11.1 Conic sections (conics)


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1 . Coni setions onis Coni setions re formed the intersetion of plne with right irulr one. The tpe of the urve depends on the ngle t whih the plne intersets the surfe A irle ws studied in lger in se.. We will disuss the remining 3 onis.. Prol Definition: Given line L lled the diretri nd point F lled fous, prol is the set of points P tht re equidistnt from the line L nd the point P, tht is the set of points P suh tht dist P,F = dist P,Line L = d Notie tht the point V, whih is hlfw etween the point F nd the diretri L is on the prol, sine distv,f = dist V, Line L. Point V is lled the verte of the prol
2 Notie lso tht the point P, smmetri to P with respet to the line FV, is lso on the prol s it stisfies P,F = dist P,Line L. Therefore, prol is smmetri with respet to the line FV. dist Here is the grph of prol. Let s summrize the properties of prol.  Verte in on the line perpendiulr to the diretri pssing through the fous  Verte is hlfw etween the fous nd the diretri  The line through the verte nd the fous is the line of smmetr of the grph  The line of smmetr is perpendiulr to the diretri Also notie tht the segment perpendiulr to the is of smmetr with one endpoint t the fous F nd the other on the prol hs the length tht is twie the length etween the fous nd the diretri: distf, A= distf,b = distf, V =. Segment AB is lled ltus retum nd determines how wide or how nrrow the prol is.
3 We would like to desrie prol in lgeri terms, tht is n eqution. Eqution of prol with verte t the origin 0,0 nd the fous F t,0 First notie tht when verte is t the origin nd the fous is t,0, then the diretri hs the eqution = . We will use the definition to derive the eqution. Let P, e point on the prol. B definition, dist P,F= disp, Q. This mens tht using the distne formul 0 After squring oth sides nd epnding the squres we get This eqution simplifies to Eqution of prol with verte t the origin 0,0 nd the fous F t 0, First notie tht when verte is t the origin nd the fous is t 0,, then the diretri hs the eqution = . We will use the definition to derive the eqution. Let P, e point on the prol. B definition, dist P,F= disp, Q. This mens tht 0 After squring oth sides nd epnding the squres we get This eqution simplifies to
4 Remrks: ian eqution of prol ontins oth nd ; one of these vriles omes squred, the other not. ii If  vrile is NOT squred, then the  is is the is of smmetr, whih mens oth verte nd the fous re on the is iii If vrile is NOT squred then the is is the is of smmetr, whih mens tht the verte nd fous re on the is. iv In oth equtions = nd =, is the nonzero oordinte of the fous, whih mens the fous is t,0, if the fous is on the is nd the fous is t 0,, when the fous in on the is. v If the fous is on the is t,0, then the diretri is perpendiulr to the is, hene it is vertil line = . If the fous is on the is t 0,, then the diretri is perpendiulr to the is, hene it is horizontl line = . vi Verte is lws midw etween the fous nd the diretri vii The ltus retum segment psses through the fous nd is perpendiulr to the is of smmetr. The length of this segment is times the length etween fous nd the verte =, with the fous eing its midpoint Emple: Find the eqution of the prol with fous t F0, nd verte t 0,0. Sine fous F is on the is, the eqution is of the form is NOT squred! =. Sine the nonzero oordinte of the fous is, =. Hene the eqution is = 8. Emple Find the eqution of prol with the diretri =  ½ nd verte t 0,0. Sine the diretri is perpendiulr to the line of smmetr nd it is vertil line = /, then the is of smmetr is the is. Therefore, in the stndrd form, will NOT e squred nd the eqution will e of the form =. Sine the verte is hlfw etween the diretri nd fous, the fous is t /,0, hene, = ½ nd the eqution is = Emple Grph the eqution = . Find the oordintes of the fous nd the eqution of the diretri. The eqution is of the form =. Therefore, the verte is t 0,0 nd the is of smmetr is the is. Fous F hs oordintes 0,, where stisfies = . This mens tht = , so F=0, . Therefore the prol will open down. To see how wide the prol is, drw the ltus retum segment. Note first, tht the distne etween verte nd fous, disv,f, is. Strting from the fous F, drw segment perpendiulr to the is of smmetr whih is the is of length distv,f = to otin point on the prol, then go in the opposite diretion the sme distne to otin the seond endpoint of ltus retum nd drw the prol.
5 Prol with the verte t h,k First rell tht when grph of funtion = f is moved to the left or right h units, the eqution eomes = f±h, tht is the in the originl eqution is repled ±h. When we move the grph k units up or down, the eqution eomes = f± k, or fter moving k to the left hnd side : k = f. This mens tht in the originl eqution is repled ± k. If then the grph of = f is moved h units to the left/right nd k units up/down, then the grph hs the eqution k = f ±h. Simpl put, the is repled ±h nd the is repled k. Tking this into ount, we n stte tht when prol = or = is shifted h units to the left/right nd k units up/down, its eqution will e k = ±h or ± h = k. Verte of the prol will e moved ordingl. Therefore, the eqution of prol with verte t h,k nd the is of smmetr prllel to the is is  k =  h, nd the eqution of prol with verte t h,k nd the is of smmetr prllel to the is is  h = k. These equtions re lled stndrd equtions of prol Emple Grph + 3 = 8. Find the oordintes of the verte, fous nd the eqution of the diretri. Rewrite the eqution in the stndrd form: 3 = 8 Determine the verte: V= h,k =,3 Determine the originl prol: The prol 3 = 8 is otined shifting the prol = 8, so its verte is t,3 shifting units to the right nd 3 units down d Drw the oordinte sstem to grph the eqution. e Drw the prol =  8. Use dshed line. The is of smmetr is the is. 8=, so =  nd hene fous F =,0. The diretri hs the eqution =  or = f Move the prol, its fous nd the diretri units to the right nd 3 units down. Determine red from the grph! new oordintes of fous +, 03 =, 3 nd the eqution of the diretri :  = or = 6.
6 Emple Grph the eqution = 0. The grph of this eqution is prol vrile is squred nd is not. It is not in the stndrd form. Write the eqution in the stndrd form = 0 Leve the terms with on the left hnd side nd move ll remining terms to the right + =  + Complete the squre on the left hnd side / =, = nd write it s squre; simplif the right hnd side + + = = Ftor out the oeffiient of on the right hnd side + = Determine the verte of the prol Eqution:  = 3, so the verte is V= , 3 3. Grph, using dshed line, the prol =  nd shift it units to the left nd 3 unit up. Prol = : verte t 0,0; is of smmetr: is; fous:  =, = ½; fous t 0,/
7 Emple Find the eqution of the prol with fous t , nd the diretri = . Drw piture showing given informtion Find the verte The line of smmetr perpendiulr to the diretri nd pssing through fous F is the line = . Sine verte is hlfw etween the fous nd the diretri nd lies on the is of smmetr, the verte is t 3, V = 3, Determine the endpoints of ltus retum nd sketh the prol Ltus retum endpoints: ,6, , d Determine the eqution The is of smmetr is prllel to the is is NOT squred; Verte is t3,; Fous is to the right of verte nd distv,f =, so =. Eqution: + = +3 or + = +3.
8 Emple A le TV reeiving dish is in the shpe of proloid of revolution. Find the lotion of the reeiver whih is pled t the fous, if the dish is feet ross t the opening nd 8 inhes deep. A proloid of revolution is otined revolving prol out its is of smmetr. The ross setion of suh proloid is prol. We n drw piture s elow. We n introdue the oordinte sstem with the origin t the verte of the prol nd find the eqution of tht prol. The stndrd eqution of suh prol is =. The given informtion llows us to determine two points on this prol:,8 nd ,8. Using the ft tht the point,8 is on the prol so = nd = 8 stisf the eqution, we n find vlue of : = 8 = /3 = 9/=.5 nd therefore the fous is t 0,.5 Therefore, the fous should e.5 inhes from the verte of the proloid..3 Ellipse Definition: An ellipse is the set of points P on the plne whose sum of the distnes from two different fied points F nd F is onstnt, tht is, distp, F + dist P, F = where is positive onstnt. d + d = Note tht the point Q, smmetri to P with respet to the line through F nd F, lso stisfies this eqution, nd hene, is on the ellipse.
9 Moreover the point Q smmetri to P with respet to the line perpendiulr to F F nd pssing through the midpoint of the segment F F is lso on the ellipse sine dist Q,F +distq, F = distp, F + dist P, F Therefore, n ellipse is smmetri with respet to the line F F s well s the line perpendiulr to F F nd pssing through point O. Point O, whih is the midpoint of the segment F F, is lled the enter of the ellipse. Here is the grph of this ellipse: The points F nd F re lled foi of the ellipse. The points V nd V re lled verties of the ellipse. The segment V V is lled the mjor is nd the segment B B is lled the minor is of the ellipse. Let s note the si properties of n ellipse:  verties, foi nd the enter of n ellipse lie on the sme line  the enter of n ellipse is hlfw etween foi  the enter of n ellipse is hlfw etween verties n ellipse is smmetri with respet to the line ontining mjor is nd smmetri with respet to the line ontining the minor is the mjor is is longer thn the minor is  the verties, foi nd the enter re on the mjor is
10 Eqution of n ellipse with the enter t the origin nd the foi long  is Suppose tht F =,0, >0. Beuse of the smmetr, F =,0 nd the length of the segment F F is. Sine in n tringle the sum of two sides must lws e greter thn the third side, in the tringle F PF, we must hve d +d > dist F,F, or >, or >. Let P =, e n point on the ellipse. Then, using the distne formul, we get 0, F P dist d nd 0, F P dist d. Therefore, sine for ellipse, distp, F + dist P, F =, we hve 0 0 or This is the eqution of n ellipse. Cn it e simplified? Let s eliminte the rdils. Rewrite: Squre oth sides: Epnd the squres: Comine the like terms: Divide oth sides nd then squre oth sides gin: Remove the prentheses on the left Comine the like terms: Sine >, then > nd therefore  is positive numer. Let or. Note tht, <.
11 With suh defined, we n rewrite the eqution of n ellipse s of this eqution nd simplifing we get This is the stndrd eqution of n ellipse with enter t the origin nd the foi on the is. How re nd relted to the grph of n ellipse?. After dividing oth sides Notie tht the interepts of the grph re the verties of the ellipse. Using the eqution we n find the interepts mke =0 nd solve for 0 Hene, the verties re V =,0, V =, 0. Similrl, we n find the interepts set = 0 nd solve for. The interepts re B =0,, nd B = 0, Our findings re summrized elow: Stndrd eqution of n ellipse with enter t 0,0 nd foi on the is:, where Center: 0,0 Foi:,0, ,0; Verties:,0,,0; = distverte, enter, = dist enter, B; = dist fous, enter ; >, > Grph:
12 Eqution of n ellipse with the enter t the origin nd the foi on the  is Work similr to the ove leds to the following fts: Stndrd eqution of n ellipse with enter t 0,0 nd foi on the is:, where Center: 0,0 Foi: 0,, 0, Verties: 0,,0, = distverte, enter, = dist enter, B; = dist fous, enter; >, > Grph: Emple Grph + =. Find the oordintes of the enter, verties nd foi.. Write the eqution in the stndrd form divide eh term :. Find the interepts: setting = 0, we get or or 3. Find the interepts: setting = 0, we get or or. Plot the interepts nd drw retngleusing dshed line ontining the interepts with the sides prllel to the  nd  is 5. Drw n ellipse n ovl shpe tht fits within the retngle
13 The enter is t the origin: 0,0 Sine the mjor is is longer nd the verties re on the mjor is, from the grph, we n red tht the verties re:,0, ,0. Sine = dist enter, verte, then =. Also, sine the minor is hs the endpoint t 0,, 0,, we onlude tht = distenter, 0,=. Sine. We get = or = 3, or 3 whih re on the mjor is hene on the is re t 3,0 nd 3,0. Therefore the foi, Emple Find the eqution of the ellipse with fous t 0,3 nd verties t 0, ±6. Sine the verties re 0,6 nd 0,6, the enter is hlfw etween the verties enter is the midpoints of the segment with endpoints 0,6 nd 0,6, hene, the enter O is t 0,0 O = ,. Sine the verties nd foi re on the is the eqution of the ellipse is of the form: 3. sine = distenter, verte = 6 nd = distenter, fous = 3, we hve Hene the eqution of the ellipse with given properties is 7 36 Eqution of n ellipse with enter t h,k If n ellipse with enter t 0,0 is shifted so its enter moves to h,k, its eqution eomes h And h k k, if the foi nd verties re on the line prllel to the is, if the foi nd verties re on the line prllel to the is
14 Emple 3 Grph. Find the oordintes of the enter, foi nd verties. 6 Note tht this is the ellipse 6, shifted, so its enter is t 3,, hene shifted 3 units to the right nd units down.. Grph, using dshed line, n ellipse 6 Center t 0,0; interepts: ; interepts: 6 Note tht = distenter, verte = ; =, = = 6 =. So = 3 nd the foi re 0, 3 nd 0, 3. Shift the enter, verties nd foi 3 units to the right nd units down. Drw the retngle nd the ellipse inside of it 3. Use the grph nd the trnsformtions to otin the oordintes of the ke points for the ellipse 3 6 Center: 3, Verties: 3,, 3,6 Foi: 3, 3 nd 3, 3
15 Emple Write the given eqution in the stndrd form. Determine the oordintes of the enter, verties nd foi = 0. Rewrite the eqution so the terms nd the terms re together; move the numer to the right hnd side = 5. From the  group, ftor out the oeffiient of ; from the group ftor the oeffiient of = Complete the squre in eh group; dd pproprite numers to the right hnd side = Write eh group s perfet squre; dd numers on the right = 6 5. Divide eh term the numer on the right hnd side 6 nd simplif This is the stndrd eqution of the ellipse. Center is t ,. Hene, this ellipse is otined shifting the ellipse 3 two units to the left nd one unit up. The numers in the denomintors re 3 nd nd sine 3 >, the verties of re on the is nd 3 = 3 nd =, or 3 nd rememer tht >. Therefore, the verties of this ellipse re t 3,0. Moreover, sine = = 3 =, =, the foi re t ±, 0. Therefore, for, the verties re 3, nd the foi re ±, or 33, nd,.
16 Emple Find the eqution of the ellipse with enter t 3,, fous t 3,3 nd verte t 3,9. Grph the ellipse.. Plot the given informtion in the oordinte sstem nd use them to determine ke vlues for the ellipse Sine = distenter, fous, then = 5 nd the seond fous is t 3, 7. Sine = distenter, verte, then = 7 nd the seond verte is t 3,5. Also, = = 75 = 95=, so = 6.Verties nd foi re on the line prllel to the is, therefore the stndrd eqution of this ellipse will e of the form h k. 3 3 Using vlues tht we determined, the eqution is or. 9 9 We n use these vlues to drw the retngle tht ontins the ellipse nd the ellipse itself.
17 . Hperol Definition: A hperol is the set of points P on the plne for whih the differene of the distnes from two different, fied points F nd F is onstnt, tht is, distp, F  dist P, F = where is positive onstnt. d d = Note tht the point Q, smmetri to P with respet to the line through F nd F, lso stisfies this eqution sine d = distq, F nd d = distq, F, nd hene, is on the hperol. Moreover the point Q smmetri to P with respet to the line perpendiulr to F F nd pssing through the midpoint of the segment F F is lso on the hperol sine dist Q,F  distq, F = distp, F  dist P, F Therefore, hperol is smmetri with respet to the line F F s well s the line perpendiulr to F F nd pssing through point O. Point O, whih is the midpoint of the segment F F, is lled the enter of the hperol. Here is the grph of this hperol:
18 The points F nd F re lled foi of the hperol. The points V nd V re lled verties of the hperol. The line V V is lled the trnsverse is nd the line perpendiulr to it nd pssing through the enter O is lled the onjugte is of the hperol. Let s note the si properties of hperol: hperol onsists of two prts lled rnhes  verties, foi nd the enter of hperol lie on the sme linetrnsverse is  the enter of hperol is hlfw etween foi  the enter of hperol is hlfw etween verties  hperol is smmetri with respet to the trnsverse nd onjugte es Eqution of hperol with the enter t the origin nd the foi on the  is Suppose tht F =,0, >0. Beuse of the smmetr, F =,0 nd the length of the segment F F is. Sine in n tringle the sum of two sides must lws e greter thn the third side, in the tringle F PF, we must hve d + dist F, F > d, or dist F,F > d d, whih mens tht >, or >. Let P =, e n point on the hperol. Then, using the distne formul, we get 0, F P dist d nd 0, F P dist d. Therefore, sine for hperol, distp, F  dist P, F =, we hve 0 0 or This is the eqution of hperol. Cn this eqution e simplified? Eliminte the solute vlue first: Rewrite: Squre oth sides: Epnd the squres:
19 Comine the like terms: Divide oth sides nd then squre oth sides: Remove the prentheses on the left Comine the like terms: Rewrite: Sine >, then > nd therefore  is positive numer. Let or. With suh defined, we n rewrite the eqution of n hperol s. After dividing oth sides of this eqution nd simplifing we get This is the stndrd eqution of hperol with enter t the origin nd the foi on the is. How re nd relted to the grph of the hperol? Notie tht the interepts of the grph re the verties of the hperol. Using the eqution we n find interepts mke =0 nd solve for 0 Hene, the verties re V =,0, V =, 0. Note tht if we tr to find the interepts set =0 nd solve for, we get the eqution = , whih hs no rel solutions. Hene this hperol hs no interepts. However, the eqution llows us to disover nother propert of hperol, nmel the eistene of smptotes. Note tht we n rewrite the eqution solve for s follows Solving for gives Notie tht, if this eqution is to hve solution, then must e positive or zero onl then the rdil is defined, tht is or, whih mens or . When is lrge, then is smll numer
20 0 nd therefore. As onsequene, when is lrge. This mens tht the lines re smptotes for the hperol. Note tht the smptotes re lines tht pss through the origin nd hve the slopes, Our findings re summrized elow: Stndrd eqution of hperol with enter t 0,0 nd foi on the is:, where Center: 0,0 Foi:,0, ,0; Verties:,0,,0; = distverte, enter, = dist fous, enter ; > Asmptotes: nd Grph: Eqution of hperol with the enter t the origin nd the foi on the  is Work, similr to the ove, leds to the following fts: Stndrd eqution of hperol with enter t 0,0 nd foi on the is:, where Center: 0,0 Foi: 0,, 0, Verties: 0,,0, = distverte, enter, = dist fous, enter; >, Asmptotes: nd Grph:
21 Emple Grph =. Find the oordintes of the enter, verties nd foi. Write the equtions of smptotes.. Write the eqution in the stndrd form divide eh term :. Find the interepts: setting = 0, we get or ; there is no interepts, so nd. Find the interepts: setting = 0, we get or or ; there re interepts, so nd. The interepts re the verties of the hperol.. Plot on the is nd on the is nd drw retngle using dshed line ontining these points with the sides prllel to the  nd  is. Identif verties of the hperol interepts 5. Drw the digonls of this retngle nd etend them to form the lines. These re the smptotes for the hperol. The smptotes divide the plne into prts the grph of the hperol will e in the prts tht ontin the verties interepts 6. Drw eh rnh of the hperol, strting lose to one of the smptotes, going to verte nd ontinuing to the other smptote. The enter is t the origin: 0,0 Sine the verties of hperol re either  or interepts, we see tht the verties re: 0,, 0,no  interepts
22 Sine =, =, nd =, we get =  or = 5, or 5. Therefore the foi, whih re on the trnsverse is hene on the is re t 0, 5 nd 0, 5 The smptotes re pssing through 0,0 nd the verties of the retngle points, nd ,, respetivel, so their equtions re nd Emple Find the eqution of the hperol with verte t 3,0 nd foi t ±5, 0. Sine the foi re 5,0 nd 5,0, then the enter, whih is hlfw etween the foi the enter is the midpoints of the segment with endpoints 5,0 nd 5,0, is t 0,0. Thus, = dist enter, fous = 5. Sine the enter is t 0,0, then the seond verte is t 3, 0 nd = dist enter, verte = Sine the verties nd foi re on the  is the eqution of the hperol is of the form: 3. Sine = 3 nd = 5, we hve Hene, the eqution of the hperol with given properties is 9 6 If hperol with enter t 0,0 is shifted so its enter moves to h,k, its eqution eomes h k, if the foi nd verties re on the line prllel to the is nd k h, if the foi nd verties re on the line prllel to the is Emple 3 Grph. Find the oordintes of the enter, foi nd verties. Find the equtions of the 6 smptotes. Note tht this is the hperol, shifted, so its enter is t 3,, hene shifted 3 units to the right nd 6 units down.
23 . Grph, using dshed line, n hperol 6 Center t 0,0; interepts: ; =, = ; verties : ,0,,0; interepts: none ; = 6, = ; = + = 0, = 0 5 ; foi: 5,0, 5,0 Asmptotes: = ±. Shift the enter, verties nd foi 3 units to the right nd units down. Shift the retngle nd the smptotes. Drw the hperol. 3. Ersing the originl grph gives lerer piture of the grph. We n use the grph nd the knowledge of trnsformtions to otin the oordintes of the ke points for the 3 hperol 6
24 Center: 3, Verties:,, 5, Foi: 3 5, nd 3 5, Asmptotes: + = ±3 or fter simplifing: = 8 nd = +. Emple Write the given eqution in the stndrd form. Determine the oordintes of the enter, verties nd foi. Find the equtions of the smptotes = 0. Rewrite the eqution so the terms nd the terms re together; move the numer to the right hnd side = 3. From the  group, ftor out the oeffiient of ; from the group ftor the oeffiient of = Complete the squre in eh group; dd pproprite numers to the right hnd side = Write eh group s perfet squre; dd numers on the right = Divide eh term the numer on the right hnd side 36 nd simplif This is the stndrd eqution of the hperol. Center is t ,. Hene, this hperol is otined shifting the hperol two units to the left nd one unit 9 down. Beuse the eqution hs interepts nd no  interepts minus is in front of the  9 term!, we onlude tht = nd = 9 nd tht verties nd foi re on the is. Moreover, = + = +9 = 3, so = 3. Therefore, for the hperol, verties re 0,, 0,; foi re 0,  3, 9 0, 3 nd the smptotes re. 3 Shifting the hperol units to the left nd unit down, ields the hperol 9 9 for whih the verties re , ± or , nd , 3 ; the foi re , ± 3  or , 3  nd , 3 ; the equtions of smptotes re or 7 nd
25 Emple Find the eqution of the hperol with enter t 3,, fous t 6, nd verte t ,. Grph the hperol.. Plot the given informtion in the oordinte sstem nd use them to determine ke vlues for the hperol Sine = distenter, fous, then = 3 nd the seond fous is t 0,. Sine = distenter, verte, then = nd the seond verte is t 5,. Also, = = 3  = 9=5, so = 5.Verties nd foi re on the line prllel to the is, therefore the stndrd eqution of this hperol will e of the form h k. 3 3 Using vlues tht we determined, the eqution is or. 5 5 We n use these vlues to drw the retngle, smptotes nd the hperol itself.
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