Thực hành luyện giải đề trước kỳ thi Đại học 3 miền Bắc - Trung - Nam Toán học: Phần 2

Luy^n J/' !f :i Ihi tJH miETl Dm:, I rung, num Iuun - iiyi iigujcn run X A C D|NH C A C Y E U TO C U A T A M TRONG MAT P H A N G TQA it G I A C Qoi A ' la d i e m d o i xifng cua A qua G thi de dang tinh diTpc A ' DQ B a i Trong mat p h i n g tpa dp Oxy cho tam gidc A B C c6 A ( ; - ) , B ( l ; 5^ C(-4; -5) V i e t phiTdng trinh (PT) c^c diTcfng t h i n g sau: 1) Dircfng cao A D ( D e BC) ' /• 2) Cac di/cJng trung tuyen B B , CC, ( B , e A C , C, e A B ) *' ' 3) Cac dirfJng phan gi^c BB2, CC2 (B2 e A C , C2 e A B ) 1) V I A D d i qua A (4; - ) , vu6ng g6c v d i BC = (-5; - ) nen c6 vectd phap tuyen n = ( ; 2) o ^ C ^ = B,C B mnh7.1 ^-l=>AB,=^AC x - = -3 ! Gia sur BjCx; y), luc ta c6 h^ PT: x = l y.l= •B- Suy PT di/cJng C C , V I C , trung d i e m canh A B nen x + = 2x,, y - = 2y, Rill' H i f d n g dSn g i a i Theo tinh chat cua difcfng phan gidc, cic d i e m doi xtfng cua A qua BB2 va CC2 deu thupc BC G p i D , la d i e m d o i xtfng cua A qua CC2 thi difcfng t h i n g Vay PT A D , la X + y - = N e u A D , c i t CC2 tai H , thi tpa dp H , la nghiem x-y =1 cua h? PT •H,(2;l) x+ y= Tifdng tif neu D2(x; y ) la d i e m d o i xtfng cua A qua BB2 thi D2(-2; - ) 3-3; SuyraPTdifdngBCiaPTdifdngD,D2: ^ ^ = ^ ^ o x - y x - y - = n dp cua B la nghiem cua h? PT: T trinh hai di/5ng trung tuyen B B , : 8x - y - = 0, CC,: 14x - 13y - = Tinh tpa dp cac dinh B , C Hi^dng dSn g i a i thi du diTdc phat trien d phan sau, chung t o i tnnh hai cdch g i a i cua thi du nhy Cdch 1: (C6 sit dung trpng l a m G cua A A B C ) Dat G(x; y) thi r8x-y = il4x-13y = •B(l;5) ^8 5^ fl - y m B a i T r e n mat p h i n g tpa dp O x y cho tam gidc A B C c6 A ( ; - ) va phifdng De chuan b i cho cic i ^ V I H , la trung d i e m A D , nen D|(0; 3) V a y PT di/6ng BB2 m x - = Tifdng tif V I A C = - A B => C Cdch 2: D a t B(x; y ) va C,(x,; y , ) i h l 8x - y - = va 14x, - 13y, - = A D , qua A ( ; - ) va vuong g6c v d i CC2 nen c6 vectd phap tuyen n = ( l ; l ) ^B(l;5) dInh B, C D BC _s y ~ trinh hai di/dng phan gidc BB2: x - = 0; CC2; x - y - = T i n h tpa cac 3) Ta v i e t PT cac difcJng phan gidc b^ng each tinh tpa dO c^c d i e m B2, C2 Taluonco TiTf^ng tir tCf A ' C // B B , tinh diTdc C ( - ; - ) rx = l Biki T r e n mat p h i n g tpa dp Oxy cho lam gidc A B C c6 A ( ; - ) va phi/dng PT dircJng B B , m 8x - y - = Ti/dng l\i PT dirdng CC, la 14x - 13y - = 14x-13y = -51 ^1 Tinh toan ti/dng tif ta thu di/dc C ( - ; - ) 2) Do B, la trung d i e m canh A C nen B,(0; - ) Suy 8« C, [8x-y = rx=i Toa dp B la n g h i p m cua h? PT T ^ ^ y-5 [14x-13y = -51 ^ Do d6 PT A D : X + 2y - = ^6 A'B //// C C , nen PT A A''BB la 14x - 13y + 51 = id CC, Gii suf B(x; y ) ta c6 he PT ^ ^ 10 Khi •3'3j oa dp ciia C la nghi?m cua h$ PT: ^ ° o x=l _ y = ' [ x - y = -3 [x-y = Jx = ^ [ x - y = -3 y = -5 T r n mat p h i n g tpa dp Oxy cho tam gidc A B C c6 C(-4; + = •B(l;5) • C M ; -5) -5) va phifdng Wnh difdng cao A D : x + 2y - = 0, difdng trung tuyen B B , : 8x - y - = Tinh tpa dp cdc dinh A , B H i r d n g d i n giili X = — y=-3 l3''3j E>tfcfng t h i n g C B qua C ( - ; - ) va vuong g6c v d i A D nen c6 vec t d phap tuyen n = ( ; - ) , dB(1;5) [8x-y = [y = , Gii sil A(x; y) vk B,(x,; y,) thl x + 2y - = 8x, - y, - = Mat khic x - = 2x,, y - = 2yi => 8x - y - 33 = fx + 2y = fx=4 Toa Aia nghi^m cua hp PT ^ =>A(4;-1) [8x-y = 33 [y = - l Bai 5, Tr&n mat phing tpa dO Oxy cho tam gidc ABC c6 B(l; 5) va phifdng tr,nh dir^ng cao AD: x + 2y - = 0, difdng phan gidc CC2: x - y - = Tinh tpa dO cdc dinh A, C i Hi/dng din giai Vi BC qua B(l; 5) vh vuong g6c vdi AD n6n c6 vectd phdp tuyen n = (2; ~i) suy PT BC la 2x - y + = Vay tpa dO C la nghi^m cua hp PT [2x-y = - : Gpi B' la diem d^i xtfng ciia B qua CC2 thi BB' c6 vectd phdp tuyen II m = (1;1) nen PT BB' la''-^y X +=y 6_^.rZ;5UB'(6;0) - = Neu BB" dt CC2 tai K(x; y) thi K U 2) Do 66, PT di/dng AC la PT B'C: x - 2y - = x + 2y = x-2y = Suy A(4; -1) Bki Tren mat phJIng tpa dp Oxy cho tam giic ABC c6 A(4; -1) va phiTdng trlnh dirdng trung tuyen BB|: 8x - y - = 0, phiTdng trinh di/dng phln gi^c CC2: X - y - = Tinh tpa dO cdc dlnh B C Hi/dng dSn giai Theo bai nd'u cho A va trung tuyen BB, ta tinh dtfdc C(-4; -5) Theo bai 3, cho A va phSn giic CC2 U'nh dir(?c B(l; 5) B^i Tren mat ph^ng tpa dp Oxy cho tam gidc ABC c6 A(4; -1) va ph\iB(1; 5) [8x-y-3 = i in-,Khum; fheo thi du 3, D2(-2; -1) la diem doi xjJng cua A qua B B ndn PT dtfdng BC la 2x - y + = Gpi C(x; y) va B,(x,; y,) thi 2x - y + = va 8x, - y, - = Iviat khac x + = 2xi, y - = 2yi nen 8x - y + 27 = fx = -4 Tpa dp C la nghipm ciia hp: x - y = -3 x - y = - r ' ^ y = -5^ >CM;-5) pal 8- TrS" '"^t P^^"g ^9 Oxy cho tam giic ABC c6 B(l; 5) va phifdng trinh drf£»ng cao AD: x + 2y - = 0, difdng trung tuyen AA,: 2x + ly + = Tinh toa d6 cic dinh A, C Hifthig din giai ! Toa dO cua A la nghi$m cua h§ PT i * ^ ~ ^ => A(4- -1) [2x + lly = - Drfdng thing BC theo bai c6 phifdng trinh 2x - y + = Tpa dp diem A, (trung diem doan BC) la nghipm cua h$ PT x - y = -3 >C(-4;-5) 2x + lly=^-3 VAN DyNG TfNH CHAT CUA CAC H I N H VA CAC TINH CHAT D^C Bl|T CUA H I N H i^; Bai Cho tam gidc ABC vu6ng cSn tai A, M(l; -1) la trung di^m cua canh BC, (1 \ trpng tam tam gidc ABC la G —;0 Tim tpa dO c^c dinh ciia tam giic d6 Htfdng din giai Nhfin xet: TCf tinh chat trpng lam GA = -2GM , ta tim dtfdc tpa dp diem A Do lam gidc ABC vuong can nen trung tuyen AM cung la diTdng cao, d6 ta Viet diTdc PT di/dng thing BC Mat khdc BM = CM = AM, ta tim diTqfc tpa ^0 ciia cdc diem B va C Hoac Viet PT canh BC, va difOng thing AC, AB (di qua A tao vdi diT&ng thing AM g6c 45") A (h.7.4)Tac6 GA = -2GM ,^ n6n- K - X G = - ( X M - X G ) ^A=XG-2(XM-XC) " lyA-yG=-2(yM-yG) lyA=yG-2(yM-yG) ^4y A ^uft/ng thing BC c6 v6ctd phip tuyen GA = (1;2) di qua M(l; -1) nen c6 PT: LMV^/ truOc Ihi DH mijn Bdc, Trung, Nam Todn hoc - Ngiiyln Van Thdng l(x - 1) + 2(y + 1) = X + 2y + = Gia sic diem B(-2a - 1; a), ta c6: MB = AM ^ - + (2 +1) a = — hoac a = — > dinh diTdc phffdng trlnh Tpa dO diem A nghi^m h$ PT [x + y - = ã3*3 12x + y +1 = ' NhĐn xet: Gpi N(x; y) thuOc tia phSn giic AC (cua BAD) M ciing phia d o i la \ x - y + = la x - y + = l(x - 1) + l(y - 2) = o TiTd6 B(4;-l) D ( ^ ; 7) TSm k h a c N Mat AB va cilng phia doi vdi diTdng t h i n g AD Ttfc la x + y - 2x + y + l diTdng t h i n g chffa c a n h AC PT AD thing (x + y - ) ( l + - ) > (2x + y + l)(2 + - ) > c u a hlnh thoi la d i e m 1(0; r4 3) ^ ' x + y - = dO tiT tpa dilm A, suy tpa dp d i e m C - ; — w 3y Chu y : C6 the v i e t phiTdng trlnh c a n h BD Wi^t PT dffdng t h i n g qua M AB AD hai g d c b ^ n g b ^ n g cdeh: ^ tao v d i h a i dffdng t h i n g Ta cung c6 the tlm tpa dp d i e m B, D b l n g c d c h suf d u n g c o n g thffc tinh d i e n SABCD = 2SABD = 2(SAMB + SAMD) U'ch hlnh thoi Gpi c a n h hlnh thoi la a Khi dd d i p n tich hlnh thoi la = a(d(M; AB) + d(M; AD)) = M4t Theo bai ta c6 N c d c h d e u h a i diTcJng t h i n g AB khac SABCD =2SABD =2AB.ADsinBAD = 2a^sina , trong8ad6 a \h gdc 1.2+2.1 hai diTdng thing AB, AD diTdc xic dinh bdi cosa= r- r- =—=>sina=5 Vly ta c6 = a -6 => a = 475 S.S giffa 75 TiT dd ta cung tlm diTdc tpa dp cAc diem B D (thupc hai diTdng thing da cho each A mot khoang bkng a) Cho tam gidc ABC can tai B phffdng trlnh canh AB cd dang > ' ^ x - y - > / - tarn diTdng tron ngoai ticp tam gidc la 1(0; 2), B e Ox Tim tpa dp cdc dinh tam giac HiTdng dSn giai x^t: Tpa dp diem B xac djnh de dang Diem A cung xac djnh de dang, ™ ta sit dung tinh cha't lA = IB = R tim C, ta sur dung tinh cha't doi xffng cua C va A qua dffdng phan giac tam giac ABC can tai B) hoac viel phffdng trlnh dffdng thing chffa canh b^ng cdch xac djnh g6c giffa dffdng thing BC va true hoanh Ll^n gidi di^ ln/// < iniin Bdc, Trun^ Sam IK::: < r , ^ Tpa dp B(2; 0) Goi Nguyln VOn TMng po dinh A thupc dffdng thing x + y - = nen A(t; - t) A(&;S&-2y[3) jChi ^ vdia?t2 Ta c6: IA = IB po A cd ho^nh dp dffdng nen A ( > ^ ; - > ^ ) •y I r /\ Fa ==2 (loai) o a ^ + ( N / a - V - l ) =8 o ^ ^ ' [a = l + 73 pifa v^o hinh ve, ta de suy tpa dp dinh: V$y tpa diem A(I + >/3;3-N/S) B \ ' Do g6c giCfa IB vdi true hoinh \k 45", g6c giffa o AB vdi true ho^nh b^ng 60", suy i B C = 30" Ta c6 IC = IB o (c-2) Vay + ; k = tan30" = -|= i- ;( , \ =c 4i >/6 ;2 Hinh7.7 ' tiviiiw:; pal Cho lam giac A B C vdi dffdng cao AH cd phffdng trlnh x = 3V3 , phffdng y = —|=x + Bdn kinh dffdng trdn npi tiep tam gidc b^ng Viet phffdng v3 vdi c ?t trinh cdc canh cua tam gidc, bie't dinh A cd lung dp dffdng e=2 \ (c-2)-2 trinh hai dffdng phan giac Irong gdc ABC v^ ACB Ian Iffdt 1^ y = — x ; \/3 y = -U(x - ) v3 Vay phffdng trinh dffdng thing chffa eanh B C Gpi tpa dp diem C ^ c =B V hay gdc giffa B C vdi true ho^nh \l 30" Vay hp so g6e eua difcJng thAng B C I= = 8: (loai) c = ^/3-I Hffdng dSn giai Nhqn xet: Ta thay hai dffdng phan gidc v i dffdng cao dong quy tai mot diem, canh BC song song hoSc trilng vdi true hoanh Hai dffdng phan giac lao vdi true ho^nh hai gdc b^ng nen tam gidc n^y can tai A c(V3-l;I->/3) (h.7.8) Do dffdng cao AH cd Bai Cho tarn giic A B C c6 phiTdng trlnh canh BC la y = dinh A thupc di/tlnJ thing x + y - = v a dipn tich tam giac la Tim tpa dp edc dinh cu| tam gidc ABC, biet A c6 ho^nh dp dffdng ' Htfdng d i n giai Nh§n xet: Tff gid thiet tam gidc ABC deu va dipn tich da biet, ta xac djnl difdc dp d^i canh cua tam gidc, tff dd tinh dffdc dp d^i dffdng cao AH phffdng trinh \ 3\f3 nen y = dffdng thing B C song song hoac tr&ng vdi true ho^nh Hai dffdng phan giac tao vdi true hoanh hai gdc b^ng b k g 30" k = ± , nen khdc, AH bSng khoang cdch tff dinh A den dffdng thing B C , tff dd ta tfn^ tam giac A B C deu dffdc tpa dp dinh A Do B C song song vdi true ho^nh va AH vuong go*- Tam dffdng iron npi tiep la l(3>/3;3) Khoang each tff I den B C bing 3, nen true ho^nh nen dffa v^o dp dii canh da biet ta de d^ng xac dinh dffdc cuaB,C (h.7.7) Gpi a la dp d^i canh tam gidc deu ABC Ta CO dipn tich tam giac 1^ S^BC ^ * 41 "4" =>a = , '3 \ Hinh 7.8 phffdng trinh dffdng thing B C la y = hoSic y = Neu phffdng trlnh dffdng * l n g B C m y = 6, thi lung dp cua diem A Ik -3 (loai) ^ay phffdng trinh canh B C 1^ y = Toa dp cic diem B va C IJl B(0; 0), C(6N^;0) dffdng thing A B cd hp so g6ck = yf3,vh '• dffdng thing C A cd h^ so' gdc P ' = - , vay phffdng trlnh cua chiing Ian lff(?tm y = V3x; y = -V3x + i8 AH = a : ^ = x/2 Luy?n gi&i 3S IfUOt Jg> IM UH J mten-BiB^TTimgrnam lomnpc- %huyht BA M lyguym van i nong piSu ki$n di dudng thdng tigp xiic vdi elip IS: DtfdNG C N I C ( E U P , H Y P E R B O L , P A R A B O L ) I (Gidi tfafch: t f a e o chiftfng tdnh mdi) T M T A T Li T H U Y E T A.EUP ' M / ' ^ ^ -f^:' V / Dfnh nghta Elip: - ' ^hi:AV + B V = C^ w , Tr6n mat p h i n g cho hai d i e m co dinh F, va F , v d i F F = c > Tap hdp v Trong mat p h i n g cho hai d i e m co dinh F i , F vdi F F = c > Tap hpp d i e m M cua m5t p h i n g cho ta luon c6 M F i + M F = 2a (a > c) (a la h^n so) goi Ik mpt elip ' c^c diem M cua mat p h i n g cho | M F , - M F J ] = 2a (trong a la mot so difdng khong d d i nho hdn c) gpi la mpt hypebol F I va F gpi la cac tieu d i e m cua hypebol ^ Khoang each F F = 2c gpi la tieu ciT cua hypebol - Hai d i e m F,, F g p i Ik tieu d i e m cua elip - 2c g p i 1^ tieu c\i cua elip N^u M n^m tren hypebol, thi MF|, M F gpi la ckc ban kinh qua tieu diem N e u M n^m tren elip thi M F i , M F gpi 1^ cdc bdn kinh qua tieu d i l m cua M cuaM - '^^^ " ' ' ^ ' ^ f)inhnghia hypebol ^a ,!:,.; ' pi/^jng t h i n g A x + By + C = la t i ^ p tiiyen da elip n d i trdn (1) k h i vk chi Phuang tiinh chtnh tdc vdcdcyiutdcim Phuang trinh chtnh tdc cua elip - Chpn he tpa dp Oxy cho Fi vh F c6 c^c tpa dp F i ( - c ; 0), F ( c ; 0) Khi d( Chpn he tpa dp Oxy cho F|(c; 0) k h i phi/dng trinh cua hypebol la - ^ - ^ =1 a^ b^ ( ) g p i \k phiTcfng trinh chinh t^c cua elip \ Elip c6 bon dinh A , ( - a ; ) A ( a ; 0), B , ( ; - b ) , ( ; b) vdib' = c'-a' phi/dng trinh tren gpi la phiTdng trinh chinh t i c cua hypebol i j uv, t^ - Cic diem A i ( - a ; 0) va A ( a ; 0) gpi la cac dinh ciia hypebol Ox gpi la true thiTc, Oy gpi la true ao cua hypebol (2) (do no khong c i t true Oy); F|(-c; 0), F ( c ; 0) la hai tieu d i e m cua elip Fi(-c; 0) va F ( c ; 0) gpi la hai tieu diem cua hypebol, 2c gpi la tieu ciT Ta CO cong thufc sau de t i m cic bdn kinh qua tieu: Ne'u M ( X o ; yo) n^m tren hypebol; •2a gpi la dp dai true thiTc, c6n 2b la dp dai cua true a o ; ' ^0 MFj = a - a y = — x , y = - — X la hai dirdng t i p m can cua hypebol; a a D a i liTdng e = — g o i la tam sai cua elip NhiT vay < e < a H i n h chff nhat g i d i han b d i ckc difdng t h i n g x = ±a, y = ±b g p i 1^ hinh chff sdciia hypebol c j i i , vc nhat cd sd cua elip Cac cong thtfc ban kinh qua tieu Elip c6 hai di/dng chuan: + DiTcJng chuan x = - la difcfng chuan tfng v d i tieu d i e m F ( c ; 0) e a DiTcfng chuan x = - - la di/dng chuan uTng v d i tieu d i e m F i ( - c ; 0) e Dinh It N e u M(x,); y„) thupc elip K i hi?u M H : , M H tiTdng tfng Ik ckc khodng 0) Ik tiep tuyen cua parabol y^ = 2px dieu Ici^ncan vadu ia:pB^ = 2AC Chti y: Vdi parabol c6 phiTdng trlnh dang y^ = -2px, (p > 0) thi dieu kipn ticp xdc la-pB^ = 2AC PHl/CfNG TRiNH C A C D l / d N G C d N I C Bai 1- Trong mat phdng vdi h? toa dp Oxy, cho Elip (E) co phiTdng trlnh im^^-^U'ifi'^i^^^ C.PARABOL - Phirpng :-2py jgu diem F g o i 1^ diTdng c h u a n c u a h y p e b o l ttfdng uTng v d i ij^^ „2 y2 + — = X a c d j n h tpa d p c d c tieu d i e m , tinh d o dai 25 16 ciJa elip (E) c a c true vk t a m sai Hifdng din giai x^ a^ +b^ Phifdng trlnh chinh tac cua (E) c6 dang: — i— = ( a > b > ) Theo de ra, ta c6: a = 5, b = ^ c = Va^ - b^ = ''^ Tpa dp cde tieu diem: F,(-3; 0); F2(3; 0) Dp dai true Idn: 2a = 10 Dp dai true bd: 2b = TSm sai: e = - = - a Bii Trong mat phdng vdi hp tpa dp Oxy, lap phiTdng trlnh chinh tic cua Elip (E) c6 dp dai true Idn b^ng 4>/2 , cic dinh nkm tren tryc nho vk cac tieu diem cua (E) ciing n^m tren mpt difdng lr6n I HUdng din giai x^ Elip (E): ^ + i - = l ( a > b > 0) Theo gi5 thi^t a = 2V2 , cdc dinh trdn Oy la a b B,(0; -b); B2(0; b); F,(-c; 0) Ttf gikc F,B,F2B2 la hlnh thoi theo gid ihict dinh Cling nkm tren duOng tr6n nen F,B|F2B2 trd hlnh vu6ng =>h = c ma a^ = b^ +c^ = 2b^ => b = c = Suy phifPng trlnh ciSa EUp (E) la: — + — = Trong mat phdng vdi hp tpa dp Oxy, cho Elip (E) c6 phi/dng trlnh Ox ^ +Oy Ian = 1.lirpt VicttaiphiTdng A, B saotrlnh chotiep AO tuy^n = 2B0.d cua (E) biet d d l hai true toa dp Hi/dng din giai * Do tinh cha't doi xiJng ciia clip (E), ta chi can xet triTcJng hdp x > 0, y > Goi A (2m; 0), B(0; m) la giao diem cua tiep tuye'n cua (E) vdi true tao (j^ (m>0) ' •i ii PhiTdng trinh diTcJng th^ng AB : X V > ' Vay phi/tfng trinh tiep tuye'n la: x + 2y - 10 = ' V i tinh chat doi xtfng nen ta c6 tiep tuyen la: X + 2y - 10 = 0; X + 2y + 10 = 0; H v 'MS ^ ^ - 2y - 10 = 0; x-"2y + 10 = X ^ l3 J — + 7t phi/dng trinh o X + 2y - 2m = 0, AB tiep xiic vdi (E) o 64 + 4.9 = 4m' o 4m^ = 100 j o m ^ = o m = 5(dom>0) '2 37t 7t + +2 0ai 6- '^''""S '"^^ phing vdi he tpa dp Decac vuong g6c Oxy, cho Elip (E) c6 ^ +—=1 2m m =>S2 = - + =^• ' ^ i ^ ' " M chuyen dpng tren tia Ox va diem N chuyen dpng tren tia Oy cho diTcfng thing M N luon tiep xuc vdi (E) Xac dinh tpa dp cua M , N de doan M N c6 dp dai nho nhat Tinh gia tri nho nha't Hi^dng dSn giai Cdch I : Gia suT M ( m ; 0) va N ( ; n) \k hai diem chuyen dpng tren hai tia Ox Oy s Bai Trong mat phing vdi h§ tpa dp Oxy, hay viet phiTOng trinh chinh ta'c cila va hinh ciia chOr nhat cd sd cua (E) co Elip (E) biet (E) c6 tam sai blng ^ ,m; MN^ = m^ + n^ = (m^ + n^) Hifdng dSn giai Goi phi/dng trinh chinh t^c cua Elip (E)lk: = 49 — + ^ = 1, a > b > a b => M N > r iA^_9_^ 16n^ ^ 9m^ Ding thiJc xay m 11- t = 25 + + ^ m = Bai Parabol y^ = 2x chia diOn tich hinh tron x^ + y^ = theo ti so nao? i grcux ioJj « ' Ke't luan: Vdi >0,n > M(2N/7;0) , N(0;V'2T) ^« A ' thi M N dat gid tri nho nhat vk gia tri Cdch 2: Gia sijT M (m; ) va N(0; n) la hai diem chuyen dpng tren hai tia Ox v4 Oy Dirdng thing M N c6 phiTdng trinh — + ^ -1 = Hifdng dSn giai Hinh tron x^ + y^ = c6 R = 2>/2 , E^irdng thing tiep xuc vdi (E) \k chi khi: 16|( -— CO dien tich la nR^ = 871 -xV2x -X \ iit?i'^f _2 Q+SquallronOAB ~ " J +99 ( -Ar ^ ] =1 + "Theo ba't ding thtfc Bunhia-copxki, ta c6: S Ta can tinh tl so — (hinh ve) do: > + x/r6:9 nho nhat la ( M N ) = y Vay phiTdng trinh chinh tic cua (E) IS: — + ^ S| = | ( > / x - x ) d x + S „ , „ r t n O A B m n m'+n^ = o m = 2V7,n = N/2T qs •I: = ,2 a 2(2a + 2b) = 20 Suy a = 3; b = X fr ^5 TCr gia thiet, ta c6 he phiTdng trinh: +9 Theo bat d i n g thtfc Cosi ta c6: Mil chu vi b i n g 20 Di/dng thing tiep xuc vdi (E) va chi khi: 16 " \ r 16 ( 3^ MN^ = m^ + n^ = (m^ + n^) > m.— + n — = o M N > Im' n^J ^ m m: — = m: — a;nfjilq I lit!.: , m n ^^ng thij-c xay + n ^ = o m = V n = V2T nni'i m >0,n >0 /7;0) N(0;V2r) thl M N dat gid trj nh6 nhaft H i r i n g dSn giai gia trj Elip (E): ^ nh6nha'tla(MN) = CdcT 3; Phircing trinh tiep tuyd'n tai diem (x„; y.,) thuoc — 16 + — 9l va N f n0;— Suy toa dp cua M va N la M 2 Q2 \ (H) c6 cung tieu diem vdi elip, suy phiTdng trinh cua Hypebol (H) c6 dang fffis'l' ,: a ng thurc xSy o Bki T.Trong mat phing vdi b s ^0 9'] Si T i r ( l ) va (2)suyraa^ = 2;b^ = x^ Vay phiTdng trinh Hypebol (H) yl) SuT dung baft d^ng thtfc Co-si hoSc Bunhia-c6pxki (nhiT cdch \k 2) ta c6 MN^ ^ ^ = c6 hai tieu diem la F,(-VrO;0), F2(Vl0;0) Hypebol Hypebol (H) c6 hai tipm can la y = ±2x = ± - x o - = o b = 2a fl6^ ^ y9oJ V ll6 o' =>MN^ = -1• +^ x„ = -j-; yo = (2) H , ^ , ' = ^ >M : :)ni;an£l rin, ' Bai Trong mat ph^ng vdi hp tpa dp Decac vu6ng gdc Oxy cho Elip (E): —j- true toa dp Oxy, cho d i l m C(2; 0) Ik Elip (E): y2 - - - — = 1, va cdc diem M ( - ; 3), N(5, n) Viet phi/dng trinh cdc di/dng thing di, dj qua M va tiep xiic vdi (E) Tim n de so cdc tiep tuyen cua E di qua N c6 mpt tiep tuyen song song vdi d| hodc d = T i m toa dp cdc diem A, B thupc (E) biet r^ng hai d i l m A, B doi xiJng qua true hoanh v^ tarn gidc ABC Ik tarn gidc deu Hvtdng dSn giai Hifdng dSn giai x^ v^ Gia suf A(x„; y„) Do A, B doi xtfng qua Ox nen B(x„, -y,)) Ta CO (E): ^ A B ' = yl vk AC^ = (x„ - 2)' + yl (1) V i A B = ACncn(xo-2)'+y?,=4y?, (2) = l,M(-2;3),N(5;n) Mif Nhan xet (E) cd hai tiep tuyen thing duTng x = ±a = ±2 va dp x = - la tiep luyc'n cua (E) di qua M ViA6(E)ncn ^ + y?,=l=>y?,=l ii4 +^ ,j Gpi dj la difilng thing qua M cd hp s6 goc k dj: y = k(x + 2) + o kx - y + 2k + = x„=2 dj: tiep xuc (E) « a ^ A ^ + b^B^ = C^ /3 Vdi X o = - thay vao (1) la CO y „ = ± — • ^ ==> A / / d = > k A = k d = Vay A (2 4S] V ; B (2 4^1 hoac A (2 / AS' , B i - \ (2 AS] [r x^ Bai Trong mat phJng vdi h? true Oxy, eho ehp (E) • • J , 2 A qua N (5; n) cd hp so goc k a = — j , A : y = - j (x - 5) + n =^ -2x - 3y + 10 + 3n = 0; A tiep xuc (E) o 4(-2)^ + l ( - ) ' = (10 + 3n)^ y_=i j ' phUdng trinh hypebol (H) e6 hai diT^ng ti^m can la y = ±2x va c6 hai diem la tieu diem cua elip (E) - | 3n^ + 20n + 25 = o n = -5 n= V n= - | loai vl A = dz Do dd N(5; - ) , ' 'i ^'^ *' Luyen giii dS trade thi DH man Bdc Trung, Nam Todn hoc - Nguyln Van Thdng COng(y TNIIIIMr\Klu^n, Vie, la chu v i thie't d i $ n th^ng la dp dai canh ben K H I D A DI$N, THfi TfC H K H I CH6P, K H I L A N G TRV Lang t r u diirng: , T H E TfC H C A CK H I T R N X O A Y T M T A T L i T H U Y E T piachuviddy v Sxq = Ph ' h la chieu cao Hinh hpp chiynhSt: A H i N H L A N G TRg Dinh nghla: S,p = 2(ab + be + ca) V; a, b, c la k i c h thirdc cua hinh hpp chff nh§t: H i n h lang tru la hinh da di?n c6 mat song song g p i 1^ day, cac canh Thi ttch khong thuoc day song song v d i r Cdc canh ben song song va b i n g ""^^ - Cac mat ben, mat ch 8x - y - 33 = fx + 2y = fx=4 Toa Aia nghi^m cua hp PT ^ =>A(4 ;-1 ) [8x-y = 33 [y = - l... [8x-y -3 = i in-,Khum; fheo thi du 3, D2 ( -2 ; -1 ) la diem doi xjJng cua A qua B B ndn PT dtfdng BC la 2x - y + = Gpi C(x; y) va B,(x,; y,) thi 2x - y + = va 8x, - y, - = Iviat khac x + = 2xi, y -. .. diT&ng thing AM g6c 45") A (h.7.4)Tac6 GA = -2 GM ,^ n6n- K - X G = - ( X M - X G ) ^A=XG -2 ( XM-XC) " lyA-yG = -2 (yM-yG) lyA=yG -2 ( yM-yG) ^4y A ^uft/ng thing BC c6 v6ctd phip tuyen GA = (1 ;2) di qua