CONIC SECTIONS. MODULE-IV Co-ordinate Geometry OBJECTIVES. Conic Sections

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1 Conic Sctions 16 MODULE-IV Co-ordint CONIC SECTIONS Whil cutting crrot ou might hv noticd diffrnt shps shown th dgs of th cut. Anlticll ou m cut it in thr diffrnt ws, nml (i) (ii) (iii) Cut is prlll to th s (s Fig.16.1) Cut is slnting ut dos not pss through th s (s Fig.16.) Cut is slnting nd psss through th s (s Fig.16.3) Fig Fig Fig Th diffrnt ws of cutting, giv us slics of diffrnt shps. In th first cs, th slic cut rprsnt circl which w hv studid in prvious lsson. In th scond nd third css th slics cut rprsnt diffrnt gomtricl curvs, which w shll stud in this lsson. OBJECTIVES Aftr studing this lson, ou will l to : rcognis circl, prol, llips nd hprol s sctions of con; rcognis th prol, llips nd hprol s crtin loci; idntif th concpt of ccntricit, dirctri, focus nd vrt of conic sction; idntif th stndrd qutions of prol, llips nd hprol; find th qution of prol, llips nd hprol givn its dirctri nd focus. MATHEMATICS 353

2 MODULE-IV Co-ordint EXPECTED BACKGROUND KNOWLEDGE Bsic knowldg of coordint Vrious forms of qution of stright lin Eqution of circl in vrious forms 16.1 CONIC SECTION Conic Sctions In th introduction w hv noticd th vrious shps of th slic of th crrot. Sinc th crrot is conicl in shp so th sction formd r sctions of con. Th r thrfor clld conic sctions. Mthmticll, conic sction is th locus of point P which movs so tht its distnc from fid point is lws in constnt rtio to its prpndiculr distnc from fid lin. Th fid point is clld th focus nd is usull dnotd S. Th fid stright lin is clld th Dirctri. Th stright lin pssing through th focus nd prpndiculr to th dirctri is clld th is. Th constnt rtio is clld th ccntricit nd is dnotd. Wht hppns whn (i) 1 (ii) 1 (iii) 1 In ths css th conic sction otind r known s llips, prol nd hprol rspctivl. In this lsson w shll stud out llips, prol, nd hprol. 16. ELLIPSE Rcll th cutting of slics of crrot. Whn w cut it oliqul, slnting without ltting th knif pss through th s, wht do w osrv? You might hv com cross such shps whn ou cut oild gg vrticll. Th slic thus otind rprsnts n llips. Lt us dfin th llips mthmticll s follows: An llips is th locus of point which movs in pln such tht its distnc from fid point rs constnt rtio to its distnc from fid lin nd this rtio is lss thn unit STANDARD EQUATION OF AN ELLIPSE Lt S th focus, ZK th dirctri nd P moving point. Drw SK prpndiculr from S on th dirctri. Lt th ccntricit. Divid SK intrnll nd trnll t A nd A' (on KS producd) rpctivl in th rtio : 1, s < MATHEMATICS

3 Conic Sctions SA. AK (1) nd SA. AK () Sinc A nd A' r points such tht thir distncs from th focus rs constnt rtio ( < 1) to thir rspctiv distncs from th dirctri nd so th li on th llips. Ths points r clld vrtics of th llips. MODULE-IV Co-ordint Z 1 B L P Z M ' K' A' S' (-,0) C N S(,0) K A B' ' Fig L' Lt AA' qul to nd C its mid point, i.., CA = CA' = Th point C is clld th cntr of th llips. Adding (1) nd (), w hv SA SA. AK. AK A or AA ( CK CA AC CK ) or. CK or Sutrcting (1) from (), w hv CK (3) SA SA ( AK AK) or ( SC CA) ( CA CS). AA or CS. or CS (4) Lt us choos C s origin, CAX s -is nd CY, lin prpndiculr to CX s -is. Coordints of S r thn (, 0) nd qution of th dirctri is Lt th coordints of th moving point P (, ). Join SP, drw PMZK. B dfinition SP. PM or SP. PM or SN NP.(NK) or ( CN CS) NP.( CK CN) MATHEMATICS 355

4 MODULE-IV Co-ordint or ( ) or (1 ) (1 ) Conic Sctions or 1 (1 ) [On dividing (1 ) ] Putting ( 1 ), w hv th stndrd form of th llips s, 1 Mjor is : Th lin joining th two vrtics A' nd A, i.., A'A is clld th mjor is nd its lngth is. Minor is : Th lin pssing through th cntr prpndiculr to th mjor is, i.., BB' is clld th minor is nd its lngth is. Principl is : Th two s togthr (mjor nd minor) r clld th principl s of th llips. Ltus rctum : Th lngth of th lin sgmnt LL' is clld th ltus rctum nd it is givn Eqution of th dirctri : Eccntricit : is givn 1 Empl 16.1 Find th qution of th llips whos focus is (1, 1), ccntricit = 1 nd th dirctri is 3. Solution : Lt P (h,k) n point on th llips thn th dfinition, its distnc from th focus =. Its distnc from dirctri or SP.PM (M is th foot of th prpndiculr drwn from P to th dirctri). or 1 h k 3 ( h 1) ( k 1) 4 11 or 7( h k ) hk 10h 10k 7 0 Th locus of P is, 7( ) which is th rquird qution of th llips. 356 MATHEMATICS

5 Conic Sctions Empl 16. Find th ccntricit, coordints of th foci nd th lngth of th s of th llips Solution : Th qution of th llips cn writtn in th following form, On compring this qution with tht of th stndrd qution of th llips, w hv 4 nd 3, thn MODULE-IV Co-ordint (i) (ii) coordints of th foci r (1,0) nd ( 1,0) [Th coordint r (, 0)] (iii) Lngth of th mjor s 4 nd lngth of th minor is = 3 3. CHECK YOUR PROGRESS Find th qution of th llips rfrrd to its cntr () whos ltus rctum is 5 nd whos ccntricit is 3 () who s minor is is qul t o th distnc twn t h foci nd whos ltus rctum is 10. (c) whos foci r th points (4,0) nd ( 4,0) nd whos ccntricit is Find th ccntricit of th llips, if its ltus rctum qul to on hlf its minor is PARABOLA Rcll th cutting of slic of crrot. Whn w cut oliqul nd ltting th knif pss through th s, wht do w osrv? Also whn tsmn hits th ll in ir, hv ou vr noticd th pth of th ll? Is thr n proprt common to th dg of th slic of th crrot nd th pth trcd out th ll in th mpl citd ov? Ys, th dg of such slic nd pth of th ll hv th sm shp which is known s prol. Lt us dfin prol mthmticll. "A prol is th locus of point which movs in pln so tht its distnc MATHEMATICS 357

6 MODULE-IV Co-ordint Conic Sctions from fid point in th pln is qul to its distnc from fid lin in th pln." STANDARD EQUATION OF A PARABOLA Lt S th fid point nd ZZ' th dirctri of th prol. Drw SK prpndiculr to ZZ'. Bisct SK t A. Sinc SA = AK, th dfinition of th prol A lis on th prol. A is clld th vrt of th prol. Tk A s origin, AX s th -is nd AY prpndiculr to AX through A s th -is. Z Y M P (, ) K A L S (,0) N X L' P' Z' Fig Lt KS AS AK Th coordints of A nd S r (0,0) nd (,0) rspctivl. Lt P(,) n point on th prol. Drw PN AS producd AN nd Join SP nd drw NP PM ZZ B dfinition of th prol SP = PM or SP = PM or ( ) ( 0) ( ) [ PM NK NA AK ] or ( ) ( ) or 4 which is th stndrd qution of th prol. 358 Not : In this qution of th prol (i) Vrt is (0,0) (ii) Focus is (,0) (iii) Eqution of th is is = 0 (iv) Eqution of th dirctri is + = 0 (v) Ltus rctum = 4 MATHEMATICS

7 Conic Sctions OTHER FORMS OF THE PARABOLA Wht will th qution of th prol whn (i) focus is (,0) nd dirctri is 0 (ii) focus is (0,) nd dirctri is + = 0, (iii) focus is (0, ) nd dirctri is 0? MODULE-IV Co-ordint It cn sil shown tht th qution of th prol with ov conditions tks th following forms: (i) 4 (ii) 4 (iii) 4 Th figurs r givn low for th ov qutions of th prols. z X' A K ' z' = 4 X (i) z ' A K ' = 4 z ' z' (ii) K A ' = 4 z' (iii) Fig Corrsponding rsults of ov forms of prols r s follows: Forms Coordints of vrt (0,0) (0,0) (0,0) (0,0) Coordints of focus (,0) (,0) (0,) (0, ) Coordints of dirctri Coordints of th is lngth of Ltus rctum Empl 16.3 Find th qution of th prol whos focus is th origin nd whos dirctri is th lin 1 0. MATHEMATICS 359

8 MODULE-IV Co-ordint Conic Sctions Solution : Lt S (0,0) th focus nd ZZ' th dirctri whos qution is 1 0 Lt P(, ) n point on th prol. Lt PM prpndiculr to th dirctri (S Fig. 16.5) B dfinition SP PM or SP PM ( 1) or 1 or or Empl 16.4 Find th qution of th prol, whos focus is th point (, 3) nd whos dirctri is th lin = 0. Solution : Givn focus is S(,3); nd th qution of th dirctri is As in th ov mpl, ( ) ( 3) CHECK YOUR PROGRESS Find th qution of th prol whos focus is (, ) nd whos dirctri is 1.. Find th qution of th prol whos focus is (,3) nd whos dirctri is HYPERBOLA Hprol is th locus of point which movs in pln such tht th rtio of its distnc from fid point to its distnc from fid stright lin in th sm pln is grtr thn on. In othr words hprol is th conic in which ccntricit is grtr thn unit. Th fid point is clld focus nd th fid stright lin is clld dirctri. Eqution of Hprol in Stndrd from : 360 MATHEMATICS

9 Conic Sctions M 1 B M P(, ) MODULE-IV Co-ordint S A Z 1 C Z A N S Lt S th focus nd ZM th dirctri. Drw SZ prpndiculr from S on dircti w cn divid SZ oth intrnll nd trnll in th rtio : 1 ( > 1). Lt th points of division A nd A s shown in th ov figur. Lt C th mid point of AA. Now tk CZ s th -is nd th prpndiculr t C s -is. Lt Now AA = SA AZ SA = ( > 1) nd AZ = ( > 1). i.. SA = AZ...(i) i.. SA = AZ...(ii) Adding (i) nd (ii) w gt SA + SA = (AZ + AZ) (CS CA) + (CS + CA) = AA Hnc focus point is (, 0). Sutrcting (i) from (ii) w gt CS =. ( CA = CA) CS = SA SA = ( AZ AZ) i.. AA = ( CZ CA) ( CA CZ) i.. i.. B 1 Fig.16.7 AA = [CZ] ( CA = CA) = (CZ) CZ = Eqution of dirctri is =. Lt P(, ) n point on th hprol, PM nd PN th prpndiculrs from P on MATHEMATICS 361

10 MODULE-IV Co-ordint th dirctri nd -is rspctivl. Thus, SP PM (SP) = (PM) = SP = PM Conic Sctions i.. ( ) ( 0) = i.. = i.. i.. = ( 1) = ( 1) i.. ( 1) = 1 Lt ( 1) = = 1 Which is th qution of hprol in stndrd from. Now lt S th img of S nd ZM th img of ZM w.r.t -is. Tking S s focus nd ZM s dirctri, it cn sn tht th corrsponding qution of hprol is dirctrics. Hnc for vr hprol, thr r two foci nd two 1. W hv ( 1) nd > 1 = If w put = 0 in th qution of hprol w gt = = ± Hprol cuts -is t A(, 0) nd A(, 0). If w put = 0 in th qution of hprol w gt 36 MATHEMATICS

11 Conic Sctions = = 1. i Which dos not ist in th crtsin pln. Hprol dos not intrsct -is. AA =, long th -is is clld trnsvrs is of th hprol nd BB =, long -is is clld conjugt is of th hprol. Notic tht hprol dos not mt its conjugt is. As in cs of llips, hprol hs two foci S(, 0), S(, 0) nd two dirctrics. C is clld th cntr of hprol. Ltus rctum of hprol is lin sgmnt prpndiculr to th trnsvrs is through n of th foci nd whos nd points li on th hprol. As in llips, it cn provd tht th lngth of th ltus rctum of hprol is Hprol is smmtric out oth th s.. Foci of hprol r lws on trnsvrs is. It is th positiv trm whos dnomintor givs th trnsvrs is. For mpl is nd lngth of trnsvrs is is 6 units. Whil long -is of lngth 10 unit. 1 hs trnsvrs is long hs trnsvrs is 5 16 Th hprol whos trnsvrs nd conjugt s r rspctivl th conjugt nd trnsvrs is of givn hprol, is clld th conjugt hprol of th givn MODULE-IV Co-ordint hprol. This qution is of th form 1. In this cs : Trnsvrs is is long -is nd conjugt is is long -is. Lngth of trnsvrs is =. Lngth of conjugt is = S Lngth of ltus rctum =. Equtions of dirctrics. Vrtics (0, ± ) A 1 B 1 B A Foci (0, ± ) Cntr (0, 0) S 1 Eccntricit () =. Fig.16.8 MATHEMATICS 363

12 MODULE-IV Co-ordint RECTANGULAR HYPERBOLA : Conic Sctions If in hprol th lngth of th trnsvrs is is qul to th lngth of th conjugt is, thn th hprol is clld rctngulr hprol. Its qution is or = ( = ) In this cs = or i.. th ccntricit of rctngulr hprol is. Empl 16.5 For th hprol , find th following (i) Eccntricit (ii) Foci (iii) Vrtics (iv) Dirctrics (v) Lngth of trnsvrs is (vi) Lngth of conjugt is (vii) Lngth of ltus rctum (viii) Cntr. Solution : Hr = 16 nd = 9, = 4 nd = 3. (i) Eccntricit () = (ii) Foci = 4.5 (, 0),0 ( 5,0) 4 (iii) Vrtics = (±, 0) = (± 4, 0) (iv) Dirctrics = ± = (v) Lngth of trnsvrs is = = 4 = 8. (vi) Lngth of conjugt is = = 3 = 6 (vii) Lngth of ltus rctum = (viii) Cntr = (0, 0) Empl 16.6 Find th qution of hprol with vrtics (±, 0) nd foci (± 3, 0) Solution : Hr = nd = 3. = 3/. W know tht = ( 1) = Eqution of hprol is MATHEMATICS

13 Conic Sctions Empl 16.7 For hprol 1, find th following : 9 7 (i) Eccntricit (ii) Cntr (iii) Foci (iv) Vrtics (v) Dirctrics (vi) Lngth of trnsvrs is (vii) Lngth of conjugt is (viii) Ltus rctum. Solution : Hr = 9 nd = 7 = 3 nd = 3 3. (i) (ii) Cntr = (0, 0) 9 (iii) Foci = (0, ) (0, 3.) (0, 6). (iv) Vrtics = (0, ± ) = (0, ± 3). MODULE-IV Co-ordint (v) Dirctrics, 3. (vi) Lngth of trnsvrs is = = 3 = 6 (vii) Lngth of conjugt is = = 3 3 = 6 3 (viii) Lngth of ltus rctum = CHECK YOUR PROGRESS (i) Trnsvrs is of th hprol is long (ii) Eccntricit of th hprol 1 is (iii) Eccntricit of rctngulr hprol is... (iv) Lngth of ltus rctum of hprol (v) Foci of th hprol 1 is t... (vi) Eqution of dirctrics of hprol (vii) Vrtics of th hprol 1. For th hprol (i) Eccntricit () =... (ii) Cntr = is... 1 is... r t..., complt th following. MATHEMATICS 365

14 MODULE-IV Co-ordint (iii) Foci =... (iv) Vrtics =... (v) Equtions of dirctrics, =... (vi) Lngth of ltus rctum =... (vii) Lngth of trnsvrs is =... (viii) Lngth of conjugt is =... (i) Trnsvrs is is long... () Conjugt is is long... Conic Sctions 1 A C % + LET US SUM UP Conic Sction "A conic sction is th locus of point P which movs so tht its distnc from fid point is lws in constnt rtio to its prpndiculr distnc from fid stright lin". (i) Focus : Th fid point is clld th focus. (ii) Dirctri : Th fid stright lin is clld th dirctri. (iii) Ais : Th stright lin pssing through th focus nd ppndiculr to th dirctri is clld th is. (iv) Eccntricit : Th constnt rtio is clld th ccntricit. (v) Ltus Rctum : Th doul ordint pssing through th focus nd prlll to th dirctri is known s ltus rctum. (In Fig.16.5 LSL' is th ltus rctum). Stndrd Eqution of th Ellips is : 1 (i) Mjor is = (ii) Minor is = (iii) Eqution of dirctri is (iv) Foci : (,0) (v) Eccntricit, i.., is givn 1 vi Ltus Rotm Stndrd Eqution of th Prol is : 4 (i) Vrt is (0,0) (ii) Focus is (,0) (iii) Ais of th prol is = 0 (iv) Dirctri of th prol is 0 (v) Ltus rctum = MATHEMATICS

15 Conic Sctions OTHER FORMS OF THE PARABOLA ARE (i) 4 (concv to th lft). MODULE-IV Co-ordint (ii) 4 (concv upwrds). (iii) 4 (concv downwrds). Eqution of hprol hving trnsvrs is long -is nd conjugt is long - 1. is is For this hprol (i) =. (ii) Cntr = (0, 0) (iii) Foci = (±, 0) (iv) Vtrics = (±, 0)(v) Lngth of ltus rctum = (vi) Lngth of trnsvrs is = (vii) Lngth of conjugt is = (viii) Equtions of dirctris r givn. Equtions of hprol hving trnsvrs is long -is nd conjugt is long - is is 1. For this hprol : (i) Vrtics = (0, ± )(ii) Cntr = (0, 0) (iii) Foci = (0, ± ) (iv) = (v) Lngth of ltus rctum =. (vi) Lngth of trnsvrs is =. (vii) Lngth of conjugt is =. (viii) Equtions of dirctris r givn = ±. MATHEMATICS 367

16 MODULE-IV Co-ordint SUPPORTIVE WEB SITES Conic Sctions TERMINAL EXERCISE 1. Find th qution of th llips in ch of th following css, whn () focus is (0, 1), dirctri is + = 0 nd = 1. () focus is ( 1,, 1), dirctri is + 3 = 0 nd = 1.. Find th coordints of th foci nd th ccntricit of ch of th following llipss: () () Find th qution of th prol whos focus is ( 8, ) nd dirctri is Find th qution of th hprol whos foci r (± 5, 0) nd th lngth of th trnsvrs is is 8 units. 5. Find th qution of th hprol with vrtics t (0, ± 6) nd = Find th ccntricit, lngth of trnsvrs is, lngth of conjugt is, vrtics, foci, qutions of dirctrics, nd lngth of ltus rctum of th hprol (i) (ii) Find th qution of th hprol with foci (0, 10), nd pssing through th point (, 3). 8. Find th qution of th hprol with foci (± 4, 0) nd lngth of ltus rctum MATHEMATICS

17 Conic Sctions ANSWERS MODULE-IV Co-ordint CHECK YOUR PROGRESS () () 100. (c) CHECK YOUR PROGRESS ( ) CHECK YOUR PROGRESS (i) -is (ii) (iii) (iv) (v) (±, 0) (vi) (vii) (±, 0) 5 3. (i) (ii) (0, 0) (iii) (0, ± ) (iv) (0, ± ) (v) (vi) (vii) (viii) (i) -is () -is TERMINAL EXERCISE 1. () () () ,0 ; , 1 ; 5 MATHEMATICS 369 ()

18 MODULE-IV Co-ordint Conic Sctions 6. (i) Eccntricit = 34 3, lngth of trnsvrs is = 6, lngth of conjugt is = 10, vrtics (± 3, 0), Foci ( 34,0), qutions of dirctrics = , ltus rctum 34 (ii) Eccntricit = 5, lngth of trnsvrs is = 1, lngth of conjugt 1 is = 1, vrtics 0, 4, Foci 5 0, 4, qutions of dirctrics, 1, 4 5 ltus rctrum = MATHEMATICS

SOLVED EXAMPLES. be the foci of an ellipse with eccentricity e. For any point P on the ellipse, prove that. tan

SOLVED EXAMPLES. be the foci of an ellipse with eccentricity e. For any point P on the ellipse, prove that. tan LOCUS 58 SOLVED EXAMPLES Empl Lt F n F th foci of n llips with ccntricit. For n point P on th llips, prov tht tn PF F tn PF F Assum th llips to, n lt P th point (, sin ). P(, sin ) F F F = (-, 0) F = (,