《微积分》课程教学课件（Calculus）11. Parametric Equation and Polar Coordinates

CALCULUSTHOMAS'NTALCALCULUSEARLYTRANSCENDENTAEChapter 11JAMESSTEWARTParametric Equations andPolar CoordinatesERMSESlide 3-21.DefinitionofPolarCoordinates11.3PolarCoordinatesIn this section we study polar coordinates and their relation to Cartesian coordinates. Youwill see that polar coordinates are very useful for calculating many multiple integrals stud-ied in Chapter 15.Slide 3-3Slide 3- 4中小间8/6To definc polar coordinates, we first fix an origin O (called the pole) and an initialry7e/6t1rfrom O (Figure 11.18). Then each point P can be located by assigning to it a polar coordi-natepair ()inwichgivesthediddistanefomOoPandgivsthedirecedangle from the initial ray to ray OP.0~g/8Pr,0)准-(司)Origin (pole)FIGURE11.20PobFIGURE11.19polihave ncpuive r-saluesuniquclitid rayAs in trigonometry, e is positive when measured counterclockwise andnegative wberFIGURE 11.18To define polarmeasured clockwise. The angle associated with a given point is not unique. While a point incoordinates for the plane,we stant with anthe plane has just one pair of Cartesian coordinates, it has infinitely many pairs of polar coorigin,called e poe and an initial yordinates. For instance, the point 2 units from the origin along the ray 9 = w/6 has polarcoordinates r = 2, 9 =/6, It also has coordinates r = 2, 8 =11/6 (Figure 11.19).In some situations we allow rto be negative, That is why we use directed distance in defin-Polar Coordinatesing P(r, 0),. The point P(2, 7r/6) can be reached by turning 7w/6 radians counterelock-P(r,0)wise from the initial my andgoing forward2 units (Figure 11.20). It can also he reached byturning w/6 radians counterckockwise fromn the initial ray and going hachwng 2 units. SoDirecdtod distaDirectedthe point also has polar coordinates r = 2, 8 =w/6fromOtn/mitialraytoOpslide 3-8

2016/11/15 1 Slide 3 - 2 Chapter 11 Parametric Equations and Polar Coordinates Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 3 - 3 11.3 Polar Coordinates Slide 3 - 4 1. Definition of Polar Coordinates Slide 3 - 5 Slide 3 - 6

EXAMPLE1Find all the polar coordinates ofthe point P(2, w/6)SolutionWe skctch the initialray ofthe coordinatesystem, draw the riy from the ori-gin that makes an angle of w/6 nadians with the initial ray, and mark the point (2, w/6)(Figure 11.21). We then find the angles for the other coordinate pairs of P in which r 2andr 2.6)-(2-%)For 2, the complete list of angkes i-(2号) ± 2#.*4m，± 6m....7w/62.PolarEquations and GraphsForr.2, the angles areSTSn5Sm± 2m,±4m，6± 6m.ad nThe corresponding coordinate pairs of P are5m/6(2-号 + 2m=) 0, ±1, ±2,**FIGURE11.21. The point P(2 /6) hasmd(-2 + 2mm)infinitcly many polar eooedinate pairs# 0, ±1, ±2,...(Example 1).When a 0, the fomulas give (2, m/6) and (2, Sm/6), When # 1, thy give(2, 13r/6) and (2, 7/6), and so onlide 3-7Slide 3-8If we hold r fixed at a constant valoe r. = α + 0, the point P(r, o) will lie a|units fromEXAMPLE2the origin O. As varies over any interval of length 2xr, P then traces a circle of radiusa(a) r = 1 and r = l are equations for the circle of radius I centered at O.sentered at O (Figure 11.22)f=a(b)8=/6,8=7/6,and9=5/6areequations forthelinclaf.+*FIGURE11.22The polar equation for acircle is r a.If we hold fised it a constant value @ = , and let r vary between 0 asd o0thepoinP()trasthlinethroghthat makesanangleofmeasureO,withthinitial rayGraphEquationr-aCircle of radius[a centered at O0-00Line through O making an angle O, with the initial raySlide 3- 10EXAMPLE3Grapb the sets of points whose polar eoondinates satisfy the followingconditions.0≤≤号18--(b) 3 ≤r±2(a) I ≤r≤ 2andand(宁≤Os告(no restriction on r)6(b)P8-5.Solution-35/52人3.RelatingPolarandCartesian(a)15r52050s1Coordinates2m(c)Oo*3slide 3-12Slide3-11

2016/11/15 2 Slide 3 - 7 Slide 3 - 8 2. Polar Equations and Graphs Slide 3 - 9 Slide 3 - 10 Slide 3 - 11 Slide 3 - 12 3. Relating Polar and Cartesian Coordinates

When we use both polar and Cartesian coondinates in a plane, we place the two originsEXAMPLE4Here are some equivalent equations expressed in terms of both polattogether and take the initial polar ray as the positive x-axis. The nay = w/2, r > 0.coordinates and Cartesian coordinates.becomes the positive y-axis (Figure 11.24), The two coondinate systems are then related byPolar equationCartesian equivalentthe following equations.rcos0 2Rys-5x-29 = 4Pcos0 sing=4a,y = Po,0)r-P=1P'0o0-Psin'8=1-r--4-1-0r =1 +2rcosef++2r*+2+2m2-y*-0r-1-cosoarSomeusaremoresimplyepresedwithpolarcoondinates others arenoFIGURE11.24Theum way torelanepolarand CaesiancoodinmEquations Relating Polar and Cartesian Coordinatestung=#X = Fcos 6,y = rsin o,r=x+y.slide 3- 14EXAMPLE6Replace the following polar equations by equivalent Cartesian equationsEXAMPLE5Find a polar equation for the circle r2 + (y - 3) = 9and identify their graphs.>(b) r2 4r cos 2+0-3-9(a) rcos0 -4(c) r.=2 cos 6 sin eC=6singSolutionWe use the substitutions r cos e = x, rsin o = y,r? = x? + y7.(a) rcos 6 =- 4(0, 3)rcoso=-4The Cartesian equation:x = =4*xThe graph:Vertical line through x = 4 on the x-axis0(b) μ2 = 4rcos 8SolutionWe apply the equations relating polar and Cartesian coordinates:r2 = 4r cos6* + ( 3) = 9The Cartesian equationx+y2 - 6y +9 = 9Eixpand (j.3)12 + yj2 = 4rx+y2-6y=0*2 4r + y* = 0Caocellaticnr2-6rsing=0+yerr24r +4+y2-4Campeing die squreor r-6sine=0r=0(x = 2) +y2 = 4+=6singIncludes bothi posxibiliticsThe graph:Cirele, nadius 2, center (h, k) = (2, 0)eu a- 16Slide 3- 154Polar CoordinatesExercises(c) r=2cos8-8ing Which polar cordinate pains abel the same poin?2 (3,0)h. (3, 0)e. (2. 2m/3)The Cartesian equation:r(2 cos8 sin 0) = 4d. (2, 7m/3)e. (3, m)r. (2, #/3)2r cos8 rsin o 4& (3, 2m)h. (2, m/3)2# y = 43, Plot the fllowing points (given in polar coordinates) Then findy = 2r 4althe polar coordinates of each pointThe graph:Line, slope m = 2, y-intercept b = 4a, (2, #/2)h. (2, 0)e. (2, #/2)d. (2, 0)6. Find the Cartesian coordinates of the following points (given inpolar coordinates).s (V2.m/4)b. (1, 0)d. (V2, #/4)e, (0, m/2)Slide 3-17slide 3-18

2016/11/15 3 Slide 3 - 13 Slide 3 - 14 Slide 3 - 15 Slide 3 - 16 Slide 3 - 17 Slide 3 - 18 Exercises

Cartesian to Polar CoordinatesExercisesExercises Polar to Cartesian Equations7. Find the polar coordinates, 0≤ o < 2 and r ≥ 0, of the fol-Replace thc polar equations in Exercises 2752 withequivalent Cartelowing points given in Cartesian coordinates.sian equations. Then describe or identify the graph.a. (1,1)b. (3, 0)27. rcos.8 228.rsino-1c. (Vi, 1)d. (3,4)29.rsino030. rco50 09. Find ithe polar coordinates, 0 ≤ 8 < 2w and r ≤ 0, of the fol.31.r=4csco32r=-3secolowing points given in Cartesian coondinates.33.r.cos9+.rsino=134. r sing = rcos oa. (3,3)b. (1, 0)35. r = 136. . =. 4rsin 0e. (-1, V5)d. (4, 3)Graphing in Polar CoordinatesGraph the sets of points whose polar coordinates satisfy the equationsand inequalities in Exercises 1126.11.7=212.0/5213, r≥ 114. 1≤r≤215.0≤0≤/6,r≥o16. 8 = 2#/3,r≤217. 0 = #/3, I ≤r≤318. 0 = 11#/4,r ≥ 1Slide 3-19Slide 3-20Exercises37. r :38. r* sin 29 = 2sin f 2 cos f39. r = cot o csc.040. r..4 tan g scc o41. F = csc 0 eraos42. rsino = Inr.+ In cos8Cartesian to Polar EquationsReplacethe Cartesian equations in Exercises5366 withequivalent11.4polar equations.53. X = 754. y= 155, x- y57. x* + y2 = 456. x y = 358, y2 = 1GraphinginPolarCoordinates“+F-60. xy = 261. j2 4r62. +x +y2163. + (y 2)2 464. (r - 5)P + y2 -2565. (r 3)P + (y + 1) - 466. (r + 2)P + (y 5) 16Slide 3-21Slide 3- 22Symmetry Tests for Polar Graphs1., Symmeby abowr the X-axis: Irthe point (r, e) lies on the graph, then the point(r, g) or (r, r 6) lies on the graph (Figure 11:26a).2.Symeybout theyaxis he point (r,) liesanthegph,tnthepoint(r, 8) or(-, 0) lics on the graph (Figure 11.26b).3Symeyboutthogi thepoint (r,) lies oegph,thnthepoint(r, e) or (r, 8 + w) lies on the graph (Figure 11.26c).1.Symmetry(/, F = 9)(.8(r.o-r(f,=0)(-r,oyoe(r.e+m)(a) Ahout he .s-atih(c) About the origin(b) Aboit the y-aisSlide 3- 23Slide 3-24

2016/11/15 4 Slide 3 - 19 Exercises Slide 3 - 20 Exercises Slide 3 - 21 Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 3 - 22 11.4 Graphing in Polar Coordinates Slide 3 - 23 1. Symmetry Slide 3 - 24

EXAMPLE2Identify thesymmetriesofthecurve=4cos8EXAMPLE1Identify the symmetries of the curve r. 1 cosThe equation 2 4 cos 8 requires cose ≥ 0, so we get the entire graph hySolutionrunning e from s/2 to w/2, The curve is symmetric about the x-axis becauseSolurtionThe cunve is symmetric about the x-axis because(r.0)on the graph ,2 = 4cos (r; 8) on the graph =.r = 1 cos 6 r = 4 cos(0)(050-0010)= r= 1- cos(-0)g0s(m) (r, 8) on the graph. (r, 8) on the graph.The curve is also symmetricabout the origin beauise(r, 0) on.the gnaph p2 4 cos 0 ()2-40080 (r, 0) on the graphTogether, these two symmetries imply symmetry about the y-axis,side 3- 25Sslide 3- 28ExercisesIdentify the symmetries of the curves in Exereises 112.3. r.= 1 sin g5. r = 2 + sin g10. 72. = sin 03. r=-1-sin(t-6) symmetry about y-axis2.Slope5.r=2+sin(tr-8)---symmetryabouty-axis10. (-ry)=sine, r*=sin(tr-8) and (-r)2=sin(rr-8)- symmetry about theorigin, x-axis and y-axsSlide 3-27side 3- 2aThe slope of a polar curve.r f(e) in the xy-plane is still given by dy/dx, which is notr' = df/d8. To see why, think of the graph of as the graph of the parametrie equationsSlope of the Curver = fto)y rsin o = f(o) sin g.x = rcoso f(0) cos.t,(0)sing +(0)stIf f is a differentiable function of e, then so are x and y and, when dx/do 0, we can calle.nf(6) cos 9 f(6) sin 6culate dy/dr from the parametric formulaprovided dk/de + 0 at (r,0).d_d/aSasie 112.E4 (0)/awihiEo（10)c.0)dfo rocooProduct Rale for derintinedfcoso0- (0)slineTherefore we seethat ady/dxis not the same as dj/de.Slide 3- 29slide 3- 30

2016/11/15 5 Slide 3 - 25 Slide 3 - 26 Slide 3 - 27 Exercises 3. r=1-sin(π-θ) - symmetry about y-axis 5. r=2+sin(π-θ) - symmetry about y-axis 10. (-r)2=sinθ, r 2=sin(π-θ) and (-r)2=sin(π-θ)- symmetry about the origin, x-axis and y-axis Slide 3 - 28 2. Slope Slide 3 - 29 Slide 3 - 30

ExercisesSlopes of Polr CurvesFind the slopes of the curves in Exercises 17-20 at the given points17.-1,117, CardioidF = + cos0; 0=±n/219, Four-leaved rose (r = sin 28; o = ±#/4, ±3#/419.-1,1,1, -111.5Areasand LengthsinPolar CoordinatesSlide 3-31Slide 3-32Theregion OTS inFigure 1130isbounded by theryseando βanmecureFflo),Weapproximatetheregion withnonoverlapping fan-shaped circularsectors based on a partition Pof angle TOS, The typical sector has radius n = (o,) andcentral angle of radian measure Aoy, Its area is e/2rr times the area of a circle ofradius rorF-8(r(oho)1.AreainthePlaneThe area ofregion OrS is approximately4-21(r0) s0.-fePPU,9FIGURE1.31The area differcatiald.4Slide 3- 33Slide 3- 34for thecurver = fo)EXAMPLE1Find the area of the region in the plane enclosed by the cardoidI /iseomtinuous, we expet the approximatios to improseasthe om othe parti= 2(1. + cos.0),tionPgoestozero,where thenormofPithelargestvalucofe,Weare then led totSolutionWe graph thecardioid (Figure 11.32) and determine that theradius OP sweepsfollowing formula defining the regionls arexout the region exactly once as 9 runs from O to 2,The area is thereforeicrom4-limC-o201+cm8["1(r0)'aP)2(1+ 2cos0 + 0P 0) d0n,2g口(2+400+21+pm2) aArea of the Fan-ShRegion Between the Origin and the CurvF-reasesg(3+ 4s+cos20)do["lraA-[30 +4i0 +]Thisisteinegalftheareadiffereial (Figure131) 6# 0.= 6#d(r(o)amSlide 3- 35slide 3-38

2016/11/15 6 Slide 3 - 31 Exercises 17. -1, 1 19. -1, 1, 1, -1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 3 - 32 11.5 Areas and Lengths in Polar Coordinates Slide 3 - 33 1. Area in the Plane Slide 3 - 34 Slide 3 - 35 Slide 3 - 36

Find theaoeregiothatlisinsidete cireled outsideEXAMPLE2cardioidr=1-cose.the area ofthe regionwhichWesketch the regionto determine itsboundaries and find the limits of integraSolutlonliesbetweentwopolartion (Figure 11.34), The outer curve is r2= 1, the inner curve is n =1cos e, and curves r,=r,(e) and r2-r(e)runs from -/2to w/2.The area, from Equation (1),isfromg=atoe=β1(rr2)do44P(c2-)d0SymmeiryFIGURE11.33The areaofthe shadedregion is.calculated by subtracting the arcs(1 (1 2cos8 + cos 0) dooftheregian botwen n und iiecoriginSqanrenfrom theareaofthergicnbetwemandthe ocipin(2 0os 0 coe2 0) dLomarlini20s6 L+gm20)Area of the Reglou 05rifo)5SnasosAld-''do-'()ao(b)8in 20 /r/28ino-号=2-弄Slide 3-37Slide 3-38Finding Polar AreasFind the areas of the rogions in Exercises 1-8,4, Inside the cardioid r = a(1 + cos.e), a > 0ExercisesExercises1. Boended by the spinl r = e for o s 8 s 5. Inside one leaf ofthe four-lened rose r = cos 206, Inside one leaf of the three-leaved rose r m cos 38(6)rCm3e(n, w)2, Bounded by the cireler = 2 sin 9 for /4 S 9≤ /2动ide 3- 39Slide 3- 4014, Insido the circle r 3a cos and outside the cardioidr=a(1 +00s8),a>ExercisesExercises15, lsside the circle r = 2oos @ ad outside the circler =Find the areas of the regions in Exercises 9-1616, laside the circle r = 6 sbove the tiner 3esee9, Shared by the circles r.= 2 cos and r = 2 sin g17, tnside the circler= 4coseand to tho rightofthe vertical linc10, Shareod by the circles r = 1 and r. 2 sin 8r=secd11. Shared by the circle r = 2 and the candioid r = 2(1 cos e)18. Inside the circle r= 4 sin & and bekow the horisontal lincr-3esc812. Shared by the cardioids r = 2(1 + cos 0) and r = 2(1 cos 0)19, a, Find the area of thparyitg figure13, Inside the lemmiscater = 6 cos 28 and outide the circler =. V-OC9.(2snordo+（200=/4)2sin 0de+2cos* 0d0=I(1cos20)do+J(1+cos 20)d0V2/20c--{0-2-0+2--0h. ltlooks as ifthe graph of r =tan g, e/2<0<w/2,could be asymplotic tothe linesx 1andx 1.is it?Give nesons for your answer.Slide 3-41slide3-42

2016/11/15 7 Slide 3 - 37 the area of the region which lies between two polar curves r1=r1 (θ) and r2=r2 (θ) from θ=α to θ=β Slide 3 - 38 Slide 3 - 39 Exercises Slide 3 - 40 Exercises Slide 3 - 41 Exercises /4 /2 2 2 0 /4 /4 /2 /4 /2 2 2 0 /4 0 /4 /4 /2 0 /4 1 1 9. (2sin ) (2cos ) 2 2 2sin 2cos (1 cos 2 ) (1 cos 2 ) 1 1 sin 2 sin 2 1 2 2 2 d d d d d d                                                               Slide 3 - 42 Exercises

We canobtainapolar oordinate fomula for the lenghofacune= (e),a βby parametrizing the curve asxr.cos9 (e)cos 6.y = rsino = (o)sin,asosp,(2)Theparametriclengthfomula, Equation (3) from Section11.2, thengives the lengtha-IV()(").2. Length of a Polar CurveThis equation becomest-'Ve+()when Equations (2) are substiuted for x and y (Exercise 29).Slide 3-43Slide 3-44EXAMPLE3Find the length of the cardioid r =1 cos e.SolutionWesketchthe cardioid to determine the limits ofintegration (Figure 11.35), TheLength of a Polar Curvepoint P(r,e) traces the cune once, counterelockwise as runs from Oto 2r,so these areIfrJ(o) has a continuous first derivative for a S e ≤β and if the pointthe values we take for ar and B.Withr. 1 c0s8,osie,P(rf) tracesthe curverf(o) exactlyonce as fruns from a to β,then thewe banvelength of the curve is()(1cos e)+ (sing)F = 1 = 00s t-I'V+() a.(3)Pr,d=.12cou8+co0+s.0-2-2comtandVe(）-V-2"-'mgFIGURE 11-35 Calculaing the length ofa eandioid (Example 3)-I'2mgma/ha0hr t5c-[-4] -44- x.Slide 3- 45Slide 3- 48ExercisesFinding Lengths of Polar CurvesFind the lengths of the curves in Exercises 21-28.21,Thespiralr=0,osesV522. The spiralre/V2,0sesm11.623, The cardioid r = 1 + cos e24.Thecurvermasin(0/2)0≤@≤,a>025. The parabolic segment r = 6/(1 + cos e), 0 ≤ 8 ≤ /2Conic Sections26. The parabolic segment r. = 2/(1 cos 0),n/2≤8≤#Slide 3-47Slide 3-48

2016/11/15 8 Slide 3 - 43 2. Length of a Polar Curve Slide 3 - 44 Slide 3 - 45 Slide 3 - 46 Slide 3 - 47 Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 3 - 48 11.6 Conic Sections

In thisectionwedefine ad revew pubolas, elipsesad hyperbolasgemtricaly andderive thcir standard Cartesian equaticns, These curves are calked comic.sections or cowicsbecause they are formed by cutting a double cone with a plane (Figure 11.36).1.Parabolasr ditsCnt fae mpolieaPtrhepeedDaXCegenerale canesoctiomssiveadShek Sac pasdSlide3-49Slide 3-50If fhe parnbola opers dowuwwed with ins focus at (0, p) and ies directrix the Tiney..p,thenBqoations (1)becomDEFINITIONSA set that consists of all thepoints in a plane equidistant from-4yy--5anda given fixed point and a given fixed line in the plane is a parabola, The fixedBy i eymittiofbolas openingpoint is the focus of the parabola. The fixed tine is the directrix.tothenight ortothe lkeft (Figute11.38)Apnbohsismpksopuimwhnafoosaddiroctissrallemeefteeppose hue the ficustFn,ghes teForerpk11.37telPF-V-P+G-pP-VF+(-pPo-Vr-+ty-(-plp-Vy+p)Whee weequale these.indsimplity, we gctnllatTrtianari一’ = 4iy.(3)Teseeqations revesdl thepaabelassyumtryabotie yasis Wealltheyaxis theor the paneaais h tesertes.The verkoxerchwheeahob24ylesatteons (Figml37,ThepostiresumberptopacbobhfcalkngthIGURE 11.38(o) The paraboba y 4px. (b) The pwholb y/ -ApxSlide 3- 51Slide 3- 52IdentifyingGrapthsEXAMPLE1Find thefocus and directrix ofthe parabolay1rExercisesMatch the parabolas in Exercises 14 with the following equationsSolutianWe find the value of p in the standardf equation y2. 4pr:=2yX=-6g,y=8,=-4r24p10,50Then find each parabolax foctus and directrix.2L.Thnwefindhanddirectrithiseo2=-43P=8x(n.0) - (8.0)Focex:Directrix:X=-号3.X =2yr'=-bySlide3-53Slide 3-54

2016/11/15 9 Slide 3 - 49 degenerate cone sections Slide 3 - 50 1. Parabolas Slide 3 - 51 Slide 3 - 52 Slide 3 - 53 Slide 3 - 54 Exercises 2 x y  2 2 x y 6 2 y x  8 2 y x 4

ExercisesParabolasExercises916givequationsofparabolas.Findachparabolasfcus and directrix Thensketchthe parabola Includethe focus anddirectrix in your sketch.9. y2 = 12x10. x2 = 6y11. y2 = 8y2.Ellipses13. y = 4r214. y = =8r212. y2 = 2x15. x = 3y216. r = 2y/2slide 3-55Slide 3-58+ PF) in deoted by 2absofapein.PceitbeitlipestistyfeaqutinVu+e+y+Vu-c+yl=aDEFINITIONSAnellipse ithet ofpoints insplawhose distanofrom tuo fixed poists in the plate have a constiasf sum.The two fised poiatsare the fociofthe ellipeto Pre ru-hand sidk, sare, inigadical,sadsqunrigan,buighthfocflipellipfocalatis halfwayhctwectnthe fociictheceteThsuhome the focal axiand ellipse cros are thellipek vertkces (Figue 1139)++hFi/(ytTheellinsdefiredwthraaflaly Fiacandranqution PF, + PF, =Zu is the.graph teateifnLe+ (°/) = 16-VC-23Peal asiRn a^andiqution (2) tkos fhe fomsH5+5-1FIGURE11.30.Potints om the focatlasisofan llipse.Slide 3-s7Slide 3- saEXAMPLE2Thie ellipe五+闯(Figee I1L41) has+-Serminajer ati:Semiminor oxis:Cestur-to-fos dstancFocVarticestinnPF+PFCester+(/) + (/- 1heawhere =22ibEtendkmalethetheewithirespect te the origin and botadu(wcoodinte axes,bes icidethy the lines s e ond y ±h. Ilates atte poits (a,0) ad (0, ±),The tngerts at these poists areperdester teses++茶一(6)牛CnsnL vith the foci and verThenajoraxis ofthisisBowverticalinidcadofhorroifsOandinfiniteifynth5Thercisoconfusion inanalyzing Equatioms (5)and (6) Ifwe findtthepons(emramt of kngth 2h jeining the pointintercpesothcooninatwewiliknowichwythmajorxisunsbauso,b),Thenmberiefistemimajpr ai, the mnber tbe semininer asisis the langer of the two axesation (3)a-VT-Pisthe centetafiteeis acircleaslide 3-60Slide 3-59

2016/11/15 10 Slide 3 - 55 Exercises Slide 3 - 56 2. Ellipses Slide 3 - 57 Slide 3 - 58 Slide 3 - 59 Slide 3 - 60