ( p)) +( p) = ( + p) +( p) = ( + p) square both sides + p + p = +p + p epand quantities = p gather like terms Solving for ields = p, which is a quadratic function of the form found in Equation. with a = p and verte (0, 0). We know from previous eperience that if the coefficient of is negative, the parabola opens downwards. In the equation = p this happens when p<0. In our formulation, we sa that p is a directed distance from the verte to the focus: if p>0, the focus is above the verte; if p<0, the focus is below the verte. The focal length of a parabola is p. If we choose to place the verte at an arbitrar point (h, k), we arrive at the following formula using either transformations from Section.7 or re-deriving the formula from Definition 7.. Equation 7.. The Standard Equation of a Vertical a Parabola: The equation of a (vertical) parabola with verte (h, k) and focal length p is ( h) =p( k) If p>0, the parabola opens upwards; if p<0, it opens downwards. a That is, a parabola which opens either upwards or downwards. Notice that in the standard equation of the parabola above, onl one of the variables,, is squared. This is a quick wa to distinguish an equation of a parabola from that of a circle because in the equation of a circle, both variables are squared. Eample 7... Graph ( +) = 8( ). Find the verte, focus, and directri. Solution. We recognize this as the form given in Equation 7.. Here, h is +soh =, and k is sok =. Hence, the verte is (, ). We also see that p = 8 sop =. Since p<0, the focus will be below the verte and the parabola will open downwards. 6
7. Parabolas 07 The distance from the verte to the focus is p =, which means the focus is units below the verte. From (, ), we move down units and find the focus at (, ). The directri, then, is units above the verte, so it is the line =. Of all of the information requested in the previous eample, onl the verte is part of the graph of the parabola. So in order to get a sense of the actual shape of the graph, we need some more information. While we could plot a few points randoml, a more useful measure of how wide a parabola opens is the length of the parabola s latus rectum. The latus rectum of a parabola is the line segment parallel to the directri which contains the focus. The endpoints of the latus rectum are, then, two points on opposite sides of the parabola. Graphicall, we have the following. the latus rectum F V It turns out that the length of the latus rectum, called the focal diameter of the parabola is p, which, in light of Equation 7., is eas to find. In our last eample, for instance, when graphing ( +) = 8( ), we can use the fact that the focal diameter is 8 = 8, which means the parabola is 8 units wide at the focus, to help generate a more accurate graph b plotting points units to the left and right of the focus. Eample 7... Find the standard form of the parabola with focus (, ) and directri =. Solution. Sketching the data ields, D The verte lies on this vertical line midwa between the focus and the directri No, I m not making this up. Consider this an eercise to show what follows.
08 Hooked on Conics From the diagram, we see the parabola opens upwards. (Take a moment to think about it if ou don t see that immediatel.) Hence, the verte lies below the focus and has an -coordinate of. To find the -coordinate, we note that the distance from the focus to the directri is ( ) =, which means the verte lies units (halfwa) below the focus. Starting at (, ) and moving down / units leaves us at (, /), which is our verte. Since the parabola opens upwards, we know p is positive. Thus p =/. Plugging all of this data into Equation 7. give us ( )( ( ) = ( ( ) = 0 + ) ( )) If we interchange the roles of and, we can produce horizontal parabolas: parabolas which open to the left or to the right. The directrices of such animals would be vertical lines and the focus would either lie to the left or to the right of the verte, as seen below. D V F Equation 7.. The Standard Equation of a Horizontal Parabola: The equation of a (horizontal) parabola with verte (h, k) and focal length p is ( k) =p( h) If p>0, the parabola opens to the right; if p<0, it opens to the left. plural of directri
7. Parabolas 09 Eample 7... Graph ( ) = ( + ). Find the verte, focus, and directri. Solution. We recognize this as the form given in Equation 7.. Here, h is +soh =, and k is sok =. Hence, the verte is (, ). We also see that p =sop =. Since p>0, the focus will be the right of the verte and the parabola will open to the right. The distance from the verte to the focus is p =, which means the focus is units to the right. If we start at (, ) and move right units, we arrive at the focus (, ). The directri, then, is units to the left of the verte and if we move left units from (, ), we d be on the vertical line =. Since the focal diameter is p =, the parabola is units wide at the focus, and thus there are points 6 units above and below the focus on the parabola. 8 7 6 As with circles, not all parabolas will come to us in the forms in Equations 7. or 7.. If we encounter an equation with two variables in which eactl one variable is squared, we can attempt to put the equation into a standard form using the following steps. To Write the Equation of a Parabola in Standard Form. Group the variable which is squared on one side of the equation and position the nonsquared variable and the constant on the other side.. Complete the square if necessar and divide b the coefficient of the perfect square.. Factor out the coefficient of the non-squared variable from it and the constant. Eample 7... Consider the equation + +8 =. Put this equation into standard form and graph the parabola. Find the verte, focus, and directri. Solution. We need a perfect square (in this case, using ) on the left-hand side of the equation and factor out the coefficient of the non-squared variable (in this case, the ) on the other.
0 Hooked on Conics + +8 = + = 8 + + + = 8 + + complete the square in onl ( +) = 8 +8 factor ( +) = 8( ) Now that the equation is in the form given in Equation 7., we see that h is soh =,and k is +sok =. Hence, the verte is (, ). We also see that p = 8 sothatp =. Since p<0, the focus will be the left of the verte and the parabola will open to the left. The distance from the verte to the focus is p =, which means the focus is units to the left of, so if we start at (, ) and move left units, we arrive at the focus (, ). The directri, then, is units to the right of the verte, so if we move right units from (, ),we dbeonthevertical line =. Since the focal diameter is p is 8, the parabola is 8 units wide at the focus, so there are points units above and below the focus on the parabola. 6 In studing quadratic functions, we have seen parabolas used to model phsical phenomena such as the trajectories of projectiles. Other applications of the parabola concern its reflective propert which necessitates knowing about the focus of a parabola. For eample, man satellite dishes are formed in the shape of a paraboloid of revolution as depicted below.
7. Parabolas Ever cross section through the verte of the paraboloid is a parabola with the same focus. To see wh this is important, imagine the dashed lines below as electromagnetic waves heading towards a parabolic dish. It turns out that the waves reflect off the parabola and concentrate at the focus which then becomes the optimal place for the receiver. If, on the other hand, we imagine the dashed lines as emanating from the focus, we see that the waves are reflected off the parabola in a coherent fashion as in the case in a flashlight. Here, the bulb is placed at the focus and the light ras are reflected off a parabolic mirror to give directional light. F Eample 7... A satellite dish is to be constructed in the shape of a paraboloid of revolution. If the receiver placed at the focus is located ft above the verte of the dish, and the dish is to be feet wide, how deep will the dish be? Solution. One wa to approach this problem is to determine the equation of the parabola suggested to us b this data. For simplicit, we ll assume the verte is (0, 0) and the parabola opens upwards. Our standard form for such a parabola is =p. Since the focus is units above the verte, we know p =,sowehave =8. Visuall, units wide (6,)? 6 6 Since the parabola is feet wide, we know the edge is 6 feet from the verte. To find the depth, we are looking for the value when = 6. Substituting = 6 into the equation of the parabola ields 6 =8 or = 6 8 = 9 =.. Hence, the dish will be. feet deep.
Hooked on Conics 7.. Eercises In Eercises - 8, sketch the graph of the given parabola. Find the verte, focus and directri. Include the endpoints of the latus rectum in our sketch.. ( ) = 6. ( + 7 ( ) ) = +. ( ) = ( +). ( +) =. ( ) =( +) 6. ( +) = 0( ) 7. ( ) = 8( ) 8. ( + ( ) ) = 7 + 9 In Eercises 9 -, put the equation into standard form and identif the verte, focus and directri. 9. 0 7 + = 0 0. +0 + =0. + 8 + 9 = 0. + + 8=0. 0 + + = 0. 7 + + =0 In Eercises - 8, find an equation for the parabola which fits the given criteria.. Verte (7, 0), focus (0, 0) 6. Focus (0, ), directri = 7. Verte ( 8, 9); (0, 0) and ( 6, 0) are points on the curve 8. The endpoints of latus rectum are (, 7) and (, 7) 9. The mirror in Carl s flashlight is a paraboloid of revolution. If the mirror is centimeters in diameter and. centimeters deep, where should the light bulb be placed so it is at the focus of the mirror? 0. A parabolic Wi-Fi antenna is constructed b taking a flat sheet of metal and bending it into a parabolic shape. If the cross section of the antenna is a parabola which is centimeters wide and centimeters deep, where should the receiver be placed to maimize reception?. A parabolic arch is constructed which is 6 feet wide at the base and 9 feet tall in the middle. Find the height of the arch eactl foot in from the base of the arch.. A popular novelt item is the mirage bowl. Follow this link to see another startling application of the reflective propert of the parabola.. With the help of our classmates, research spinning liquid mirrors. To get ou started, check out this website. This shape is called a parabolic clinder.
7. Parabolas 7.. Answers. ( ) = 6 Verte (, 0) Focus (, ) Directri = Endpoints of latus rectum (, ), (, ) 6 7 8 9 0. ( + ) 7 ( ) = + Verte ( 7, ) Focus ( 7, ) Directri = Endpoints of latus rectum ( 0, ), (, ). ( ) = ( +) Verte (, ) Focus ( 6, ) Directri =0 Endpoints of latus rectum ( 6, 8), ( 6, ) 8 7 6 7 6
Hooked on Conics. ( +) = Verte (0, ) Focus (, ) Directri = Endpoints of latus rectum (, ), (, 6) 6 7 8. ( ) =( +) Verte (, ) Focus (, ) Directri = Endpoints of latus rectum (, ), (, ) 6. ( +) = 0( ) Verte (, ) Focus (, 0) Directri =0 Endpoints of latus rectum (, 0), (8, 0) 0 9 8 7 6 0 8 6 6 8 7. ( ) = 8( ) Verte (, ) Focus (, ) Directri = Endpoints of latus rectum (, ), (, ) 9 7 6 7
7. Parabolas 8. ( + ) ( ) = 7 + 9 Verte ( 9, ) Focus (, ) Directri = Endpoints of latus rectum (, ), (, ) 9. ( ) = 7( ) Verte (, ) Focus (, ) Directri =. ( +) =8( 6) Verte (, 6) Focus (, 8) Directri = 0. ( + ) = ( ) Verte (, ) Focus (, 9 ) 0 Directri = 0. ( +) = ( 0) Verte (0, ) Focus ( 79 8, ) Directri = 8 8. ( ) = ( ). ( 9 ) = ( ) Verte (, ) Verte (, 9 ) Focus (, ) Focus (, 9 ) Directri = Directri = 7. = 8( 7) 6. ( ) =0 ( ) 7. ( +8) = 6 9 ( + 9) 8. ( ) =6 ( + 7 ( ) = 6 ( + 9. The bulb should be placed 0.6 centimeters above the verte of the mirror. (As verified b Carl himself!) 0. The receiver should be placed.06 centimeters from the verte of the cross section of the antenna.. The arch can be modeled b = ( 9) or =9. One foot in from the base of the arch corresponds to either = ±, so the height is =9 (±) = feet. ) ) or
6 Hooked on Conics 7. Ellipses In the definition of a circle, Definition 7., we fied a point called the center and considered all of the points which were a fied distance r from that one point. For our net conic section, the ellipse, we fi two distinct points and a distance d to use in our definition. Definition 7.. Given two distinct points F and F in the plane and a fied distance d, an ellipse is the set of all points (, ) in the plane such that the sum of each of the distances from F and F to (, ) isd. ThepointsF and F are called the foci a of the ellipse. a the plural of focus (, ) d d F F d + d = d for all (, ) on the ellipse We ma imagine taking a length of string and anchoring it to two points on a piece of paper. The curve traced out b taking a pencil and moving it so the string is alwas taut is an ellipse. The center of the ellipse is the midpoint of the line segment connecting the two foci. The major ais of the ellipse is the line segment connecting two opposite ends of the ellipse which also contains the center and foci. The minor ais of the ellipse is the line segment connecting two opposite ends of the ellipse which contains the center but is perpendicular to the major ais. The vertices of an ellipse are the points of the ellipse which lie on the major ais. Notice that the center is also the midpoint of the major ais, hence it is the midpoint of the vertices. In pictures we have,
7. Ellipses 7 Minor Ais Major Ais V V F C F An ellipse with center C; focif, F ; and vertices V, V Note that the major ais is the longer of the two aes through the center, and likewise, the minor ais is the shorter of the two. In order to derive the standard equation of an ellipse, we assume that the ellipse has its center at (0, 0), its major ais along the -ais, and has foci (c, 0) and ( c, 0) and vertices ( a, 0) and (a, 0). We will label the -intercepts of the ellipse as (0,b)and(0, b) (We assume a, b, andc are all positive numbers.) Schematicall, (0,b) (, ) ( a, 0) ( c, 0) (c, 0) (a, 0) (0, b) Note that since (a, 0) is on the ellipse, it must satisf the conditions of Definition 7.. Thatis,the distance from ( c, 0) to (a, 0) plus the distance from (c, 0) to (a, 0) must equal the fied distance d. Since all of these points lie on the -ais, we get distance from ( c, 0) to (a, 0) + distance from (c, 0) to (a, 0) = d (a + c)+(a c) = d a = d
8 Hooked on Conics In other words, the fied distance d mentioned in the definition of the ellipse is none other than the length of the major ais. We now use that fact (0,b) is on the ellipse, along with the fact that d =a to get distance from ( c, 0) to (0,b) + distance from (c, 0) to (0,b) = a (0 ( c)) +(b 0) + (0 c) +(b 0) = a b + c + b + c = a b + c = a b + c = a From this, we get a = b + c,orb = a c, which will prove useful later. Now consider a point (, ) on the ellipse. Appling Definition 7., weget distance from ( c, 0) to (, ) + distance from (c, 0) to (, ) = a ( ( c)) +( 0) + ( c) +( 0) = a ( + c) + + ( c) + = a In order to make sense of this situation, we need to make good use of Intermediate Algebra. ( + c) + + ( c) + = a ( + c) + = a ( c) + ( ( ) + c) + = (a ) ( c) + ( + c) + = a a ( c) + +( c) + a ( c) + = a +( c) ( + c) a ( c) + = a c a ( c) + = a c ( a ) ( c) + = ( a c ) a ( ( c) + ) = a a c + c a a c + a c + a = a a c + c a c + a = a a c ( a c ) + a = a ( a c ) We are nearl finished. Recall that b = a c so that ( a c ) + a = a ( a c ) b + a = a b a + b =
7. Ellipses 9 This equation is for an ellipse centered at the origin. To get the formula for the ellipse centered at (h, k), we could use the transformations from Section.7 or re-derive the equation using Definition 7. and the distance formula to obtain the formula below. Equation 7.. The Standard Equation of an Ellipse: For positive unequal numbers a and b, the equation of an ellipse with center (h, k) is ( h) ( k) a + b = Some remarks about Equation 7. are in order. First note that the values a and b determine how far in the and directions, respectivel, one counts from the center to arrive at points on the ellipse. Also take note that if a>b, then we have an ellipse whose major ais is horizontal, and hence, the foci lie to the left and right of the center. In this case, as we ve seen in the derivation, the distance from the center to the focus, c, can be found b c = a b. If b>a, the roles of the major and minor aes are reversed, and the foci lie above and below the center. In this case, c = b a. In either case, c is the distance from the center to each focus, and c = bigger denominator smaller denominator. Finall, it is worth mentioning that if we take the standard equation of a circle, Equation 7., and divide both sides b r,weget Equation 7.. The Alternate Standard Equation of a Circle: The equation of a circle with center (h, k) and radius r>0is ( h) ( k) r + r = Notice the similarit between Equation 7. and Equation 7.. Both equations involve a sum of squares equal to ; the difference is that with a circle, the denominators are the same, and with an ellipse, the are different. If we take a transformational approach, we can consider both Equations 7. and 7. as shifts and stretches of the Unit Circle + = in Definition 7.. Replacing with ( h) and with ( k) causes the usual horizontal and vertical shifts. Replacing with a and with b causes the usual vertical and horizontal stretches. In other words, it is perfectl fine to think of an ellipse as the deformation of a circle in which the circle is stretched farther in one direction than the other. Eample 7... Graph (+) 9 + ( ) =. Find the center, the lines which contain the major and minor aes, the vertices, the endpoints of the minor ais, and the foci. Solution. We see that this equation is in the standard form of Equation 7.. Here h is + so h =, and k is sok =. Hence, our ellipse is centered at (, ). We see that a =9 so a =,andb =sob =. This means that we move units left and right from the center and units up and down from the center to arrive at points on the ellipse. As an aid to sketching, we draw a rectangle matching this description, called a guide rectangle, and sketch the ellipse inside this rectangle as seen below on the left. This was foreshadowed in Eercise 9 in Section 7..
0 Hooked on Conics 7 7 6 6 Since we moved farther in the direction than in the direction, the major ais will lie along the vertical line =, which means the minor ais lies along the horizontal line, =. The vertices are the points on the ellipse which lie along the major ais so in this case, the are the points (, 7) and (, ), and the endpoints of the minor ais are (, ) and (, ). (Notice these points are the four points we used to draw the guide rectangle.) To find the foci, we find c = 9= 6 =, which means the foci lie units from the center. Since the major ais is vertical, the foci lie units above and below the center, at (, ) and (, 6). Plotting all this information gives the graph seen above on the right. Eample 7... Find the equation of the ellipse with foci (, ) and (, ) and verte (0, ). Solution. Plotting the data given to us, we have From this sketch, we know that the major ais is horizontal, meaning a>b. Since the center is the midpoint of the foci, we know it is (, ). Since one verte is (0, ) we have that a =,soa =9. All that remains is to find b. Since the foci are unit awa from the center, we know c =. Since a>b,wehavec = a b,or= 9 b,sob = 8. Substituting all of our findings into the equation ( h) a + ( k) b =, we get our final answer to be ( ) 9 + ( ) 8 =.
7. Ellipses As with circles and parabolas, an equation ma be given which is an ellipse, but isn t in the standard form of Equation 7.. In those cases, as with circles and parabolas before, we will need to massage the given equation into the standard form. To Write the Equation of an Ellipse in Standard Form. Group the same variables together on one side of the equation and position the constant on the other side.. Complete the square in both variables as needed.. Divide both sides b the constant term so that the constant on the other side of the equation becomes. Eample 7... Graph + + + = 0. Find the center, the lines which contain the major and minor aes, the vertices, the endpoints of the minor ais, and the foci. Solution. Since we have a sum of squares and the squared terms have unequal coefficients, it s a good bet we have an ellipse on our hands. We need to complete both squares, and then divide, if necessar, to get the right-hand side equal to. + + + = 0 + + = + ( +6 ) = ( + ) + ( +6 +9 ) = ++(9) ( ) +( +) = ( ) +( +) = ( ) +( +) = ( ) ( +) + = Now that this equation is in the standard form of Equation 7., we see that h is soh =, and k is + so k =. Hence, our ellipse is centered at (, ). We see that a =soa =, and b =sob =. This means we move units left and right from the center and unit up and down from the center to arrive at points on the ellipse. Since we moved farther in the direction than in the direction, the major ais will lie along the horizontal line =, which means the minor ais lies along the vertical line =. The vertices are the points on the ellipse which lie along the major ais so in this case, the are the points (, ) and (, ), and the endpoints of the minor ais are (, ) and (, ). To find the foci, we find c = =, which means The equation of a parabola has onl one squared variable and the equation of a circle has two squared variables with identical coefficients.
Hooked on Conics the foci lie units from the center. Since the major ais is horizontal, the foci lie units to the left and right of the center, at (, ) and ( +, ). Plotting all of this information gives As ou come across ellipses in the homework eercises and in the wild, ou ll notice the come in all shapes in sizes. Compare the two ellipses below. Certainl, one ellipse is more round than the other. This notion of roundness is quantified below. Definition 7.. The eccentricit of an ellipse, denoted e, is the following ratio: e = distance from the center to a focus distance from the center to a verte In an ellipse, the foci are closer to the center than the vertices, so 0 <e<. The ellipse above on the left has eccentricit e 0.98; for the ellipse above on the right, e 0.66. In general, the closer the eccentricit is to 0, the more circular the ellipse; the closer the eccentricit is to, the more eccentric the ellipse. Eample 7... Find the equation of the ellipse whose vertices are (±, 0) with eccentricit e =. Solution. As before, we plot the data given to us
7. Ellipses From this sketch, we know that the major ais is horizontal, meaning a>b. With the vertices located at (±, 0), we get a =soa =. We also know that the center is (0, 0) because the center is the midpoint of the vertices. All that remains is to find b. To that end, we use the fact that the eccentricit e = which means e = distance from the center to a focus distance from the center to a verte = c a = c = from which we get c =.Togetb,weusethefactthatc = a b,so = b from which we get b = 7 ( h) 6. Substituting all of our findings into the equation a + ( k) b =, ields our final answer + 6 7 =. As with parabolas, ellipses have a reflective propert. If we imagine the dashed lines below representing sound waves, then the waves emanating from one focus reflect off the top of the ellipse and head towards the other focus. F F Such geometr is eploited in the construction of so-called Whispering Galleries. If a person whispers at one focus, a person standing at the other focus will hear the first person as if the were standing right net to them. We eplore the Whispering Galleries in our last eample. Eample 7... Jamie and Jason want to echange secrets (terrible secrets) from across a crowded whispering galler. Recall that a whispering galler is a room which, in cross section, is half of an ellipse. If the room is 0 feet high at the center and 00 feet wide at the floor, how far from the outer wall should each of them stand so that the will be positioned at the foci of the ellipse? Solution. Graphing the data ields
Hooked on Conics 0 units tall 00 units wide It s most convenient to imagine this ellipse centered at (0, 0). Since the ellipse is 00 units wide and 0 units tall, we get a =0andb = 0. Hence, our ellipse has the equation + =. 0 0 We re looking for the foci, and we get c = 0 0 = 900 = 0, so that the foci are 0 units from the center. That means the are 0 0 = 0 units from the vertices. Hence, Jason and Jamie should stand 0 feet from opposite ends of the galler.
7. Ellipses 7.. Eercises In Eercises - 8, graph the ellipse. Find the center, the lines which contain the major and minor aes, the vertices, the endpoints of the minor ais, the foci and the eccentricit.... 7. 69 + =. 9 + = ( ) + ( ) + 0 ( +) + 6 ( +) 9 ( ) ( ) 0 =. = 6. = 8. ( +) + 6 ( ) + 9 ( ) + 8 ( ) ( +) ( ) 8 = = = In Eercises 9 -, put the equation in standard form. Find the center, the lines which contain the major and minor aes, the vertices, the endpoints of the minor ais, the foci and the eccentricit. 9. 9 + 0 9 = 0 0. + 0 +9=0. +8 0 +7 + 7 = 0. + +=0. 9 + 8 = 0. 6 + +0 +=0 In Eercises - 0, find the standard form of the equation of the ellipse which has the given properties.. Center (, 7),Verte(, ), Focus (, ) 6. Foci (0, ±), Vertices (0, ±8). 7. Foci (±, 0), length of the Minor Ais 0 8. Vertices (, ), (, ); Endpoints of the Minor Ais (8, ), (8, 0) 9. Center (, ),Verte(0, ), eccentricit 0. All points on the ellipse are in Quadrant IV ecept (0, 9) and (8, 0). (One might also sa that the ellipse is tangent to the aes at those two points.). Repeat Eample 7.. for a whispering galler 00 feet wide and 7 feet tall.. An elliptical arch is constructed which is 6 feet wide at the base and 9 feet tall in the middle. Find the height of the arch eactl foot in from the base of the arch. Compare our result with our answer to Eercise in Section 7..
6 Hooked on Conics. The Earth s orbit around the sun is an ellipse with the sun at one focus and eccentricit e 0.067. The length of the semimajor ais (that is, half of the major ais) is defined to be astronomical unit (AU). The vertices of the elliptical orbit are given special names: aphelion is the verte farthest from the sun, and perihelion is the verte closest to the sun. Find the distance in AU between the sun and aphelion and the distance in AU between the sun and perihelion.. The graph of an ellipse clearl fails the Vertical Line Test, Theorem., so the equation of an ellipse does not define as a function of. However, much like with circles and horizontal parabolas, we can split an ellipse into a top half and a bottom half, each of which would indeed represent as a function of. With the help of our classmates, use our calculator to graph the ellipses given in Eercises - 8 above. What difficulties arise when ou plot them on the calculator?. Some famous eamples of whispering galleries include St. Paul s Cathedral in London, England, National Statuar Hall in Washington, D.C., and The Cincinnati Museum Center. With the help of our classmates, research these whispering galleries. How does the whispering effect compare and contrast with the scenario in Eample 7..? 6. With the help of our classmates, research etracorporeal shock-wave lithotrips. It uses the reflective propert of the ellipsoid to dissolve kidne stones.
7. Ellipses 7 7.. Answers. 69 + = Center (0, 0) Major ais along =0 Minor ais along =0 Vertices (, 0), (, 0) Endpoints of Minor Ais (0, ), (0, ) Foci (, 0), (, 0) e =. 9 + = Center (0, 0) Major ais along =0 Minor ais along =0 Vertices (0, ), (0, ) Endpoints of Minor Ais (, 0), (, 0) Foci (0, ), (0, ) e = ( ) ( +). + = 9 Center (, ) Major ais along = Minor ais along = Vertices (, 0), (, 6) Endpoints of Minor Ais (0, ), (, ) Foci (, + ), (, ) e = 6 ( +) ( ). + = 6 Center (, ) Major ais along = Minor ais along = Vertices ( 9, ), (, ) Endpoints of Minor Ais (, ), (, ) Foci ( +, ), (, ) e = 9 8 7 6
8 Hooked on Conics ( ) ( ). + = 0 Center (, ) Major ais along = Minor ais along = Vertices (, + ), (, ) Endpoints of the Minor Ais ( 0, ), ( + 0, ) Foci (, ), (, ) e = 6 ( ) ( +) 6. + = 9 Center (, ) Major ais along = Minor ais along = Vertices (, ), (, ) Endpoints of the Minor Ais (, ), (, ) Foci ( +, ), (, ) e = ( +) ( ) 7. + = 6 0 Center (, ) Major ais along = Minor ais along = Vertices (, + ), (, ) Endpoints of the Minor Ais ( 6, ), (, ) Foci (, 7), (, ) e = 0 9 8 7 6 6
7. Ellipses 9 ( ) ( ) 8. + = 8 8 Center (, ) Major ais along = Minor ais along = Vertices (, + ), (, ) Endpoints of the Minor Ais (, ), ( +, ) Foci (, + 0), (, 0) e = 7 6 6 7 ( ) ( ) 9. + = 9 Center (, ) Major Ais along = Minor Ais along = Vertices (8, ), (, ) Endpoints of Minor Ais (, ), (, ) Foci (7, ), (, ) e = ( ) 0. + = Center (0, ) Major ais along =0 Minor ais along = Vertices (0, ), (0, + ) Endpoints of Minor Ais (, ), (, ) Foci (0, ), (0, 8) e = ( ) ( +). + = 8 Center (, ) Major ais along = Minor ais along = Vertices (, ), ( +, ) Endpoints of Minor Ais (, + ), (, ) Foci (, ), ( +, ) e = 6 6 ( ) ( ). + = 6 8 Center (, ) Major Ais along = Minor Ais along = Vertices (, ), (, ) Endpoints of Minor Ais (, + ), (, ) Foci ( +, ), (, ) e =
0 Hooked on Conics.. 7. + ( ) = 9 Center ( 0, ) Major Ais along = 0 (the -ais) Minor Ais along = Vertices (0, ), (0, ) Endpoints of Minor Ais (, ( ),, ( Foci e = 0, + ( ) + 9 ), ( ( 7) 0, ) = 6. ) ( ) ( +). + = 6 Center (, ) Major Ais along = Minor Ais along = Vertices (, + 6 ),(, 6) Endpoints of Minor Ais (, ), ( ) +, Foci (, ), (, ) e = 6 6 9 + 6 = + = 8. ( 8) ( ) + ( ) ( ) ( 8) ( +9) 9. + = 0. + = 7 6 8. Jamie and Jason should stand 00 7.86 feet from opposite ends of the galler.. Thearchcanbemodeledbthetophalfof 9 + 8 =. One foot in from the base of the arch corresponds to either = ±. Plugging in = ± gives = ± and since represents a height, we choose = 6.7 feet.. Distance from the sun to aphelion.067 AU. Distance from the sun to perihelion 0.98 AU. =
7. Hperbolas 7. Hperbolas In the definition of an ellipse, Definition 7., we fied two points called foci and looked at points whose distances to the foci alwas added to a constant distance d. Those prone to sntactical tinkering ma wonder what, if an, curve we d generate if we replaced added with subtracted. Theanswerisahperbola. Definition 7.6. Given two distinct points F and F in the plane and a fied distance d, a hperbola is the set of all points (, ) in the plane such that the absolute value of the difference of each of the distances from F and F to (, ) isd. ThepointsF and F are called the foci of the hperbola. (, ) F F (, ) In the figure above: and the distance from F to (, ) the distance from F to (, ) = d the distance from F to (, ) the distance from F to (, ) = d Note that the hperbola has two parts, called branches. The center of the hperbola is the midpoint of the line segment connecting the two foci. The transverse ais of the hperbola is the line segment connecting two opposite ends of the hperbola which also contains the center and foci. The vertices of a hperbola are the points of the hperbola which lie on the transverse ais. In addition, we will show momentaril that there are lines called asmptotes which the branches of the hperbola approach for large and values. The serve as guides to the graph. In pictures,
Hooked on Conics Transverse Ais F V V F C A hperbola with center C; focif, F ; and vertices V, V and asmptotes (dashed) Before we derive the standard equation of the hperbola, we need to discuss one further parameter, the conjugate ais of the hperbola. The conjugate ais of a hperbola is the line segment through the center which is perpendicular to the transverse ais and has the same length as the line segment through a verte which connects the asmptotes. In pictures we have V Conjugate Ais C V Note that in the diagram, we can construct a rectangle using line segments with lengths equal to the lengths of the transverse and conjugate aes whose center is the center of the hperbola and whose diagonals are contained in the asmptotes. This guide rectangle, much akin to the one we saw Section 7. to help us graph ellipses, will aid us in graphing hperbolas. Suppose we wish to derive the equation of a hperbola. For simplicit, we shall assume that the center is (0, 0), the vertices are (a, 0) and ( a, 0) and the foci are (c, 0) and ( c, 0). We label the
7. Hperbolas endpoints of the conjugate ais (0,b)and(0, b). (Although b does not enter into our derivation, we will have to justif this choice as ou shall see later.) As before, we assume a, b, andc are all positive numbers. Schematicall we have (0,b) (, ) ( c, 0) ( a, 0) (a, 0) (c, 0) (0, b) Since (a, 0) is on the hperbola, it must satisf the conditions of Definition 7.6. That is, the distance from ( c, 0) to (a, 0) minus the distance from (c, 0) to (a, 0) must equal the fied distance d. Since all these points lie on the -ais, we get distance from ( c, 0) to (a, 0) distance from (c, 0) to (a, 0) = d (a + c) (c a) = d a = d In other words, the fied distance d from the definition of the hperbola is actuall the length of the transverse ais! (Where have we seen that tpe of coincidence before?) Now consider a point (, ) on the hperbola. Appling Definition 7.6, weget distance from ( c, 0) to (, ) distance from (c, 0) to (, ) = a ( ( c)) +( 0) ( c) +( 0) = a ( + c) + ( c) + = a Using the same arsenal of Intermediate Algebra weaponr we used in deriving the standard formula of an ellipse, Equation 7., wearriveatthefollowing. It is a good eercise to actuall work this out.
Hooked on Conics ( a c ) + a = a ( a c ) What remains is to determine the relationship between a, b and c. To that end, we note that since a and c are both positive numbers with a<c,wegeta <c so that a c is a negative number. Hence, c a is a positive number. For reasons which will become clear soon, we re-write the equation b solving for / to get ( a c ) + a = a ( a c ) ( c a ) + a = a ( c a ) a = ( c a ) a ( ( c a ) c a ) ( c a ) = a As and attain ver large values, the quantit (c a ) 0sothat (c a ). B setting a b = c a we get b. This shows that ± b a a as grows large. Thus = ± b a are the asmptotes to the graph as predicted and our choice of labels for the endpoints of the conjugate ais is justified. In our equation of the hperbola we can substitute a c = b which ields ( a c ) + a = a ( a c ) b + a = a b a b = The equation above is for a hperbola whose center is the origin and which opens to the left and right. If the hperbola were centered at a point (h, k), we would get the following. Equation 7.6. The Standard Equation of a Horizontal a Hperbola For positive numbers a and b, the equation of a horizontal hperbola with center (h, k) is ( h) ( k) a b = a That is, a hperbola whose branches open to the left and right If the roles of and were interchanged, then the hperbola s branches would open upwards and downwards and we would get a vertical hperbola. Equation 7.7. The Standard Equation of a Vertical Hperbola For positive numbers a and b, the equation of a vertical hperbola with center (h, k) is: ( k) ( h) b a = Thevaluesofa and b determine how far in the and directions, respectivel, one counts from the center to determine the rectangle through which the asmptotes pass. In both cases, the distance
7. Hperbolas from the center to the foci, c, as seen in the derivation, can be found b the formula c = a + b. Lastl, note that we can quickl distinguish the equation of a hperbola from that of a circle or ellipse because the hperbola formula involves a difference of squares where the circle and ellipse formulas both involve the sum of squares. Eample 7... Graph the equation ( ) =. Find the center, the lines which contain the transverse and conjugate aes, the vertices, the foci and the equations of the asmptotes. Solution. We first see that this equation is given to us in the standard form of Equation 7.6. Here h is soh =,and k is so k = 0. Hence, our hperbola is centered at (, 0). We see that a =soa =,andb =sob =. This means we move units to the left and right of the center and units up and down from the center to arrive at points on the guide rectangle. The asmptotes pass through the center of the hperbola as well as the corners of the rectangle. This ields the following set up. 7 6 6 6 7 Since the term is being subtracted from the term, we know that the branches of the hperbola open to the left and right. This means that the transverse ais lies along the -ais. Hence, the conjugate ais lies along the vertical line =. Since the vertices of the hperbola are where the hperbola intersects the transverse ais, we get that the vertices are units to the left and right of (, 0) at (0, 0) and (, 0). To find the foci, we need c = a + b = + = 9. Since the foci lie on the transverse ais, we move 9 units to the left and right of (, 0) to arrive at ( 9, 0) (approimatel (.9, 0)) and ( + 9, 0) (approimatel (7.9, 0)). To determine the equations of the asmptotes, recall that the asmptotes go through the center of the hperbola, (, 0), as well as the corners of guide rectangle, so the have slopes of ± b a = ±. Using the point-slope equation
6 Hooked on Conics of a line, Equation., ields 0=± ( ), so we get = and = +. Putting it all together, we get 7 6 6 7 6 7 Eample 7... Find the equation of the hperbola with asmptotes = ± and vertices (±, 0). Solution. Plotting the data given to us, we have This graph not onl tells us that the branches of the hperbola open to the left and to the right, it also tells us that the center is (0, 0). Hence, our standard form is =. Since the vertices a b are (±, 0), we have a =soa =. In order to determine b,werecallthattheslopesofthe asmptotes are ± b a. Since a = and the slope of the line = is, we have that b =,so b = 0. Hence, b = 00 and our final answer is 00 =.
7. Hperbolas 7 As with the other conic sections, an equation whose graph is a hperbola ma not be given in either of the standard forms. To rectif that, we have the following. To Write the Equation of a Hperbola in Standard Form. Group the same variables together on one side of the equation and position the constant on the other side. Complete the square in both variables as needed. Divide both sides b the constant term so that the constant on the other side of the equation becomes Eample 7... Consider the equation 9 6 = 0. Put this equation in to standard form and graph. Find the center, the lines which contain the transverse and conjugate aes, the vertices, the foci, and the equations of the asmptotes. Solution. We need onl complete the square on : 9 6 = 0 9 ( +6 ) = 0 9 ( +6 +9 ) = 0 (9) 9 ( +) = 9 ( +) = Now that this equation is in the standard form of Equation 7.7, we see that h is + so h =, and k is so k = 0. Hence, our hperbola is centered at (, 0). We find that a =soa =, and b = 9 so b =. This means that we move unit to the left and right of the center and units up and down from the center to arrive at points on the guide rectangle. Since the term is being subtracted from the term, we know the branches of the hperbola open upwards and downwards. This means the transverse ais lies along the vertical line = and the conjugate ais lies along the -ais. Since the vertices of the hperbola are where the hperbola intersects thetransverseais,wegetthattheverticesare of a unit above and below (, 0) at ( (, ) and, ). To find the foci, we use 0 c = a + b = 9 += Since ( the foci lie on the transverse ais, we move units above and below (, 0) to arrive at ) ( ), 0 and, 0. To determine the asmptotes, recall that the asmptotes go through the center of the hperbola, (, 0), as well as the corners of guide rectangle, so the have slopes of ± b a = ±. Using the point-slope equation of a line, Equation., weget = + and =. Putting it all together, we get 0
8 Hooked on Conics 6 Hperbolas can be used in so-called trilateration, or positioning problems. The procedure outlined in the net eample is the basis of the (now virtuall defunct) LOng Range Aid to Navigation (LORAN for short) sstem. Eample 7... Jeff is stationed 0 miles due west of Carl in an otherwise empt forest in an attempt to locate an elusive Sasquatch. At the stroke of midnight, Jeff records a Sasquatch call 9 seconds earlier than Carl. If the speed of sound that night is 760 miles per hour, determine a hperbolic path along which Sasquatch must be located. Solution. Since Jeff hears Sasquatch sooner, it is closer to Jeff than it is to Carl. Since the speed of sound is 760 miles per hour, we can determine how much closer Sasquatch is to Jeff b multipling 760 miles hour hour 9 seconds =.9 miles 600 seconds This means that Sasquatch is.9 miles closer to Jeff than it is to Carl. In other words, Sasquatch must lie on a path where (the distance to Carl) (the distance to Jeff) =.9 This is eactl the situation in the definition of a hperbola, Definition 7.6. In this case, Jeff and Carl are located at the foci, and our fied distance d is.9. For simplicit, we assume the hperbola is centered at (0, 0) with its foci at (, 0) and (, 0). Schematicall, we have GPS now rules the positioning kingdom. Is there still a place for LORAN and other land-based sstems? Do satellites ever malfunction? We usuall like to be the center of attention, but being the focus of attention works equall well.
7. Hperbolas 9 6 Jeff Carl 6 6 6 We are seeking a curve of the form = in which the distance from the center to each focus a b is c =. As we saw in the derivation of the standard equation of the hperbola, Equation 7.6, d =a, sothata =.9, or a =0.9 and a =0.90. All that remains is to find b. To that end, we recall that a + b = c so b = c a = 0.90 =.097. Since Sasquatch is closer to Jeff than it is to Carl, it must be on the western (left hand) branch of 0.90.097 =. In our previous eample, we did not have enough information to pin down the eact location of Sasquatch. To accomplish this, we would need a third observer. Eample 7... B a stroke of luck, Kai was also camping in the woods during the events of the previous eample. He was located 6 miles due north of Jeff and heard the Sasquatch call 8 seconds after Jeff did. Use this added information to locate Sasquatch. Solution. Kai and Jeff are now the foci of a second hperbola where the fied distance d can be determined as before 760 miles hour hour 8 seconds =.8 miles 600 seconds Since Jeff was positioned at (, 0), we place Kai at (, 6). This puts the center of the new hperbola at (, ). Plotting Kai s position and the new center gives us the diagram below on the left. The second hperbola is vertical, so it must be of the form ( ) (+) =. As before, b a the distance d is the length of the major ais, which in this case is b. We get b =.8 sothat b =.9 andb =.6. With Kai 6 miles due North of Jeff, we have that the distance from the center to the focus is c =. Sincea + b = c,wegeta = c b =9.6 =.9. Kai heard the Sasquatch call after Jeff, so Kai is farther from Sasquatch than Jeff. Thus Sasquatch must lie on the southern branch of the hperbola ( ).6 (+).9 =. Looking at the western branch of the