CHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 1 CHNG I - I CNG V VCT A: TOM TAT LY THUYET - Vect la oan thang co dnh hng Ky hieu :AB; CD hoaca ; b- Vect khong la vect co iem au trung iem cuoi : Ky hieu0 - Hai vect cung phng la hai vect co gia song song hoac trung nhau - Hai vect cung phng th hoac cung hnghoac ngc hng- Hai vect bang nhau neu chung cung hng va cung o dai TNG V HIU HAI VECT -nh ngha: ChoAB a = ;BC b = . Khi oAC a b = + - Tnh chat : * Giao hoan :a b +=b a +* Ket hp ( a b + ) +c=( a b ++c ) * Tn h chat vect khonga +0=a- Quy tac 3 iem : Cho A, B ,C tuy y, ta co :AB+ BC = AC- Quy tac hnh bnh hanh . Neu ABCD la hnh bnh hanhth AB+ AD = AC- Quy tac ve hieu vec t : Cho O , B ,C tuy y ta co :CB OC OB = TCH CA VECT VI MT S - Cho keR , ka la 1 vect c xac nh: * Neuk > 0 thkacung hng via; k < 0 th k angc hng via* o dai vect kabang k . a- Tnh chat : a) k(ma ) = (km)a b) (k + m)a= k a+ ma c) k( a+b)= ka+ kb d)ka=0 k = 0 hoaca= 0 - b cung phnga ( a=0) khi va ch khi co so k thoa b =ka- ieu kien can va u eA , B , Cthang hang la co so k sao choAB =k AC- Chob khong cungphnga , xluon c bieu dienx = ma+ nb ( m, n duy nhat ) www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 2 I - CC BI TP LIN QUAN N VCT 1)Rt gn cc biu thc sau: a)OM ON + AD + MD + EK EP MD AB MN CB PQ CA NM + + + +2)Chng minh rnga)AB +CD=AD+CB b)AC +BD=AD+BC c)AB +CD+EA=ED+CB d)AD+BE+CF =AE +BF +CD =AE +BD+CE e)AB +CD+EF +GA=CB+ED+GF3)Chohnh bnh hnh ABCD tm O. CMR :AO BOCO DO O + + + = , Vi I bt k4 IA IB IC ID IO + + + =4)Cho tam gic C a im MN v P n ttrung imC C CMR: MN BP = ;MA PN = . 5)Cho t giac ABCD, goi M, N, P, Q lan lt la trung iem AB, BC, CD, DA.Chng minh:; MN QP NP MQ = =6)Cho tam giac ABC co trc tam H va O tam la ng tron ngoai tiep . Goi B la iem oi xng B qua O . Chng minh:C B AH ' = . 7)ChohnhbnhhanhABCD.DngBC PQ DC NP DA MN BA AM = = = = , , , . Chng minh O AQ =8)Cho 4 im btMNPQChng minh cc ng thc sau: a.PQ NP MN MQ + + = ;c)NP MN QP MQ + = + ; b.MN PQ MQ PN + = + ; 9)Cho ng gic C Chng minh rng: a.0 AD BA BC ED EC + + = ; b.AD BC EC BD AE + =10)Cho 6 im M N P Q R S Chng minh: a)PN MQ PQ MN + = + .b) RQ NP MS RS NQ MP + + = + + . 11)Cho 7 iem A ; B ; C ; D ; E ; F ; G . Chng minh rang : a.AB +CD +EA =CB +ED b.AD +BE +CF =AE+BF +CD c.AB +CD + EF+GA =CB +ED + GFd.AB -AF +CD -CB +EF-ED =0 12) Cho hnh bnh hnh C c tm O CMR:0 OA OB OC OD + + + = . thc trung im Cho 2 imv13) Cho Mtrung imCMR vi imbt:2 IA IB IM + = . 14) Vi N sao cho2 NA NB = CMR vibt:2 3 IA IB IN + =15) Vi P sao cho3 PA PB = CMR vibt:3 2 IA IB IP = 16) thc trng tm Cho tam gic C c trng tm : CMR:0 GA GB GC + + =Vibt:3 IA IB IC IG + + = . www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 3 M thuc onv M14GA . CMR2 0 MA MB MC + + =17) thc hnh bnh hnh Cho hnh bnh hnh C tm O CMR: a)0 OA OB OC OD + + + = ; b vibt:4 IA IB IC ID IO + + + = . 18) Gi G l trng tm tam gic C chng minh rng : a)0 GA GB GC + + =b) ( )13AG AB AC = +19) Gi n t trng tm ca tam gic ABC v C aChng minh rng :AA' ' ' 3 ' BB CC GG + + =b)Gi M,N,P l cc im tho: 1 1 1, ,3 3 3MA MB NB NC PC PA = = =Chng minh rng cc tam gic ABCv tam gic MNP c cng trng tm 20) Cho hnh bnh hnh C v mt im M ty Chng minh rng : MA MC MB MD + = +21) Cho hai hnh bnh hnh ABCD v ABEF. Dng cc vect EHv FG bng Chng minh rng CDGH l hnh bnh hnh22) Cho tam gic ABC ni tip trong ng trn O trc tm ca tam gica)Gi D l im i xng ca A qua tm O Chng minh rng C b)Gi K l trung im ca AH v I l trung im ca Cchng minh OK = IH 23) Cho hnh bnh hnh C iv F n ttrung im ca hai cnhv C ng cho B n t ct F v C ti M v N chng minh rng : DM = MN = NB 24) Gi G l trng tm ca tam gic ABC. Dng AD = GC v DE = GB Chng minh rng025) T imnm ngoi ng trn O ta 2 tip tuynv C vi O i giao im ca O v CTrn ng trung trc ca ony 1 im MT M tip tuyn M vi O Chng minh rng : |MA| = | MF | 26) Cho tam gic C n ngoi ca tam gic ta v cc hnh bnh hnh J CPQ CRS Chng minh rng :0 RJ IQ PS + + =27) Cho tam gic C c trung tuyn MTrn cnh C y hai imv F sao cho FFC i Ngiao im ca M v Tnh tngAF AE AN MN + + +28) Cho hnh bnh hnh CTrn ng cho C y im OQua O cc ng thng song song vi cc cnh ca hnh bnh hnh ctv C ti M v N ctv C tiv F Chng minh rng : a)OA OC OB OD + = +b)BD ME FN = +29) Cho tam gic u ni tip ng trn tm O ay xc nh cc im M N P sao cho: OM=OA +OB;ON=OB+OC; OP=OC+OA b)Chng minh rng OA +OB+OC=030) Cho tam gic ABC. Gi im i xng viqua; im i xng vi C qua;C im i xng vi A qua C . Chng minh rng vi mt im O bt k ta c : ' ' ' OA OB OC OA OB OC + + = + +31) Cho n im trn mt phng n An k hiu chng l A1, A2 n. Bn Bnh k hiu chng l B1, B2 n.www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 4 32) Chng minh rng :A1B1+ A2B2+...+ AnBn= 0 33) Cho ngu giac eu ABCDE tam O Chng minh: O OE OD OC OB OA = + + + +34) Cho luc giac eu ABCDEF co tam la O . CMR : a)OA+OB+OC+OD+OE +OF=0b)OA+OC+OE=0c)AB+AO+AF =AD d)MA+MC+ME =MB+MD+MF ( M tuy y ). 35) Cho tam giac ABC ; ve ben ngoai cac hnh bnh hanh ABIF ; BCPQ ; CARS Chng minh rang : RF +IQ +PS =036) chotgicCiJnttrungimCvitrungimJCMR: 0 EA EB EC ED + + + = . 37) Cho tam gic C vi M N Ptrung imC C CMR: a) 0 AN BP CM + + = ;b) AN AM AP = + ; c)0 AM BN CP + + = . 38) Cho hnh thang Cy n C y nhgi trung imCMR: EA EB EC ED DA BC + + + = + . 39) Cho 6 im A, B, C, D, E, F. CMR : (bng nhiu cch khc nhau) a)AB CD AD CB + = + b)AB CD AC DB = + c)AD BE CF AE BF CD + + = + +40) Cho tam gic ABC vi M, N, P l trung im cc cnh AB, BC, CA. Chng minh rng : a)AN BP CM O + + = b)AN AM AP = + c)AM BN CP O + + =41) Cho hai im A, B. Cho M l trung im A, B. Chng minh rng vi im I bt k ta c : 2 IA IB IM + = . 42) Vi im N sao cho2 NA NB = . CMR vi I bt k :2 3 IA IB IN + =43) Vi im P sao cho3 PA PB = . CMR vi I bt ki :3 2 IA IB IP = .Tng qut tnh cht trn. 44) Cho tam gic ABC v G l trng tm ca tam gic.Chng minh rngAG BG CG O + + = . Vi I bt k ta c :3 IA IB IC IG + + = . MthuconAGv 14MG GA = .CMR:2MA MB MC O + + = .ViIbki 2 4 IA IB IC IM + + = . 45) Cho t gic ABCD. Gi M, N ca AB v CD . CMR : a)2 AD BC MN + =b)2 AC BD MN + =c) Tm v tr im I sao cho IA IB IC ID O + + + =d) Vi M bt k, CMR : 4 MA MB MC MD MI + + + =46) (Khi nim trng tm ca h n im v tm t c ca h n im) Cho n im 1 2, ,...,nAA A . GiGlimthomn 1 2...nGA GA GA O + + + = .CMRvibkiM: 1 2...nMA MA MA nMG + + + = . Gi I l im tho mn 1 1 2 2...n nn IA n GA n GA O + + + = . CMRvi M bt k :

1 1 2 2 1... ( .. )n n nn MA n MA n MA n n MG + + + = + +47) Cho lc gic u ABCDEF. CMR hai tam gic ACE v BDF cng trng tm. www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 5 48) Cho lc gic ABCDEF. Gi M, N, P, Q, R, S ln lt l trung im ca AB, CD, EF, BC, DE, FA. CMR hai tam gic MNP v QRS cng trng tm. 49) Cho hai tam gic ABC v ABC l cc im thuc BC, CA, AB sao cho : ' ' ' ' ' ', , AB kAC BC kB A CA kCB = = =v1 k = . CMR hai tam gic ABC v ABC cng trng tm. 50) Cho t gic li ABCD. Gi M, N , P, Q l trung im AB, BC, CD, DA. CMR hai tam gic ANP v CMQ cng trng tm. (Mt s ng thc v trc tm, trng tm, tm ng trn ngoi tip, tm ng trn ni tip) 51) Cho tam gic ABC, G, H, O, I l trng tm, trc tm, tm ng trn ngoi tip v tm ng trn ni tip. a)3OG OA OB OC = + + b)OH OA OB OC = + + c)2HO HA HB HC = + +d)aIA bIB cIC O + + = e)A tan Tan HA TanBHB CHC O + + =f) Gi M l im bt k nm trong tam gic ABC. CMR : BCM ACM ABMS IA S IB S IC O + + =(M nm ngoi th khng cn ng). 52) (Nhn mnh bi ton v m rng ra nhiu trng hp). Cho tam gic ABC. Gi M l trung im AB v N l mt im trn cnh AC sao cho NC = 2NA. Gi K l trung im MN. a) CMR : 1 14 6AK AB AC = + .b) D l trung im BC. CMR :1 14 3KD AB AC = +53) Cho tam gic ABC aXc nh im I sao cho :2 0 IA IB + =bXc nh im K sao cho :2 KA KB CB + =Cho tam gic ABC a)Tm im M tho mn :0 AM MB MC + = b)Tm im N tho mn :BN AN NC BD = + +c)Tm im K tho mn :0 BK BA KA CK + + + =d)Tm im M tho mn :2 0 MA MB MC + =e)Tm im N tho mn :2 0 NA NB NC + + =f)Tm im P tho mn :2 0 PA PB PC + =54) Cho hnh bnh hnhABCD. Tm im M tho mn: 4AM AB AC AD = + +55) Cho lc gic ABCDEF .Tm im O tho mn : OF 0 OA OB OC OD OE + + + + + =56) ChoABC A . Tm M sao cho a/2 3 0 MA MB MC + + =b/2 4 0 MA MB MC + = 57) Cho t gic C Tm M sao cho a/2 2 0 MA MB MC MD + + =b/2 5 2 0 MA MB MC MD + + =58) Cho tam gic ABC aXc nh cc im D,E tho mn:4 0 ; 2 0 DA DB EA EC = + =b)Tm qu tch im M tho mn: 4 2 MA MB MA MC = +59) Cho hai im phn bit A,B a)Hy xc nh cc im P,Q,R tho: 2 3 0; 2 0; 3 0 PA PB QA QB RA RB + = + = + =www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 6 60) Cho tam gic ABC v M, N ln lt l trung im AB, AC.Gi P, Q l trung im MN v BC. CMR : A, P , Q thng hng.Gi E, Ftho mn : 13ME MN = , 13BF BC = . CMR : A, E, F thng hng. 61) Cho tam gic ABC, E l trung im AB v F thuc tho mn AF = 2FC. Gi M l trung im BC v I l im tho mn 4EI = 3FI. CMR : A, M, I thng hng. Ly N thucBC sao cho BN = 2 NC v J thuc EF sao cho 2EJ = 3JF. CMR A, J, N thng hng. Ly im K l trung im EF. Tm P thuc BC sao cho A, K, P thng hng. 62) Cho tam gic ABC v M, N, P l cc im tho mn :3 MB MC O = ,3 AN NC = ,PB PA O + = . CMR : M, N, P thng hng. (1 1 1,2 2 4MP CB CAMN CB CA = + = + ). 63) Cho tam gic ABC v L, M, N tho mn2 , LB LC =12MC MA= ,NB NA O + = . CM : L, M, N thng hng. 64) Cho tam gic ABC vi G l trng tm. I, J tho mn :2 3 IA IC O + = ,2 5 3 JA JB JC O + + = .65) CMR : M, N, J thng hng vi M, N l trung im AB v BC. 66) CMR J l trung im BI. 67) Gi E l im thuc AB v tho mnAE kAB = . Xc nh k C, E, J thng hng. 68) Cho tam gic ABC. I, J tho mn :2 ,3 2 = IA IB JA JC O = + . CMR : ng thng IJ i qua G. II H TRC TA TOM TAT LY THUYET: - Truc la ng thang tren o xac nh iem O va 1 vectico o dai bang 1.Ky hieu truc (O;i ) hoac xOx - A,B nam tren truc (O;i ) thAB = AB i . Khi oAB goi la o dai ai so cuaAB- He truc toa o vuong goc gom 2 truc Ox Oy. Ky hieu Oxy hoac (O;i ; j ) - oi vi he truc (O;i ; j ), neua =x i+y jth (x;y) la toa o cuaa . Ky hieua= (x;y) - Choa= (x;y) ; b = (x;y) ta coab = (x x;y y) k a =(kx ; ky) ; k e R b cung phnga ( a=0) khi va ch khi co so k thoax=kx vay= ky - Cho M(xM ; yM) va N(xN ; yN) ta coP la trung iem MN thxp = 2M Nx x +va yP = 2M Ny y + MN= (xM xN ; yM yN) - Neu G la trong tam tam giac ABC th xG = 3A B Cx x x + +va yG = 2A B Cy y y + + www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 7 BI TP 69) Choa= (1;3), b= (2; 5), c= (4;1) aTm tavectu= 2 ab+ 3cbTm tavectxsao chox+a=b ccTm cc sv h sao choc= k a+ hb70) Cho(2; 3); (5;1); ( 3; 2) a b c = = = .a/ Tm taca vect2 3 4 u a b c = + b/ Tm tavectxsao cho2 x a b c + =c/ Tm cc s h vsao cho c ha kb = +71) Cho cc vecta = (3;1) ,b = (2;1)c = (4;1) 72) Tm cc s xy sao cho x a + y. b+ 7 c=0Chou= 2 i 3 jvv= k i+ 4 jTm cc gi tr ca hai vectu vvcng phng 73) Cho cc vecta= ( 1;4), b = (2; 3), c = (1;6) Phn tchctheoa vb74) Cho 3 vect a= (m;m) ,b = (m 4;1) ,c = (2m + 1;3m 4Tm m a +bcng phng vi c75) Xt xem cc cp vect sau c cng phng hng?Nu cng phng th c cng hng hng? a)a= (2;3) ,b= ( 10; 15) b)a= (2;3) ,b= ( 10; 15) c)a= (0;7) ,b= (0;8) d)a= ( 2;1) ,b= ( 6;3) e)a= (0;5) ,b= (3;0) 76) Trong mt phng Oxy cho 3 im 1;-2; 3;2; C0;4 Tm taM trong mi trng hp sau: a/2 3 CM AB AC = b/2 4 AM BM CM + =c/ ABCM l hnh bnh hnh. 77) Trong mt phng Oxy cho 3 im 1;4; 3;1; C-1;2Tm taM trong mi trng hp sau: a/2 5 AM BM CM + =b/2 3 0 MA MB =\c/ ABMC l hnh bnh hnh. \d/ Tm tatrng tmca tam gic C \e/ Tm tatrung im M N P n ttrung im ca cc cnh CC 78) Trong mt phng Oxy cho tam gic 1;1; 2;4; C3;2a/ Tm tatrng tmca tam gic C b/ Tm tatrung im M N P n ttrung im ca cc cnh CC 79) Trong mt phng Oxy cho tam gic 6;-3); B(1;0); C(3;2).a/ Tm tatrng tmca tam gic C b/ Tm tatrung im M N P n ttrung im ca cc cnh CC c/ Tm Chnh bnh hnh Tm tatmca hnh bnh hnh 80) Trong mt phng Oxy cho 3 im -2;1); B(0;2); C(4;4). a/ Chng minh rng 3 im C thng hng b/ Tm tagiao imca ng thngv trc Ox c/ Tm tagiao imca ng thngv trc Oy 81) Trong mt phng Oxy cho 3;4; 2;5 a/ Tm a Ca;1 thuc ng thngb/ Tm M Ctrung im M. 82) Trong mt phng Oxy cho 1;3; 0;1; C0;3; 2;7 Chng minh// C 83) Trong mt phng Oxy cho -1;1); B(1;3); C(-2;0) www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 8 a/ Chng minh C nm trn ng thng i qua b/ Tm giao im ca ng thngv trc Oy c/ Chng minh: O hng thng hng 84) Trong mt phng Oxy cho 1;-1); B(3;1); C(y;2). a/ Tm y C thng hng b/ Tm giao im giav Ox c/ Tm giao imv Oy 85) Trong mt phng Oxy cho 4;5; C-2;1) a/ Tm tatrung imca on C b/ Chng minh: OC hng thng hng c/ Tm M OMChnh bnh hnh 86) Cho A(-1;5) , B(3;-3) a/ Tm tatrung im M cab/ Tm taN sao cho trung im N c/ Tm taP sao cho trung im P d/ ng thng i qua ct Ox ti K Tm taK e/ ng thng i qua ct Oy ti L Tm taL f/ Tm taim C sao choOC AB = . g/ Tm ta sao cho3 DA DB AB =87) Cho A(1,2); B(2; 4); C(3,-3) a/ Chng minh rng C p thnh mt tam gic b/ Xc nh trng tmca tam gic C c/ Tm ta sao cho Otrng tm tam gicd/ Tm ta Chnh bnh hnh e/ Tm taF sao cho OFhnh bnh hnh f/ Cho a 1 Xc nh ta Cthng hng g/ Xc nhK Ox e KChnh thang h/ Tm tagiao im ca ng thng i quav ng thng i qua OC 88) Cho cc im -2;1; 4;2; C-1;-2 n ttrung im cc cnh C Cca tam gic C Tm tacc nh ca tam gic C Chng minh rng trng tm tam gic C v C trng nhau. 89) Cho cc im 3;2) ,B(2;4) ,C(3; 2). aTm tatrng tm tam gic C bTm taimsao cho Ctrng tm tam gicc Tm taimsao cho Chnh bnh hnh90) Cho 3 im 2; 3) ,B(2;1) ,C(2; 1) aTm imsao cho Chnh bnh hnhbi im i xng viquaChng minh rng Chnh bnh hnh 91) Cho tam gic ABC c A( 1;1), B(5; 3 nh C nm trn trc Oy v trng tmnm trn trc Ox Tm tonh C 92) Cho tam gic C bit trng tm 1;2trung im ca C 1; 1 trung im cnh C3;4Tm tocc nh C 93) Cho cc im 2;3 9;4 Mx; 2 Tm x 3 im M thng hng 94) Cho cc im 1;1 3;2 Cm + 4;2m + 1Tm m C thng hng 95) Cho 3 im 1;8 1;6 C3;4 Chng minh rng: C thng hng 96) Cho 4 im 0;1 1;3 C2;7 0;3 Chng minh rng: hai ng thngv C song song97) Cho 4 im 2; 3) ,B(3;7) ,C(0;3), D( 4; 5 Chng minh rng: hai ng thngv C song song98) Cho cc im 4;5) , B(1;2) ,C(2; 3) aChng minh rng: ba im C to thnh mt tam gicbTm taimsao choAD = 3 BC +ACwww.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 9 cTm taimsao cho Otrng tm ca tam gic99) Cho tam gic C cc cnh C Cn t c trung imM 2;1) ,N(1; 3) ,P(2;2) aTm tacc nh C bChng minh rng: cc tam gic C v MNP c trng tm trng nhau CHNG II TCH V HNG CA HAI VCT V NG DNG 1: GIA TR LNG GIAC CUA MOT GOC BAT KY( T 00 en1800) TOM TAT LY THUYET - nh ngha : Trenna dng tron n v lay iem M thoa goc xOM = o va M( x ; y) *. sin goc o la y; ky hieu sin o = y*. cos goc o la x0; ky hieu cos o = y0

*. tang goc o layx(x = 0); ky hieu tan o = yx *. cotang goc o laxy(y = 0); ky hieu cot o = xy - Bang giatr lng giac cua cac goc ac biet BI TP 100)Tnh gia tr bieu thc A = Cos 200 + cos 800+ cos 1000+ cos1600 101)Tnh gia tr bieu thc:o00 300 450 600 900 Sin o0 21 22 231 Cos o1 23 22 210 tan o0 3313Cot o3 1 330 www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 10 A=( 2sin 300 + cos 135 0 3 tan 1500)( cos 1800 -cot 600) B= sin2900 + cos 21200- cos200- tan2600+ cot21350 102)n gian cac bieu thc: a) A= Sin 1000 + sin 800+ cos 160 + cos 1640 b) B= 2 Sin (1800-) cot - cos(1800- ) tan cot(1800- ) . (Vi 00< A ( hoac chon A => B ) Viet p/tng thang (A) oi xng vi ng thang(D1):A1x + B1y + C1 = 0 qua ng thang (D2) : A2 x + B2 y + C2 = 0 Trng hp (D1) // (D2) : - B1 : Lay iem M0 e (D1) . Tmtoa o iem M0/ oi xng vi M0 qua (D2)- B2 : Viet p/t ng thang (A) Qua M0/ va song song vi (D1) hoac (D2) Trng hp (D1) cat (D2) : CACH 1 :-B1:Tm giao iemM0(x0 ; y0)cua hai ng thang (D1) va (D2) -B2 : Layiem M1 e (D1) (M1 = M0 ) , tm toa o iem M2 oi xng vi M1 qua (D2)-B3 :Viet p/t ng thang (A ) qua hai iem M0 , M2

CACH 2 :-B1:Tm giao iemM0(x0 ; y0)cua hai ng thang (D1) va (D2)-B2 :p/t ng thang (A) qua iem M0 co dang : A(x x0) + B(y y0) = 0-B3 :Lap p/t bac hai hai an A , B : cos [ (A) ; (D2) ] = cos [ (A) ; (D1) ] . chon 1 trong hai so A hoac B tm an con laiViet phng trnh ng thang(A) i qua iem M0 (x0 ; y0) va cachiem M1(x1 ; y2) mot oan bang dPP :+ Phng trnh ng thang (A) co dang :A(x x0) + B(y y0) = 0 + Lap pt bac hai hai an A , B : d[ M1 ; (A)] = d +Giai pt (*) vi an A (hoac B) , vi tham so B (hoac A ) +Chon B => A ( hoac chon A => B ) www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 23 BI TP 222)Xt v tr tng i ca cc cp ng thng sau y nu ct nhau th tm tagiao ima) 2x + 3y + 1 = 0 v4x + 5y 6 = 0 b) 4x y +2 = 0 v-8x+2y + 1 = 0c) + =+ =t yt x2 35v x 4 2t 'y 7 3t ' d) + = =t yt x2 21v x 2 3t 'y 4 6t ' 223)Tnh gc to bi hai ng thng : x+ 2y + 4 = 0 , ( )3 31x tt Ry t= +e= +; b) 1 2x 2t x 1 3t 'D : ; (D ) : .(t,t ' )y 3t y 3 6t ' 224)Tnh hong cch t im M4;-5 n cc t sau y :a) 3x 4y + 8 = 0b) + ==t yt x3 22 225)tnh khoang cach t iem en ng thang c cho trc tng ng nh sau : a/ A(3;5) va (d1) : 4x + 3y + 1 = 0 b/B(1;2) va (d2) : 3x 4y + 1 = 0 226)Tnh hong cch gia hai ng thng :3x + 4y 50 = 0 v( )2 41 3x tty t= e= +R

227)lap phng trnh ng phan giac cua cac goc gia hai ng thang: (d1) : 2x + 4y + 7 = 0 va(d2) : x 2y 3 = 0 (d1) : x + 4y + 1 = 0 va(d2) : x y 1 = 0 228)tm phng trnh tap hp cac eu hai ng thang : (d1) : 5x + 3y 3 = 0va (d2) : 5x + 3y + 7 = 0 www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 24 229)Trong mt phng Oxy cho ng thng ng thng d:3x + 4y 12 = 0. aXc nh tocc giao im ca d vi Ox Oy bTnh tohnh chiuca gc O trn ng thng dcVit phng trnh ng thng d' i xng vi O qua ng thng d230)Trong mt phng Oxy cho 2 ng thng d1: 4x 3y 12 = 0; d2: 4x + 3y 12 = 0. aTm tocc nh ca tam gic c 3 cnh nm trn d1,d2 v trc tung bXc nh tm v bn nh ng trn ni tip tam gic ni trn 231)Lp phng trnh cc cnh ca tam gic MNP bit N2;- 1 ng cao h t M c phng trnh 3x 4y + 270 ng phn gic trong t P c phng trnhx + 2y 5 = 0. 232)Cho tam gic ABC c A(-1;3 ng caonm trn ng thng yx ng phn gic trong ca gc C nm trn ng thng x + 3y + 20 Vit phng trnh cnh C 233)Cho im M1;6 v ng thng d:2x 3y + 3 = 0. aVit phng trnh d2 qua M v vung gc vi d bXc nh tohnh chiu vung gc ca M n d 234)Vit phng trnh ng thng qua C4;-3) v ct OxOy ti 2 imsao cho tam gic OABcn. 235)Vit phng trnh ng thng qua 3;-5 v ct trc Ox Oy ti PQ sao cho trung im PQ 236)Vit phng trnh ng thng qua J4;-4 v to vi 2 trc tomt tam gic c din tchl 4 (vdt). 237)Cho im 2;1 Vit pt ng thng d quachn trn hai trc tabng nhau238)Vit phng trnh ng thng c h gc-3/4 v to vi hai trc tomt tam gic c din tch 24 239)Vit phng trnh ng thng qua 4; 1 v to vi hai hai na trc dng OxOy ti hai im MN sao cho : a.OM + ON nh nht b.din tch tam gic OMN nh nhtc.2 21 1OM ON nh nht240)Trong mt phng Oxy cho cc im 10 ; 52 v ng thngc phung trnh :2x y +1 = 0a.Xc nh giao im cavi ng thng i qua hai imb.Tm im C trn ng thngsao cho tam gic C cn ti C241)Vit pt ng thng d1)Qua N(1;-1 v to vi trc honh mt gc 60o

2)Qua I(1;-1 v to vi ng thng d: 3x y + 20 mt gc 45o3i qua im( ) 1; 2 Bv to vi hng dng ca trc Ox mt gc 030 o = . 4 i qua im( ) 3; 4 Cv to vi trc Ox mt gc 045 | = . 5 i qua im 2 ; 1 v to vi:2x + 3y + 40gc 450

242)Vit pt hai cnh gc vung ca mt tam gic vung cn bit1 Mt nh -3;2 v cnh huyn c pt : 3x + 4y 1 = 0www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 25 2 Mt nh 0;1 v cnh huyn c pt : -2x + y + 3 = 0 243)Cho C cn tipt cnh y BC : 3x y +50 ; pt cnh bnAB : x + 2y 10Lp pt cnh C bit n i qua im M1;-3)244)Tam gic C cncnh y C: x + 3y + 10cnh bn : x y + 5 = 0 . ng thng cha cnh C i qua im M-4 1Tm tanh C245)Lp phng trnh ng thngi qua P2; -1 v cng vi hai ng thng (d1) : 2x - y + 5 = 0 v (d2) : 3x + 6y - 10 to thnh mt tam gic cn c nhgiao im ca d1) v (d2)246)Cho im M2;5 v t d: x + 2y 2 = 0a Tm taim M i xng vi im M qua db Vit pt t d i xng vi d qua M247)Cho ng thng d: + = =t yt x2 32 v im -1;4 Vit pt ng thng d1) i xng ca d qua248)Cho ng thng() : 3 4 12 0 d x y + = . a Xc nh tocc giao im A, B ca d n t vi trc Ox, Oy. b Tm tohnh chiu H ca gc toO trn (d). c Vit phng trnh ca ng thng( )1di xng ca d) qua O. 249)Cho ng thng() : 2 3 3 0 d x y + =v im( ) 5;13 M . a Vit PT ng thng d1) qua M v song song vi d). b Vit PT ng thng i xng viqua d1) 250)Cho ng thng: x 1 3tty 2 tvit phng trnh ng thng : a i xng viqua 1) : 2x + y + 3 = 0 b i xng viqua 2) : 2x + 6y - 3 = 0 251)Tm qu tch cc im cch ng thng : a) (D): 2x + 5y 10 mt hong bng 3 ; b : 2x - y + 3 0 mt hong bng5252)Tm qu tch cc im cch u hai ng thnga) 5x + 3y 3 = 0 v 5x + 3y + 7 = 0 ;b) 4x 3y + 2 = 0 v y 3 = 0253)Vit pt ng thng : a Qua 2;7 v cch 1;2 mt on bng 1 b Qua 2;2 v cch u hai im 1;1C3 ; 4 c Cch u 3 im -1 ; 1) ,B(4 ; 2) , C(3 ; -1) d) qua im M2;5 v cch u hai im P-1;2) , Q(5;4)254)Cho ng d: x y + 2 + 0 nh hong cch t 3;5 n d bng 3 255)Cho tam gic C c 2;3 v C4 bit pt C: 3x + y + 10 Tnh din tch tam gic ABCwww.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 26 256)Cho hnh ch nht bit pt hai cnh 4x y + 3 = 0 , x + 4y 50 v mt nh 7;-1) 1 Tnh din tch hnh ch nht ; 2 Vit pt hai cnh cn i ca hnh ch nht 257)Cho hnh vung ABCD c A(- 4 ; 5) v ng thng cha 1 ng cho c pt: 7x y +80Lp pt cc cnh v ng cho th hai ca hvung258)Cho hnh vung ABCD c pt AB: 3x + 4y + 1 = 0 v pt CD: 3x + 4y 10 = 0 a Tnh din tch hnh vung b Vit pt hai cnh cn i nu bit 1;-1) 259)Cho hnh vung c mt nh0 ;5 v mt ng cho nm trn ng thng c phng trnh : 7x y + 80 Vit phng trnh cc cnh v ng cho th hai ca hnh vung . 260)Cho tam gic ABC c B(2,-1) , ng cao quac phng trnh 3x 4y + 27 = 0 , phn gic qua C c phng trnh 2x y + 5 = 0 a/ Vit phng trnh ng thng cha cnh C v tm tanh C b/ Lp phng trnhng thng cha cnhC261)Cho tam gic ABC c A(2,-1) v phng trnh hai ng phn gic trong ca gcv C n cdB): x 2y + 1 = 0 ,(dC): x + y + 3 = 0 . Tm phng trnh cc cnh262)Trong mt phng Oxy cho tam gic Cc -1;3), ng caoc pt:yx ; ngphn gic trong gc C ca tam gic c pt: x + 3y - 2 = 0 . bVit ptt cc cnh ca tam gic CcTm chu vi ca tam gic C 263)Lp pt cc cnh tam gic bit 2 ; - 1),ng cao : 3x 4y + 27 = 0 ;ng phn gic trong CD : x + 2y 5 = 0 264)Cho ng thng d: x y + 20 v hai im O0;02;0a CMR hai im O nm v cng mt pha i vi ng db Tm im i xng ca O qua dc Tm trn d im M sao chodi ng gp hc OM ngn nht MT S THI I HC V CAO NG 265)(HSPKT K Trong mt phng Oxy Cho tam gic C bit nh -1,2) , B(2,0) , C(-3,1)1/ Xc nh tm ng trn ngoi tiptam gic C2/ Tm im M trn ng thng C sao cho din tch tam gic M bng 1/3 din tch tam gic ABC266)(HKTQD p phng trnh cc cnh tam gic Cbit -4,5) v hai ng cao h t hai nh cn i ca tam gic c pt:5x + 3y 4 = 0 v 3x + 8y +13 = 0267)(HTCKT) cho ng cong Cm : x2 + y2 +2mx 6y +4 m = 0 1/ CMR (Cm) l ng trn vi mi mTm tp hp tm ng trn hi m thau i2/ Vi m4 hy vit phng trnh dng thng vung gc vi dng thng3x-4y+10 0 v ct ng trn ti hai im sao cho6www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 27 268)(HHH)cho M(5/2,2) v hai ng thng c phng trnh : y = x/2 ; y 2x0Lp phng trnh ng thngdi qua M v ct hai ng thng ni trn ti hai imv B sao cho M l trung im 269)(HMC)y vit phng trnh ng trn ngoi tip tam gic Cbit phng trnh AB: y x 2 = 0 , BC: 5y x +2 = 0 vAC: y+x 8 =0270)(HGTVT)cho hnh bnh hnh C c s o din tch bng 4it tanh A(1,0) , B(2,0) v giao imca hai ng cho C nm trn ng thngy xy tm tacc nh C v 271)(Hc vin QS) Tam gic C cncnh y C: x + 3y + 10cnh bn : x y + 5 = 0 . ng thng cha cnh C i qua im M-4 1Tm tanh C272)(HHng khng) Cho tam gic ABC c B(2,-1) , ng cao quac phng trnh 3x 4y + 27 = 0 , phn gic qua C c phng trnh 2x y + 5 = 0 1/ Vit phng trnh ng thng cha cnh C v tm tanh C 2/ Lp phng trnhng thng cha cnhC273)(HQS)cho A(2,-4) , B(4/3,2/3) , C(6,0) . Tm tm v bn knh ng trn ni tip tam gic ABC274)(HM bn cngcho tam gic C vi cc nh 1201C-2,1) 1/ Vit phng trnh ng thng cha cnh 2/ Lp phng trnh ng cao C ca tam gicC 3/ Lp phng trnh ng trn ngoi tip ca tam gic C275)(H An Giang KD)cho hnh thoi ABCD c A(1,3) , B(4,-1) ait cnhsong song vi trc Ox v nhc honhmTm tacc nh C v D bLp phng trnh ng trn ni tip hnh thoi C276)(H thng mi) Cho tam gic ABC c A(2,-1) v phng trnh hai ng phn gic trong ca gcv C n cdB): x 2y + 1 = 0 ,(dC): x + y + 3 = 0 . Tm phng trnh cnh C277)(H ty nguyn p phng trnh tng qut ca ng thng i qua im -2,3) v cch u hai im 5-1) , B(3,7)278)(HSP HNi KA) cho tam gic ABC c nh 11ng cao tv C n t c phng trnh : - 2x + y 8 = 0 v 2x + 3y 60Lp phng trnh ng cao h tv xc nh tanh C ca tam gic C279)(H ngoi ng) cho 3 im -1,7) ; B(4,-3) ; C(-4;1Lp phng trnhng trn ni tip tam gic280)HQG:(2000) cho Parabol (P) : y2 =4x v hai ng thng : (D): m2x +my + 1 = 0 (L): x my + m20 vi mtham s thc hc0 a.CM (D) vung gc (L) v giao im cav L di ng trn mt ng thng c nh hi m thay i b.CM (D) v (L) lun tip xc vi Pivn tcc tip im cav L vi PChng minh ng thngun i qua mt im c nh hi m thay i 281)(CCN4) Cho tam gic ABC c nh -10 hai trung tuyn xut pht tv C n lc c phng trnh : 5x+ y 9 = 0 v 4x +5y 10 = 0a.Xc nh ta trng tmca tam gic Cwww.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 28 bLp phng trnh tng qut ca ba cnh tam gic C282)(HANinh) Cho tam gic c im M-1,1) l trung im ca mt cnhcn hai cnh kia c phng trnh l : x +y 2 = 0v 2x +6y +3 = 0 . Hy xc nh tacc nh tam gic283)(CNTin hc Trong mt phng Oxy cho cc im 10 ; 52 v ng thngc phung trnh : 2x y +1 = 0 a.Xc nh giao im cavi ng thng i qua hai imBb.Tm im C trn ng thngsao cho tam gic C cn ti C284)( khi A - 2006)Trong mt phngcho ba ng thng1 2 3: 3 0; : 4 0; : 2 0. d x y d x y d x y + += = = 285)Tm toim M nm trn ng thng d3 sao cho hong cch t M nng thng d1 bng hai n hong cch t M n ng thng d2. 286)( thi khi A nm 2005)Trong mt vi h toOxy cho hai ng thng d1: x y = 0 v d2: 2x + y 1 = 0. 287)Tm tocc nh ca hnh vung C bit rng nhthuc d1, nh C thuc d2 v cc nh thuc trc honh PHNG TRNH NG TRN Phng trnh chnh tac :ng tron (C) tam I(a ; b) ; ban knh R > 0 co p/t chnh tac la : (x a)2 + (y b)2 = R2

lu y :* Neu a = b = 0 th p/t ng tron co dang : x2 + y2 = R2 - la p/t ng tron tam O ban knh R .* ng tron tam I(a ; b) va qua goc toa o O co phng trnh : (x a)2 + (y b)2 = a2 + b2 * ng tron tam I(a ; b) va tiep xuc vi truc hoanh Ox co phng trnh: (x a)2 + (y b)2 = b2 * ng tron tam I(a ; b) va tiep xuc vi truc tung Oy co phng trnh: (x a)2 + (y b)2 = a2 Phng trnh tong quat : Phng trnh co dang: x2 + y2 + 2Ax + 2By + C = 0(vi A2 + B2 C > 0 )eu xac nh motng tron (C) co tam I(A ; B) va ban knh R=2 2A B C ; c goi la phng trnh tong quat cua ng tron. Luy :+ Neu C = 0 ng tron (C) i qua goc toa o1.V tr tng oicua ngtron vi ng thang : www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 29 Cho ng tron C(I ; R) va ng thang (D) . d[I ; (C)] < R (D) cat (C) tai tai hai iem phan biet . d[ I ; (C)] > R (D) khongcat (C) d [ I ; (C) ] = R (D) tiep xuc vi (C) 4.Phng tch cua diem M(x0 ; y0 ) oi vi ng tron (C):Neu (C) : x2 + y2 +2Ax + 2By + C = 0 th : P M / (C) = f(xo ; yo) = xo2 + yo2 + 2Axo + 2Byo + C Neu (C) : ( x a)2 + (y b)2 = R2 th : P M / (C) = f(xo ; yo) = (x0 a)2 + (y0 b)2 = R2 Nhan xet:* P M / (C) > 0 M ngoai (C) * P M / (C) < 0 M trong (C) * P M / (C) = 0 M e (C) 5.Truc ang phng cua hai ng tron :cho hai ng tron (C1) va (C 2) khong ong tam lan lt co phng trnh: (C 1) : f1(x ; y) = x2 + y2 + 2A1x + 2B1y + C1

(C2 ) : f2(x ; y) = x2 + y2 + 2A2x + 2B2y + C2 Truc ang phng cua hai ng tron (C 1) va (C 2) co phng trnh : f1(x ; y) = f2(x ; y) Hay 2(A1 A2)x + 2(B1 B2)y + C1 C2 = 0 PP GIAI 1 SO DANG TOAN VE NG TRON : Van e 1 :viet phng trnh ng trn Xac nh tam va ban knh cua 1 ng tronCACH 1 :ong nhat p/t a cho vi p/t : x2 + y2 +2Ax + 2By + C = 0 .T o tm cac he so A ; B ; C , roi suy ra tam I( - A ; - B); ban knh R= 2 2A B CCACH 2 :a p/t ve dang ( x a)2 + ( y b)2 =R2 => Tam I(a ; b) ; ban knh la R Viet phng trnh ng tron (C) CACH 1 :+ Gia s p/t ng tron can tm co dang : x2 + y2 + 2Ax + 2By + C = 0(1) + da vao cac gia thiet bai toan cho lap he 3 p/t3 an A , B , C + Giai he tm A , B , C . Roi viet p/t ng tronCACH 2 :+ Tm tam I(a ; b)+ Tm ban knh :* Neu A e (C) th R = IA; * Neu ng thang (D) tiep xuc vi (C) th : R = d[ I ; (D)] + Viet p/t ng tron dang : (x a)2 + (y b)2 = R2; (2) www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 30 Ghi chu : -(C) i qua 3 iem phan biet A , B , C cho trc : The toa o 3 iemvao (1) ,tm 3 he so A , B , C-Tm toa o tam I cua (C) can lap c 1 he p/t. Vi moi gia thiet di ay se chuyen thanh 1 p/t tm tam : a)ng tron qua hai iem A , B IA2 = IB2

b)ng tron tiep xuc vi ng thang (D) tai A I thuoc ng thang (D) vuong goc vi (D) tai A (Ngha la toa o iem I thoa p/t ng thang (D) i qua A va (D) (D)c)ng tron i qua A va tiep xuc vi (D) d[ I ; (D) ] = IA d)ng tron tiep xuc vi hai ng thang song song (D) va (D) d[ I ; (D) ] = d[ I ; (D) ] e)ng tron tiep xuc vi hai ng thang cat nhau(D) va (D) d[ I ; (D) ] = d[ I ; (D') ]I ; ' D D thuocng phan giac cua f)ng tron co tam I e (D) toa o tam I thoa p/t cua (D) ac biet : Tm tam I cua ng tron noi tiep A ABC, khi a biet toa o 3 iem A, B, C*Tm toa o iem D la chan ng phan giac trong ve t A, qua he thc : .ABDB DCAC * Toa o tam I thoa he thc : BAIA IDAD

BI TP 288)Tm tm v bn nh ca cc ng trn sau a) x2 + y2 2x 2y 2 = 0 b) 16x2 + 16y2 + 16x 8y = 11c) 7x2 + 7y2 4x + 6y 1 = 0 289)Vit phng trnh ng trn ng nhvi : 1/ A(-1,1) , B(5,2) 2/ A(-1,-2) , B(2,1) 3) A(1;1), B(7;5) 4/A(1;3), B(5;1)290)Lp pt vng trn qua 3 im1/ A(1,3) , B(5,6) , C(7,0)2/ A(5,3) , B(6,2) , C(3,-1) www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 31 3/ A(1;2) , B(5;2) , C(1;-3) 3/ A(0;1), B(1;-1), C(2;0)291)Lp pt ng trn bit : 1/ Tm I(2,2) , bn knh R = 3 2/ Tm I(1,2) , v i qua A(3,1) 2) Tm I(2;-3) v qua A(1;4)3/ Tm I(-4;2 v tip xc vi d: 3x + 4y 16 = 04/Tm I(-1;2 v tip xc vi d: x 2y + 7 = 05/ Tmthuc d: x + y 1 = 0 v qua A(-2;1), B(4;2) 6/ Qua A(1,2) , B(3,1) v c tm trn (d) 7x + 3y +1 = 0 7/ i qua 3155 v c tm nm trn trc honh8/ i qua M(-13N21 v c tm nm trn ng phn gic ca gc phn t th nht 292)Lp pt ng trn tip xc vi cc trc tav1/Qua A(2,4)2/ c tm nm trn ng thng 3x 5y 8 = 0 3) Qua M(2,-1)293)Lap phng trnh cua ng tron (C) trong cac trng hp sau : a/ (C ) co tam I(-1;2) va tiep xuc vi ng thang (d) : x- 2y + 7 = 0 b/ (C ) co ng knh la AB vi A(1 ;1) va B(7 ;5) 294)trong mat phang Oxy , hay lap phng trnh cua ng tron (C ) co tam la iem I(2 ;3) va thoa man ieu kien sau : a/ (C ) co ban knh la 5 b/ (C ) i qua goc toa oc/ (C) tiep xuc vi truc Ox d/ (C) tiep xuc vi truc Oye/ (C) tiep xuc vi ng thang (d) :4x + 3y -12 = 0 295)lap phngtrnh cua ng tron (C) i qua hai iem A(1;2) va B(3;4) ong thi tiep xuc vi ng thang (d):3x + y -3 = 0. 296)Trong mt phng Oxy cho ng trn C : 297)Lp phng trnh ng trni xng vi Cqua ng thng : x-2 = 0 . 298)cho 3 iem A(1;4) ,B(-7;4) ,C(2;-5)a/ lap phng trnh ng tron (C) ngoai tiep tam giac ABC. b/ tm tam va ban knh cua (C) 299)Vit phng trnh ng trn trong cc trng hp sau : a)ng trn tip xc trc Ox ti -1,0) v qua B(3,2) b)ng trn tmv tip xc vi ng thngc)Qua 42 v tip xc 2 ung thng x - 3y - 2 = 0 v x - 3y + 18 =0d)C tm trn ng thng x5 v tip xc vi 2 ng thng 3x y + 3 = 0vx 3y + 9 = 0 e)Qua 1234 V tip xc vi d y3 3xf)Tm nm trn ng thng d: 4x + 3y 20 v tip xc vi hai ng thngg)(d1): x + y + 4 = 0, (d2): 7x y + 4 = 0 h)Tip xc vi d 3x 4y 310 ti 1 -7) v R = 5 www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 32 300)Trong mt phng vi h taOxy cho:

ng trnv ng thng. 301)Tm taimsao cho ng trn tmc bn nh gp i bn nh ng trn tip xc ngoi vi ng trn. 302)Vit phng trnh ng trn c honhtm a9bn nh R2v tip xc vi ng thng d: 2x+y-10=0 303)Trong mt phng cho tam gic i chn ng cao tM v Ntrung im ca cc cnhv C Vit phng trnh ng trn i qua cc imM N 304)Cho (d) : (1 m2)x + 2my + m2 4m +10Vit phng trnh ng trn tip xc vi d vi mi m305)Vit phng trnh ng trn ngoi tip tam gic c 3 cnh trn 3 ng thng sau 5y = x 2 , y = x + 2 , y = 8 x 306)Vit pt ng trn ni tip tam gic O bit : 1)Cho A(4,0) , B(0,3)2) A(4,0) , B(0,4) 307)Cho tam gic ABC c A(1/4, 0) ; B(2,0) ; C(-2,3)1/ Tm gc C ca tam gic2/ Lp pt ng trn ni tip tam gic C3/ Vit pttt ca ng trn ni tip Tam gic v song song C308)Cho (d1) : 4x 3y 12 = 0 , (d2) : 4x + 3y 120 Tnh tacc nh ca tam gic c 3 cnh n t nm trn cc ng thng d1) , (d2 v trc tung .Xc nh tm v bn knh ng trn ni tip ca tam gic ni trn 309)Trong mt phng vi h taOxycho tam gic C c ba gc nhnbit5 ; 4 v2 ; 7 iv Fhai ng cao ca tam gicy vit phng trnh ca ng trn ngoi tip t gic F 310)Tm m phng trnh sau l phng trnh ng trn1/ x2 + y2 + 4mx 2my +2m + 3 = 02/ x2 + y2 2(m+1)x +2my +3m2 2 = 0 311)Cho (Cm) : x2 + y2 2(m +2)x + 4my +19m 6 = 01/ Tm m Cmvng trn c bn nh R102/Tm m Cmvng trn3/Tm tp hp tm ca Cm312)Cho (Cm) : x2 + y2 + (m + 2)x (m + 4)y + m + 1 = 01/nh m Cm c bn nh nh nht 2/ Tm tp hp tm ca Cm3/CMR : (Cm) lun qua 2 im c nh vi mi m 4/ Tm tt c cc im ca Cm hng th i quaCho (Ca) : x2 + y2 + 2(1 cosa)x- 2ysina + 3 = 0 , a ] 2 , 0 [ t e

1/ Tm a Cavng trn 2/ Tm tp hp tm ng trn Ca313)Cho (Cm) : x2 + y2 2x (m-1)y + m2 4 = 01/ Tm m Cm i qua A(2,3) .Xc nh tm v bn nh ca ng trn ng m tm c2/ Tm m Cm c bn nh n nht314)Cho h ng trn Cm : x2 + y2 2mx 2(m+1)y + 2m 1 = 01/ CMR : khi m thay ih ng trn Cm un i qua 2 im c nhwww.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 33 2/ CMR : vi mi mh ng trn Cm un ct trc tung ti 2 dim phn bit315)Cho cc ng trn C x2+ y2 =1 v (Cm) x2 + y2 2(m+1)x + 4my = 51/ CMR c 2 ng trn Cm1) , (Cm2 Tip xc vi ng trn C ng vi 2 gi tr m1 , m2 ca m2/Xc nh pt ng thng tip xc vi c 2 ng trn C1) , (C2)316)Cho h ng cong Ct : x2 + y2 2(1+cost)x 2(sint)y + 6cost 3 = 01/ Chng t Ct unng trn thc2/ Tm qu tch tm ng trn Ct hi t thay i3/ Chng t Ct un i qua 1 im c nh4/ Trong h Ct c ng trn no bn nh1 3 hng ? Viet phng trnh tiep tuyen cua ng tron Dang 1: Viet phng trnh tiep tuyen cua t (C) tai iem M0(x0 ; y0) ( M0 e (C) ) CACH 1 : * Xac nh toa o tam I cua t(C) * Tiep tuyen la ng thang qua M0 va co vec t phap tuyen la IM CACH 2 : * DungPP phan oi toa o : + Neu ng tron (C) co p/t : x2 + y2 + 2Ax +2by + C = 0 th p/t tiep tuyen laxox + yoy + A(xo + x) + B(yo + y) + C = 0 + Neu ng tron (C) co p/t : (x a)2 + (y b)2 = 0 th p/t tiep tuyen la :(xo a) (x a) + (yo b) (y b) = R2

Dang 2:Viet p/t tiep tuyen cua (C) co phng cho trc ( tc la biet he so goc k cua tiep tuyen , hoac tiep tuyen song song , hoac tiep tuyen vuong goc vi ng thang cho trc )www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 34 Phng phap : Xac nh tam I va ban knh R cua ng tron (C)Viet p/t tiep tuyen (A) cua (C) : -Neu tiep tuyen (A) co he so goc k th p/t co dang : y = kx + b ( vi he so b cha biet )-Neu (A) // (D) : Ax + By + C = 0 th p/t (A) co dang : Ax + By + C = 0( vi C cha biet) -Neu (A) (D) : Ax + By + C = 0 th p/t (A) co dang : Bx Ay + C = 0( voi C cha biet )Dung / k : (A) tiep xuc vi (C) d[ I ; (A) ] = R. e tm he so cha bietDang 3:Viet phng trnh tiep tuyen cua (C) xuat phat t iem A(x0 ; y0)( Ae (C) )Phng phap : CACH 1 : Xac nh tam I va ban knh R cua ng tron (C)p/t tiep tuyen co dang : A(x -x0 ) + B (y y0) = 0 ( Vi he so A , B cha biet )Dung /k: tiep xuc d[ I ; (A) ] = R . lap p/t an so A , vi tham soB(hoac an B tham so A) .Giai p/t tren , roi chon B => A ( hoac chon A => B)CACH 2 :Xac nh tam I va ban knh R cua ng tron (C)p/t tiep tuyen co dang : y = k(x x0) + y0 kx y kx0 + y0 = 0Dung /k : d[ I ; (A) ] = R lap p/t bac hai 1 an k . Roi giai tm kGhi chu : Neu tm c 2 gia tr k th co hai tiep tuyen vi (C) i qua ANeu tm c 1 gia tr k thcan xet trng hp ng thang (A) qua A(x0;y0) song song vi Ox cop/t:x = x0 x x0 = 0 ,co phai la tiep tuyen cua (C) khong . Bang cach kiem tra ?; d I R , neu ung thtieptuyen th hai la ng thang x = x0

V tr tng oi cua hai ng tron Tiep tuyen chung cua hai ng tron1.V tr tng oi cua hai ng tron :-Xac nh tam I1 ; I2; ban knh R1 ; R2cua hai ng tron (C1) va (C2)www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 35 -Tnh d = I1I2 ( oan noi hai tam)- So sanh : Neu 1 21 2R R d R Rth (C1) cat (C2) Neu d = R1 + R2 th (C1) va (C2) tiep xuc ngoai Neu d = 1 2R Rth (C1) va (C2) tiep xuc trongNeu d > R1 + R2 th (C1) va (C2) nam ngoai nhauNeu d < 1 2R Rth (C1) va (C2) ng nhau2.Tiep tuyen chung cua hai ng tron : ( ch xet trng hp R1 = R2 )CACH 1 : -Xac nh v tr tng oi cua hai ng tron-Xet cac trng hp:Trng hp 1: Neu (C1) va (C2) nam ngoai nhau th co 4 tiep tuyen chung : + Tm giao iem M cua hai tiep tuyen chung ngoai qua he thc :11 22RMI MIR + Tm giao iem N cua hai tiep tuyen chung trong qua he thc : 11 22RNI NIR + Viet p/t tiep tuyencua (C1) hoac (C2) i qua iem M va N Trng hp 2:Neu (C1) va (C2)cat nhau th co hai tiep tuyen chung ngoai. Tm giong nh TH1 , oi vi tiep tuyen chung ngoaiTrng hp 3 : Neu (C1) va (C2) tiep xuc ngoai co hai tiep tuyen chung ngoai ( tm giong TH1 ) ; va 1tiep tuyen chung trong la trucang phngcua hai ng tron Trng hp 4 : Neu (C1) va (C2) tiep xuc trong co 1 tiep tuyen chung trong la truc ang phng cua hai ng tronwww.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 36 Trng hp 5 : Neu (C1) va (C2) ng nhau th hai ng tron khong co tiep tuyen chungCACH 2 :- Xac nh tam I1 , I2 va ban knh R1 , R2 cua hai ng tron , va suy ra VTT cua hai ng tron- P/t tiep tuyen chung(A) co dang y = kx + b hay kx y + b = 0 -Dung ieu kien :(A) la tiep tuyen chung cua hai ng tron ;1 1*;2 2dI RdI R .-Giai he pt tm k & b -Xet trng hp (A) : x + c = 0 dunghe (*) tm c neu coCACH 1 :- P/t (A) co dang : Ax + By + C = 0(A2 + B2 = 0) - (A) la tiep tuyen chung ;1 1*;2 2dI RdI R . Ta kh an C T hai p/t , c p/t bac hai vi an A (hoac B) . Giai p/t tman A theo B ( hoac B theo A) , roi chon B => A ( hoac chon A => B )Ghi chu :Neu R1 = R2 th hai tiep tuyen chung ngoai song song va cach eu ng noi tam I1I2mot khoang la R1 . Con tiep tuyen chung trong la truc ang phng cua hai ng tron ******** www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 37 BI TP 317)vietphngtrnhtieptuyencuangtron(C): ( ) ( )2 21 2 25 x y + + = taiiem M(4;2) thuoc ng tron (C). 318)lapphngtrnhtieptuyenvingtron 2 24 2 0 x y x y + = bietrangtiep tuyen i qua iem A (3;-2) 319)vietphngtrnhtieptuyen(d)vingtron 2 24 6 3 0 x y x y + + + = bietrang (d) song song vi ng thang (d1) : 3x y + 2010 = 0 . 320)cho ng tron (C) : 2 27 0 x y x y + =va ng thang (d):3x + 4y 3 = 0a/ tm toa o giao iem cua (d) va (C). b/ lap phng trnh tiep tuyen vi (C) tai cac giao iem o. c/ tm toa o giao iem cua hai tiep tuyen. 321)lap phng trnh tiep tuyen (d) cua ng tron (C) :2 26 2 0 x y x y + + =biet rang (d) vuong goc vi ng thang (d1): 3x y + 4 = 0 322)cho ng tron (C) : 2 26 2 6 0 x y x y + + + =va iem A(1;3) a/ chng to rang iem A nam ngoai ng tron (C). b/ lap phng trnh tiep tuyen vi (C) xuat phat t A. lapphngtrnhngtron(C)iquahaiiemA(1;2),B(3;4)vatiepxucvi ng thang (d) : 3x + y 3 = 0323)Trong mt phng vi h taOxy cho ng trn C :v im Xc nh tatm v bn nh ca ng trn C Vit phng trnh cc tip tuyn ca ng trn C t im324)Cho ng trn T c phng trnh : a Xc nh tm v bn nh ca T b Vit phng trnh tip tuyn ca T bit tip tuyn ny vung gc vi ng thng d c phng trnh 12x - 5y + 2 = 0. 325)Trong mt phng vi h taOxycho ng trn C :v ng thng d : 3x - 4y + 230Vit phng trnh tip tuyn ca ng trn C bit tip tuyn ny vung gc vi ng thng d 326)Trong mt phng Oxy cho ng trn Lp phng trnh tip tuyn vi ng trn C bit rng tip tuynqua 327)Trong mt phng Oxy cho ng trn C :Vit phng trnh cc tip tuyn ca C i qua im F 0; 3 www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 38 328)Trong mt phng vi h taOxy cho ng trn C c phng trnh : a Vit phng trnh tip tuyn ca ng trn bit cc tip tuyn ny vung gc vi ng thng. b Tm iu in ca m ng thngtip xc vi ng trn329)Trong mt phng vi h taOxy cho tam gic C bit4; - 2) , B (- 2; 2) , C (- 4 ; - 1Vit phng trnh ng trn C ngoi tip tam gic C v phng trnh tip tuyn vi C ti330)Trong mt phng vi h taOxy cho ng trn. Tm tt c cc tip tuyn casong song vi ng thng. 331)Trong mt phng vi h tacc vung gc Oxycho imv ng trn (O) : 1 Chng minh rng mt im nm ngoi ng trn O 2 Vit phng trnh cc ng thng i qua imv tip xc vi ng trn O 332)Trong mt phng vi h tacc vung gc Oxy cho ng thng v hai im 1 Vit phng trnh ng trni quav c tm. 2 Vit phng trnh ng tip tuyn tivi ng trn. 3 Vit phng trnh cc tip tuyn vi bit tip tuyn i qua Tm tatip im . 333)Trong mt phng vi h taOxy cho im - 2; 1 v ng thng d : 3x - 4y = 0 a Vit phng trnh ng trn C c tmv tip xc vi ng thng d b Vit phng trnh tp hp cc im m qua cc imv c hai tip tuyn n C sao cho hai tip tuyn vung gc vi nhau 334)Cho ng trn V ng thng a. Chng minh rnghng ct www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 39 b. T im M thuc cc tip tuyn M M ti Ccc tip im Chng minh rng hi M thay i trnthun i qua mt im c nh 335)Cho h ng trnc phng trnh:Tm tp hp tm cakhithay i 336)Vit phng trnh ng trn i qua 10 v tip xc vi hai ng thng 337)Trong mt phng tacho ng trnv mt im . Vit phng trnh ng thng i quav cttheo mt dy cung cdi 8 338)Trong mt phng vi h cc trc chun cho ng trnv ng thng Chngminh rng tmt imM bt trnta un chai tip tuynphn bit ti (C). a.i s hai tip tuyn t M ti C c cc tip im vChng minh rng hi M chy trnng thngun i qua mt im c nh 339)Cho ng trnv ng thng( tham s a Chng minh rnglun ctti hai im phn bit. b. Tmdi onun t gi tr n nht nh nht 340)Cho h ng trnc phng trnh:Chng minh rngun tip xc vi hai ng thng c nh 341)Trong mt phng tachoc phng trnhVit phng trnh cc tip tuyn t imn. 342)Cho hai ng trn c tm n t v 1 Chng minhtip xc ngoi viv tm tatip im. 2 i mt tip tuyn chung hng i quacav Tm tagiao im cav ng thng. 3.Vit phng trnh ng trongi quav tip xc vi hai ng trnvti . www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 40 343)Trong mt phng vi h tovung gc Oxy xt h ng trn c phng trnh ( tham s 344)Xc nh taca tm ng trn thuc h cho m tip xc vi trc Oy 345)Cho h ng trnc phng trnh:Timtip xc vi 346)Cho h ng trnc phng trnh:Tmtip xc vi ng trn 347)Cho ng trn c phng trnh:Vit phng trnh tip tuyn ca ng trn i qua. 348)Tm cc gi tr ca a h sau c ng hai nghim 349)Trong mt phng Oxy cho ng trn C :v ng thngc phng trnh : 350)Tm taim T trnsao cho qua T c hai ng thng tip xc vi C ti hai imv 351)Trong mt phng vi h taOxy cho ng trn :v imiv cc tip im ca cc tip tuyn tn Vit phng trnh ng thng. 352)Trong mt phng vi h taOxy cho ng trn C :v ng thng d: Tm taim M nm trn d sao cho ng trn tm M c bn nh gp i bn nh ng trn C tip xc ngoi vi ng trn C 353)Trong mt phng vi h ta0xy cho hai im2; 0 v6; 4 Vit phng trnh ng trn C tip xc vi trc honh ti imv hong cch t tm ca C n im bng 5 354)Cho hai ng trn : 1.Xc nh cc giao im cav.2. Vit phng trnh ng trn i qua 2 giao imv im 0; 1 355)Trong mt phng vi h tacc vung gc Oxy cho ng trn C : www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 41 v ng thng d :. Vit phng trnh ng trn C' i xng vi ng trn C qua ng thng d Tm ta cc giao im ca C v C'356)Cho ng trn C : Lp phng trnh ng trn C' i xng vi ng trn C qua ng thng d:. 357)Tmdi dy cung xc nh bi ng thng 4x + 3y - 80 v ng trn tm2; 1 tip xc vi ng thng 5x - 12y + 15 = 0. 358)Trong mt phng Oxy cho hai ng thng . Vit phng trnh ng trnquav tip xc vi ng thngti giao im cavi trc tung 359)Trong mt phng vi h tacc vung gc Oxy Vit phng trnh ng thng i quav tip xc vi ng trn 360)Trong mt phng vi h tacc vung gc Oxy cho cc im Xc nhtaim tm ng trn ni tip tam gic . 361)Trong mt phng vi h taOxy cho parabol (P) :v im. Vit phng trnh ng trnc tmv tip xc vi tip tuyn ca ti. 362)Trong mt phng vi h taOxy cho hai im 2;0 v 6;4 Vit phng trnh ng trn C tip xc vi trc honh ti imv hong cch t tm ca C n im bng 5 363)Trong mt phng vi h tacac vung gc Oxy cho ng trn v ng thng Vit phng trnh ng trn C' i xng vi ng trn C qua ng thng dTm ta cc giao im ca C v C' 364)Cho ng trnv imVit phng trnh ng thng i qua M ct ng trn ti 2 imsao cho Mtrung im ca on365)Trong mt phng Oxy cho h ng trn: Chng minh rng hcun tip xc vi hai ng thng c nh 366)Trong mt phng Oxy cho h ng trn: .Tm m ct ng trnti hai im phn bitv Chng minh rng hing thngc phng hng i 367)Trong mt phng taOxy cho hai ng thng : www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 42 1 Tm tacc nh ca tam gic c ba cnh n t nm trn cc ng thng v trc tung2 Xc nhtm v bn nh ng trn ni tip ca tam gic ni trn 368)Lp phng trnh ng thng qua gc tav ct ng trn : thnh mt dy cung cdi bng 8. 369)Cho vng trn (C) :v im3; 5y tm phng trnh cc tip tuyn tn vng trn i s cc tip tuyn tip xc vi vng trn ti M, N. y tnhdi MN 370)Cho h vng trn : 1 Chng minh rng hi m thay i h vng trn un un i qua hai im c nh2 Chng minh rng vi mi m h vng trn un un ct trc tung ti hai im phn bit371)Trong mt phng cho ng trn : Tm m tn ti duy nht mt im P m t c 2 tip tuyn PP ti C cc tip im sao cho tam gic P u 372)Trong mt phng Oxy cho ng trn C :v ng thng D) c phng trnh :Vit phng trnh ng thng vung gc viv tip xc vi ng trn 373)Trong mt phng Oxy cho ng trn C :v ng thngc phng trnh :Vit phng trnh ng thng song song viv ct ng trn ti hai im M N sao chodi MN bng 2 374)Trong mt phng vi h taOxy cho ng trn C :. y vit phng trnh cc tip tuyn ca C bit cc tip tuynvung gc vi ng thng x + y0 375)Cho ba im 0 ; 1 ; 2 ; 0 ; C3 ; 2 Tp hp cc im Mx ; y sao cho : 376)Cho 1; 1 v 2 ; 3tp hp cc im M sao cho : 377)Cho hai ng trn C :v C :, M l im di sao chodi tip tuyn t M ti C gp hai ndi tip tuyn t M ti C Tm qu tch MVi gi tr no ca m thdi tip tuyn pht xut t 5 ; 4 n ng trn C :bng 1? www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 43 378)Trong mt phng vi h trc taOxy cho 2;1 v 2 ng thngvVit PT ng trntip xctiv c tm thuc. 379)Mt ng trn qua im 3;5 v ct Oy ti im 0;4 v im 0;-2Vit phng trnh ng trn cho bit tm v bn nh 380)Cho hai ng thng(d) v (c phng trnh n t: 2x-y+2=0 v 2x+y-4=0 . Vit phng trnh ng trn C c bn nh R nm trong gc nhn ca hai ng thng d v v tip xc vi chng 381)Trong hng gian Oxy cho 2 ng trn : Lp phng trnh tip tuyn chung ca 2 ng trn 382)Trong mt phng toOxy cho im M6;2 v ng trn C : 383)Lp phng trnh ng thng d qua M v ct C ti 2 im ; sao cho 384)Trong mt phng Oxyp phng trnh ung trn qua 1;2 ; 3;1 v c tmthuc ng thng : 7x+3y+10 385)Trong mt phng toOxy cho h ng cong : a Chng minh rng h ng trn v tn ti 1 ng thngtrc ng phng ca tt c cc ng trn b Chng minh rng cc ng trn ca hun tip xc vi nhau ti 1 im c nh Tm im386)Cho 2 ng trn 0 v 0' tip xc ngoi ting gc C vung trong thuc O v C thuc O'Tm qu tch trung imca BC. 387)Trong mt phng Oxy p phng trnh ng trn C tip xc vi ng thng : x-y-20 ti im M 3;1 v tmthuc ng thng : 2x-y-2=0 . 388)Trong mt phng Oxy cho 2 ng trn :a Chng minh rng ;vct nhau ti 2 im phn bitvb Vit phng trnh ng trn quav tip xc vi ng thng ; x-2y+4=0 389)Cho ng trn O;R 2 ng nhMN Tip tuyn tict M tict N ti K PQ n t trung im cav K Chng minh tm ng trn ngoi tip tam gic PQ di chuyn trn mt ng thng c nh vic nh 390)Cho ng trn C c phng trnh:v im 4;7 a Lp phng trnh ng trn C' tip xc vi C bit C' i qua imwww.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 44 b Trong trng hp C' tip xc ngoi C hy tm trn C im M trn C' im N sao cho tam gic MN c din tch n nht Vi tm ca ng trn C 391)Cho ng trn C: x2 + y2 + 4x - 4y - 10; im0;1 v ng thng : x - y = 0. 1 Vit phng trnh tng qut ca cc tip tuyn d1);(d2 ca ng trn C di qua 2 Tnh cosin cc gc nhn to bin t vi d1),(d2).392)Cho ng trn C: Vit cc phng trnh tip tuyn ti cc im c to nhng s nguyn thuc ng trn 393)Cho hai imv 1 Tm qu tch cc imsao cho 2 Tm qu tch cc imsao chotrongmt s cho trc 394)Cho 2 h ng trnn t c phng trnh: Tm trc ng phng ca Chng minh rng hi m thay icc trc ng phngun i qua 1 im c nh www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 45 E LP : -PHNG TRNH:

-nh tren truc ln :-nh tren truc nho:-o dai2 truc -Tieu iem : - Tieu c : -Tam sai : -Ban knh tieu -PT ng chuan : -PT tiep tuyen tai iem M0(x0 ; y0) : -k tiep xuc cua (E) vi

2 212 2x ya b

(b2 = a2 c2; a,b,c > 0 ) A1( -a ; 0) ; A2 (a ; 0) B1(0 ; -b ) ; B2( 0 ; b) 2a ; 2b F1(-c ; 0) ; F2 (c ; 0) 2c e = ca 0 ) A1(0 ; -a) ; A2(0 ; a) B1(- b ; 0) ; B2(b ; 0) 2a ; 2b F1(0 ; - c) ; F2(0 ; c) 2c e = ca < 1 MF1 = a +ey; MF2 = a ey2a aye c www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 46 PP GIAI MOT SO DANG TOAN VE E LPVane1 : - Xac nh cac yeu to cua E lp - Lap PT chnh tac cuaE lp - Tm iem tren E lp( Da vao gia thiet bai toan ap dung cac tnh chat cua E lp co lien quan , e giai bai toan ) Vane2 :-VietPT tiep tuyen cua E Lp (E) :2 212 2x ya b Dang 1:Tiep tuyen tai iem M0(x0 ; y0)e(E)(Dung cong thc phan oi toa o)Dang 2:Tiep tuyen (D) co phng cho trc( Tiep tuyen co he so goc k cho trc ; Tiep tuyen song song hoac vuong goc vi ng thang cho trc ) CACH 1 : - Viet dang p/t tiep tuyen (D) ( Nh bai ng tron dang 2) Dung k tiep xuc e tm he so con lai cua tiep tuyen . Roi viet p/t tiep tuyenCACH 2 :( Tm toa o tiep iem , roi viet p/t tiep tuyen ) - Goi I (x0 ; y0) la tiep iem cuatieptuyen (D) vi (E )=>p/t (D) co dang :0 012 2xx yya b -Dung k co cung he so goc hoac tch hai he so goc bang 1 , e lap phng trnh bac nhat hai an x0 , y0. (V du : (D) song song vi ng thang Ax + By + C = 0 0 02 20x ya bA B ) (A):Ax+By+C=0A2a2 + B2 b2 = C2 0 012 2xx yyb a (cong thc phan oi toao) A2 b2 + B2 a2 = C2 www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 47 -T p/t tren ket hp vi p/t Bac hai 2 an x0 , y0 : 2 20 012 2x ya b (v M0e (E) ) lap he . Giai tm x0 , y0

-Lap p/t tiep tuyen vi x0 , y0 va tm cDang 3: Viet p/t tiep tuyen qua iem M (x1 ; y1) ( M e (E ) ) CACH 1 : - Xet ng thang (D) qua iem M va co he so goc k : y = k(x x1) + y1 - Dung k tiep xuc e giai tm k * Neu tm c hai gia tr k , ta viet c hai phng trnh tiep tuyen cua ( E ) qua M * Neu tm c mot gia tr k , Ta xetng thang qua M co p/t : x x1 = 0 . Kiem tra qua k tiep xucxem ng thang co phai la tiep tuyen cua (E ) khongCACH 2 :- Goi M(x0 ; y0) la tiep iem cua (D) va (E ) , khi o p/t tiep tuyen co dang : 0 012 2xx yya b

- Ta co he : 2 20 0102 2. .1 0 1 012 2x yM Ea bx x y yM Da b . Giai he tm x0 , y0 ; roi viet p/t tiep tuyenVane3 :Tap hp iem M (x ; y) la E lpAp dung :Tap hp iem M la E Lp(E ) neu thoa 1 trong 2 tnh chat sau : 1. Tong khoang cach t M en hai iem co nh F1 ; F2 la 1 hang so bang 2a ( a > 0 ) . Ngha la : (E ) = { M / F1M + F2M = 2a }. trong o F1 , F2 la hai tieu iem cua (E ) , va F1F2 = 2c( a > c ) www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 48 2.Ty so gia khoang cach t iem M en iem co nh F, vi khoang cach t M en 1 ng thang co nh (A) la 1 hang so dng nho hn 1 ( Ky hieu la e < 1, va goi la tam sai ) . Ngha la : (E ) : {M / [ , ]MFed M} . Vi F goi la tieu iem , (A) goi la ng chuan . BI TP 395)Cho elip( )2 2: 9 4 36 E x y + = . 1) Tm tocc nh, tiu im, tnh tm sai ca (E). 2) Cho( ) 1;1 M , lp PT ng thng qua M v ct (E) ti hai im A, B :MA MB = . 396)Lp PT chnh tc cuae elip (E) , bit: 1) (E) i qua cc im ( ) ( )3 3; 2 , 3; 2 3 M N . 2) Hai tiu im( ) ( )1 22;0 , 2;0 F F v a) trc ln cdi bng 4. b) (E) i qua gc to . 397)Cho elip( )2 2:16 25 100 E x y + = . 1) Tm tocc nh, tiu im, tnh tm sai ca (E). 2) Tm toca im( ) M E e , bit2Mx = . Tnh khong cch t M n hai tiu im cuae (E). Tm tt c cc gi tr ca b ng thngy x b = +c im chung vi (E). Cho elip( )2 2: 4 9 36 E x y + = . 1) Tm tocc nh, tiu im, tnh tm sai ca (E). 2) Cho( ) 1;1 M , lp PT ng thng qua M v ct (E) ti hai im A, B :MA MB = . 398)Trong h toOxy cho hai im( ) ( ) ( )1 24;0 , 4;0 0;3 vF F A . 1) Vit PT chnh tc ca elip (E) i qua A v nhn 1 2; FFlm cc tiu im. 2) Tm taim( ) M E esao cho 2 12 MF MF = . www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 49 399)Vit PT chnh tc cuae elip (E), bit: 1) Trc ln thuc Ox,di trc ln bng 8; trc nh thuc Oy cdi bng 6. 2) Trc ln thuc Oy cdi bng 10, tiu c bng 6. 3) Hai tiu im thuc Ox; trc ln cdi bng 26, tm sai 1213e = . 4) (E) i qua cc im( ) ( ) 4;0 , 0;3 M N . 5) Hai tiu im:( ) ( )1 21;0 , 5;0 F F ; tm sai 35e = . 6) (E) c tm( ) 1;1 I , tiu im( )11;3 F , trc nh cdi bng 6. 400)Tm tm sai ca elip (E) ,bit: 1) Cc nh trn trc nh nhn on thng ni hai tiu im di mt gc vung. 2) di trc ln bng hai lndi trc nh. 3) Khong cch gia hai nh, mt nh trn trc ln v nh kia thuc trc nh bng tiu c ca (E). 401)Chng t rng PT: 2 20 . 0, . 0 viAx By F AB AF + + = > 1) L PT ca mt elip nu 2 204 4c da ea c| |+ > |\ .. Tm tocc tiu im ca elip. 2) L mt im nu 2 204 4c dea c+ = . 403)Cho elip( )2 2: 4 9 36 E x y + = . 1) Vit (E) di dng chnh tc, txc nh tocc nh, cc tiu im v tnh tm sai ca (E). 2) Tm tt c cc gi tr ca m ng thng() : 2 0 d x y m =tip xc vi (E). 3) Tm tt c cc gi tr ca m ng thng (d) ct (E) ti hai im A,B:1 AB = . . TIP TUYN CA ELIP. 404)Vit PT tip tuyn ca elip( )2 2: 116 9x yE + = , bit: 1) Tip tuyn i qua im( ) 4;0 A . 2) Tip tuyn i qua im( ) 2; 4 B . 3) Tip tuyn song song vi ng thng() : 2 6 0 x y A + = . 4) Tip tuyn vung gc vi ng thng() : 0 x y A = . www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 50 405)Vit PT tip tuyn ca elip( )2 2: 19 4x yE + =bit tip tuyn to vi ng thng () : 2 0 x y A =mt gc 045 o = . 406)Vit PT tip tuyn chung ca hai elip sau: ( ) ( )2 2 2 21 2: 1, : 19 4 4 9x y x yE E + = + = . 407)Vit PT cc ng thng cha cc cnh ca hnh vung ngoi tip elip 2 213 6x y+ = . 408)Cho elip( )2 2: 19 4x yE + = . Vit PT tip tuyn vi (E) i qua im( ) 3; 2 A . Tm toca tip im ? 1) Vit PT ca elip( ) Ec tiu c bng 8, tm sai 45e =v cc tiu im nm trn Ox, i xng nhau qua trc Oy. 2) Vit PT cc tip tuyn ca (E) i qua im ( )150;4A . 3) Tnh din tch hnh phng chn bi (E) v hai tip tuyn ni trn. 409)Cho elip( )2 2: 19 5x yE + = . Mt hnh ch nht c gi l ngoi tip eipnu mi cnh ca hnh ch nht u tip xc vi (E). Trong tt c cc hnh ch nht ngoi tip (E), hy xc nh: 1) Hnh ch nht c din tch nh nht. 2) Hnh ch nht c din tch nh nht.410)Vit PT cc cnh ca hnh vung ngoi tip elip( )2 2: 124 12x yE + = . QU TCHI VIELIP. 411)(H Hu_96) Cho elip( )2 22 2: 1x yEa b+ = . Gi 1 2A Al trc ln ca (E). K cc tip tuyn 1 1 2 2, At A tca (E). Mt tip tuyn qua im( ) M E e , ct 1 1 2 2At A t v theo th t ti 1 2T T v. 1) CMR: Tch s 1 1 2 2. ATATkhng ph thuc vo v tr im M . 412)Cho h elip( ) ( )22: 2 0 1xE y x mm= < < . 1) a (E) v dng chnh tc, xc nh toca tm, cc tiu im 1 2, F Fv cc nh 1 2, A Athuc trc ln ca (E). 2) Tm qu tch cc nh 1 2, A Akhi m thay i. 3) Tm qu tch cc tiu im 1 2, F Fkhi m thay i. www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 51 413)CMR: iu kin cn v ng thng() : 0 d Ax By C + + =( )2 20 A B + >tip xc vi elip( )2 22 2: 1x yEa b+ =l : 2 2 2 2 2C A a B b = + . 414)CMR: iu kin cn v ng thng() : d y kx m = +tip xc vi elip ( )2 22 2: 1x yEa b+ =l : 2 2 2 2m ka b = + . BI TP HYPEBOL 415)Lp pt chnh tc ca ypebobit: 1 Na trc thc bng 4 tiu c bng 10 2 Tiu c bng13 2 mt tim cn :x y32=3, tm sai5 = e , (H) qua) 6 ; 10 ( M4di trc o bng 12 tm sai 45= e5 1 nh-10; 0 v mt tim cn :x y52=6i qua 2 im) 5 2 ; 2 5 ( Av) 40 ; 45 ( B7qua M24;5 v mt tim cn:x y125 =8 gc gia 2 tim cn bng 600, (H) qua M(6;3) 9i qua) 3 ; 2 4 ( Mc tiu im trng vi tiu im ca : 2x2+7y2=70 10, (H) qua)59;534 4( M v M nhn 2 tiu im di 1 gc vung 416)Tm tm sai ca :12222= byax bit: 1 hong cch gia tiu im v nh trn trc o bngdi trc thc 2 nh trn trc o nhn F1, F2 di 1 gc 1200 3 ypebo c hai tim cn vung gc giypebo vung 4di trc thc gp 2 ndi trc o 417)Cho Hypebol (H):12222= byax Mmt im tu trnd1); (d2 n tcc ng thng qua M v song song vi 2 tim cn caCMR din tch hnh bnh hnh to bi d1), (d2 v tim cn hng i www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 52 418)Lp phng trnh chnh tc ca ypebo bit tngdi 2 bn trc: a+b7 v phng trnh hai tim cn:.43x y =419)Cho h ng cong Cm):1252222=+mymx( 5 ; 0 = = m m )1 Tu theo gi tr ca m hy xc nh hi no Cm) l elip, khi no l Hypebol? 2 i s 1 im tu trn ng thng x1 vhng thuc trc honh CMR qua c 4 ng cong ca h Cm i qua Trong 4 ng y c bao nhiu elip, bao nhiu Hypebol? 420)Trong mp toOxy cho -2; 0); B(2; 0) v M(x; y).1 Xc nh toM bit M nm pha trn trc honh v s o gc 090 = AMB ; s o gc MAB = 300. 2 Khi x y thay i sao cho s o gc M gp i s o gc M i M chy trn ng cong no? 421)Cho Elip (E): x2 + 3y2 = 9 i hypebo c cc tiu im trng vi 2 nh trn trc n ca ; 2 tim cn cha 2 ng cho ca hnh ch nht c s caVit pt ca422)Cho Hypebol (H): 5x2 y2 4 = 0. Tm cc nh tiu im tm sai c phng trnh cc tim cn Cho (H): 9x2 16y2 144 = 0 1 Tm cc tiu im cc nh tm sai v phng trnh cc tim cn ca2 Vit pt chnh tc ca ip c cc tiu im trng vi cc tiu im cav ngoi tip hnh ch nht c s ca423)Cho (H):19 42 2= y x 1 Xc nh tocc nh cc tiu im tm sai v cc tim cn caV2 Tm n ng thng ynx 1 c im chung vi424)Cho hypebol (H): ) 0 ( 12222> > = b abyax Cho mt s thc dng Xt cc ng thng d1): y = kx; (d2):xky1 =a. Hy tm k sao cho (d1) v (d2 u ctb iv C n tgiao im ca d1 vi nm trong gc phn t th nht ivn tgiao im ca d2 vi nm trong gc phn t th haiy tm sao cho hnh thoi C c din tch nh nht 425)Cho Hypebol (H): 116 42 2= y x. www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 53 1 Tm phng trnh ng cho ca hnh ch nht tm O c 4 nh thucsao cho h s gc cc ng chos nguyn 2 i mt ng thng bt ct hai tim cn ti P Q; ctti R S Chng minh rng: PRQS 3 Ly im K thucT K hai ng thng n t song song vi hai tim cn Chng minh din tch hnh bnh hnh gii hn bi hai ng thngv hai tim cn c din tch hng i 4 i s hai ng thng qua tm O v vung gc vi nhau ctti 4 im to thnh mt hnh thoi Vit phng trnh hai ng thnghi hnh thoi c din tch nh nht 5. Cho A(-20 Tm hai imC thuc nhnh phi casao cho tam gic Ctam gic u 426)Trong mt phng Oxy cho Hyperbol (H) :19 42 2= y x i dng thng qua O c h s gc ; d'ng thng qua O v vung gc vi d a Tm iu in ca d v d' u ctb Tnh theodin tch ca hnh thoi vi 4 nh4 giao im ca d;d' v ). 427)Trong mt phng taOxy cho yperboc phng trnh : 9x2- 16y2= 144 a Tm tanhtiu imtm sai tim cn cab Lp phng ng trn Cng nh F1F2 Vi F1F2hai tiu im ca428)Trong mt phng Oxy cho hypeboc phng trnh : 12222= byax Tiu im F1Tm im M trnsao chodi MF1 ngn nhtdi nht 429)Trong mt phng Oxy cho ip:14 92 2= +y x V Hyperbol (H) :14 12 2= y x www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 54 Lp phng trnh ng trn i qua giao im ca ip v yperbo 430)Trong mt phng toOxy cho yperboc phng trnh :20x2- 25y2 = 100a Tnh hong cch t im c honh 10 = x n 2 tiu imb Tm b phng trnh ng thng : yx+b c im chung vi yperbol trn. 431)Trong mt phng Oxy cho yperboc phng trnh : (H):12222= byax a Tnh hong cch t im M thucn tim cn ca n b T im M cc ng thng song song vi 2 tim cn v ct chng ti P;QTnh din tch t gicOPMQ 432)Cho Hypebol (H):19 162 2= y x Lp pt chnh tc eipbitc 2 tiu im2 tiu im cavi qua cc nh ca hnh ch nht c s ca433)Cho Hypebol (H):12222= byax CMR: Tch hong cch t 1 im M0 bt trnn 2 ng tim cn1 s hng i 434)Cho Hypebol (H):13 62 2= y x Tm toim M thucsao cho M nhn 2 tiu im F1, F2 cadi 1 gc 900. 435)Cho Elip (E): 4x2 + 9y236 Lp pt ypebo c 2 tiu im trng vi 2 tiu im cua v i qua) 1 ; 2 2 ( A436)Cho (H):112 42 2= y x Tm M trnsao cho hong cch t M n tiu im tri gp i hong cch t M n tiu im phi ca437)Cho A(-a; 0; a;0 iCng trn thay i qua; MMng nh ca C un song song vi Ox Tm qu tch M v M 438)Cho M( ) tan ;cost bta, ( )2) 1 2 (t+ = k tTm qu tch im M www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 55 Tm qu tch tm cc ng trn chn trn Ox v Oy hai on thng cdi n t2a v 2b 439)Cho Hypebol (H):12222= byax. CMR: vi mi M trnta c: 1, OM2 MF1.MF2 = a2 b2. 2, (MF1 + MF2)2 = 4(OM2 + b2) 440)Cho (H):15 42 2= y x Tm im M trnnhn 2 tiu im di 1 gc 1200. 441)Cho (H):116 92 2= y x Mt ng thng d qua tiu im F1 vung gc vi trc thc ct ti M N Tnh MN 442)Cho (E): 16x2 + 9y2 = 144 1 Vit pt chnh tc cac cng hnh ch nht c s vi2 Tm M trnnhn 2 tiu im di 1 gc 600. 3, Tm M trn (H) sao cho 2 bn nh qua tiu vung gc vi nhau 443)Cho F4; 0 v ng thng d: 4x 9 = 0. Tm tp hp cc im M trn mp tosao cho t s hong cch t M n F v t M n d bng 34 444)Tnh din tch hnh ch nht c nh nm trn :116 202 2= y x hai cnh i qua 2 tiu im v song song vi Oy 445)Cho (H):14 252 2= y xv ng thng d: 2x + 15y 10 = 0 1 CMR: d un ctti 2 im phn bit xA>0). Tnh AB 2 Tm C trnsao cho tam gic C cn A. 446)Cho (H): x2 2y2 6 v M3;1 Lp pt ng thng d qua M ctti 2 imv B sao cho MA = MB. 447)Cho (H): x2 4y2 32 v ng thng d: x + 6y0 1 CMR: d un ctti 2 im phn bit Tnh2 Tm im C nm trnsao cho tam gic C c din tch bng 30

448)Cho A(4; 1) v (H):14 22 2= y x. Tm M trnsao cho M ngn nht www.VNMATH.comCHUYN BI TP HNH HC LP 10 ( c s dng ti liu t cc ngun khc). BIN SON : TRN MAI SANG 56 www.VNMATH.com