# 10.2 The Ellipse and the Hyperbola

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1 CHAPTER 0 Conic Sections Solve. 97. Two surveors need to find the distnce cross lke. The plce reference pole t point A in the digrm. Point B is meters est nd meter north of the reference point A. Point C is 9 meters est nd meters north of point A. Find the distnce cross the lke, from B to C. 98. A ridge constructed over ou hs supporting rch in the shpe of prol. Find n eqution of the prolic rch if the length of the rod over the rch is 00 meters nd the mimum height of the rch is 40 meters. 00 m 40 m C 0 m A (0, 0) B Use grphing clcultor to verif ech eercise. Use squre viewing window. 99. Eercise Eercise Eercise Eercise The Ellipse nd the Hperol S Define nd Grph n Ellipse. Define nd Grph Hperol. Grphing Ellipses An ellipse cn e thought of s the set of points in plne such tht the sum of the distnces of those points from two fied points is constnt. Ech of the two fied points is clled focus. (The plurl of focus is foci.) The point midw etween the foci is clled the center. An ellipse m e drwn hnd using two thumtcks, piece of string, nd pencil. Secure the two thumtcks in piece of crdord, for emple, nd tie ech end of the string to tck. Use our pencil to pull the string tight nd drw the ellipse. The two thumtcks re the foci of the drwn ellipse. Focus Center Focus Ellipse with Center (0, 0) The grph of n eqution of the form + = is n ellipse with center (0, 0). The -intercepts re (, 0) nd -, 0, nd the -intercepts re (0, ), nd 0, -. The stndrd form of n ellipse with center (0, 0) is + =.

2 Section 0. The Ellipse nd the Hperol EXAMPLE Grph 9 + =. Solution The eqution is of the form + =, with = nd = 4, so its grph is n ellipse with center (0, 0), -intercepts (, 0) nd -, 0, nd -intercepts (0, 4) nd 0, (0, 4) (, 0) (, 0) 4 (0, ) Grph + 4 =. EXAMPLE Grph 4 + = 4. Solution Although this eqution contins sum of squred terms in nd on the sme side of n eqution, this is not the eqution of circle since the coefficients of nd re not the sme. The grph of this eqution is n ellipse. Since the stndrd form of the eqution of n ellipse hs on one side, divide oth sides of this eqution = = 4 4 Divide oth sides = Simplif. We now recognize the eqution of n ellipse with = 4 nd =. This ellipse hs center (0, 0), -intercepts (4, 0) nd - 4, 0, nd -intercepts (0, ) nd 0, (0, ) (, 0) (4, 0) 4 (0, ) Grph =. The center of n ellipse is not lws (0, 0), s shown in the net emple. Ellipse with Center (h, k) The stndrd form of the eqution of n ellipse with center h, k is - h - k + =

3 4 CHAPTER 0 Conic Sections + - EXAMPLE Grph + =. Solution The center of this ellipse is found in w tht is similr to finding the center of circle. This ellipse hs center -,. Notice tht = nd =. To find four points on the grph of the ellipse, first grph the center, -,. Since =, count units right nd then units left of the point with coordintes -,. Net, since =, strt t -, nd count units up nd then units down to find two more points on the ellipse. (, ) ( ) ( ) - 4 Grph =. CONCEPT CHECK In the grph of the eqution 4 + =, which distnce is longer: the distnce etween the -intercepts or the distnce etween the -intercepts? How much longer? Eplin. Focus Center Focus Grphing Hperols The finl conic section is the hperol. A hperol is the set of points in plne such tht the solute vlue of the difference of the distnces from two fied points is constnt. Ech of the two fied points is clled focus. The point midw etween the foci is clled the center. Using the distnce formul, we cn show tht the grph of - = is hperol with center (0, 0) nd -intercepts (, 0) nd -, 0. Also, the grph of - = is hperol with center (0, 0) nd -intercepts (0, ) nd 0, -. Hperol with Center (0, 0) The grph of n eqution of the form - = is hperol with center (0, 0) nd -intercepts (, 0) nd -, 0. Answer to Concept Check: -intercepts, 4 units

4 Section 0. The Ellipse nd the Hperol The grph of n eqution of the form - = is hperol with center (0, 0) nd -intercepts (0, ) nd 0, -. The equtions - = nd - = re the stndrd forms for the eqution of hperol. Helpful Hint Notice the difference etween the eqution of n ellipse nd hperol. The eqution of the ellipse contins nd terms on the sme side of the eqution with sme-sign coefficients. For hperol, the coefficients on the sme side of the eqution hve different signs. (, ) (, ) (, ) (, ) Grphing hperol such s - = is mde esier recognizing one of its importnt chrcteristics. Emining the figure to the left, notice how the sides of the rnches of the hperol etend indefinitel nd seem to pproch the dshed lines in the figure. These dshed lines re clled the smptotes of the hperol. To sketch these lines, or smptotes, drw rectngle with vertices (, ), -,,, - nd -, -. The smptotes of the hperol re the etended digonls of this rectngle. EXAMPLE 4 Grph - =. Solution This eqution hs the form - =, with = 4 nd =. Thus, its grph is hperol tht opens to the left nd right. It hs center (0, 0) nd -intercepts (4, 0) nd -4, 0. To id in grphing the hperol, we first sketch its smptotes. The etended digonls of the rectngle with corners (4, ), 4, -, -4,, nd -4, - re the smptotes of the hperol. Then we use the smptotes to id in sketching the hperol. (, ) (4, ) 4 (, ) 4 (4, ) 4 Grph 9 - =.

5 CHAPTER 0 Conic Sections EXAMPLE Grph 4-9 =. Solution Since this is difference of squred terms in nd on the sme side of the eqution, its grph is hperol s opposed to n ellipse or circle. The stndrd form of the eqution of hperol hs on one side, so divide oth sides of the eqution. 4-9 = 4-9 = Divide oth sides. 9-4 = Simplif. The eqution is of the form - =, with = nd =, so the hperol is centered t (0, 0) with -intercepts (0, ) nd 0, -. The sketch of the hperol is shown Grph 9 - =. Although this is eond the scope of this tet, the stndrd forms of the equtions of hperols with center (h, k) re given elow. The Concept Etensions section in Eercise Set 0. contins some hperols of this form. Hperol with Center (h, k) Stndrd forms of the equtions of hperols with center (h, k) re: - h - - k = - k - - h = Grphing Clcultor Eplortions To grph n ellipse using grphing clcultor, use the sme procedure s for grphing circle. For emple, to grph + =, first solve for. = - 0 Y Y = - = { B Net, press the Y = ke nd enter Y = nd Y = -. B B (Insert two sets of prentheses in the rdicnd s - / so tht the desired grph is otined.) The grph ppers s shown to the left.

6 Section 0. The Ellipse nd the Hperol 7 Use grphing clcultor to grph ech ellipse =. + =. 0 + = = = = 0.8 Voculr, Rediness & Video Check Use the choices elow to fill in ech lnk. Some choices will e used more thn once nd some not t ll. ellipse 0, 0, 0 nd -, 0 0, nd 0, - focus hperol center, 0 nd -, 0 0, nd 0, -. A(n) is the set of points in plne such tht the solute vlue of the differences of their distnces from two fied points is constnt.. A(n) is the set of points in plne such tht the sum of their distnces from two fied points is constnt. For eercises nd ove,. The two fied points re ech clled. 4. The point midw etween the foci is clled the.. The grph of - = is (n) with center nd -intercepts of.. The grph of + = is (n) with center nd -intercepts of. Mrtin-G Interctive Videos Wtch the section lecture video nd nswer the following questions. 7. From Emple, wht informtion do the vlues of nd give us out the grph of n ellipse? Answer this sme question for Emple. 8. From Emple, we know the points (, ),, -, -,, nd -, - re not prt of the grph. Eplin the role of these points. See Video Eercise Set Identif the grph of ech eqution s n ellipse or hperol. Do not grph. See Emples through.. + =. 4-4 =. - = =. - + =. + = Sketch the grph of ech eqution. See Emples nd = = = =. 9 + =. + 4 =. 4 + = = Sketch the grph of ech eqution. See Emple = = = =

7 8 CHAPTER 0 Conic Sections Sketch the grph of ech eqution. See Emples 4 nd = 0. - =. - = =. - 4 = =. - =. 4 - = 00 MIXED Grph ech eqution. See Emples through. 7. = - 8. = = = =. + = = = 9 MIXED SECTIONS 0., 0. Identif whether ech eqution, when grphed, will e prol, circle, ellipse, or hperol. Sketch the grph of ech eqution. If prol, lel the verte. If circle, lel the center nd note the rdius. If n ellipse, lel the center. If hperol, lel the - or -intercepts = 4. = = = = = = 4. + = 4. = = = = = 48. = = 0. = REVIEW AND PREVIEW Perform the indicted opertions. See Sections. nd CONCEPT EXTENSIONS The grph of ech eqution is n ellipse. Determine which distnce is longer, the distnce etween or the distnce etween How much longer? See the Concept Check in this section.. + = = = = 9. If ou re given list of equtions of circles, prols, ellipses, nd hperols, eplin how ou could distinguish the different conic sections from their equtions. 0. We know tht + = is the eqution of circle. Rewrite the eqution so tht the right side is equl to. Which tpe of conic section does this eqution form resemle? In fct, the circle is specil cse of this tpe of conic section. Descrie the conditions under which this tpe of conic section is circle. The orits of strs, plnets, comets, steroids, nd stellites ll hve the shpe of one of the conic sections. Astronomers use mesure clled eccentricit to descrie the shpe nd elongtion of n oritl pth. For the circle nd ellipse, eccentricit e is clculted with the formul e = c d, where c = 0-0 nd d is the lrger vlue of or. For hperol, eccentricit e is clculted with the formul e = c d, where c = + nd the vlue of d is equl to if the hperol hs -intercepts or equl to if the hperol hs -intercepts. Use equtions A H to nswer Eercises 70. A. - D. - 9 = B. 4 + = C. 4 + = = E = F. + = G. - = H =. Identif the tpe of conic section represented ech of the equtions A H.. For ech of the equtions A H, identif the vlues of nd.. For ech of the equtions A H, clculte the vlue of c nd c. 4. For ech of the equtions A H, find the vlue of d.. For ech of the equtions A H, clculte the eccentricit e.. Wht do ou notice out the vlues of e for the equtions ou identified s ellipses? 7. Wht do ou notice out the vlues of e for the equtions ou identified s circles? 8. Wht do ou notice out the vlues of e for the equtions ou identified s hperols? 9. The eccentricit of prol is ectl. Use this informtion nd the oservtions ou mde in Eercises, 7, nd 8 to descrie w tht could e used to identif the tpe of conic section sed on its eccentricit vlue. 70. Grph ech of the conic sections given in equtions A H. Wht do ou notice out the shpe of the ellipses for incresing vlues of eccentricit? Which is the most ellipticl? Which is the lest ellipticl, tht is, the most circulr? 7. A plnet s orit out the sun cn e descried s n ellipse. Consider the sun s the origin of rectngulr coordinte sstem. Suppose tht the -intercepts of the ellipticl pth of the plnet re {0,000,000 nd tht the -intercepts re {,000,000. Write the eqution of the ellipticl pth of the plnet.

8 Integrted Review 9 7. Comets orit the sun in elongted ellipses. Consider the sun s the origin of rectngulr coordinte sstem. Suppose tht the eqution of the pth of the comet is -,78,000,000 -,400,000.4 * * 0 = Find the center of the pth of the comet. 7. Use grphing clcultor to verif Eercise Use grphing clcultor to verif Eercise. For Eercises 7 through 80, see the emple elow. Emple - - Sketch the grph of - =. 9 Solution This hperol hs center (, ). Notice tht = nd = Sketch the grph of ech eqution. (, ) ( ) ( ) = = = = = = Integrted Review GRAPHING CONIC SECTIONS Following is summr of conic sections. Conic Sections Stndrd Form Grph Prol = - h + k (h, k) 0 0 (h, k) Prol = - k + h (h, k) 0 0 (h, k) Circle - h + - k = r (h, k) r Ellipse center (0, 0) + = Hperol center (0, 0) - = Hperol center (0, 0) - =

9 0 CHAPTER 0 Conic Sections Identif whether ech eqution, when grphed, will e prol, circle, ellipse, or hperol. Then grph ech eqution = 4. = + 4. = =. 9-9 =. - 4 = = 8. + = 9. = = = = = 4. = = 0. Solving Nonliner Sstems of Equtions S Solve Nonliner Sstem Sustitution. Solve Nonliner Sstem Elimintion. In Section 4., we used grphing, sustitution, nd elimintion methods to find solutions of sstems of liner equtions in two vriles. We now ppl these sme methods to nonliner sstems of equtions in two vriles. A nonliner sstem of equtions is sstem of equtions t lest one of which is not liner. Since we will e grphing the equtions in ech sstem, we re interested in rel numer solutions onl. Solving Nonliner Sstems Sustitution First, nonliner sstems re solved the sustitution method. EXAMPLE Solve the sstem e - = - = Solution We cn solve this sstem sustitution if we solve one eqution for one of the vriles. Solving the first eqution for is not the est choice since doing so introduces rdicl. Also, solving for in the first eqution introduces frction. We solve the second eqution for. - = Second eqution - = Solve for. Replce with - in the first eqution, nd then solve for. - = First eqution \$%& - - = Replce with = - + = = 0 = or = Let = nd then let = in the eqution = - to find corresponding -vlues. Let =. Let =. = - = - = = - = - = 0 The solutions re (, ) nd (, 0), or the solution set is,,, 0. Check oth solutions in oth equtions. Both solutions stisf oth equtions, so oth re solutions

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