REVIEW OF CONIC SECTIONS

REVIEW OF CONIC SECTIONS In this section we give geometric definitions of parabolas, ellipses, and hperbolas and derive their standard equations. The are called conic sections, or conics, because the result from intersecting a cone with a plane as shown in Figure. ellipse parabola hperbola FIGURE Conics PARABOLAS ais focus verte FIGURE F(, p) F parabola directri P(, ) A parabola is the set of points in a plane that are equidistant from a fied point F (called the focus) and a fied line (called the directri). This definition is illustrated b Figure. Notice that the point halfwa between the focus and the directri lies on the parabola; it is called the verte. The line through the focus perpendicular to the directri is called the ais of the parabola. In the 6th centur Galileo showed that the path of a projectile that is shot into the air at an angle to the ground is a parabola. Since then, parabolic shapes have been used in designing automobile headlights, reflecting telescopes, and suspension bridges. (See Challenge Problem.4 for the reflection propert of parabolas that makes them so useful.) We obtain a particularl simple equation for a parabola if we place its verte at the origin O and its directri parallel to the -ais as in Figure. If the focus is the point, p, then the directri has the equation p. If P, is an point on the parabola, then the distance from P to the focus is O =_p p PF s p p and the distance from P to the directri is. (Figure illustrates the case where p.) The defining propert of a parabola is that these distances are equal: FIGURE s p p We get an equivalent equation b squaring and simplifing: p p p p p p p 4p Thomson Brooks-Cole copright 7 An equation of the parabola with focus, p and directri p is 4p

REVIEW OF CONIC SECTIONS If we write a 4p, then the standard equation of a parabola () becomes a. It opens upward if p and downward if p [see Figure 4, parts (a) and (b)]. The graph is smmetric with respect to the -ais because () is unchanged when is replaced b. (, p) =_p (, p) =_p =_p ( p, ) (p, ) =_p (a) =4p, p> (b) =4p, p< (c) =4p, p> (d) =4p, p< FIGURE 4 If we interchange and in (), we obtain += 5 _, FIGURE 5 = 5 4p which is an equation of the parabola with focus p, and directri p. (Interchanging and amounts to reflecting about the diagonal line.) The parabola opens to the right if p and to the left if p [see Figure 4, parts (c) and (d)]. In both cases the graph is smmetric with respect to the -ais, which is the ais of the parabola. EXAMPLE Find the focus and directri of the parabola and sketch the graph. SOLUTION If we write the equation as and compare it with Equation, we see that 4p, so p 5. Thus the focus is p, ( 5, ) and the directri is 5. The sketch is shown in Figure 5. ELLIPSES F F FIGURE 6 FIGURE 7 P P(, ) F (_c, ) F (c, ) An ellipse is the set of points in a plane the sum of whose distances from two fied points F and F is a constant (see Figure 6). These two fied points are called the foci (plural of focus). One of Kepler s laws is that the orbits of the planets in the solar sstem are ellipses with the Sun at one focus. In order to obtain the simplest equation for an ellipse, we place the foci on the -ais at the points c, and c, as in Figure 7 so that the origin is halfwa between the foci. Let the sum of the distances from a point on the ellipse to the foci be a. Then P, is a point on the ellipse when PF PF a that is, s c s c a or s c a s c Squaring both sides, we have c c 4a 4as c c c Thomson Brooks-Cole copright 7 which simplifies to as c a c We square again: a c c a 4 a c c which becomes a c a a a c

REVIEW OF CONIC SECTIONS (_a, ) (_c, ) FIGURE 8 + = a@ b@ (, b) b a (a, ) c (c, ) (, _b) From triangle F F P in Figure 7 we see that c a, so c a and, therefore, a c. For convenience, let b a c. Then the equation of the ellipse becomes b a a b or, if both sides are divided b a b, a b Since b a c a, it follows that b a. The -intercepts are found b setting. Then a, or a, so a. The corresponding points a, and a, are called the vertices of the ellipse and the line segment joining the vertices is called the major ais. To find the -intercepts we set and obtain b, so b. Equation is unchanged if is replaced b or is replaced b, so the ellipse is smmetric about both aes. Notice that if the foci coincide, then c, so a b and the ellipse becomes a circle with radius r a b. We summarize this discussion as follows (see also Figure 8). (, a) 4 The ellipse a b a b (_b, ) FIGURE 9 + =, a b b@ a@ (, ) (, c) (, _c) (, _a) (b, ) has foci c,, where c a b, and vertices a,. If the foci of an ellipse are located on the -ais at, c, then we can find its equation b interchanging and in (4). (See Figure 9.) 5 The ellipse b a has foci, c, where c a b, and vertices, a. EXAMPLE Sketch the graph of 9 6 44 and locate the foci. SOLUTION Divide both sides of the equation b 44: a b (_4, ) {_œ 7, } (4, ) {œ 7, } (, _) 6 9 The equation is now in the standard form for an ellipse, so we have a 6, b 9, a 4, and b. The -intercepts are 4 and the -intercepts are. Also, c a b 7, so c s7 and the foci are (s7, ). The graph is sketched in Figure. FIGURE 9 +6 =44 EXAMPLE Find an equation of the ellipse with foci, and vertices,. SOLUTION Using the notation of (5), we have c and a. Then we obtain b a c 9 4 5, so an equation of the ellipse is 5 9 Thomson Brooks-Cole copright 7 Another wa of writing the equation is 9 5 45.. Like parabolas, ellipses have an interesting reflection propert that has practical consequences. If a source of light or sound is placed at one focus of a surface with elliptical cross-sections, then all the light or sound is reflected off the surface to the other focus (see

4 REVIEW OF CONIC SECTIONS Eercise 59). This principle is used in lithotrips, a treatment for kidne stones. A reflector with elliptical cross-section is placed in such a wa that the kidne stone is at one focus. High-intensit sound waves generated at the other focus are reflected to the stone and destro it without damaging surrounding tissue. The patient is spared the trauma of surger and recovers within a few das. HYPERBOLAS FIGURE P is on the hperbola when PF - PF =a P(, ) F (_c, ) F (c, ) A hperbola is the set of all points in a plane the difference of whose distances from two fied points F and F (the foci) is a constant. This definition is illustrated in Figure. Hperbolas occur frequentl as graphs of equations in chemistr, phsics, biolog, and economics (Bole s Law, Ohm s Law, suppl and demand curves). A particularl significant application of hperbolas is found in the navigation sstems developed in World Wars I and II (see Eercise 5). Notice that the definition of a hperbola is similar to that of an ellipse; the onl change is that the sum of distances has become a difference of distances. In fact, the derivation of the equation of a hperbola is also similar to the one given earlier for an ellipse. It is left as Eercise 5 to show that when the foci are on the -ais at c, and the difference of distances is, then the equation of the hperbola is 6 PF PF a a b (_a, ) b =_ a (_c, ) FIGURE - = a@ b@ b = a (a, ) (c, ) where c a b. Notice that the -intercepts are again a and the points a, and a, are the vertices of the hperbola. But if we put in Equation 6 we get b, which is impossible, so there is no -intercept. The hperbola is smmetric with respect to both aes. To analze the hperbola further, we look at Equation 6 and obtain a b s a This shows that a, so. Therefore, we have a or a. This means that the hperbola consists of two parts, called its branches. When we draw a hperbola it is useful to first draw its asmptotes,which are the dashed lines ba and ba shown in Figure. Both branches of the hperbola approach the asmptotes; that is, the come arbitraril close to the asmptotes. a =_ b (, c) a = b 7 The hperbola a b has foci c,, where c a b, vertices a,, and asmptotes ba. (, a) (, _a) If the foci of a hperbola are on the -ais, then b reversing the roles of and we obtain the following information, which is illustrated in Figure. Thomson Brooks-Cole copright 7 FIGURE - = a@ b@ (, _c) 8 The hperbola a b has foci, c, where c a b, vertices, a, and asmptotes ab.

REVIEW OF CONIC SECTIONS 5 (_5, ) =_ 4 (_4, ) (4, ) = 4 (5, ) EXAMPLE 4 Find the foci and asmptotes of the hperbola 9 6 44 and sketch its graph. SOLUTION If we divide both sides of the equation b 44, it becomes 6 9 FIGURE 4 9-6 =44 which is of the form given in (7) with a 4 and b. Since c 6 9 5, the foci are 5,. The asmptotes are the lines and 4 4. The graph is shown in Figure 4. EXAMPLE 5 Find the foci and equation of the hperbola with vertices, and asmptote. SOLUTION From (8) and the given information, we see that a and ab. Thus, b a and c a b 5 4. The foci are (, s5) and the equation of the hperbola is 4 SHIFTED CONICS We shift conics b taking the standard equations (), (), (4), (5), (7), and (8) and replacing and b h and k. EXAMPLE 6 Find an equation of the ellipse with foci,, 4, and vertices,, 5,. SOLUTION The major ais is the line segment that joins the vertices,, 5, and has length 4, so a. The distance between the foci is, so c. Thus, b a c. Since the center of the ellipse is,, we replace and in (4) b and to obtain 4 as the equation of the ellipse. EXAMPLE 7 Sketch the conic -=_ (-4) and find its foci. 9 4 7 8 76 SOLUTION We complete the squares as follows: Thomson Brooks-Cole copright 7 (4, 4) (4, ) (4, _) -= (-4) FIGURE 5 9-4 -7+8+76= 4 9 8 76 4 9 8 6 76 4 44 4 9 4 6 9 4 4 This is in the form (8) ecept that and are replaced b 4 and. Thus, a 9, b 4, and c. The hperbola is shifted four units to the right and one unit upward. The foci are (4, s) and (4, s) and the vertices are 4, 4 and 4,. The asmptotes are 4. The hperbola is sketched in Figure 5.

6 REVIEW OF CONIC SECTIONS EXERCISES 8 Find the verte, focus, and directri of the parabola and sketch its graph... 4. 4 4. 5. 8 6. 5 7. 8. 9 Find an equation of the parabola. Then find the focus and directri. 9.. 6 Find the vertices and foci of the ellipse and sketch its graph... 9 5. 4 6 4. 4 5 5 5. 9 8 4 7 6. A 5 6 Click here for answers. _ 6 4 7 64 7 8 Find an equation of the ellipse. Then find its foci. S Click here for solutions.. 4. 4 8 6 9 64 9 5 5 Identif the tpe of conic section whose equation is given and find the vertices and foci. 5. 6. 7. 4 8. 8 6 6 9. 4. 4 4 48 Find an equation for the conic that satisfies the given conditions.. Parabola, verte,, focus,. Parabola, verte,, directri 5. Parabola, focus 4,, directri 4. Parabola, focus, 6, verte, 5. Parabola, verte,, ais the -ais, passing through (, 4) 6. Parabola, vertical ais, passing through,,,, and, 9 7. Ellipse, foci,, vertices 5, 8. Ellipse, foci, 5, vertices, 9. Ellipse, foci,,, 6 vertices,,, 8 4. Ellipse, foci,, 8,, verte 9, 4. Ellipse, center,, focus,, verte 5, 4. Ellipse, foci,, passing through, 4. Hperbola, foci,, vertices, 44. Hperbola, foci 6,, vertices 4, 7. 8. 45. Hperbola, foci, and 7,, vertices, and 6, 46. Hperbola, foci, and, 8, vertices, and, 6 47. Hperbola, vertices,, asmptotes 48. Hperbola, foci, and 6,, asmptotes and 6 Thomson Brooks-Cole copright 7 9 Find the vertices, foci, and asmptotes of the hperbola and sketch its graph. 9.. 6 44 5 6. 4. 9 4 6 49. The point in a lunar orbit nearest the surface of the moon is called perilune and the point farthest from the surface is called apolune. The Apollo spacecraft was placed in an elliptical lunar orbit with perilune altitude km and apolune altitude 4 km (above the moon). Find an equation of this ellipse if the radius of the moon is 78 km and the center of the moon is at one focus.

REVIEW OF CONIC SECTIONS 7 5. A cross-section of a parabolic reflector is shown in the figure. The bulb is located at the focus and the opening at the focus is cm. (a) Find an equation of the parabola. (b) Find the diameter of the opening CD, cm from the verte. V A B 5 cm cm F 5 cm 5. In the LORAN (LOng RAnge Navigation) radio navigation sstem, two radio stations located at A and B transmit simultaneous signals to a ship or an aircraft located at P. The onboard computer converts the time difference in receiving these signals into a distance difference PA PB, and this, according to the definition of a hperbola, locates the ship or aircraft on one branch of a hperbola (see the figure). Suppose that station B is located 4 mi due east of station A on a coastline. A ship received the signal from B microseconds (s) before it received the signal from A. (a) Assuming that radio signals travel at a speed of 98 fts, find an equation of the hperbola on which the ship lies. (b) If the ship is due north of B, how far off the coastline is the ship? C D 56. (a) Show that the equation of the tangent line to the parabola 4p at the point, can be written as p. (b) What is the -intercept of this tangent line? Use this fact to draw the tangent line. 57. Use Simpson s Rule with n to estimate the length of the ellipse 4 4. 58. The planet Pluto travels in an elliptical orbit around the Sun (at one focus). The length of the major ais is.8 km and the length of the minor ais is.4 km. Use Simpson s Rule with n to estimate the distance traveled b the planet during one complete orbit around the Sun. 59. Let P, be a point on the ellipse a b with foci F and F and let and be the angles between the lines PF, PF and the ellipse as in the figure. Prove that. This eplains how whispering galleries and lithotrips work. Sound coming from one focus is reflected and passes through the other focus. [Hint: Use the formula to show that tan tan. See Challenge Problem..] å P(, ) F F + = a@ b@ m m tan m m A coastline 4 mi sending stations P B 6. Let P, be a point on the hperbola a b with foci F and F and let and be the angles between the lines PF, PF and the hperbola as shown in the figure. Prove that. (This is the reflection propert of the hperbola. It shows that light aimed at a focus F of a hperbolic mirror is reflected toward the other focus F.) 5. Use the definition of a hperbola to derive Equation 6 for a hperbola with foci c, and vertices a,. å P 5. Show that the function defined b the upper branch of the hperbola a b is concave upward. F F 54. Find an equation for the ellipse with foci, and, and major ais of length 4. 55. Determine the tpe of curve represented b the equation k k 6 P Thomson Brooks-Cole copright 7 in each of the following cases: (a) k 6, (b) k 6, and (c) k. (d) Show that all the curves in parts (a) and (b) have the same foci, no matter what the value of k is. F F

8 REVIEW OF CONIC SECTIONS ANSWERS S.,,, 8.,,,, Click here for solutions. ( 8, ), 8., 4, (, s) 4. ( 5, ), 4 _ ( s, ) =_ 8.,, (, 6), 6 4.,,,, = 6, _ 6 _4 5.,, (, s5) 6., and 5,, (s, ) (, ) (,_) 5.,,, 5, 6., 5, ( 5 4, 5), 4 (_, 5) = 7.,foci (, s5) 4 9 8.,foci ( s5, ) 9 4 9.,,,,., 4, (, s), 5 5 = 7.,, 5,, 8.,, (, 5 8 ), 7 8 (_5, _) (_, _).,, (, s),.,, (s, ), = = Thomson Brooks-Cole copright 7 9.,focus ( 4, ), directri 4.,focus (,, directri 5 ).,,,.,,, 6 _ œ 5 _œ 5 _. ( s6, ), ( s5, ), (s6) (, ) {+œ 5, }

REVIEW OF CONIC SECTIONS 9 4. 5, 5 and, 5, 4. 44. 6 8 7, 5 and, 5, 5 4 4 45. 4 5 46. 47. 9 6 9 6 5. Parabola,,, (, 4) 6. Hperbola,,, (s, ) 7. Ellipse, (s, ),, 8. Parabola,, 4, (, 4) 9. Hperbola,,,, ; (, s5). Ellipse, (, ), (, s ). 8. 4. 4. 6 5. 6 6. 4 7. 8. 9. 44 69 4 4. 4. 5 9 9 4. 9 s7 s7 5 4 6 5 48. 4 49.,76,6,75,96 5. (a) p 5, (b) s 5. (a) (b) 48 mi,5,65,9,75 54. 8 55. (a) Ellipse (b) Hperbola (c) No curve 56. (b) 57. 9.69 58..64 km Thomson Brooks-Cole copright 7

REVIEW OF CONIC SECTIONS SOLUTIONS. = =. 4p =,sop = 8.The verte is (, ),thefocusis 8,,andthe directri is = 8.. 4 + = = 4. 4p = 4,so p =. Theverteis(, ),thefocusis(, ), and the directri is =.. 4 = = 4. 4p = 4,so p =.Theverteis(, ),thefocusis 6, 6, and the directri is =. 6 4. =. 4p =,sop =.Theverteis(, ), the focus is (, ), and the directri is =. 5. ( +) =8( ). 4p =8,sop =.The verte is (, ),thefocusis(, 5),andthe directri is =. 6. =( +5). 4p =,sop = 4.Theverteis (, 5),thefocusis 5 4, 5, and the directri is = 4. Thomson Brooks-Cole copright 7

REVIEW OF CONIC SECTIONS 7. + + +5= + += 4 ( +) = ( +). 4p =,sop =. The verte is (, ),thefocusis( 5, ),and the directri is =. 8. + =6 = 6 ( 6 +9)= 6 + 8 ( ) = + ( ) = ( +). 4p =,sop =.Theverteis(, ),thefocus 8 is, 5 8, and the directri is = 7 8. 9. The equation has the form =4p, wherep<. Since the parabola passes through (, ), wehave =4p( ),so4p = andanequationis = or =. 4p =,sop = 4 and the focus is 4, while the directri is = 4.. The verte is (, ), so the equation is of the form ( ) =4p( +),wherep>. The point (, ) is on the parabola, so 4=4p() and 4p =. Thus, an equation is ( ) =( +). 4p =,sop = and the focus is, while the directri is = 5.. 9 + 5 = a = 9=, b = 5, c = a b = 9 5=. The ellipse is centered at (, ), with vertices at (±, ). The foci are (±, ).. 64 + = a = =, b = 64 = 8, c = a b = 64 = 6. The ellipse is centered at (, ), with vertices at (, ±). The foci are (, ±6). Thomson Brooks-Cole copright 7

REVIEW OF CONIC SECTIONS. 4 + =6 4 + 6 = a = 6 = 4, b = 4=, c = a b = 6 4=.Theellipseis centered at (, ), with vertices at (, ±4). The foci are, ±. 4. 4 +5 =5 5/4 + = 5 a = = 5, b = =, 4 c = a b 5 = = =.The 4 4 ellipse is centered at (, ), with vertices at ± 5,. The foci are ±., 5. 9 8 +4 =7 9( +)+4 =7+9 9( ) +4 ( ) =6 + = a =, b =, 4 9 c = 5 center (, ), vertices (, ±),foci, ± 5 6. 6 + +4 = 7 6 +9+( + +)= 7+9+ ( ) +( +) =4 ( ) ( +) + = a =, b = =c center 4 (, ), vertices (, ) and (5, ),foci ±, 7. The center is (, ), a =,andb =, so an equation is 4 + 9 =. c = a b = 5, so the foci are, ± 5. 8. The ellipse is centered at (, ), witha =and b =.Anequationis c = a b = 5,sothefociare ± 5,. ( ) 9 + ( ) 4 =. Thomson Brooks-Cole copright 7 9. 44 5 = a =, b =5, c = 44 + 5 = center (, ), vertices (±, ),foci(±, ), asmptotes = ± 5. Note: It is helpful to draw a a-b-b rectangle whose center is the center of the hperbola. The asmptotes are the etended diagonals of the rectangle.

REVIEW OF CONIC SECTIONS. 6 = a =4, b =6, 6 c = a + b = 6 + 6 = 5 =. The center is (, ), the vertices are (, ±4),thefociare, ±,andthe asmptotes are the lines = ± a b = ±.. =4 4 4 = a = 4==b, c = 4+4= center (, ), vertices (, ±), foci, ±, asmptotes = ±. 9 4 =6 4 9 = a = 4=, b = 9=, c = 4+9= center (, ), vertices (±, ),foci ±,,asmptotes = ±. 4 + = 8 ( +) ( 4 +4)= 8+ ( ) ( ) = 8 ( ) 6 ( ) 9 = a = 6, b =, c = 5 center (, ), vertices ± 6,,foci ± 5,,asmptotes =± 6 ( ) 6 or =± ( ) 4. 6 +64 9 9 = 5 6( +4 +4) 9( + +5)=5+64 5 6( +) 9( +5) ( +) ( +5) =44 = 9 6 a =, b =4, c =5 center (, 5), vertices ( 5, 5) and (, 5),foci( 7, 5) and (, 5), asmptotes +5=± 4 ( +) Thomson Brooks-Cole copright 7 5. = + =( +).Thisisanequationofaparabola with 4p =,sop = 4.Theverteis(, ) and the focus is, 4.

4 REVIEW OF CONIC SECTIONS 6. = + =.Thisisanequationofahperbola with vertices (±, ). The foci are at ± +, = ±,. 7. =4 + 4 = +( +)= +( ) = ( ) + =. This is an equation of an ellipse with vertices at ±,. The foci are at ±, =(±, ). 8. 8 =6 6 8 +6=6 ( 4) =6.Thisisanequationofaparabola with 4p =6,sop =.Theverteis(, 4) and the focus is, 4. 9. + =4 + + +=4 +4 ( +) 4 =4 an equation of a hperbola with vertices (, ± ) = (, ) and (, ). The foci are at, ± 4+ =, ± 5. ( +) 4 =.Thisis. 4 +4 + = 4 + + 4 + = 4 + + = + + =.This /4 is an equation of an ellipse with vertices, ± =, ±.Thefociareat, ± =, ± 4 /.. Theparabolawithverte(, ) and focus (, ) opens downward and has p =, so its equation is =4p = 8.. The parabola with verte (, ) and directri = 5 opens to the right and has p =6, so its equation is =4p( ) = 4( ).. The distance from the focus ( 4, ) to the directri =is ( 4) = 6, so the distance from the focus to the verte is (6) = and the verte is (, ). Since the focus is to the left of the verte, p =. An equation is =4p( +) = ( +). 4. The distance from the focus (, 6) to the verte (, ) is 6 =4. Since the focus is above the verte, p =4.An equation is ( ) =4p( ) ( ) = 6( ). 5. The parabola must have equation =4p,so( 4) =4p() p =4 =6. 6. Vertical ais ( h) =4p( k). Substituting (, ) and (, ) gives ( h) =4p( k) and ( h) =4p( k) ( h) =( h) 4+4h + h = h h = =4p( k). Substituting (, 9) gives [ ( )] =4p(9 k) 4=4p(9 k). Solving for p from these equations gives p = 4( k) = 9 k +4 +=. 4( k) =9 k k = p = ( +) = ( ) 8 Thomson Brooks-Cole copright 7 7. The ellipse with foci (±, ) and vertices (±5, ) has center (, ) and a horizontal major ais, with a =5and c =,sob = a c =.Anequationis 5 + =.

REVIEW OF CONIC SECTIONS 5 8. The ellipse with foci (, ±5) and vertices (, ±) has center (, ) and a vertical major ais, with c =5and a =,sob = a c =.Anequationis 44 + 69 =. 9. Since the vertices are (, ) and (, 8), the ellipse has center (, 4) with a vertical ais and a =4. The foci at (, ) and (, 6) are units from the center, so c =and b = a c = 4 =. An equation is ( ) ( 4) + = ( 4) + =. b a 6 4. Since the foci are (, ) and (8, ), the ellipse has center (4, ) with a horizontal ais and c =4.Theverte (9, ) is 5 units from the center, so a =5and b = a c = 5 4 = 9.Anequationis ( 4) ( +) + = a b ( 4) 5 + ( +) 9 =. 4. Center (, ), c =, a = b = 5 9 ( ) + 5 ( ) = 4. Center (, ), c =, major ais horizontal a + b =and b = a c = a 4. Since the ellipse passes through (, ),wehavea = PF + PF = 7 + a = 9+ 7 and b = + 7,sothe ellipse has equation 9+ 7 + + 7 =. 4. Center (, ), vertical ais, c =, a = b = 8= 8 = 44. Center (, ), horizontal ais, c =6, a =4 b = 5 6 = 45. Center (4, ), horizontal ais, c =, a = b = 5 4 ( 4) 5 ( ) = 46. Center (, ), vertical ais, c =5, a = b =4 9 ( ) 6 ( ) = 47. Center (, ), horizontal ais, a =, b a = b =6 9 6 = 48. Center (4, ), horizontal ais, asmptotes = ±( 4) c =, b/a = a = b c =4=a + b =a a = ( 4) ( ) = 49. In Figure 8, we see that the point on the ellipse closest to a focus is the closer verte (which is a distance a c from it) while the farthest point is the other verte (at a distance of a + c). So for this lunar orbit, (a c)+(a + c) =a = (78 + ) + (78 + 4),ora = 94;and(a + c) (a c) =c = 4, or c =. Thus, b = a c =,75,96,andtheequationis,76,6 +,75,96 =. 5. (a) Choose V to be the origin, with -ais through V and F. Then F is (p,), A is (p, 5), so substituting A into the equation =4p gives 5 = 4p so p = 5 and =. (b) = = CD = Thomson Brooks-Cole copright 7 5. (a) Set up the coordinate sstem so that A is (, ) and B is (, ). PA PB = ()(98) =,76, ft = 45 b = c a =,9,75,5,65,9,75 =. 5 mi =a a =,andc = so (b) Due north of B = ()(),5,65,575 = = 48 mi,9,75 59

6 REVIEW OF CONIC SECTIONS 5. PF PF = ±a ( + c) + ( c) + = ±a ( + c) + = ( c) + ± a ( + c) + =( c) + +4a ± 4a ( c) + 4c 4a = ±4a ( c) + c a c + a 4 = a c + c + c a a = a c a b a = a b (where b = c a ) a b = 5. The function whose graph is the upper branch of this hperbola is concave upward. The function is = f() =a + b = a b + b,so = a b b + / and = a b b + / b + / = ab b + / > for all,andsof is concave upward. 54. We can follow eactl the same sequence of steps as in the derivation of Formula 4, ecept we use the points (, ) and (, ) in the distance formula (first equation of that derivation) so ( ) +( ) + ( +) +( +) =4will lead (after moving the second term to the right, squaring, and simplifing) to ( +) +( +) = + +4, which, after squaring and simplifing again, leads to + =8. 55. (a) If k>6, thenk 6 >, and k + =is an ellipse since it is the sum of two squares on the k 6 left side. (b) If <k<6,thenk 6 <,and k + =is a hperbola since it is the difference of two squares k 6 on the left side. (c) If k<,thenk 6 <, andthereisno curve since the left side is the sum of two negative terms, which cannot equal. (d) In case (a), a = k, b = k 6,andc = a b =6, so the foci are at (±4, ). Incase(b),k 6 <,so a = k, b =6 k,andc = a + b =6, and so again the foci are at (±4, ). 56. (a) =4p =4p = p, so the tangent line is = p ( ) =p( ) 4p =p p =p( + ). (b) The -intercept is. 57. Use the parametrization = cos t, = sin t, t π to get L =4 π/ (d/dt) +(d/dt) dt =4 π/ 4sin t +cos tdt=4 π/ sin t +dt Thomson Brooks-Cole copright 7 Using Simpson s Rule with n =, t = π/ = π,andf(t) = sin t +,weget L 4 π f() + 4f π +f π + +f 8π +4f 9π + f π 9.69

REVIEW OF CONIC SECTIONS 7 58. The length of the major ais is a, soa = (.8 )=5.9 9. The length of the minor ais is b,so b = (.4 )=5.7 9. An equation of the ellipse is a + b =,orconvertingintoparametric equations, = a cos θ and = b sin θ. So L =4 π/ (d/dθ) +(d/dθ) dθ =4 π/ a sin θ + b cos θdθ Using Simpson s Rule with n =, θ = π/ = π,andf(θ) = a sin θ + b cos θ,weget L 4 S =4 π.64 km f() + 4f π +f π + +f 8π +4f 9π + f π 59. a + = b a + = = b b a ( 6= ). Thus, the slope of the tangent line at P is b. TheslopeofF P is a + c and of FP is. B the formula from Problems Plus, we have c tan α = + c + b a b a ( + c) = a + b ( + c) a ( + c) b = a b + b c c + a c using b + a = a b and a b = c = b (c + a ) c (c + a ) = b c and tan β = c b a b a ( c) So α = β. 6. The slopes of the line segments F P and F P are implicitl, a = = b b a Problems Plus, and tan α = + = a b ( c) a ( c) b = a b + b c c a c = b (c a ) c (c a ) = b c b a + c b a ( + c) = b (c + a ) c (c + a ) + c and c,wherep is (, ). Differentiating the slope of the tangent at P is b a, so b the formula from = b ( + c) a a ( + c)+b using /a /b = and a + b = c = b c Thomson Brooks-Cole copright 7 So α = β. tan β = + b a + c b a ( c) = b ( c)+a a ( c)+b = b (c a ) c (c a ) = b c