PHNG PHP TO TRONG MT PHNG Vect & To Ta ca im va vect: M(x;y) OM = x.i + y.j a = (a1;a2) a = a1.i+ a2.j trong i = (1;0), j = (0;1) l cc vect n v ca trc. Gi sa = (a1;a2) v b = (b1;b2). Vect bng nhau To vect tng, hiu, tch vect vi 1 s: a = b a1 = b1, a2 = b2 a b= (a1 b1;a2 b2) ka = (ka1;ka2) To ca AB: AB = (xB xA;yB yA). di ca vect: ,a ,= a ,2+a _2 Khong cch gia 2 im: AB = ,AB ,= (xBxA)2+(yByA)2 To im M chia on AB theo t s k 1: MA = kMB xM = ABxkx 1k , yM = AByky 1k To trung im M ca on AB: xM = ABxx 2+ , yM = AByy 2+ To trng tm G ca ABC: xG = ABCxxx 3++ , yG = ABCyyy 3++ Vect cng phng: a , b a = kb 1212aab b=(b1b2 = 0) Tch v hng ca 2 vect:a.b = ,a , .,b , .cos(a,b) a b a.b = 0 a.b = a1b1 + a2b2. a2 = |a|2 Gc ca 2 vect: cos(a,b ) = a.b| a | . | b |G GG G= 112222221212a bab aa.bb+ ++. 1/ Cho 3 im A(3;5), B(3;3), C(1;2). . Chng minh rng A, B, C l cc nh ca 1 tam gic. Tm to im D sao cho ABDC l hnh bnh hnh. . Tm to im E sao cho AE = 2AB 3AC. . Tnh chu vi v din tch ABC. . Tm to trng tm G, to trc tm H ca ABC, to tm I cang trn ngoi tip ABC. Chng minh I, H, G thng hng. . Tm giao im ca ng phn gic ngoi gc A vi BC. -2- GV: Nguyn Vit Lc2/ Cho ABC vi A(1;0), B(4;0), C(0;m) vi m 0. Tnh to trng tm G ca ABC theo m. Xc nh m GAB vung ti G. 3/ChoABCvungcntiA.BitM(1;1)ltrungimcacnhBCv G(_;0) l trng tm ABC. Tm to cc nh A, B, C. 4/ Cho 2 im A(1;3), B(3;1). Tm to im C sao cho ABC u. 5/ . Cho ABC vung ti A, AB = c, AC = b. Gi M l im trn cnh BC sao cho CM = 2BM, N l im trn cnh AB sao cho BN = 2AN. Tm h thc lin h gia b v c sao cho AM CN. . Cho ABC vung cn ti A, tnh gc t to bi cc trung tuyn ca tam gic k t B v C. | Phng trnh ng thng. ]| Phng trnh ng thng (d) qua im M(xo;yo) vung gc vi n = (a;b): a(x xo) + b(y yo) = 0(a2 + b2 = 0) ]} Phng trnh tng qut ca ng thng: ax + by + c = 0 (a2 + b2 = 0), trong vect php tuyn n = (a;b), vect ch phng a = (b;a), h s gc k = a/b. ]~ Phng trnh ng thng i qua im M(xo;yo) c h s gc k ( Oy): y yo = k(x xo) ] ng thng c h s gc k, tung gc b (ct Oy ti B(0;b)): y = kx + b ] Phng trnh ng thng theo on chn (i qua 2 im A(a;0), B(0;b)): ] ng thng (d) i qua im M(xo;yo) c vect ch phng a = (a1;a2): {o1o2xxtayyta=+ =+ : phng trnh tham s. oo12xxyyaa=(a1.a2 = 0): phng trnh chnh tc. } V tr tng i ca hai ng thng. * Cho 2 ng thng (d1): a1x + b1y + c1 = 0, (d2): a2x + b2y + c2 = 0 t D = 1122ab ab , Dx = 1122b cb c , Dy = 1122ca ca + (d1) ct (d2) a1:a2 b1:b2 D 0(x = Dx:D, y = Dy:D) + (d1) , (d2) a1:a2 = b1:b2 c1:c2 D = 0, Dx 0 hay Dy 0 + (d1) (d2) a1:a2 = b1:b2 = c1:c2 D = Dx = Dy = 0 (E). Gi H l hnh chiu vung gc ca M ln trc Oy. Gi s AH ct OMtiP.ChngminhrngkhiMthayitrn(E)thPlunlunchytrn1 ng cong (C) c nh. V th ng cong (C). -3- 6/ Vit phng trnh tham s, phng trnh chnh tc, phng trnh tng qut ca ng thng i qua im Mo nhn a lm vect ch phng: . Mo(1;2), a = (3;1). Mo(2;1), a = (0;1). Mo(5;3), a = (2;0) 7/ Vit phng trnh ng thng i qua im Mo vung gc vi n: .Mo(1;2), n = (2;2). Mo(2;1), n = (2;0). Mo(2;1), n = (2;1) 8/ . Vit phng trnh tham s, chnh tc, tng qut ca ng thng i qua 2 im A(3;1), B(2;5). . Vit phng trnh tham s ca ng thng 2x + 3y 2 = 0. 9/ Cho ng thng d:{x 22t y3t=+ =+ v 2 im A(2;5), B(1;7). . Tm trn d im M cch im A mt khong bng 4. . Tm trn d im C sao cho ABC cn. < 10>Cho 2 im A( 4;3), B(1;5). Tm trn ng thng d: x 2y 3 = 0 im M sao cho MA2 + MB2 nh nht. < 11>. Vit phng trnh ng thng D i qua im A(2;3) ct hai .thng d:{x 72t y3t= = + v d:{x 54m y73m= + = + ti B, C sao cho A l trung im ca BC. . Lp phng trnh ng thng i qua im P(3;0) ct hai ng thng 2x y 2 = 0, x + y + 3 = 0 ti A, B sao cho P l trung im ca on AB. < 12>Cho ABC c phng trnh hai cnh x + y 2 = 0, 2x + 6y + 3 = 0, trung im mt cnh l M(1;1). Tm to cc nh. < 13>Lp phng trnh ng thng nu im P(2;3) l chn ng vung gc h t O xung ng thng ny. < 14> ChoP(2;3),Q(1;0).LpphngtrnhngthngquaQvvunggc vi PQ. < 15>Cho ABC vi A(2;1), B(1;1), C(3;2).. Lp phng trnh ng cao BH ca tam gic. . Lp phng trnh ng thng i qua nh A v song song vi BC. < 16> Chongthng(d):2x+3y+4=0.Lpphngtrnhngthngi qua im M(2;1) v: . song song vi (d).. vung gc vi (d). < 17>Cho M(2;1), N(5;3), P(3; 4) l trung im ca cc cnh AB, BC, CA ca tam gic ABC. Lpphng trnh cc cnh ca tam gic. < 18> HnhchnhtABCDchainhA(5;1)vC(0;6),mtcnhcphng trnh x + 2y 12 = 0. Tm phng trnh cc cnh cn li. < 19>Tm to trc tm ca tam gic nu bit cc cnh c phng trnh: AB: 4x y 7 = 0, BC: x + 3y 31 = 0, CA:x + 5y 7 = 0. < 20>Tm im chiu ca im P( 6;4) ln ng thng 4x 5y + 3 = 0. -4- < 21>Tm im B i xng vi im A(5;13) qua ng thng 2x 3y 3 = 0. < 22>Tm im chiu ca im P(8;12) ln ng thng i qua 2 im A(2;3), B(5;1). < 23>Tm im N i xng vi im M(8;9) qua ng thng i qua hai im A(3; 4), B(1;2). < 24>Lp phng trnh ng thng i xng ca ng thng (D): x 2y 5= 0 qua A(2;1). < 25> ChoABCviB(3;5),C(4;3),phngictronggcA:x+2y8=0. Tm phng trnh cc cnh ABC. < 26> Lpphngtrnhccngthng(d1),(d2)theothtiquaccim A(0;4), B(5;0) bit ng thng (): 2x 2y + 1 = 0 l ng phn gic ca gc to bi (d1) v (d2). < 27>Cho ng thng phng trnh (a + 2)x + (a2 9)y + 3a2 8a + 5 = 0. Vi gi trno ca a th ng thng ny : . song song trc honh.. song song trc tung.. i qua O. < 28>Lp phng trnh cc cnh v trung tuyn ca ABC nu A(3;2), B(5;2), C(1;0). < 29> ChoABCviA(3;3),B(2;1),C(11;2).Vitphngtrnhngthng qua A v chia ABC thnh hai phn c t s din tch bng 2. < 30>Cho ABC c phng trnh ng cao CH: 2x + y + 3 = 0, phng trnh phngicAD:xy=0,cnhACquaimM(0;1)vAB=2AM.Vit phng trnh cc cnh ca tam gic. < 31> ChoABCviA(3;1),B(5;7)vtotrctmN(4;1).Lpphng trnh cc cnh ca tam gic. < 32>Cho ABC c A( 4;1), B(3;2), C(8;5) v im L trn cnh AC sao cho LC = 3LA, CE l trung tuyn k t C. Tm to giao im ca BL v CE. < 33>Cho 2 im A(3;2), B(3;1) v ng thng d: x + y 4 = 0. Tm phng trnh ng thng song song vi d v ct on AB ti im M sao cho MB = 2MA. < 34>Cho ABC c phng trnh cnh AB: 5x 3y + 2 = 0, phng trnh ng cao AH: 4x 3y + 1 = 0 v BK: 7x + 2y 22 = 0. Lp phng trnh hai cnh cn li v ng cao th ba. < 35>Lp phng trnh cc cnh ca ABC bit nh B( 4;5) v phng trnh hai ng cao AH: 5x + 3y 4 = 0, CK: 3x + 8y + 13 = 0. < 36> LpphngtrnhcccnhcaABCbitnhA(5;2)vphngtrnh hai trung tuyn BM: 5x + 4y = 0, CN: 3x y = 0. < 37> LpphngtrnhcccnhcaABCbitnhA(4;1)vphngtrnh hai phn gic BD: x 1 = 0, CE: x y 1 = 0. < 38> LpphngtrnhcccnhcaABCbitnhB(2;6)vphngtrnh ng cao AH: x 7y + 15 = 0, phng trnh phn gic gc A: 7x + y + 5 = 0. GV: Nguyn Vit Lc-5- < 39> LpphngtrnhcccnhcaABCbitnhB(2;1)vphngtrnh ng cao AH: 3x 4y + 27 = 0, phng trnh phn gic CE: x + 2y 5 = 0. < 40> LpphngtrnhcccnhcaABCbitnhC(4;1)vphngtrnh ng cao AH: 2x 3y + 12 = 0, phng trnh trung tuyn AM: 2x + 3y = 0. < 41> LpphngtrnhcccnhcaABCbitnhB(2;7)vphngtrnh ng cao AH: 3x + y + 11 = 0, phng trnh trung tuyn CM: x + 2y + 7 = 0. < 42> LpphngtrnhcccnhcaABCbitnhC(4;3)vphngtrnh phn gic AD: x + 2y 5 = 0, phng trnh trung tuyn AM: 4x +13y 10 = 0. < 43> LpphngtrnhcccnhcaABCbitnhA(3;1)vphngtrnh phn gic BD: x 4y+10 = 0, phng trnh trung tuyn CM: 6x +10y 59 = 0. < 44>Cho phng trnh 2 phn gic AD: y + 4 = 0, BE: 7x + 4y + 5 = 0 ca 1 tam gic v phng trnh cnh AB: 4x + 3y = 0. Vit ph.trnh 2 cnh cn li. < 45>Lp phng trnh ng thng i qua im M(3;7) v chn trn cc trc to nhng on khc 0 bng nhau. < 46> LpphngtrnhngthngiquaimM(2;3)vchntrncctrc to nhng on bng nhau. < 47> LpphngtrnhngthngquaimM(4;3)vcttrchonh,trc tung ln lt ti A, B tho 5AM 3MB = 0. < 48>Lp phng trnh ng thng i qua im C(1;2) to vi gc to mt tam gic c din tch bng 2 vdt. < 49>Lp phng trnh ng thng qua im M(3;1) v ct hai na trc Ox, Oy ti A, B sao cho: . Din tch OAB nh nht.. OA + OB nh nht. < 50>Xc nh gi tr m v n hai .thng mx + 8y + n = 0 v 2x + my 1 = 0 . song song.. trng nhau.. vung gc. < 51> .Xcnhgitrmhaingthng(m1)x+my5=0,mx+ (2m 1)y + 7 = 0 ct nhau ti 1 im trn trc honh. . nh m ba ng thng (1): 2x + y 1 = 0, (2): x + 2y + 1 = 0, (3): mx y 7 = 0 ng quy ti mt im. < 52>Tm giao im 2 ng thng trong cc trng hp sau: . x + 5y 35 = 0, 3x + 2y 27 = 0.. {x 2t y3t= = , {x 3t1 y6t3=+ =+ . . 3x + 5y 4 = 0, 6x + 10y 7 = 0..x y2 = 0, x2 2y = 0. < 53>Cho a2 + b2 > 0 v 2 .thng (d1): (a b)x + y = 1, (d2): (a2 b2)x + ay = b. Xc nh giao im ca (d1) v (d2), tm iu kin i vi a, b giao im nm trn trc honh. < 54>Chng minh rng ng thng (d): (m 2)x + (m 1)y + 2m 1 = 0 lun i qua 1 im c nh. -6- ~ Khong cch t mt im n mt ng thng. ]| Khong cch t im M(xo;yo) n ng thng (): ax + by + c = 0: d(M,) = oo 22axbycab+++t F(x,y) = ax + by + c. * M1(x1;y1), M2(x2;y2) nm v khc pha i vi () F(x1,y1).F(x2,y2) < 0 * M1(x1;y1), M2(x2;y2) nm v cng pha i vi () F(x1,y1).F(x2,y2) > 0 ]} Phng trnh ng phn gic gc to bi 2 ng thng (d1), (d2): 11122222221122a xbycaxbycabab++++= ++< 55>Tnh khong cch t im A n ng thng (d): . A(2;1), (d): 4x + 3y + 10 = 0.. A(0;3), (d): 5x 12y 23 = 0. . A(2;3), (d): 3x 4y 2 = 0.. A(1;1), (d): {x 12t y2t= + =+ . < 56>Tnh khong cch gia 2 ng thng song song: . 3x 4y 10 = 0, 6x 8y + 5 = 0. . 5x 12y + 26 = 0, 5x 12y 13 = 0. < 57>Tnh din tch hnh vung c 1 cnh nm trn ng thng x 2y 7 = 0 v nh A(2;5). < 58>Cho ABC c ph.trnh cc cnh AB: x +21y 22 = 0, BC: 5x 12y +7 =0, CA: 4x 33y +146 = 0. Tnh khong cch t trng tm tam gic n cnh BC. < 59>Tnh din tch hnh vung c hai cnh nm trn hai ng thng 5x 12y 65 = 0, 5x 12y + 26 = 0. < 60>Khong cch t im M n cc ng thng 5x 12y 13 = 0 v 3x 4y 19 = 0 bng tng ng l 3 v 5. Tm to im M. < 61> VitphngtrnhngthngiquaimM(2;4)cchimA(0;3)khong cch bng 1. < 62> VitphngtrnhngthngiquaimA(2;5)cchimB(5;1)khong cch bng 3. < 63> VitphngtrnhngthngiquaimM(1;2)ckhongcchnim A(2;5) bng na khong cch n im B(1;8). < 64> Tmphngtrnhngthng(d) cchimA(1;1)mtkhongbng2, cch im B(2;3) mt khong bng 4. < 65>Tm phng trnh .thng (d) c h s gc bng 2, cch im A(2;3) mt khong bng 4. < 66>Vit phng trnh ng thng vung gc vi .thng 2x + 6y 3 = 0 v cch im A(5;4) khong cch l 10. GV: Nguyn Vit Lc-7- < 67>Lp phng trnh ng thng song song vi ng thng 3x 4y 10 = 0 v cch ng thng ny khong cch bng 3. < 68>Vit phng trnh ng thng i qua gc to to vi hai ng thngx y + 12 = 0, 2x + y + 9 = 0 mt tam gic c din tch bng 1,5 vdt. < 69>Cho 2 im A(2;2), B(5;1). Tm trn ng thng (d): x 2y + 8 = 0 im C sao cho dt(ABC) = 17 vdt. < 70>Cho ABC c A(2;3), B(3;2) v dt(ABC) = 15 vdt. Tm im C bittrng tm G nm trn ng thng (d): 3x y 8 = 0. < 71> imE(1;1)ltmmthnhvungcmtcnhnmtrnngthng x 2y + 12 = 0. Lp phng trnh cc cnh cn li. < 72>Lp ph.trnh cc cnh hnh vung c 2 nh lin tip A(2;0), B(1;4). < 73> LpphngtrnhcccnhcahnhvungbitnhA(5;1)vphng trnh mt cnh l 4x 3y 7 = 0. < 74> Lpphngtrnhtphpnhngimckhongcchnngthng 8x 15y 25 = 0 bng 2. < 75>Lp phng trnh tp hp nhng im cch u 2 ng thng song song: . 3x y + 7 = 0, 3x y 3 = 0.. 5x 2y 6 = 0, 10x 4y + 3 = 0. < 76>Tm trn ng thng 2x y 5 = 0 im P sao cho tng khong ccht n hai im A(7;1) v B(5;5) l nh nht. < 77>Tm trn ng thng 3x y 1 = 0 im P sao cho hiu khong ccht n hai im A(4;1) v B(0;4) l ln nht. < 78>Cho ng thng (): mx + y m 2 = 0 v 2 im A(2;1), B(4;2). . nh m () ct on AB. . nh m khong cch t gc to O n () l ln nht. < 79>Lp ph.trnh cc ng phn gic gc nhn v t to bi 2 ng thng: . x 2y 3 = 0, 2x + 4y + 7 = 0.. 3x + 4y 1 = 0, 5x + 12y 2 = 0. . x 3y + 5 = 0, 3x y 2 = 0 < 80>Cho 2 ng thng d1: x + 2y 11 = 0 v d2: 3x 6y 5 = 0. Lp phng trnh ng phn gic ca gc to bi d1 v d2 m c cha im M(1;3). < 81> LpphngtrnhngthngiquaimP(2;1)tovihaingthng (d1): 2x y + 5 = 0, (d2): 3x + 6y 1 = 0 mt tam gic cn c nh l giao im ca (d1) v (d2). Gc gia hai ng thng.cos(d1,d2) = |cos(n1, n2)| = 1212 22221122| aabb|ab.ab+++. < 82>Tm gc gia 2 ng thng: . 5x y + 7 = 0, 3x + 2y = 0.. 3x y + 5 = 0, 2x + y 7 = 0. < 83> VitphngtrnhngthngiquaimM(2;1)tovingthng (L): x = 1 + t, y = 2 3t 2 gc 45o. -8- < 84>nh m gc to bi 2 .thng 4x + my 20 = 0, 2x 3y + 6 = 0 l 45o. < 85>Cho phng trnh cc cnh ca tam gic: 3x + 4y 1 = 0, x 7y 17 = 0, 7x + y + 31 = 0. Chng minh rng tam gic ny cn. < 86> imA(4;5)lnhcamthnhvungcmtngchonmtrn ng thng 7x y + 8 = 0. Lp phng trnh cc cnh v ng cho th hai ca hnh vung. < 87>Mt hnh vung c tm I(2;3) v 1 cnh c phng trnh: x 2y 1 = 0. Lp phng trnh cc ng cho v cc cnh cn li ca hnh vung. < 88>Lp phng trnh cnh hnh vung c 2 nh i din A(1;3), C(6;2). < 89>Cho ABC cn nh C(4;3), phng trnh cnh AC: 2x y 5 = 0, phng trnh cnh AB: x y = 0. Vit phng trnh cnh BC. < 90>Cho ABC cn ti A, phng trnh cnh BC: x + 2y 3 = 0, cnh AB: 2x + y 3 = 0. Lp phng trnh cnh AC bit AC i qua im M(2;1). < 91>Cho ABC vung cn ti A(4;1), ph.trnh cnh huyn 3x y + 5 = 0. Tm phng trnh 2 cnh gc vung. < 92>Cho 2 ng thng (d1): x + y 1 = 0, (d2): x 3y + 3 = 0. Vit phng trnh ng thng (d) i xng ca (d1) qua (d2). ng trn. ]| Phng trnh ng trn (C) tm I(a;b) bn knh R: (x a)2 + (y b)2 = R2 x2 + y2 2ax 2by + c = 0 vi R2 = a2 + b2 c > 0 ]} Phng trnh tip tuyn ca ng trn: a. Tip tuyn vi (C) ti im M(xo;yo) (C): Tip tuyn qua M vung gc vi IM = (xo a;yo b) do c phng trnh: (xo a)(x xo) + (yo b)(y yo) = 0. b. Tip tuyn vi (C) i qua im M(xo;yo) e (C):+ Phng trnh tip tuyn () c dng: a(x xo) + b(y yo) = 0 ax + by axo byo = 0 (a2 + b2 0) + () tip xc (C) d(I,) = R, t tm c a, b. c. Phng trnh tip tuyn vi (C) c h s gc k: * Nu tip tuyn c h s gc k: phng trnh tip tuyn c dng y = kx + m * Nu tip tuyn song song ng thng ax + by + c = 0: phng trnh tip tuyn c dng ax + by + m = 0. * Nu tip tuyn vung gc ng thng ax + by + c = 0: phng trnh tip tuyn c dng bx ay + m = 0. + () tip xc (C) d(I,) = R, t tm c m. < 93> Lp phng trnh ng trn trong cc trng hp sau: . Tm O bn knh R = 3. . Tm I(2;3) bn knh R = 7. . i qua gc O v c tm I(6;8).. i qua A(2;6) c tm I(1;2). GV: Nguyn Vit Lc-9- . Tm O v tip xc vi ng thng 3x 4y + 20 = 0. . Tm I(1;1) v tip xc vi ng thng 5x 12y + 9 = 0. . i qua im B(9;9) v tip xc vi trc honh ti A(6;0). . i qua A(3;1), B(1;3), tm I nm trn ng thng 3x y 2 = 0. . i qua 3 im A(1;3), B(0;2), C(1;1). !0. i qua im M(1;2) v tip xc vi cc trc to . !1. i qua 2 im A(1;3), B(4;2) v tip xc vi ng thng x y + 1 = 0. !2. Qua A(1;1), B(0;2) v tip xc vi ng trn (x 5)2 + (y 5)2 = 16. !3. Bn knh R = 5, tip xc vi .thng x 2y 1 = 0 ti im A(3;1). !4. i qua A(2;1), tip xc vi hai .thng 2x + y 5 = 0, 2x + y + 15 = 0. !5. i qua M(1;2), tip xc vi hai .thng 7x y 5 = 0, x + y + 13 = 0. !6. Tm I(3;1), 1 dy cung di bng 6 nm trn .th. 2x 5y + 18 = 0. !7. Tm nm trn ng thng 2x + y = 0 v tip xc vi 2 ng thng 4x 3y + 10 = 0, 4x 3y 30 = 0. !8. Tm nm trn .thng 4x 5y 3 = 0 v tip xc vi 2 ng thng 2x 3y 10 = 0, 3x 2y + 5 = 0. !9. Tip xc vi 3 ng thng: 4x 3y 10 = 0, 3x 4y 5 = 0, 3x 4y 15 = 0. < 94>Phng trnh no xc nh 1 ng trn. Tm tm v bn knh. . x2 + y2 4x + 6y 3 = 0.. x2 + y2 8x = 0. . x2 + y2 + 4y = 0.. x2 + y2 2x + 4y + 14 = 0 . x2 + y2 + 4x 2y + 5 = 0.. x2 + y2 + 6x 4y + 14 = 0. < 95> Binlunstnggiaocangthng(D):mxy2m+3=0v ng trn (C): x2 + y2 2x + ' = 0. < 96>Lp phng trnh ng knh ca ng trnx2 + y2 + 4x 6y 17 = 0bit rng ng knh vung gc vi ng thng 5x + 2y 13 = 0. < 97>. Lp phng trnh tip tuyn vi ng trn (x + 2)2 + (y 3)2 = 25 ti im A(5;7). . Lp phng trnh tip tuyn vi ng trn x2 + y2 + 2x 19 = 0 bit tip tuyn i qua im A(1;6). Vit phng trnh ng thng qua 2 tip im. . Lp phng trnh tip tuyn vi ng trn x2 + y2 + 10x 2y + 6 = 0 bit tuyn song song vi ng thng 2x + y 7 = 0. . Lp phng trnh tip tuyn vi ng trn x2 + y2 2x + 4y = 0 bit tiptuyn vung gc vi ng thng x 2y + 9 = 0. -10. im M(25;2) nm trn Elip v trc nh ca n bng 6. . im M(2;2) nm trn Elip v na trc ln bng 4. . Hai im M(4; 3 ), N(22;3) nm trn Elip. . im M(15;1) nm trn Elip v tiu c bng 8. !0. im M(2; ` ) nm trn Elip v tm sai bng _. !1.imM(8;12)nmtrnElipvbnknhquatiuimbntrica im M l r1 = 20. !2. i qua im M( 32 4 ;_) v M nhn on ni 2 tiu im di gc . !3. Khong cch gia 2 ng chun bng 5, tiu c bng 4. !4. Khong cch gia 2 ng chun bng 16, trc ln bng 8. !5. Khong cch gia 2 ng chun bng 13, trc nh bng 6. !6. Khong cch gia 2 ng chun bng 32, tm sai bng . !7. im M( 5;2) nm trn Elip v kh.cch gia hai .chunbng 10. < 99>Lp phng trnh Elip bit: . Phng trnh cc cnh hnh ch nht c s l x 4 = 0, y 3 = 0. . Tm O, hnh ch nht c s c 1 cnh nm trn ng thng x 2 = 0 v c ng cho bng 6. . Mt nh l (5;0), phng trnh ng trn ngoi tip hnh ch nht cs l x2 + y2 = 41. < 1= 00>Tm di cc trc, nh, tiu im, tm sai ca (E): . x2 + 25y2 = 25.. x2 + 5y2 = 15.. 4x2 + 9y2 = 25. . 9x2 + 25y2 = 1.. x2 + 4y2 = 1. < 1= 01>Tnh din tch t gic c 2 nh l 2 tiu im, 2 nh cn li l 2 nh trn trc nh ca Elip x2 + 5y2 = 20. < 1= 02> KimchngrngimM(4;- ' - )nmtrnElip16x2 +25y2 =400,tnh cc bn knh qua tiu im ca im M. < 1= 03>Tm cc im M trn Elip 7x2 +16y2 = 112 sao cho F1M = 2,5. < 1= 04>Xt v tr tng i ca ng thng v elip: . 2x y 3 = 0, 22xy 169+= 1,. 2x + y 10 = 0, 22xy 94+= 1, . 3x + 2y 20 = 0, x2 + 4y2 = 40. < 1= 05>Tm tm sai ca Elip trong cc trng hp: . di trc ln bng k ln di trc nh (k > 1). . Mi tiu im nhn trc nh di gc 2. . nh trn trc nh nhn 2 tiu im di gc 2. (i = 1, 2) < 98>Lp phng trnh chnh tc ca Elip nu: . Trc ln bng 10, tiu c bng 8. . Trc nh bng 24, tiu c bng 10. . Tiu c bng 8, tm sai bng ` .. Trc ln bng 20, tm sai bng ` . . Trc nh bng 10, tm sai bng ' . GV: Nguyn Vit Lc-20- . nh m (Cm) l ng trn c bn knh bng 1. Gi ng trn ny l (C). Vit phng trnh ng thng (d) tip xc vi ng trn (C) ti im A(1 + ;1 ). . Vit ph.trnh tt c cc tip tuyn vi (C) bit chng vung gc (d). < 54>Cho ng trn (Cm): x2 + y2 12mx 2(m + 1)y + (m + 1)2 = 0. . Bin lun theo m s giao im ca (Cm) vi trc tung. . nh m (Cm) tip xc ngoi vi ng trn (C): x2 + y2 = 1. < 55>Cho h ng trn (Cm): x2 + y2 2mx + 2my + 2m2 1 = 0. . Chng minh rng (Cm) lun l mt ng trn c bn knh khng i. . Tm tp hp tm I ca h (Cm). Suy ra (Cm) lun tip xc vi 2 ng thng c nh m ta phi vit phng trnh. < 56>Cho h ng trn (Cm): x2 + y2 + (m + 2)x (m + 4)y + m + 1 = 0 . nh m (Cm) l ng trn c bn knh nh nht. . Tm tp hp tm cc ng trn (Cm). < 57>Cho h ng trn (Cm): x2 + y2 (m 2)x + 2my 1 = 0 . Chng t rng cc ng trn ny u qua 2 im c nh khi m thay i. Vit phng trnh trc ng phng ca h ng trn ny. . Vit phng trnh cc tip tuyn ca (C2) k t A(0;1) khim = 2. < 58>Cho 2 ng trn (C1): x2 + y2 x 6y + 8 = 0, (C2): x2 + y2 2mx 1 = 0. nh m (C1) tip xc (C2). Ch r loi tip xc. < 59>Cho 2 ng trn(C): x2 + y2 1 = 0, (Cm): x2 + y2 2(m + 1)x + 4my 5 = 0 . Tm tp hp tm cc ng trn (Cm) khi m thay i. . Chng minh rng c hai ng trn (Cm) tip xc vi (C) ng vi 2 gi tr ca m. Vit phng trnh cc tip tuyn chung ca 2 ng trn (Cm) . < 60>Cho ng trn (C): x2 + y2 = 1 v ng thng (d): ax + by +1 = 0. . Tm iu kin i vi a, b (d) tip xc vi (C). . M v N l hai im trn (C) vi xM = 1, yN = 1. Xc nh a, b tng khong cch t M v N n (d) nh nht vi gi thit (d) tip xc vi (C). < 61>Cho h ng cong ph thuc tham s m, c phng trnh : F(x,y) = x2 + y2 2m(x a) = 0 trong a l mt s dng cho trc (c nh). .Vigitrnocam,phngtrnhtrnlphngtrnhcang trn?K hiu Cm l ng trn ng vi gi tr m. . Chng t rng on thng ni im O vi im A(2a;0) lun ct Cm. < 62>Cho h phng trnh {22(2m1)x2my5m80 xy6x8y0+++=++= (*) . Gii h phng trnh (*) khi m = 2. -22- .XthnhchnhtPQRSnitiptrong(E)vccccnhsongsong vi cc trc ca (E). Tm to cc nh hnh ch nht sao cho n c din tch ln nht. < 70>Cho elip (E): 4x2 + 9y2 = 36 v im M(1;1). . Lp phng trnh ng thng (D) qua M ct (E) ti 2 im M1, M2 sao cho MM1 = MM2. . ng thng () qua M ct (E) ti 2 im P, Q. Tm tp hp cc trung im I ca on PQ. < 71>Cho elip (E): 4x2 + 9y2 = 36 v 2 .thng (D): ax by = 0, (D): bx + ay =0 (a2 + b2 > 0). Gi M, N l cc giao im ca (D) vi (E), P, Q l cc giao im ca (D) vi (E). . Tnh din tch t gic MPNQ theo a v b. . Tm iu kin i vi a, b din tch t gic MPNQ nh nht. < 72>Cho hypebol (H): 9x2 4y2 = 36. . Xc nh to cc nh, to cc tiu im, tm sai v cc tim cn ca (H). V hypebol cho. . Tm cc gi tr ca n .thng y = nx 1 c im chung vi hypebol. < 73>Cho hypebol (H): 16x2 9y2 = 144. .Tmditrcthc,trco,tiuc,tiuim,nh,tmsaiv phng trnh cc ng tim cn ca (H). . Tm im M (H) sao cho khong cch t O n M bng na tiu c. < 74>Cho hypebol (H): 9x2 16y2 = 144 c tiu im F1, F2. . Tm im M (H) sao cho F1M F2M. . Gi s M(xo;yo) (H). Tnh OM2 F1M.F2M v (F1M + F2M)2 4OM2. < 75>Cho hypebol (H): 5x2 4y2 20 = 0 v ng thng (D): x y + m = 0. .Chngminhrng(D)lunct(H)ti2imM,N(xM< xN)thuc 2 nhnh khc nhau ca (H). . nh m sao cho 3F1M = F2N, vi F1, F2 l tiu im ca (H). < 76>Cho hypebol (H): 9x2 16y2 = 144. . Tm nhng im trn (H) nhn 2 tiu im di 1 gc 120o. . A l nh trn trc thc c honh dng. Tm to nhng im M, N trn (H) sao cho AMN u. < 77>Cho hypebol (H): 4x2 y2 = 4. . Tm nhng im trn (H) c to nguyn. Bi Tp n -19- < 45>Cho 2 ng trn (C1): x2 + y2 2x 2y + 1 = 0, (C2): x2 + y2 12x 12y + 36 = 0. Chng minh rng (C1) v (C2) ngoi nhau. Tm khong cch ngn nht v di nht ni 1 im ca (C1) vi 1 im ca (C2). < 46>Cho 2 ng trn c tm ln lt l I, J (C1): x2 + y2 4x + 2y 4 = 0, (C2): x2 + y2 10x 6y + 30 = 0 . Chng minh (C1) tip xc ngoi vi (C2) v tm to tip im H. . Gi (D) l 1 tip tuyn chung khng i qua H ca (C1) v (C2). Tm to giao im K ca (D) v ng thng IJ. Vit phng trnh ng trn (C) i qua K v tip xc vi (C1) v (C2) ti H. < 47>Cho 3 im A(3;1), B(0;7), C(5;2). . Chng minh rng ABC vung v tnh din tch ca n. . im M chy trn ng trn ngoi tip ABC. Chng minh rng khi trng tm G ca MBC chy trn 1 ng trn, vit ph.trnh ng trn < 48>. Cho 2 im A(4;0), B(0;3). Vit phng trnh ng trn ngoi tip v ni tip OAB. .TmtphptmccngtrntipxcviOxvctOytiim A(0;1). < 49>Cho ng trn (C): x2 + y2 8x 6y + 21 m2 = 0 v im I(5;2). . Chng minh rng I nm trong ng trn (C). . Tm ph.trnh ng thng ct (C) ti 2 im nhn I lm trung im. < 50> ChoA,Bl2imtrntrchonhchonhlnghimcaphng trnh: x2 2(m + 1)x + m = 0 v im E(0;1). . Vit phng trnh ng trn ng knh AB. . Vit phng trnh ng trn ngoi tip EAB. < 51>Cho ng thng (d): 2x + my + 1 2 = 0 v hai ng trn(C1): x2 + y2 2x + 4y 4 = 0, (C2): x2 + y2 + 4x 4y 56 = 0. .GiIltmngtrn(C1).Tmmsaocho(d)ct(C1)ti2im phn bit A v B. Vi gi tr no ca m th din tch IAB ln nht v tnh gi tr ln nht . . Chng minh (C1) tip xc vi (C2). Vit phng trnh tng qut ca tt c cc tip tuyn chung ca (C1) v (C2). < 52>Cho h ng trn c phng trnh x2 + y2 2mx + 2(m 2)y + 3 = 0 . Tm gi tr ca m phng trnh trn xc nh 1 ng trn (C). . Tm tp hp tm ng trn (C) khi m thay i. . Tm cc ng trn (C) tip xc vi ng thng x = 3. < 53>Cho h ng cong (Cm): x2 + y2 2x 2y + m = 0. . Vi iu kin no ca m th (Cm) l ng trn? Xc nh tm v bn knh ca (Cm) trong trng hp ny. Bi Tp n -21- . Xc nh m h phng trnh (*) c 2 nghim (x1;y1), (x2;y2) sao cho biu thc E = (x1 x2)2 + (y1 y2)2 t gi tr ln nht. < 63>Cho elip (E): 4x2 + 9y2 = 36 c tiu im F1, F2. . Tm cc im M(E) tho MF1 = 2MF2. . Chng minh rng vi mi im M (E) ta u c 2 < OM < 3 . Gi s M(xo;yo) (E). Tnh F1M.F2M + OM2 v 4OM2 (F1M F2M)2. < 64>Cho elip (E): 3x2 + 4y2 48 = 0. . Tm nhng im M trn (E) sao cho F1M:F2M = 3:5. . ng thng (D) qua I(2;1) c h s gc l ct (E) ti B v C. Tm im A trn (E) sao cho ABC c din tch ln nht. < 65>Cho elip (E): 2 2 y x 10036+= 1. . Tm to cc tiu im F1, F2, tm sai, phng trnh cc ng chun. . Qua F1, dng 1 dy AB ca (E) v vung gc vi trc honh. Tnh di on AB. Tm im M trn (E) sao cho di F1M nh nht. < 66>Cho elip (E): 3x2 + 5y2 = 30. . Xc nh to cc nh, to cc tiu im v tm sai ca elip. . ng thng i qua tiu im F2(2;0) ca (E), song song vi trc tung, ct (E) ti 2 im A v B. Tnh khong cch t A v t B ti tiu im F1. < 67>Cho elip (E): 9x2 + 25y2 225 = 0. .Tmtung caimthuc(E)chonhx = 33 5vtnhkhong cch t im n hai tiu im. . Tm cc gi tr ca b ng thng y = x + b c im chung vi (E). . Gi I, J l 2 im trn (E) sao cho OI OJ. Tnh 2211OIOJ+. < 68>Cho elip (E): x2 + 4y2 4 = 0 v im A(2;0). . Gi (D) l ng thng qua M(3;1) v c h s gc k. Bin lun theo k s giao im ca (D) v (E). . Tm 2 im B, C trn (E) sao cho ABC l tam gic u. . Mt gc vung xAy quay quanh nh A ct (E) ti 2 im E, F. Chng minh rng ng thng EF lun i qua 1 im c nh. < 69>Cho elip (E): 9x2 + 4y2 = 36. . Vit phng trnh ng thng (D) qua A(1;3) v song song vi ng phn gic I. Tm to giao im ca (D) v (E). -18- . Tm im M trn (C) sao cho MAB cn ti M. . Vit phng trnh tip tuyn vi (C) song song ng thng AB. < 35>Cho ng trn (C): (x + 1)2 + (y 2)2 = 13 v .thng (d): x 5y 2 = 0. . Tm to giao im A, B ca (d) v (C). . Tm im M trn (C) sao cho MAB vung. < 36>Cho im A(2;1) v ng trn (C): (x 1)2 + (y + 3)2 = 9. . Chng minh rng A nm ngoi ng trn. . Qua A v 2 tip tuyn AT1 v AT2 n (C) (T1, T2 l 2 tip im). Vit phng trnh ng thng T1T2 v tnh di T1T2. < 37>Cho im A(3;0) v ng trn (C): x2 + y2 + 2x 4y 20 = 0. . Chng minh rng im A trong ng trn. . Vit phng trnh dy cung qua A sao cho dy cung ngn nht. < 38>Cho 3 im A(1;2), B(2;1), C(1;5). . Chng minh rng tp hp cc im M sao cho MA2 + MB2 = MC2 l 1 ng trn (T) m ta phi xc nh tm v bn knh. .Vitphngtrnhngthng(D)quaAct(T)ti2imE,Fsao cho on EF ngn nht. < 39>Cho 2 .trn (C1): x2 + y2 2x 2y 2 = 0, (C2): x2 + y2 8x 2y +16 = 0 . Chng minh rng (C1) v (C2) tip xc nhau. . Vit phng trnh tip tuyn chung ca (C1) v (C2). < 40>Tm phng trnh tip tuyn chung ca 2 ng trn (C1): x2 + y2 10x + 24y = 56, (C2): x2 + y2 2x 4y = 20. < 41>Cho ng thng (): x y 3 = 0 v im M(2cos2t;2(1 + sintcost)). . Chng minh rng tp hp cc im M l 1 ng trn (C). . Vit phng trnh tip tuyn ca (C) vung gc vi (). < 42>Cho 3 im A(0;4), B(3;0), C(3;0). . Vit phng trnh ng trn (T) tip xc vi ng thng AB ti B v tip xc vi ng thng AC ti C. . Gi M l 1 im bt k trn (T), d1, d2, d3 ln lt l khong cch t M ti cc ng thng AB, AC, BC. Chng minh rng: d1.d2 = d3. < 43> Chongtrn(C):x2+ y22x+ 4y+ 4=0. Vitphng trnh ng thng () song song vi ng thng (D): 3x + 4y 1 = 0 v chia ng trn (C) thnh 2 cung c t s di bng 2. < 44>Cho .thng (D): 3x 2y 1 = 0 v .trn (C): x2 + y2 2x + 4y + 3 = 0. . Xc nh v tr tng i ca (D) v (C). . Tm im M(xo;yo) (D) sao cho xo + yo t gi tr nh nht. .Tmcc imN(x1;y1)(C)saochox1+ y1tgitrlnnhthoc gi tr nh nht. Bi Tp n -17- < 25>Cho ABC u c nh A(1;0), phng trnh cnh BC: 2x 4y + 3 = 0. . Tnh chiu cao AH v dt(ABC). . Vit phng trnh hai cnh cn li ca tam gic. . Tm im M trn BC sao cho MA + MO nh nht. < 26>Cho 2 im A(1;4), B(4;1). Tm im C trn trc Ox sao cho ABC c chu vi nh nht. < 27> Cho2imA(2;5),B(2;3)v.thng(d):x4y+4=0.Chngminh rng (d) ct ng thng AB ti im M ngoi on AB. Tnh t s MA:MB. < 28>Cho ng thng (): {x 3t y24t= =+ v 2 im A(1;2), B(2;3). . Chng minh rng () ct on AB. . Tm ph.trnh ng thng (D),() v (D) cch () mt khong bng 3. < 29>Cho 2 ng thng (d1): x 2y + 3 = 0, (d2): 2x + y 1 = 0. . Chng minh rng (d1) (d2). . Tm trn Ox cc im cch u (d1) v (d2). < 30>Cho 2 im A(1;2), B(3;2) v 2 ng thng(d1): 2x + 3y 6 = 0, (d2): 2x + 3y 12 = 0. . Tnh khong cch gia (d1) v (d2). . nh m ng thng (): mx + y + 1 = 0 ct on AB. . Tm phng trnh ng thng (D) qua A v ct (d1), (d2) ti E, F sao cho EF = 3. < 31>Cho 2 im B(2;3), C(1;0) v ng thng(): (m 2)x + (m 1)y + 2m 1 = 0. . Chng minh rng () lun i qua 1 im c nh A. . nh m () ct on BC. . nh m khong cch t B n () ln nht. < 32>.Trnmtphngto,vitphng trnhngtrn(T)tmQ(2;1) bnknhr=10.Chngminhrng(khngdnghnhv)imA(0;3)nm ngoi ng trn (T). .VitphngtrnhccngthngiquaimA(0;3)vkhngc im chung vi ng trn (T). < 33>Cho ng trn (C) tm I(1;2) v bn knh R = 3. . Vit phng trnh tng qut ca ng trn (C). . Vit phng trnh ng thng cha dy cung ca ng trn (C) nhn gc to O(0;0) lm trung im. < 34>Cho ng trn (C): x2 + y2 2x 9 = 0 v 2 im A(1;3), B(4;0). -16- < 12>Cho ABC c nh C(2; 4), trng tm G(0;4) v M l trung im ca BC . Gi s M(2;0). Xc nh to A v B. .GisMdingtrnngthng(D):x+y2=0,tmqutch im B. Xc nh M di cnh AB l ngn nht. < 13> ChoABCvungtiA,phngtrnhcnhBC:3xy3=0,cc nhA,Bthuctrchonhvbn knhngtrnnitip bng2.Tmto trng tm G ca ABC. < 14>Cho ABC vung ti C vi CA = a, CB = b. Vit phng trnh tp hp cc im M sao cho MA2 + MB2 = 2MC2. < 15>Cho ABC cn c phng trnh cnh y BC v cnh bn AB ln lt l: x + 2y = 0, x y + 6 = 0. Vit phng trnh ng thng qua B v song song vi AC. < 16>Cho ABC bit A(1;2), B(2;0), C(3;1). . Xc nh tm ng trn ngoi tip ABC. . Tm im M trn ng thng BC sao cho SABM = _SABC. < 17>Vit phng trnh ng thng (d) i qua im A(1;2) v cch u 2 im M(2;3), N(4;5) . < 18> VitphngtrnhngthngiquaimB(1;3)cchimA(2;5) khong cch bng 8. < 19>Cho 2 ng thng (d1): x 3y + 6 = 0, (d2): 2x y 3 = 0. Tm phng trnh ng thng (d) i xng ca (d2) qua (d1). < 20>Cho ng thng (d): 3x + 4y 12 = 0. Vit phng trnh ng thng (d) i xng vi (d) qua O. < 21>Cho A(1;3), B(3;3), I(6;2) v ng thng (D): 4x + 3y 3 = 0. . Vit phng trnh ng thng qua I v vung gc vi (D). . Vit phng trnh ng thng qua I v cch u 2 im A,B. . Tm im C trn (D) sao cho dt(IBC) = - ', - `vdt. < 22>Cho hnh ch nht ABCD c tm I(;0), phng trnh ng thng AB l x 2y + 2 = 0 v AB = 2AD. Tm to cc nh A, B, C, D bit rng nh A c honh m. < 23> ChoM(2;6),gi(d)lngthngquaMctOxtiA,OytiB.Tm phng trnh ng thng (d) bit 3OA 4OB = 0 v M trn on AB. < 24>Cho ABC vi A(0;1), B(1;2), C(5;1) . Tm phng trnh cnh BC v ng cao AH. .Gi(D)lngthngiquaAchsgcm.nhm(D)ct BC ti mt im nm pha ngoi on thng BC. -10-Phng Php To Trong Mt Phng Elip. Hypebol: nh Ngha: nh ngha (E) = {M / F1M + F2M = 2a}(H) = {M / F1M F2M = 2a} K hiu a l na trc ln, K hiu a l na trc thc, b l na trc nhb l na trc o : ElipHypebol Phng trnh chnh tc + = 1 = 1 Tiu c F1F2 = 2cc2 = a2 b2c2 = a2 + b2 Tiu imF1(c;0), F2(c;0)F1(c;0), F2(c;0) Tm saie = c/ae = c/a Bn knh qua tiu im r1 = F1M = a + exMr1 = F1M = ,exM + a, ca im M(E) r2 = F2M = a exMr2 = F2M = ,exM a, nh A1(a;0), A2(a;0)A1(a;0), A2(a;0) B1(0;b), B2(0;b) Phng trnh cc cnhx = ax = a hnh ch nht c sy = by = b Phng trnh cc tim cny = _x Ph.trnh cc ng chun (1,2): x = + ea = +ca2(1,2): x = + ea = +ca2 Tnh cht iiFM d(M,) = e(i = 1, 2) iiFM d(M,) = eA1 A2OF1F2MB2 B12c 2a 2b x 2a2/c (2) (1) xy F2 F1A1O 2a 2a2/c2c (2)(1)2bM -14-Phng Php To Trong Mt Phng < 1= 26>Trn parabol y2 = 16x tm cc im c bn knh qua tiu im bng 13. < 1= 27>Xt v tr tng i ca ng thng x y + 2 = 0 v parabol y2 = 8x. Cnic nh Ngha: Cnic l tp hp cc im M ca mt phng c t s khong cch t n ti 1 im c nh F v 1 .thng c nh () (khng i qua F) bng mt hng s e. e: Tm sai, F: Tiu im, (): ng chun ng vi tiu im F * Nu e < 1: cnic l Elip. * Nu e > 1: cnic l Hypebol. * Nu e = 1: cnic l Parabol. < 1=28>Lp phng trnh cnic nu bit: . Tm sai _, tiu im F(2;1) v ng chun tng ng x 5 = 0. . Tm sai bng ` , tiu im F(5;0) v ng chun tng ng 5x 16 = 0 . Tm sai , tiu im F( 4;1) v ng chun tng ng y + 3 = 0. .Tmsaibng' ` ,tiuimF(0;13)vngchuntngng13y 144 = 0. . Tm sai , tiu im F(3;0) v ng chun tng ngx + y 1 = 0. .Tmsaibng5,tiuimF(2;3)vphngtrnhngchun tng ng 3x y + 3 = 0. .imA(3;5)nmtrncnic,tiuimF(1; 4),ph.trnhng chun tng ng x 2 = 0. . im A(3;5) nm trn cnic, tiu im F(2;3), phng trnh ngchun tng ng x + 1 = 0. .imM(2;1)nmtrncnic,tiuimF(1;0),phngtrnhngchun tng ng 2x y 10 = 0. !0.imM(1;2)nmtrncnic,tiuimF(2;2),phngtrnhngchun tng ng 2x y 1 = 0. -12-Phng Php To Trong Mt Phng . Khong cch gia 2 nh trn 2 trc bng k ln tiu c (k > ). < 1= 06>Tm sai Elip bng 10, bn knh qua tiu im ca im M trn Elip bng 10. Tnhkhong cch t M n ng chun cng pha vi tiu im ny. < 1= 07>Tm sai Elip bng , khong cch t im M trn Elip n ng chun bng20.TnhkhongcchtMntiuimcngphavingchun ny. < 1= 08>Mt Elip c tm sai bng _, tiu im F(2;0). Tnh khong cch t mt imMtrnElipchonhbng2nngchuncngphavitiu im cho. < 1= 09> MtElipctmsaibng,mtngchuncphngtrnhx=16. Tnh khong cch t 1 im M trn Elip c honh bng 4 n tiu im cng pha vi ng chun cho. < 1= 10>Tm tm sai ca Elip bit khong cch gia 2 ng chun = k ln tiu c < 1= 11>Lp phng trnh chnh tc ca hypebol nu: . Tiu c bng 10 v trc o bng 8. . Tiu c bng 6, tm sai bng . . Trc thc bng 16, tm sai bng ` . . Phng trnh tim cn y = ' x, tiu c bng 20. . Cc im M(6;1), N(8;22 ) nm trn hypebol. . Tm sai bng 2, im M(5;3) nm trn hypebol. . Phng trnh tim cn y = _x, im M( " ;1) nm trn hypebol. . Tng 2 bn trc a + b = 7, phng trnh 2 tim cn y = x .Mtnhl(3;0),phngtrnhngtrnngoitiphnhchnht c s x2 + y2 16 = 0 !0. Qua im M vi xM = 5 v F1M = " , F2M = - - ' . !1. Khong cch gia 2 ng chun bng - - ` - `, tiu c bng 26. !2. Khong cch gia 2 ng chun bng - ` - , trc o bng 6. !3. Khong cch gia 2 ng chun bng ` , tm sai bng . !4. Phng trnh tim cn y = x, khong cch gia 2 ng chun = - - . !5. Phng trnh ng chun x = ' , im M(3;` ) nmtrn hypebol. !6. Phng trnh tim cn y = , phng trnh ng chun x = - ' - . < 1= 12>Tm di cc trc, tiu im, tm sai, phng trnh cc tim cn: Bi Tp n -23- . ng thng (D) qua A(1;4) ct (H) ti 2 im phn bit M, N sao cho A l trung im ca MN. Tm to M, N. < 78>Cho parabol (P): y2 = 4x. .TmimMtrn(P)cbnknhquatiuimbng10vtung dng. Tm im N trn (P) sao cho OMN vung ti O. . Tm 2 im A, B trn (P) sao cho OAB u. < 79>Cho parabol y2 = 2x v ng thng (d): 2x 2my 1 = 0. Chng t rng vimim,(d)luniquatiuimFca(P)vct(P)tihai imM,N. Tm tp hp cc trung im I ca MN khi m thay i. < 80>Cho parabol (P): y2 = 8x v im I(2;4) nm trn (P). Xt gc vung thay iquayquanhimIv2cnhgcvungct(P)ti2imM,N(/I). Chng minh rng ng thng MN lun lun i qua 1 im c nh. < 81>Cho parabol (P): y2 = x v gi F l tiu im ca (P). Gi s ng thng (d) i qua F ct (P) ti 2 im M1 v M2. . Tnh M1M2 khi (d) , Oy. .Gis(d)Oy.Giklhsgcca(d).TnhM1M2theok.Xc nh cc im M1, M2 sao cho M1M2 ngn nht. < 82>Cho 3 ng thng D1: 3x + 4y 6 = 0, D2: 4x + 3y 1 = 0, D3: y = 0. Gi A = D1iD2, B = D2iD3, C = D3iD1. . Vit phng trnh phn gic trong ca gc A ca ABC v tnh SABC. . Vit phng trnh ng trn ni tip ABC. < 83>Cho hai im A(3;3 ), B(3; 3). . Vit phng trnh ng trn ngoi tip ABO. . Lp phng trnh chnh tc ca parabol i qua 2 im A, B. < 84>Cho hai im A(2;3) v B(2;1). .Vitphngtrnh ngtrn iqua2imA,Bvctmnmtrn trc honh. . Vit phng trnh chnh tc ca parabol i qua im A. V ng trn v parabol tm c trn cng mt h trc to . < 85>Cho hai im A(5;0) v B(4;32). . Lp phng trnh ng trn nhn AB lm ng knh. Tm to cc giao im ca ng trn v trc honh. . Lp phng trnh chnh tc ca elip i qua A v B.