L F'(-c, 0) F(c, 0) M' D' x + d = 0 x - d = 0

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1 Hperol Reference: Advnced Level Pure Mheics S.L. Green (4 h Ediion) Chper VII p.9 - p.0 Edied Mr. Frncis Hung Ls upded: Deceer 7, 0 M D N' N P(, ) L F'(-c, 0) F(c, 0) A'(-,0) B' B A(,0) L' M' D' + d 0 - d 0. Definiion F is he focus, -is is he principl is. DD is he direcri. PF A vrile poin P(, ) oves so h e ( consn clled he eccenrici > ). PN where N is he foo of perpendiculr fro P ono DD. In priculr, when P oves o A (eween B nd F) on he principl is, AF eab Produce FB furher o poin A such h A F ea B; hen A is on he curve. (Noe h A nd A re on he opposie sides of DD.) Bisec AA (0, 0) (clled he cenre) nd le AA, hen A (, 0), A (, 0). Le DD e d. c e( d) () c + e( + d) () [() + ()] : c e (3) [() ()] : de (4) Now le P (, ), PF epn ( c) + e( d ) ( e) + e ( ) (3) nd (4) e e + e + e e + (e ) (e ) (given h e > e ( e ) > 0) Le e > 0, hen he equion of he locus ecoes nd (e) c + c (6) Replce nd in (5) respecivel, here is no chnge. The locus is sericl ou he -is nd -is (7) (5) hp:// Pge

2 Hperol Noes Mr. Frncis Hung. For n poin P(, ) on he hperol, P (, ) is he ige of P refleced long -is. B (7), he locus is seric ou -is P (, ) lso lies on he curve. Le F ( c, 0) e he ige of F(c, 0) refleced long -is. MM : d is he ige of he direcri DD refleced long -is. F' P' Then e (ll dshes re iges refleced long -is) P' N' P is he ige of P, which lies on he curve. F' P e for n poin P on he ellipse. PN' There re wo foci F(, 0), F (, 0) nd wo direcrices DD : d 0 nd EE : + d The equion gives + Fro which or. A(, 0) nd A (, 0) re he verices of he hperol. I follows h here is no poin eween o. The curve hs wo disinc rnches. 4. The lus recu is he line segen LL hrough F(c, 0) nd is perpendiculr o he principl is. To find he coordines of L nd L : e B susiuion, ± e L ( e, e ), L ( e, e ) LL e LL (8) e hp:// Pge

3 Hperol Noes Mr. Frncis Hung 5. Geoericl proper Le P, Q e wo poins on he se rnch of hperol. N is he foo of perpendiculr drwn fro P ono he line DD. K is he foo of perpendiculr drwn fro Q ono he line DD. The chord PQ is produced o R on DD. Join RF, he line joining P, F is produced o ee DD S. Le QFR α, RFS β, QRF θ hen PF epn, QF eqk definiion PF PN (9) QF QK ΔPNR ~ ΔQKR (equingulr) M D N K R Q F S + d 0 - d 0 M' D' P PF PN PR PF PR (0) QF QK QR QF QR PF sin θ Appl sine lw on ΔPRF: () PR sin β QF sin θ ΔFQR, () QR sin α () () (0) α β RF is he eerior ngle isecor of PFQ. Le P, Q e wo poins on he differen rnches of hperol. N is he foo of perpendiculr drwn fro P ono he line DD. K is he foo of perpendiculr drwn fro Q ono he line DD. The chord PQ cus DD R. Join RF. Q Le QFR α, PFR β, PRF θ hen PF epn, QF eqk definiion PF PN (3) QF QK ΔPNR ~ ΔQKR (equingulr) PF PN PR PF PR (4) QF QK QR QF QR PF sin θ Sine lw on ΔPRF: (5) PR sin β QF sin θ ΔFQR, (6) QR sin α (5) (6) (4) α β RF is he inerior ngle isecor of PFQ. M N R K θ β α + d 0 - d 0 M' D B D' F P hp:// Pge 3

4 Hperol Noes Mr. Frncis Hung 6. Vriion (0, 0) is he cenre of he curve. The curve is hperol which opens upwrd nd downwrd. The foci re F(0, c), F (0, c). c e (3) The direcri is DD whose equion is d nd EE whose equion is d, F(0,c) where d (4) e I is cler h + c (6) D' d D E' -d E F'(0,-c) If he cenre of hperol is rnsled o V(h, k), hen is new equion is ( h) ( k) (7) Eple Consider he equion I cn e rnsfored ino sndrd equion copleing he squre: ( + ) 3( 4) 4 0 ( + + ) 3( 4 + 4) ( ) ( + ) 6 ( ) ( + ) 3 I is hperol which opens upwrd nd downwrd wih cenre (, ),, 3. c c 5 c e 5, d e Preric equions secθ (8) re he preric equions of, where θ is preer. n θ Siilrl n θ (9) re he preric equions of. secθ hp:// Pge 4

5 Hperol Noes Mr. Frncis Hung 8. Equion of chords using preers. Le A( sec α, n α), B( sec β, n β) e poins of he hperol. Then he equion of AB is nβ secβ nβ n α nβ secβ secα secβ ( ) sin α sinβ cosα cosβ cosαcosβ ( ) cosαcosβ cosα cosβ ( sin αcosβ cosαsinβ ) ( cosβ cosα) ( α β) nβ secβ nβ sin secβ sin α+β sin α β α β α β nβ sin cos α+β α β secβ sin sin α β nβ cos α+β secβ sin α + β nβ α + β α β secβ α β sin sin cos cos α β α + β α β α + β cos sin secβcos nβsin α β α + β sec β cos sinβsin β β sec β cos β α + sinβsin α + α + β α + β secβ cosβcos cos α β α + β α + β The equion of chord is cos sin cos (0) 9. Equion of ngen θ. As β α θ, equion of chord AB ecoes ngen P wih preer θ. The equion of ngen θ is given sinθ cosθ sec θ n θ () If ( 0, 0 ) lies on he hperol, hen 0 sec θ, 0 n θ. 0 0 () hp:// Pge 5

6 Hperol Noes Mr. Frncis Hung 0. Suppose he ngen P( sec θ, n θ) cus he direcri T, hen PT isecs FPF nd PFT 90. M D Equion of ngen P is sec θ n θ secθ To find T: le, hen co θ e e Q P( sec θ, n θ) T, cscθ( ecosθ) e e α β e cscθ( ecosθ) cscθ( ecosθ) F'(-c, 0) T TF e e ( e ) G F(c, 0) n θ sin θ PF secθ e ( ecos θ) cscθ( ecosθ) sin θ TF PF ( e ) ( ecos θ) + d 0 - d 0 c e M' D' ( e ) ( e ) ( e ) TF PF Le F PT α, TPF β, PT is produced o cu -is G, PGF φ. To find G, pu 0 in sec θ n θ G ( cos θ, 0). GF cos θ + e (cos θ + e), FG e cos θ (e cos θ) PF ( secθ + e) + ( n θ) sec θ + esecθ + e + ( e ) n θ ( sec θ n θ) + esecθ + e + e ( + n θ) + esecθ + e + e sec θ ( + esecθ) PF ( secθ e) + ( n θ) ( + e sec θ) (e sec θ ) (Q e >, sec θ >, PF > 0) FG sinβ Appl sine lw on ΔPGF, PF sin φ ( e cosθ) sinβ sinβ cos θ (3) ( esecθ ) sin φ sin φ F' G sin α Appl sine lw on ΔPGF, PF' sin φ ( cosθ + e) sin α sin α cos θ (4) ( + esecθ) sin φ sin φ Copre () nd () α β PT isecs FPF.. Le P( sec θ, n θ) e poin on, hen he difference of he focl disnces is consn. PF ( secθ e) + ( n θ) (e sec θ ) PF ( secθ + e) + ( n θ) (e sec θ + ) PF PF (e sec θ + ) (e sec θ ) hp:// Pge 6

7 Hperol Noes Mr. Frncis Hung. Condiion for ngenc. Le l + + n 0 e ngen, hen i is proporionl o sec θ n θ. secθ n θ i.e. l n sec θ l n, n θ n sec θ n l θ n n (l) () n () + n (l) (5) 3. Equion of ngens, given slope. Le + k e ngen, hen i is proporionl o sec θ n θ. secθ n θ i.e. k sec θ, n θ k k sec θ n θ k k () k k ± Given slope, equion of ngens re ± (6) Noe h when 0, i.e. or, wo ngens cn e found. When < <, no ngens cn e drwn. 4. The locus of he fee of perpendiculrs fro he foci o ngen is he uilir circle wih cenre (0, 0) nd rdius. Equion of ngen: ± (7) ngen The perpendiculr line hrough F is + e (8) The perpendiculr line hrough F is + e (9) (7) + (8) : ( + )( + ) e + + c + + ( + ) Siilrl, eliining fro (7) nd (9) gives +. R P( sec θ, n θ) F'(-c, 0) F(c, 0) Q hp:// Pge 7

8 Hperol Noes Mr. Frncis Hung 5. The produc of perpendiculrs fro he foci is. Using he forul (6) in secion 3, equion of ngen is ± e ± e ± + + B disnce forul, QF, RF ( ) ( ) e + c, e c +, c + 6. The spoes. Clerl he hperol hs no vericl spoes. ± ± (30) Le + k e n olique spoe. li li ± ± (Siilrl, li ± ) k li ( ) li li 0 + k li ( + ) li + li The wo spoes re (3) nd (3) B(0,) F'(-c,0) A'(-,0) A(,0) F(c,0) B'(0,-) AA is clled he rnsverse is (). BB is clled he conjuge is (), i hs no rel inersecion wih he curve. hp:// Pge 8

9 Hperol Noes Mr. Frncis Hung 7. Recngulr hperol If, he spoes re he lines + 0 nd 0, which re righ ngles. The hperol is hen sid o e recngulr or equilerl. To find he eccenrici, + c e e e The equion e (33) or (34) F(0,c) P(,) A(0,) F'(-c,0) F(c,0) A'(-,0) N A(, 0) A'(0,-) F'(0,-c) Le P(, ) e poin on (33) Le N e he foo of perpendiculr fro P on he line + 0. N + PN N cos 45 + PN sin 45 PN N PN sin 45 N cos 45 N + PN PN N (33): ( ) ( ) N PN If we ke he spoe N s he new X-is, nd he oher spoe ( 0) s he new Y-is, hen he k new equion of he recngulr hperol, referred o is spoes, is in he for: XY k, where k. Wih his sse, he spoes re given X 0 or Y hp:// Pge 9

10 Hperol Noes Mr. Frncis Hung 8. Preric equions k k (35) re he preric equions of k. 9. Chord, ngen nd norl. Le A k k,, B k k, e wo poins on he recngulr hperol k. k k k ( ) k Equion of chord AB is given : k k k k k + k 0 ( ) The equion of chord is: + k( + ) 0 (36) As B A, he equion of ngen A is + k 0 (37) k Equion of norl is given : k which is equivlen o 3 k 4 + k 0 (38) 0. Eple Find he preer of he poin in which he norl ee he curve gin. Le he required preer e. Copre he norl 3 k 4 + k 0 (38) nd chord: + k( + ) 0 (36) Since he re idenicl, 3 4 k + k k( + ) 3 Eple 3 Find he locus of id poins of chords k which re prllel o he dieer. Le he chord e + k( + ) 0 (36) I is prllel o. k The id-poin is ( ) (40) (39): k + (39), + The locus is. I is clled he conjuge dieer. Eple 4 If ( 0, 0 ) lie on k, hen 0 k, 0 k. The equion of ngen is : + k 0 k + (k) k 0 ( ) k k + (40) hp:// Pge 0

11 Hperol Noes Mr. Frncis Hung Eple 5 If PP is dieer of recngulr hperol nd he norl P ee he curve gin Q, show h PQ suends righ ngle P. PP psses hrough (0, 0), PQ norl P. Tr o show h PP Q 90 k 6 4 P() Le he preer of P e. Then since PP psses hrough, preer of P Q - B eple, preer of Q is P' 3 k PP QP k + k 3 k ( ) PP QP. There re os 4 norls pss hrough given poin (p, q). k The norl k, is 3 k k I psses hrough (p, q): 3 p q k 4 + k 0 6 k 4 p 3 + q k 0 4 I is polnoil equion of degree 4, which hs, in generl os 4 (rel) roos in (,, 3, 4 ) If,, 3, 4 re rel, 3 lies on one rnch nd -6 one on he oher, (s),, 3 is posiive nd 4 is negive or,, 3 is negive nd he 4 is posiive (p,q) hp:// Pge

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