Cẩm nang hướng dẫn luyện thi Đại học - Hình học: Phần 1

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Phần 1 tài liệu Cẩm nang luyện thi Đại học - Hình học cung cấp cho người đọc các kiến thức: Phương pháp tọa độ trong mặt phẳng (viết phương trình đường thẳng, viết phương trình đường tròn, phương trình các đường Cônic,...), hình học không gian tổng hợp. Mời các bạn cùng tham khảo.

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516.0076 C120N Th.S NGUYEN TAT THU (Nhom giao vien chuyen luyen thi Dai hoc) LUYEN TH DAI HOC HINH HOC SACH D1\ CHO HOC SINH LUitN THI DAI HOC, CAO DANG DVL.013508 HA XUAT BAN DAI m QUQJIGJA H A J O L
JION Th.S NGUYEN TAT THU 1^ (Nhom giao vien chuyen luyen thi Dai hoc) Mm mum LUYEN THI DAI HOC liiNU nee SACHDANHCHOHOCSINHHJY$NTHIDAIHOC, CAODJNG (:. NHA XUAT BAN DAI HQC QUOC GIA HA Npi
L6I NOI D A U Jldfi ikiiL f Cac em hpc sinh than men! * Theo cau tnic de thi Dai hpc - Cao dSng ciia Bp Giao d\ic thi trong de thi Trong nhung nam gan day, de thi Dai hpc luon c6 3 cau thupc ve phan mon vao cac truong Dai hpc - Cao dSng c6 3 diem Hinh hpc dupe chia thanh 3 Hinh hpc so cap. Cac bai toan thupc chu de hinh hpc rat da dang va sang tao cau, moi cau 1 diem nhu sau: , trong loi giai da gay khong it kho khan cho cac thi sinh. Hon nua, nhung chu de Cau 5. Gom cac van de ve hinh hpc khong gian tong hpp Hinh hpc so cap cac em hpc sinh dupe hpc 6 ca ba khoi lop nen luong kien thuc Noi dung ciia cau 5 trong de thi thuong gom hai y: Tinh the ti'ch khoi da dien ciia cac em hpc sinh ciing bj roi rac. Voi niem dam me va nhieu nam kinh (khoi chop va khoi lang try) va y thii hai thuong xoay quanh cac va'n de nghi^m giang day danh cho bp mon toan, thay Nguyen Tat Thu da danh thoi +) Chung minh quan he vuong goc, quan h ^ song song; gian va tam huye't vie't tap sach "Cam nang luy^n thi Dai hpc Hinh hpc so cap" nham khoi day niem yeu thich toan hpc, ren luyfn ky nang tir lam bai, t u on tap. +) Tinh goc giiia hai duong thang cheo nhau; Npi dung ciia bp sach dupe chia lam 3 chuong ^ ^ ^ +) Tinh khoang each t u mot diem den mat phang; i .• ' Chuong 1. Phuong phdp toa do trong mat phdng * +) Tinh khoang each giiia hai duong thang cheo nhau. * Chuong 2. Hinh hoc khong gian Cau 7a, 7b: Gom cac van de ve phuong phap tpa dp trong mat phang. Chii yeu Chuong 3. Phuong phdp toa do trong khong gian ' xoay quanh cac van de sau Voi lo'i vie't khoa hpc, sinh dpng, bp sach giiip cac em tiep can mon toan mot +) Lap phuong trinh duong thang, duong tron, Elip, Hypebol; ^ each nh? nhang, t u nhien, khoi nguon cam hung khi t u hpc. +) Xac dinh tpa dp ciia mot diem. Kien thuc trpng tam ngan gpn, day dii bao gom ly thuyet, phuong phap giai Cau 8a, 8b: Gom cac van de ve phuong phap tpa dp trong khong gian, chii yeu cac dang bai d i t u co ban den mo rpng va chuyen sau, qua do giiip cac em hieu xoay quanh cac chii de ro ban chat, phan tich, lap luan nhuan nhuyen de chu dpng tim ra phuong phap giai quyet bai toan. +) Lap phuong trinh duong thang, phuong trinh mat phang, phuong trinh mat cau; V i du minh hpa trong tung phan dupe phan loai, s^p xep chpn Ipc tir de den kho nham dan dat cac em den nhirng dang bai thi Dai hpc. Loi giai vira chi tiet +) Chung minh v i t r i tuong doi giCra duong thang, mat phSng va mat cau; vua gpi mo de cac em tung buoc vira phan tich vua tim toi ra each giai chinh +) Xac djnh tpa dp ciia mpt diem thoa man tinh chat cho truoc. xac va thu vj nhat. Loi binh va nhan xet cua tac gia sau moi bai giai la kinh Sau day, chiing ta di phan tich de tim loi giai cac bai toan dupe trich trong nghi^m quy bau cho cac em. cac de thi Dai hpc nam 2013. , ,^, Lam nhieu bai tap de nang cao nang luc t u duy, do la mot each hpc toan * K h o i A-2013 h i f u qua nhat. Cac em se hung thii hon khi dupe thu sue voi nhieu bai t^p khac nhau va tinh huo'ng da dang c6 kem huong dan giai. Cau 5. Cho hinh chop S.ABC c6 day ABC la tam giac vuong tai A, ABC = 30". De su dung bp sach hi^u qua va mang lai ket qua cao nhat, cac em can kien Tam giac SBC deu canh a va mat ben SBC vuong goc voi day. Tinh theo a tri tim hieu de nam chSc ly thuyet, cham chi ren luy^n ky nang lam bai thong the tich ciia khoi chop S.ABC va khoang each t u C den (SAB). qua cac v i du, bai tap trinh bay trong bp sach. N^y ( ^ Phan tich: Yeu cau ciia bai toan c6 2 y: Tinh the tich khoi chop va tinh Hi vpng, tap sach: " Cam nang luyen thi Dai hoc Hinh hpc so cap" se tiep khoang each t u diem C den mat p h i n g (SAB). , j .^ tyc la nguon tai li^u bo ich cho cac em hpc sinh trong ky thi D ^ i hpc - Cao +) Tinh the tich khoi chop S.ABC : dang toi. De tinh the tich ciia khoi chop, truoc he't phai xac dinh dupe chan duong Mac du tac gia da danh nhieu tam huye't cho cuon sach, song sy sai sot la cao ciia hinh chop? Theo de bai, mat ph5ng (SBC) vuong goc voi day nen dieu kho tranh khoi. Chung toi rat mong nhan dupe su phan bi^n va gop y quy duong cao ciia tam giac SBC ke t u S chinh la duong cao ciia hinh chop. bau cua quy doc gia de nhung Ian tai ban sau cuon sach dupe hoan thi^n hon. Hon nij-a, tam giac SBC deu nen chan duong cao ciia hinh chop chinh la Nguyen Tat Thu trung diem BC.
Nen the tich cua khoi chop dugc tinh nhu sau: 3V AB 1 HK . W Mat khac HK 1 S M , do do H K 1 (SAB). Trong tarn giac vuong BND, ta co I N = IB = l A t u day ta tim dupe x hay tim dupe diem C va I . Khi do ta lap dupe phuong trinh duong thang BN (di Trong tarn giac SHN ta c6: qua N va vuong goc voi AC). Dua vao tinh chat IB = l A ta tim dupe tpa dp ' 1 __L^ 1 _ 4 20 aV39 diem B. Vay ta co loi giai nhu sau: HK^ SH^ HM^ 3a^ a^ 3a^ 26 L o i giai. V i C € d => C(x;-2x - 5). Goi I la giao diem cua AC va BD, ta co I Ta co: d(C,(SAB)) = ^ . d ( H , ( S A B ) ) = 2.HK = la trung diem AC nen I x - 4 -2x + 3
hangVt^t Trong tam giac vuong BND, ta c6 IN = IB = lA => IN^ = lA^ Tu gia thiet ta c6 M(0;a),a > 0 . 2 r - 2 x + i i ^ 2 r x + 4 ] 2 r2x + 13^ Trong tam giac vuong MAI, ta c6: I 2 ) + I 2 J I 2 J + I 2 J . 0 . De xac dinh tpa dp diem M ta can thiet lap mpt phuong thang A . trinh, gpi H la giao diem ciia AB va IM, ta xet tam giac MAI vuong tai A va +) Tim tpa dp diem M: Vi M thupc A nen tpa dp cua M chi c6 1 an (an t). Do AH la duong cao, AH = — = 2V2. AM = 2V3O nen tu day ta tim dupe t, tu do suy ra M. Vay ta c6 loi giai sau: Tij- day, su dung cong thuc duong cao trong Loi giai. Vi (P) 1 A => n(-3;-2;l) la vec to phap tuyen ciia (P) do do: tam giac vuong ta tim dupe AM, suy ra MI Phuong trinh mat phang (P) la : -3(x -1) - 2(y - 7) + l(z - 3) = 0 va MH. Mat khac MH chinh la khoang each Hay 3 x - 2 y - z - 1 4 = 0. , ^ tu M den AB (tuc la duong thang d). Tu day j ta tim dupe a va tpa dp diem M, do do ta c6 Ta CO M e A=>M(6-3t;-l-2t;-2 + t) phuong trinh MI, hon niia dp dai MI tinh Ta CO A M = 2N/30 A M ^ = 120 dupe nen ta tim dupe tpa dp diem I. Vay ta I CO loi giai nhu sau: o ( 5 - 3 t f + ( - 8 - 2 t f + ( - 5 + t f =120
C t y TNHH MTV DVVH Khang Vi^t Lum m»s mym m vn mnn IH)L - isguyen n n - n m - Cau 8b. Trong khong gian voi he tpa dp Oxyz, cho mat phang: The tich khoi chop S.ABCD la: V = I s H . S . o r n = a^ - (P): 2x + 3y + z - 1 1 = 0 va mat cau (S): + + - 2x + 4y - 2z - 8 = 0. Chung minh (P) tiep xiic voi (S). Tim tpa dp tiep diem cua (P) va (S). Gpi M la trung diem CD va K la hinh chieu vuong goc ciia H len SM Ta c6 Phan tich: CD 1 (SHM) => CD 1 H K +) De chung minh mat phSng (P) tiep xiic voi m|it cau (S) c6 tam I , ban kinh R Dodo H K l ( S C D ) . • ta Chung minh d(I,(P)) = R. * Mat khac AH//(SCD) => d(A, (SCD)) = d(H, (SCD)) = HK = SH.HM ^ aV21 +) De tim tpa dp tiep diem H ciia mat cau (S) va mp(P) ta viet phuong trinh I H VSH^+HM^ 7 di qua I vuong goc voi (P). Khi do, tpa dp H la giao diem cua duong thang Cau 7a. Trong mat phMng voi h? tpa dp Oxy, cho hinh thang can ABCD c6 hai I H voi (P). Ta CO loi giai nhu sau: duong cheo vuong goc voi nhau va A D = 3BC. Duong th^ng BD c6 Lai giai. Mat cau (S) c6 tam la l ( l ; - 2 ; l ) , R = >yi4 . phuong trinh x + 2y - 6 - 0 va tam giac ABD c6 true tam la H ( - 3 ; 2 ) . Tim 2(1)+ 3(-2) +1-11 T a c o : d(I,(P)) = = 714 =R. tpa dp cac dinh C va D . rs Phan tich: Gpi I la giao diem cua AC va BD. Tu de bai, ta tha'y A I vuong goc Vay (P) tiep xuc voi (S). ' voi BD nen H thupc A I va I la hinh chieu ciia H len BD, t u day ta tim dupe Phuong trinh duong thSng d d i qua I va vuong goc voi (P) la : diem I va viet dupe phuong trinh AC va BD, khi do tpa dp ciia C va D chi x-1 y+2 ^ z-i CO 1 an. Lai c6, tam giac IBC vuong can tai I va tam giac CBH vuong tai B 2 ~ 3 " 1 Vi H e d = > H ( l + 2 t ; 3 t - 2 ; l + t ) . nen ta c6 IC = I H = IB, tu do ta tim dupe tpa dp C, B. Dua vao ID = -316 ta xac djnh dupe tpa dp diem D . Do H € ( P ) nen 2 ( l + 2t) + 3 ( 3 t - 2 ) +1 + t - 1 1 = 0 « t = 1 . Vay P ( 3 ; 2 ; l ) . Vay ta eo loi giai nhu sau: * KhoiB-2013 Loi giai. Gpi I la giao diem ciia hai duong cheo AC va BD, ta c6 A I vuong Cau 5. Cho hinh chop S.ABCD c6 day la hinh vuong canh a, mat ben SAB la goc voi BD nen H thupc A I . tam giac deu va nam trong mat phSng vuong goc voi mat phang day. Tinh Phuong trinh H I : 2x - y + 8 = 0 . theo a the tinh cua khoi chop S.ABCD va khoang each t u diem A den m|t p h i n g (SCD). Tpa dp diem I la nghiem ciia h^ < ^ ^~^o • ^~ => l f - 2 - 4) Phan tich: [2x-y + 8 = 0 [y = 4 ^ ' ' +) De tinh the tich khoi chop, trudc het ta phai d i xac djnh chan duong cao ciia Ta CO Tam giac BIC vuong can tai I nen goc CBI = 45". Mat khac, ACBH hinh chop. V i mat phling (SAB) vuong vuong tai H nen BI phan giac ciia goc CBH I la trung diem CH, do do goc voi mat phang day nen chan IC = IB = I H . duong cao ciia hinh chop chinh la Vi C € l H = * C ( c ; 2 c + 8)=^IC = I H « ( c + 2)^+(2e + 4 f = 5 o c = - l , c = - 3 . chan duong cao cua tam giac SAB ve tu S hay do chinh la trung diem canh Suy ra C ( - l ; 6 ) . AB (do tam giac SAB deu). +) Ta tha'y A H song song voi (SCD) nen Tuong t u B(6 - 2b; b) ^ IB = I H o (8 - 2b) + (4 - b) = 5 o b = 5,b = 3. d(A,(SCD)) = d(H,(SCD)). Ta c6 loi +) Voi b ^- 3 ^ B(0;3). Ta CO =^ = 1 ^ ID = -316 , suy ra D ( - 8 ; 7 ) giai nhu sau Lai giai. Gpi H la trung diem canh AB, suy ra +) Voi b = 5=> B(-4;5), tuong tu ta c6 D(4;1) . S H 1 A B = > S H I (ABCD) va SH = ^ .
•-ty iiMtitiTvii V uvvH Khang Vi^t Cau 7b. Trong mat phMng vdi tpa dp Oxy, cho tarn giac ABC c6 chan duong Ta cd H(3 + 2t;5 + 3t;-t)e(P)=^2(3 + 2t) + 3(5 + 3t) + t - 7 = 0 ^ t = - l = > H ( l ; 2 ; l ) cao ha tu dinh A la H 17 , chan duong phan giac trong cvia goc A la Vi H la trung diem A A' nen A ' ( - l ; - l ; 2 ) . D (5; 3) va trung diem ciia canh AB la M (0; 1). Tim tpa dp dinh C. Phan tich: Tir de bai ta c6 phuong trinh Cau 8b. Trong khong gian vdi h? tpa dp Oxyz, cho cac diem A(l; - 1 ; l), BC, suy ra phuong trinh ciia AH. Do B (-1; 2; 3) va duong thSng A: ^ _ y ^ _ z - 3 .... do, tpa dp ciia A chi c6 1 an, lay doi -2 1 3 . Viet phuong trinh xung qua M ta c6 tpa dp diem B. Dua duong thing di qua A, vuong gdc vdi hai duong thang qua AB va A. vao A H vuong goc vdi BH ta tim dupe Phan tich: De viet phuong trinh duong thang ta can tim mpt diem di qua va tpa dp diem A va B. Lay doi xung M mot VTCP. De tim VTCP, ta thudng tim hai vecto khong cimg phuong va qua AD ta dupe dir'm N thuoc duong cung vuong gdc vdi dudng thang dd. Khi dd , tich cd hudng ciia hai vecto dd la VTCP ciia dudng thang. thang AC, tu do ta viet dupe phuong trinh AC. Dua vao C la giao diem ciia Trong de bai, dudng thang d can viet phuong trinh vuong gdc vdi AB va A BC va AC ta tim dupe C. Vay ta c6 loi nen AB A U^^ la VTCP ciia d. Vay ta cd loi giai nhu sau: giai nhu sau: Loi giai. Gpi d la dudng thang can lap phuong trinh. Loi giai. Ta c6, phuong trinh BC: 2x - y - 7 = 0; phuong trinh A H : x + 2y - 3 = 0 Ta cd AB = (-2; 3; 2) va u = (-2;1;3) la VTCP ciia dudng thing A . Vi A € A H ^ A ( 3 - 2 a ; a ) ^ B ( 2 a - 3 ; 2 - a ) . Vi d vuong gdc vdi A va AB nen a = AB A u = (6; 2; 4) la VTCP ciia d . Vi AH.HB = 0 nentaco =^a = 3 ^ A ( - 3 ; 3 ) , B ( 3 ; - 1 ) . Vay phuong trinh d : -• j - l _ y + 1 _ z - 1 Phuong trinh AD : y = 3 => N (0; 5) la diem doi xung ciia M qua AD 3 1 2 eAC * KhoiD-2013 => Phuong trinh AC : 2x - 3y + 15 = 0 va phuong trinh BC : 2x - y - 7 = 0 Cau 5. Cho hinh chdp S.ABCD cd day ABCD la hinh thoi canh a, canh ben SA Vay C (9; 11). vuong gdc vdi day, BAD = 120" . M la trung diem canh BC va SMA = 45*^. Cau 8a. Trong khong gian vdi h^ tpa dp Oxyz, cho diem A (3; 5; 0) va mat Tinh theo a the tich khdi chdp S.ABCD va khoang each tu Dden (SBC). phSng (P) : 2x + 3y - z - 7 = 0. Viet phuong trinh duong thing di qua A Phan tich: De bai cho SA la dudng cao ciia hinh chdp nen ta di tinh SA dua vuong goc vdi (P). Tim tpa dp diem doi xung ciia A qua (P). vao tarn giac SAM vuong tai A . Phan tich: Duong thang d can viet phuong trinh vuong goc vdi (P) nen Vi AD//(SCB) nen d(D,(SBC)) = d(A,(SBC)) . Vaytacdldi giai nhu sau: nhan VTPT ciia (P) lam VTCP, tu do ta viet dupe phuong trinh d. Gpi A' doi xung vdi A qua (P) va H la giao diem ciia d va (P), ta cd H la Loi giai. trung diem AA'. Do do, xac dinh dupe H ta se cd diem A'. Vay ta cd loi giai Loi giai. Vi B'AD = 120" => ABC = 60" nhu sau: Loi giai. Ta cd n = (2;3;-l) la VTPT ciia (P). Suy ra duong thSng d di qua A => AABC deu, suy ra AM 1 BC va AM = va vuong gdc vdi (P) nhan n lam VTCP. Tarn giac SAM vuong tai M va SMA = 45° • SA = AM = aVi , x = 3 + 2t • Phuong trinh d : y = 5 + 3t, t e R . D z = -t Ma S A B C D = 2 S ^ B C = 2 . - Gpi H la giao diem ciia d vdi (P) va A' la diem doi xung vdi A qua (P).
Cant nang luyen thi DH Hinh Hoc - Nguyen latThti 1 1 a V s a^S _3a^ Ta c6: IC^ = l A ^ (7 - 2c)^ + (c - i f = 25 « c^-6c +5=0 c = l,c = 5 The tich k h o i chop S.ABCD la: V = - S A . S A B C D " 2 • ~ 2 ~ ~ 2 ~8~' SuyraC(4;l). V i AD//BC ^ AD//(SBC) d(D,(SBC)) = d(A,(SBC)) . Voi a = -5=> A ( - 5 ; - 2 ) , B ( - 4 ; 5 ) ^ B H = ( 2 ; - l ) Ve A H 1 S M . BCl AM Phuongtrinh A C : 2 x - y + 8 = 0=>C(c;2c + 8). ' " ' ' • BC1(SAM)=>BC1 AH ri Do BC 1 S A T a c o : I C ' = I B 2 =>(c + l ) +(2c + 7) = 25 « c^ + 6c + 5 = 0 c =-l,c =-5 T u d o suy ra A H 1 (SBC) => d(A,(SBC)) = A H . Suy ra C ( - l ; 6 ) . 1 1 1 _8_ aV6 Ma >AH = Vay C ( 4 ; l ) hoac C ( - l ; 6 ) . j,,, - +• AH^ AM^ SA^ AM^ 3a' Cau 7b. T r o n g m a t phang v o i he tpa dp Oxy, cho d u o n g t r o n (C): : r < (x -1)^ + (y - 1)'^ = 4 va d u o n g thang A : y - 3 = 0 . T a m giac M N P c6 true tam t r u n g v o i tam cua (C), cac d i n h N va P thupc A , d i n h M va t r u n g d i e m Vay d(D,(SBC)) = ciia canh M N thupc (C). T i m tpa dp d i e m P. «- 9 3 Phan t i c h : Ta thay (C) va A tiep xuc v o i nhau tai T, ma tam I la true tam C a u 7a. T r o n g m a t p h 5 n g O x y cho tam giac A B C c6 M la t r u n g d i e m '2'2 nen M la giao cua T I v o i (C). A B , d i e m H ( - 2 ; 4 ) , I ( - 1 ; 1 ) Ian l u o t la chan d u o n g cao ve t u B va t a m ducmg Goi J la t r u n g d i e m M N , suy ra IJ la d u o n g t r u n g b i n h nen IJ song song v o i t r o n ngoai tiep t a m giac ABC. T i m toa dp d i n h C. A va J thupc (C) nen ta t i m d u p e tpa dp d i e m J, lay d o i x u n g ta c6 d i e m N . Phan tich: V i M va I la t r u n g d i e m canh A B va t a m d u o n g t r o n ngoai tiep V i P thupc A nen tpa d p cau P chi c6 1 an, dua vao N I v u o n g goc v o i M P ta t a m giac A B C nen I M v u o n g goc v o i AB, t u day ta c6 p h u o n g t r i n h cua A B . t i m dupe P. Vay ta c6 l o i giai n h u sau: Ta goi toa d o cua A (chi c6 1 an) lay d o i x u n g qua M ta c6 toa d o ciia B . L a i giai. D u o n g tron (C) c6 tam 1(1; 1), R = 2. D u a vao A H va B H v u o n g goc ta t i m d u p e tpa d p ciia A va B. Ta C O d ( l , A ) = K nen suy ra Co A , H nen ta c6 p h u o n g t r i n h A C . D u a vao l A = IC ta t i m d u p e tpa dp A tiep xuc (C) tai T. d i e m C. Do 1 la true tam tam giac 7 1 P M N nen M I v u o n g goc A, L o i giai. Ta c6 I M = . V i I M 1 A B nen p h u o n g t r i n h A B : 2'2 A suy ra x^^ - x, =^1 . 7x - y + 33 = 0 . M a M thupc (C) nen M ( l ; -1) Suy ra A ( a ; 7a + 33) H A = (a + 2; 7a + 29) Gpi J la t r u n g d i e m M N suy ra B ( - a - 9;-7a - 30) ^ B H = (a + 7;7a + 34). IJ la d u o n g t r u n g b i n h cua tam giac M T N , suy ra y j = y , = 1 Do BH 1 A H BH.HA = 0 o (a + 2)(a + 7) + (7a + 29)(7a + 34) = 0 M a J thupc (C) nen J(3; 1) hay J(-l; 1). +) V o i J(3; ] ) t h i N(5; 3). Gpi P(t; 3) thupc A . Ta c6 M 4/MP =^ t = -1 =^ P ( - l ; 3). a^ + 9a + 20 = 0 P(3; 3). . Voi a = -4 A ( - 4 ; 5), B ( - 5 ; -2) ^ BH = (3; 6 C a u 8a. T r o n g k h o n g gian v o i h^ tpa dp Oxyz, cho cac d i e m A ( - l ; -1; -2), P h u o n g t r i n h A C : x + 2y - 6 = 0 C ( 6 - 2c;c). B(0; 1; 1) va mat ph^ng ( P ) : x + y + z - l = 0 . T i m tpa d p h i n h chieu
Ldm ridHg lU^fH THI UH HMH Hl?e - Nguyen i ai nur vuong goc cvia A tren (P). Viet phuong trinh mat phang di qua A, B va vuong goc voi (P). Phan tich: De lap phuong trinh mat phang ta can tim mot diem di qua va f suy ra SH = MH. tan 60° = | . Mat khac tam giac ABD deu canh a VTPT. De tim VTPT ta thuang tim hai vecto khong cimg phuong c6 gia nen S^BCD = ^ S ^ B D = 2 . - j — = - y - song song hoac nam trong mat phing do. Voi bai toan tren, mat phang can viet di qua A,B va vuong goc voi (P) nen AB A np la VTPT. Vay ta c6 loi The tich khoi chop S.ABCD la giai sau: V = -3S H . S^^^^ ^ 3 e D3 = 2- - - 2- ^ = ^ Loi giai. Gpi (a) la mat phang can lap. Ta c6 n = (l;l;l) la VTPT cua (P). Ta CO AB//(SCD) Vi (a) di qua A , B va vuong goc voi (P) nen n ' = AB A n = (-l;2;-l) la r:>d(AB,SC) = d(AB,(SCD)) VTPT ciia (a). = d(B,(SCD)) . Phuong trinh (a) la: x - 2y + z - 1 = 0. = |d(H,(SCD)) ^ Cau 8b. Trong khong gian voi he tQa do Oxyz, cho diem A(-l; 3; -2) va mat phang (P): X - 2y - 2z + 5 = 0. Tinh khoang each tu A den (P). Viet phuong Gpi N, K theo thu hx la hinh chieu cua H len CD va SN, khi do d(H,(SCD)) = HK. trinh mat ph^ng di qua A va song song voi (P). , . - l - 6 + 4 + 5| 2 ViHN=2d(B,CD)=2£V3^aV3 HK= ^"'^^ Khoang each tu A den mat phang (P): d( A,(P)) = — , =— \/l + 4 + 4 3 ' ' ^ 3 VSH^+HN^ 7 • Goi (Q) la mat phang can tim. Vay d(AB,SC) = ^ ^ 11 (Q) di qua A va c6 mpt vecto phap tuyen la n = (l;-2;-2) Cau 7a. Trong mat ph^ng Oxy cho hinh vuong ABCD c6 A(1;1), AB = 4. Goi = > ( Q ) : X - 2y - 2z +3 = 0. De thi thu truang THPT Chuyen Luong The Vinh nam 2014. M la trung diem canh BC, K - ; — la hinh chieu vuong goc ciia D len * KhoiA .6 Cau 5, Cho hinh chop S . A B C D c6 day A B C D la hinh thoi canh a va AM . Tim toa do cac dinh con lai ciia hinh vuong, biet Xg < 2 . BAD = 60^'. Hinh chieu ciia S len mat phing ( A B C D ) la trpng tam tam Loi giai. giac A B C . Goc giiia mat phSng ( A B C D ) va ( S A B ) bSng 60° . Tinh the tich Gpi N la giao diem ciia DK va AB. khoi chop S . A B C D va khoang each giua hai duong thang SC va A B . Khi do ADAN = AABM AN = BM Loi giai. => N la trung diem canh AB. -— r 4 8^ CQ'I la trong tam tam giac A B C , suy ra SH 1 ( A B C D ) . H Taco AK= , phuong trinh Ke MH vuong goc voi AB, M thuoc AB. Ta CO SMH la goc giua hai mat phSng ( S A B ) va ( A B C D ) , do do V^ ^J AM:2x + y - 3 = 0, D K : x - 2 y - 3 = 0. SMH = 60". Vi N G DK ^ N(2n + 3;n) =:> AN = (2n + 2;n - 1 ) . HB 1 . AT.\s ax/s
Cty TNHH MTV DWHKhang Vi^t « n = -l,n = - l . Phuong trinh mat cau (S): (x -1)^ + (y -1)^ + (z -1)^ = 4 . I-) Voi t = 5=^l(-5;7;13), R = 12. +) Voi n = - i = ^ X B = 2 X N =^>2 (loai). 5 Phuong trinh mat cau (S): (x + 5)^ + (y - jf + (z -13)^ = 144 . +) Voi n = - l = > X B = l < 2 , y B = - 3 ^ B ( l ; - 3 ) . Cau 8b. Trong khong gian Oxyz cho hai duong thing duong thing Phuang trinh BC: y = -3 => C (5; -3). d : ^ =^ = | , ^ • ^ = Z^ = ^^ va diem A(2;3;3). Viet phuong Phuongtrinh C D : X = 5 = > D ( 5 ; 1 ) . trinh mat cau (S) di qua A , c6 tam nam tren duong thing A va tiep xiic voi Cau 7b. Trong mat phSng Oxycho tarn giac ABCco true tarn H(-6;7), tarn duong thing d. duong tron ngoai tiep l ( l ; l ) va D(0;4) la hinh chieu vuong goc ciia A len Lai giii. ' duong thSng BC . Tim tpa do dinh A . Gpi I la tam cua mat cau (S), ta c6 l(2 + t;-t;3 + 2t). Loi giai. Suy ra AI = (t;-t-3;2t)=> lA = V6t^+6t+ 9 . Ta CO HD = (b;-3), suy ra phuong trinh B C : 2 x - y + 4 = 0. Duong thing d di qua B(-1;-1;0) va c6 u = (-2;l;2) la VTCP. Phuong trinh D H : x + 2y - 8 = 0 . BI = (t + 3; -t +1; 2t + 3) => BIA u = (-4t -1; -6t -12; -t + 5) Goi M la trung diem canh B C , ta c6 I M = d(l,BC) = N/S . Kc duong kinh BB', khi dcS AHB'C la hinh binh hanh BIAU ^(4t + l)2+(6t + 1 2 f + ( t - 5 f V53t2+142t + 170 Dodo d(l,d) = non A H = B ' C - 2 1 M - 2 N / 5 . Vi A G D H A ( 8 ~ 2a;a) ^ A H = {2a -14;7 - a). Theo de bai, ta c6 d{l,d) = lA 53t^ + 142t +170 = 54t^ + 54t + 81 Suy ra (2a -14^ + ( a - 7 ^ = 2 0 ^ ( a - 7 ^ = 4 => a = 9,a = 5. o t ^ - 8 8 t - 8 9 = 0 o t = - l , t = 89. Vay A (2; 5) hoac A (-10; 9). +) Voi t = - l : ^ l ( l ; l ; l ) , R = IA = 3. ^ y-2 z- 3 Cau 8a. Trong khong gian Oxyz cho duong thang ^ ' — ^ = Phuong trinh mat cau (S): (x -1)^ + (y -1)^ + (z -1)^ = 9 . mat phang (u): x + 2y + 2z + 1 = 0, (p): 2x - y - 2z + 7 = 0 . Viet phuong trinh +) Voi t = 89=>l(91;-89;18l), R = IA = 748069. mat cau (S) ccS tarn nam tren duong thing d va (S) tiep xuc voi hai mat Phuong trinh mat cau (5): (x - 9l)^ + (y + 89f + (z - 18l)^ = 48069. phang (a) va ((i). * Kh6iD Lai giai. Cau 5. Cho hinh lang try dung A B C A ' B ' C c6 tam giac ABC vuong t^i A, Goi I la tam cua mat cau (S), l e d nen l(-t;2 + t;3 + 2t). AB ^- a, BC = 2a va A A ' = 2a . G(?i M la trung diem ciia canh BB'. Tinh the Vi (S) tiep xiic voi hai mat phing (a) va (p) non d(I,(a)) = d(I,(P)) tich khoi chop BMCA' va c6 sin cua goc giua hai duong thing A'M va BC. Lai giii. 5t + l l 7t + l 5t + n = 7t + l o t = 5,t = - l 3 3 Ve duong cao A ' H ciia tam giac A ' B ' C . :—» +) Voi t = - l ^ I ( l ; l ; l ) , R = 2, Ta CO A'H 1 (BMC) va A'H = ^ ' B ' . A ' C B'C
Cty TNHH MTV DWH Khang Vir 1 AB.u, ACu- Ma S^MBC = ^BC.BM = 1.2a.a = . cos IAB; u 2 ) = COS ^ AC; U 2 ) ABl AC The tich khoi t u dien BMCA' la: 26 26a-13 \ - a = 1 (thoa man). .'.'j i h ;^ f ->i ,> ij, .a2=^ \ V--AH.S,MBC-3— ^ \ \ ^ "v^"^(2.af.(5a-3f \ \ B' y-' \ Goi N la trung diem C C , ta c6 \ N / Vay tpa dp diem A ( 1 ; - 2 ) , B ( 5 ; 4 ) . MN//BC nen goc giija hai duong N. / / N. Cau 7b. Trong mat phang Oxy cho ba duong thang d j : 4x + y - 9 = 0, thang A ' M va BC bang goc giira hai / / N N r s d 2 : 2 x - y + 6 = 0, d 3 : x - y + 2 = 0. Tim tpa dp cac dinh ciia hinh thoi duong thang A ' M va M N . A N C ABCD, biet hinh thoi A B C D c6 di^n tich bMng 15, cac dinh A , C thupc Ta CO A ' M ^ = A'B'2 + B'M^ = 2a^ dj, B thupc d j va D thupc d 2 . j^., ^ ^ A ' N ^ = C'N^ + A'C'^ = a^ + 3a^ = 4a^ Lot gidi. J2 Suy ra c o s M A ' N = - = 4 • ' Vi B D 1 A C nen phuong trinh B D : y = - x + m . ,. 2A'N.A'M y = -X + m 9-m 4m-9 B = B D n d , , suy ra B •B Vay CO sin cua goc giiia hai duong thSng A ' M va BC bang 4x+y-9=0 . 3 ''—j Cau 7a. Trong mat phang toa dp Oxy, cho hinh binh hanh ABCD c6 D ( - 6 ; - 6 ) , m-6 2m+ 6 Tuong t u D = BD n d 2 => D duong trung true cua doan DC c6 phuong trinh la Aj : 2x + 3y +17 = 0 va ~T~'~T~ 1 2m-1' duong phan giac goc BAC c6 phuong trinh la : 5x + y - 3 = 0 . Xac dinh toa Suy ra tpa dp trung diem cua BD la I 2' 2 dp cac dinh con lai cua hinh binh hanh. Lai gidi. Vi l e A C r ^ ^ - " " ^ ^ ^+2 = 0 c > m = 3.Suy ra B(2;l), D(-1;4),I f l 5 [2'2j Gpi I la trung diem ciia CD, I € Aj nen I (3a - 1 ; -2a - 5). Tarn-q ^ 15 5 25 Taco U j . D i = 0 trong do Di(3a + 5;-2a +1) va U j ( - 3 ; 2 ) la vecto chi phuong =>Ar = ciia Aj suy ra a = - 1 . 1 Ma A e dg A(a;a + 2) => A l ^ = 2 nen ta c6: Vay l ( - 4 ; - 3 ) = : > C ( - 2 ; 0 ) . a — V 2; Vi A e A2 nen tpa dp diem A c6 dang A (a; 3 - 5a). 25 a — Mat khac ABCD la hinh binh hanh tuong duong voi D A , DC khong cimg V 2j = —a = 3,a = -2. Vay tpa dp cac dinh ciia hinh thoi la: ph^o^S Y ^ . ^ = DC . ^ _^ ^ , _ ^ . i n / i-^ =a + 4 A(3; 5), B(2; 1),C(-2;0), D ( - l ; 4 ) hoac A(-2;0), B(2; 1),C(3;5),D(-l;4). XB-a = 4 Xg AB = DC « ^ .B(a + 4 ; 9 - 5 a ) . yg - 3 + 5a = 6 yB=9-5a- Cau 8a. Trong khong gian Oxyz cho duong th^ng d : - — - = = , mat 2 1 3 DA, DC khong cung phuong khi va chi khi * ~~6~~ ^ ^ Ph5ng ( a ) : 2 x + y - 2 z + 2 = 0 va hai diem A ( 0 ; - 1 ; - 2 ) , B(2;3;1) . Viet Duong th^ng A2 la phan giac goc BAC nhan vecto U 2 = (-1;5) lam vec to chi phuong trinh mat ph^ng (p) di qua hai diem A , B va cSt duong thSng d tai phuong nen C sao cho C each mat ph^ng (a) mot khoang bang 2 . , ^
Cam nang luyftt thi DH Hinh hgc - N^yen Tat Thu Lcri gidi, CHLTONG 1. Vi C 6 d nen C(5 + 2t;-2 -1;5 + 3t). P H U O N G PHAP TQA D O T R O N G M A T P H A N G T O M TAT LY T H U YET. Theodebai d(C,(a)) = 2 — =2 t = ±2 I , Tpa dp trong m|t phang. +) t = 2 ^ C ( 9 ; - 4 ; 11) ^ A B = (2;4;3), A C - (9;-3; 14) A B A A C = (65; -1;-42) • Cho u ( x j , y j ) ; v(x2;y2) va k e R . K h i d o : Suy ra n = (65; - 1 ; -42) la VTPT cua (p) . 1) u + v = (xi + X 2 ; y i + y 2 ) 2) u - v = (xj-X2;yi-y2) Phuang trinh ( p ) : 65x - y - 42z - 85 = 0 . X j - X2 3)ku = (kxi;kyj) 4)'"= + yl 5) u=v A B A A C = ( l ; l ; - 2 ) lyi=y2 6) u.v = XjX2 + y i y 2 = > u l v o u.v = 0 o x j X 2 +yiy2 =0 Suy ra n = ( l ; 1; -2) la VTPT cua ( p ) . Xj = kxj Phuang trinh ( p ) : x + y - 2z - 3 = 0. • Hai vec to u ( x j , y j ) ; v(x2;y2) cung phuang voi nhau yi=ky2 • x+3 y—4 z—3 — Cau 8b. Trong khong gian Oxyz cho duang thang A : — = = — ^ va u.v XiX2+yiy2 • Goc giiia hai vec to u ( x p y j ) ; v(x2;y2): cos(u,v) = hai diem A ( 0 ; 1 ; 2 ) , B(1;-1;2) . Viet phuang trinh mat phSng (a) di qua A , B —•— • u V 3V6 • Cho A ( x A ; y A ) ; B ( x B ; y B ) . K h i d 6 : va ck duong thSng A tai C sac cho tam giac A B C c6 di^n tich b^ng . Lai giii. 1) AB = ( x B - X A ; y B - y A ) 2) AB= AB = 7(XB - + (yg - y A ) ' Taco C G A = > C ( - 3 - t ; 4 + 2t;3 + 2t) X - ^ A + X B Suyra A B = ( l ; - 2 ; 0 ) , A C = ( - t - 3 ; 2 t + 3;2t +1) ' 2 3) trong do I la trung diem cua A B . . AB A AC - (-4t - 2;-2t - 1 ; - 3 ) : ABAAC = V20t^+20t + 14 . 3N/6 3N/6 • AB 1 CD AB.CD = 0. Ma S AABC ABAAC o 2 0 t 2 +20t + 14 = 54 • Cho tam giac ABC voi A ( x ^ ; y ^ ) , B(xB;yB)/ C ( x c ; y c ) • Khi do trpng tam o t ^ + t - 2 = 0 o t = l,t = -2. V _ XA + XB + Xc X(~ +) Vai t = 1 => AB A AC = (-6;-3;-3) => n = ( 2 ; l ; l ) la VTPT cua (a). G (x^; y(2 ) ciia tam giac ABC la : ^ 3 V -ZAIZBIZC Phuang trinh ( a ) : 2x + y + z - 3 = 0 . 3 +) Vai t - -2 AB A AC = (6;3;-3) =^n^ ( 2 ; l ; - l ) la VTPT cua (a). I I . Phuang trinh.dutmg thang ' ! I . Phuang trinh duoiig t h i n g Phuang trinh ( a ) : 2x + y - z +1 = 0 . I I . Vec to chi phuang (VTCP), vec to phap tuyen (VTPT) ciia duong t h i n g : Cho duong thSng d. • n = (a; b) 0 gpi la vec to phap tuyen ciia d neu gia ciia no vuong vai d. • u =(Uj;u2) ^ 0 goi la vec to chi phuang ciia d neu gia ciia no trung hoac song song voi duong thSng d. s'
M o t d u a n g thSng c6 v 6 so V T P T va v 6 so V T C P (Cac vec t o nay l u o n c i i n g 5, Phuong trinh duong phan giac cua goc tao boi hai duong thang p h u o n g v o i nhau) Cho hai duong thang d j : a^x + b j y + C j = 0 va d j : a j X + b j y + C2 = 0 • M o i quan h§ g i u a V T P T va VTCP: n.u = 0 . P h u o n g t r i n h p h a n giac ciia goc tao b o i hai d u o n g thang la: • N e u n = (a; b) la m o t V T P T ciia d u o n g thSng d t h i u = (b; - a ) la m o t V T C P a i x + b i y + c i _ ^ a 2 X + b 2 y + C2 sH-ii- cua d u o n g thang d . yja^ + b^ ^jal + hi -An^fH rfsU 0 j • D u o n g t h i n g A B c6 A B la VTCP. ITI Phuong trinh duong tron. , 1.2. P h u a n g trinh duong t h i n g , , , -' :rfi8 mfc'i « 1. Phuong trinh d u o n g tron : 1.2.1. P h u a n g trinh tong quat ciia duong t h a n g : Cho d u o n g t r o n (C) t a m I(a; b ) , ban k i n h R, k h i d o p h u a n g t r i n h ciia (C) la: Cho d u o n g t h i n g d d i qua d i e m A ( x o ; y o ) va c6 n = (a;b) la VTPT, k h i d o (x-a)2+(y-b)2=R2. p h u a n g t r i n h t o n g quat ciia d c6 dang: a(x - X Q ) + b(y - yg) = 0 . N g o a i ra p h u a n g t r i n h : x^ + y^ - 2 a x - 2 b y + c = 0 v o i a^ + b^ - c > 0 cung 1.2.2. P h u a n g trinh tham so cua duong t h a n g : la p h u a n g t r i n h ciia d u a n g t r o n c6 tam I(a; b ) , ban k i n h R = +b^ - c . Cho d u a n g t h a n g d d i qua d i e m A(xQ;yo) va c6 u = (a;b) la VTCP, k h i d o 2. Phuong trinh tiep tuyen : i , fx = +at XQ Cho d u o n g t r o n (C) : ( x - a ) ^ + ( y - b ) 2 =R^ p h u a n g t r i n h t h a m so' cua d u o n g thang d la: " , t€R. ^ ly = yo+bt • Tiep t u y e n A ciia (C) tai d i e m M la d u a n g thang d i qua M va v u o n g goc 2. V i tri tuong doi giua hai duong thang. voi I M . Cho hai d u o n g t h i n g d^ : a^x + b j y + c^ = 0; d j : a j X + b j Y + C j = 0 . K h i d o v i • D u o n g t h a n g A : A x + By + C = 0 la tiep t u y e n ciia (C) d ( I , A) = R t r i t u o n g d o i giira c h i i n g p h u thuoc vao so n g h i ^ m cua h ^ : • D u o n g t r o n ( C ) : (x - a)^ + (y - b)^ ^ R^ c6 hai tiep t u y e n c i i n g p h u a n g v o i ajx + b j y + c, =0 O y la X = a ± R . N g o a i hai tiep t u y e n nay cac tiep t u y e n con lai deu c6 a2X + b 2 y + C 2 = 0 dang : y = k x + m . I V . E lip ^it • • N e u (I) v 6 n g h i ^ m t h i d j / / d 2 . 1. D i n h nghia: T r o n g m a t p h i n g cho hai d i e m co d i n h Fj,F2 c6 F1F2 = 2c. Tap • N e u (I) v 6 so n g h i ^ m t h i d j = d2 h o p cac d i e m M ciia mat p h i n g sao cho M F j + M F j 2a (2a k h o n g d o i va • N e u (I) CO n g h i ^ m d u y nhat t h i d j va d2 cat n h a u v a n g h i ^ m cua h? la tpa a > c > 0 ) la m o t d u o n g e l i p . d o giao d i e m . • Fj,F2 : la hai tieu d i e m va 2c la tieu c u ciia elip. :.H 3. Goc giira hai duong thang. •'i^^' Cho hai d u o n g t h i n g d j : a^x + b^y + Cj = 0; d2 : a2X + b 2 y + C j = 0 . G p i a la • M F j , M F 2 : la cac ban k i n h qua tieu. goc n h p n tao b a i hai d u o n g t h i n g d j va d2 . 2. Phuong trinh chinh t i c ciia elip: a^ao + b i b ^ ^ + ^ = 1 voi h^=a^-c^. Taco : cosa= ' ' a^ b2 ^a^ + b ^ ^ a ^ + b ^ 4. K h o a n g each tu mpt diem den duong thang. VaydiemM(xo;yo)6(E)«^ +^ = l va b . t u M den A d u o c t i n h b o i cong thiic: d ( M , ( A ) ) = a b , ,,,, . . . , : Va^+bl • True d o i x u n g O x , O y . T a m d o i x u n g O . , ,
m m mng mygfi im vii iiinii nyi-ivauym i«t mw- • Dinh: Ai(-a;0), A 2 ( a ; 0 ) , Bi(0;-b) va 62(0; b ) . A ^ A j =2agpi la dp dai VI. Parabol 1. Dinh nghia: Parabol la tap hpp cac diem M ciia mat phang each deu mot tryc Ion, B i B j = 2b gpi la dp dai tryc be. duong thJing A co' dinh va mot diem F co djnh khong thupc A . • Tieudiem: ¥^i-c;0), F2(c;0). Q A : duong chuan; F : tieu diem va d(F, A) = p > 0 la tham so tieu. • Npi tiep trong hinh chif nhat co so PQRS 2. Phuong trinh chinh tac ciia Parabol: y^ = 2px ' CO kich thuoc 2a va 2b voi b = a - c^ . 3. Hinh dang ciia ParaboI(P): c 0 • Tarn sai: e = — = 0. V. Hypebol 1. D i n h nghia: Trong mat phMng voi h# tpa dp Oxy cho hai diem FpFj c6 §1. VIET PHl/ONG TRlNH O I / O N G THANG FjF2 =2c. Tap hpp cac diem M cua mat phSng sao cho JMF^ - M F 2 | = 2a 1. Phuong phap chung. (2a khong doi va c > a > 0) la mot Hypebol. • Fj, F2 : la 2 tieu diem va FjFj = 2c la tieu cu. De lap phuong trinh duong thSng A ta thuong diing cac each sau • MFj, MF2 : la cac ban kinh qua tieu. • Tim diem M ( X o ; y Q ) ma A di qua va mpt VTPT n = (a; b ) . Khi do phuong 2 2 ' X V 3 2 2 2. Phuang trinh chinh tac ciia hypebol: —- - ^ = 1 voi b = c - a . trinh duong thang can lap la: a(x - XQ) + b(y - y^,) = 0. a b • Gia sii duong thang can lap A : ax + by + e = 0. Dya vao dieu ki^n bai toan ta 3. T i n h chat va h i n h dang ciia hypebol (H): tim dupe a = mb, e = nb . Khi do phuong trinh A : m x + y + n = 0. Phucmg • True doi xiing Ox (true thue), Oy (true ao). Tam doi xiing O . phap nay ta thuong ap dung doi voi bai toan lien quan den khoang each va goc • Dinh: Aj(-a;0), A 2 (a;0). Dp dai tryc thue: 2a va dp dai tryc ao: 2b. • Phuong phap quy tich: M ( X Q ; y g ) e A : ax + by + c = 0 axg + byg + c = 0 . • Tieu diem Fj(-c; 0), F2( e; O). Chii y: ]) Cho duong t h i n g A : ax + by + c = 0 • Haiti^mcan: y = ±—x. • Ne'u Aj//A thi Aj : ax + by + C j = 0, c, 5* c • Ne'u A 2 1 A thi A j : bx - ay + C j = 0. • Hinh ehu nhat co so PQRS c6 kich thuoc 2a, 2b voi b^ = c^ - a^. 2) Duong thSng di qua hai diem A(a;0), B(0;b); ab^tO co phucmg trinh: . e • Tam sai: e = — = a b 3 3 • H a i d u o n g chuan: x = ± - = ± — . 2. Cac dang toan thuong gap. e c • D p d a i c a c b a n k i n h q u a tieu ciia M ( x Q ; y o ) e ( H ) : D a n g t o a n 1: V i e t p h u a n g t r i n h du&ng t h i n g k h i b i e t d i e m d i +) MF| = exo + a v a MFj =exQ - a k h i XQ > 0. qua va phucmg cua duong t h i n g ^ ^ ... +) MF^ = - C X Q - a v a M F 2 = - e x g + a k h i XQ < 0. Phuong phap giai. ^2 2 2 2 . M(xo;yo)e(E): — - i - = l o - ^ - ^ = 1 v a ta l u o n co >a. * Ne'u duong t h i n g di qua hai diem A, B thi AB la VTCP. a^ b2 Hai duong t h i n g song song thi chung co cung VTCP va ciing VTPT.
Cty TNHHMIV DWH Khang Vi$t Cam nang luy?n thl Hinh HQC - Nguyen lat iRu V i dy 1.1.1. T r o n g mat p h i n g O x y , c h o b a d i e m A ( 2 ; 2 ) , B ( 1 ; - 5 ) , C ( 8 ; - 4 ) . K h i d o ej + 6 2 = 12 la VTCP cua A D , nen n = ( 3 ; l ) la VTPT cua A D . .5V2' 5V2. 1) C h u n g m i n h rang A , B , C la ba d i n h cua m p t t a m giac. 2) Viet p h u a n g t r i n h d u a n g t r u n g tuyen A M cua t a m giac A B C . P h u o n g t r i n h A D : 3x + y - 8 = 0 . 3) Viet p h u o n g t r i n h d u o n g t r u n g tr^c canh A B . C a c h 3 : X e t E ( x ; y ) 6 A D , ta c6 d(E, AB) = d(E, A C ) 4) Viet p h u a n g t r i n h d u o n g phan giac t r o n g va phan giac ngoai goc A . M a p h u o n g t r i n h A C : 7x - y -12 = 0, A B : x + y - 4 = 0 ^ J K1 d Lot gidi. 7x-y-12 x+ y-4 7 x - y - 1 2 = 5x + 5 y - 2 0 x-3y+4=0 Nen 1) T a c o A B = ( - l ; - 7 ) , A C = ( 6 ; - 6 ) 5V^ 7 x - y - 1 2 = - 5 x - 5 y + 20 3x + y - 8 = 0 • V d i m o i d i e m F ( X ; y ) , ta k i hieu f ( F ) = x - 3y + 4 . < . , Vi - 1 - 7 nen hai vecta A B va A C k h o n g c i i n g p h u a n g , hay A , B , C la 6 —6 Ta CO f(B) = 20, f(C) = 24 => f(B).f(C) > 0 d o d o B va C nam c u n g phi'a so v d i ba d i n h cua m p t t a m giac. d u o n g thSng x - 3y + 4 0 . V a y p h u a n g t r i n h A D : 3x + y - 8 = 0 . 9 9] 2) V i M la t r u n g d i e m canh BC nen M 2 2 Dang toan 2: Viet phuang trinh ducmg th4ng di qua m6t diem 5 _13 va lien quan den khoang each va goc. Suy ra A M = nen n = ( l 3 ; 5 ) la VTPT cua d u o n g thSng A M . 2' 2 iV Phuong phap giai. V a y p h u a n g t r i n h A M la: 1 3 ( x - 2 ) + 5 ( y - 2 ) = 0 hay 13x + 5 y - 3 6 = 0 •f 3) Goi N la t r u n g d i e m A B , ta c6 N '3 1' 2'"2j D u o n g thang d i qua d i e m A(xQ;yQ) nen p h u o n g t r i n h c6 d a n g a ( x - X o ) + b ( y - y Q ) = 0 v d i a^ + b ^ > 0 . Ta CO A B = ( - l ; - 7 ) la V T P T cua d u o n g t r u n g tr^c c?inh A B nen p h u o n g V i dv 1.1.2. T r o n g mat p h a n g Oxy cho hai d i e m A ( l ; 2 ) , B ( 3 ; - l ) . Viet p h u o n g t r i n h d u o n g thJing A biet: t r i n h d u o n g t r u n g true canh A B la: x + 7y +16 = 0 . 4) G o i A D la d u o n g p h a n giac t r o n g goc A . De lap p h u o n g t r i n h A D ta c6 cac 2) D u o n g thang A d i qua A va each B m p t khoang bang . V2 each sau 3) D u o n g t h i n g A d i qua C(1;0) va khoang each iu A den A gap 2 Ian Cach 1: Ta c6 A B = 5^2, A C = 6^2, B C = ( 7 ; l ) ^ B C = 5V2 khoang each t u B den A . „ BD AB 5 BD 5 _ — 5 — Lot gidi. Suv ra = = — => = — => B D = — D C ^ C D A C 6 BC 11 11 1) Ta CO A B = ( 2 ; - 3 ) la VTPT cua d u o n g th^ng A . 35 46 X D - 1 = Suy ra p h u o n g t r i n h d u o n g thSng A la: n 11 = ^ D ^46 50^ Hay 50 11' 11 2 ( x - l ) - 3 ( y - 2 ) = 0 » 2 x - 3 y + 4 = 0. yr, + 5 = — 2) V i A d i qua A nen p h u o n g t r i n h A ed dang 24 72 a ( x - l ) + b ( y - 2 ) = 0 hay ax + b y - a - 2 b = 0 (*) v d i a ^ + b ^ > 0 . ; D o do A D = n - ( 3 ; l ) la V T P T cua A D . 11 11^ P h u o n g t r i n h A D : 3x + y - 8 = 0 . Theo de bai d(B, A) = nen ta cd: V2 Cach 2: D a t = AB r 1 . 7 ^ - AC r 1. n 3a-b-a-2b 1 2a-3b AB ' " ^ = AC =
Cty TNHHMTVDVVnmdngVt, a = b Za^ - 24ab + I Z b ^ = 0 (a - b)(7a - 17b) = 0 » p = O A + OB + A B = a + b + Va^ + b ^ > a + b + = (a + b)[ 1 + ^ a = —b 7 7i '1 1^ • a = b , ta chpn a = b = 1 thay vao (*) ta c6 x + y - 3 = 0 M I t k h a c , t u ( l ) t a c 6 1 = 2 —+ — •a + b > 8 . a b a+ b • a =yb,tachon a = 17,b = 7 thay vao (*) ta c6 17x + 7 y - 3 1 = 0 . Suy ra P > 8 1 + - = 8 + 4N/2 . 3) V i A d i q u a C(1;0) nen p h u o n g t r i n h A c6 dang D a n g thuc xay ra k h i a = b = 4 . ax + b y - a = 0 (**) v o i a^ + b^ > 0 . X v M a t khac d ( A , A) = 2d(B, A) nen ta c6 Phuong t r i n h d : — + -^ = lx + y - 4 = 0 . a + b 2a-b a + b = 2a - b a = 2b 1 a + b = 2a - b 4) T a c o d ( 0 , d ) = - a + b = -2a + b a = 0 • K h i a = 2 b , ta chpn a = 2,b = 1 thay vao (**) ta d u p e 2x + y - 2 = 0 • K h i a = 0 , thay vao (**) ta c6 y = 0 . (I n Ma - L + - L . i — + — = - nen d ( 0 , d ) < 2 V 2 . V i dy 1.1.3. T r o n g m a t p h a n g O x y cho d i e m E(2;2). Viet p h u o n g t r i n h a^ b^ 2^ a b 8 d u o n g thSng d d i qua E va cat hai tia Ox, O y tai A , B sao cho: DSng thuc xay ra k h i a = b = 4 . 1) 0 A = 2 0 B X y 2) Tarn giac A O B c6 d i ^ n tich bang 8 Suy ra p h u o n g t r i n h d: — + Y = lx + y - 4 = 0 . 3) T a m giac A O B c6 chu v i nho nhat 7 V i dv 1.1.4. T r o n g m a t phang O x y cho d i e m P ( l ; 7 ) va hai d u o n g thang 4) K h o a n g each t u O den d Ion nhat. d j : 2x - y + 1 = 0 ; d2 : X - 2y - 1 = 0 . Lai gidi. 1. Vie't p h u o n g t r i n h d u o n g t h i n g A d i qua P va cat d ^ d j tai hai d i e m A,B X V Gia s u A(a;0), B(0;b) v o i a,b > 0 . P h u o n g t r i n h d c6 dang: — + — = 1 . sao cho t a m giac A I B can tai I v o i I = d j n d 2 . a b 2. Viet p h u o n g t r i n h d u o n g t h i n g d d i qua M ( - 3 ; 4 ) cat d p d j Ian l u o t tai Vi E e d nen - + - = 1 (1). a b E,F sao cho M F = 2 M E . 3. Viet p h u o n g t r i n h d u o n g t h i n g A j d i qua* P va tao v o i d j m o t goc a 1) T a c o O A = 2 0 B < r > a = 2 b , thay vao (1) ta du(7c - + - = 1 => b = 3,a = 6 b b thoa cos a = P h u o n g t r i n h d u o n g t h i n g d : - + — = lx + 2 y - 6 = 0 . 6 3 Lai gidi. 2) Ta CO dien tich t a m giac O A B : S = - O A . O B = - a b - 8 ab = 16 hay a = — 2x-y +l = 0 fx^-l ^ 2 2 ^ b 1- Xet he p h u o n g t r i n h o-^ , suy ra d i n d , = I ( - 1 ; - 1 ) . x-2y-l =0 [y = - l ^ i 2 v . / Thay vao (1) ta dugc: - + - = ! « . b ^ - 8 b +16 = 0 < ^ b = 4 = > a = 4 . 8 b G Q I d ' la d u o n g p h a n giac ciia goc tao b o i hai d u o n g t h i n g d p d j . P h u o n g X y W n h cua d ' : P h u o n g t r i n h d : — + -^ = l < r > x + y - 4 = 0 . 2x-y + l x-2y-l 2x-y+l=x-2y-l x+y+2=0 3) Ta CO chu v i t a m giac O A B la: 2 x - y + l = - x + 2y + l
C a w ttang luyftt thi Oil Uhih hm \'i;iiiu-n \U Jill: cry 1 i\nM nTrvTJWiTKnang Vtet V i A A I B can tai I nen A ± d ' . D o do, p h u o n g t r i n h A la: P h u o n g t r i n h p h a n giac cua goc tao b o i g i i i a hai d u o n g thang d j , d 2 la: X - y + 6 = 0 hoac x + y - 8 = 0 . Ai:3x-7y-10 2x + 5y + 3 = 5 x - 2 y - 7 2. G o i E ( m ; 2 m + 1), F(2n + l ; n ) , A2 :7x + 3y + 4 = 0 ...•
Cam nang luyfn Iht DH Hinh ho: \xuifen Tat Thu Cty TNHHMTVDWH Khang Vift t u y e n c h u n g ciia (Ci) va (C2) t h i khoang each t u I i va h den d u o n g t h i n g do V i d v 1.1.9. Viet p h u a n g t r i n h cgnh A B cua h i n h c h i j nhat A B C D , biet A B , Ian l u g t bang Ri va R 2 , tuc la: BC, C D , D A Ian l u g t d i qua cac d i e m M ( - 3 ; 4 ) , N(6;-9), P(8;2), Q ( - 2 ; - 3 ) 5A-12B + C va S ABCD = 50. = 15 Lot gidi. A + 2B + C V i A B d i qua M nen p h u a n g t r i n h c6 dang: a(x + 3) + b(y - 4) = 0 = 5 (2) Phuang t r i n h B C : b(x - 6) - a(y + 9) = 0 . T u (1) va (2) ta suy ra : | 5 A - 1 2 B + C| = 3|A + 2 B + C| M l t khac: 8b + 6a 11a-2b (3a + 4 b ) ( l l a - 2 b ) Hay 5 A - 1 2 B + C = ± 3 ( A + 2B + C ) . SABCD=d(Q,BC).d(P,AB) = Va^+b^ V a ^ + b ^ T H I : 5 A - 1 2 B + C = 3 ( A + 2B + C ) C = A - 9B thay vao ( 2 ) : Theo de bai S^B(;I5 = 50 nen ta c6 p h u o n g t r i n h 2 A - 7 B | = 5 V A 2 + B 2 => 21A^ + 28AB - 24B^ = 0 => A = ~ B 8a^ + 38ab-33b2 = 0 ^=1 33a2+38ab-8b2| = 25(a2+b2) b~4 N e u ta chpn B= 21 t h i se dxxgc A = -14 ±10N/7 , C = -203 ± 10^7 . 58a2+38ab + 1 7 b 2 = 0 a___ll Vay CO hai tiep t u y e n : (-14 lloV? )x + 21y - 2 0 3 ± 10%/? = 0. b~ 2 T H 2 : 5 A - 12B + C = -3(A + 2B + C) C = thay vao (2) ta dugc: • - = - t a c h p n a = 3,b = 4 . P h u a n g t r i n h A B : 3x + 4 y - 7 = 0 . b 4 96A? + 28AB + 51B^ ^ 0 . P h u a n g t r i n h nay v 6 n g h i g m . Vi 1.1.8. T r o n g m a t phang v o l h ^ toa d p O x y cho h i n h chCr nhat A B C D • ^ = t a c h p n a = 11, b = - 2 . P h u o n g t r i n h A B : l l x - 2 y + 41 = 0 . CO d i e m 1(6; 2) la giao d i e m ciia 2 d u a n g cheo A C va B D . D i e m M ( 1 ; 5) Vi 1.1.10. T r o n g mat phang O x y cho h i n h thoi A B C D ngoai tiep d u a n g thuoc d u o n g thang A B va t r u n g d i e m E ciia canh C D thuQC d u a n g thang tron ( C ) : ( x - 1 ) ^ + ( y + l)2 =20. Diem B nam tren duong thang d : x + y - 5 = 0 . Viet p h u o n g t r i n h d u o n g thMng A B . d : 2 x - y - 5 = 0 va X B > 0 . Viet p h u a n g t r i n h canh A B cua h i n h thoi, biet Lai gidi. AC = 2BD. Vi Eed:^E(a;5-a)=:>iE = ( a - 6 ; 3 - a ) . Lai gidi. Gpi N la t r u n g d i e m cua A B , suy ra D u o n g t r o n ( C ) c6 t a m 1(1;-1), ban k i n h R = 275 . I la t r u n g d i e m ciia E N nen : Dat I B = x,x > 0 . D o A C = 2BD => l A = 2IB = 2 x . Xfj =2x, - x ^ =12-a N: Ke I H 1 A B = ^ I H = R = 2N/5. yN=2yi-yE=a-^ Trong tam giac A I B , ta c6: =i>MN = ( l l - a ; a - 6 ) . 1 1 • = —1 +IB'1 1 1 ^ + ^ => X = 5 hay IB = 5 . a = 6 IH^ IA2 20 4x2 ^2 Vi E 1 M N = * M N . I E = 0 o ( l l - a ) ( a - 6 ) + (a-6)(3-a) = 0 « . a = 7' M a t khac B e d = ^ B ( b ; 2 b - 5 ) = > i B = ( b - l ; 2 b - 4 ) nen • V o i a = 6 => M N = ( 5 ; 0 ) , suy ra p h u o n g t r i n h A B : y - 5 = 0 . "b = 4 * V o i a = 7 => M N = (4; 1), suy ra p h u a n g t r i n h A B : x - 4y +19 = 0 . IB = 5 o ( b - l ) 2 + ( 2 b - 4 ) 2 =25« 5 Do > 0 nen ta c6 b = 4 => B ( 4 ; 3 ) .
Cam nang luyftt thi DH Hinh hqc- Nguyen Tat Thu Cty TNHH MTV DWH Khang Vi(t Phuong trinh AB c6 dang: a(x - 4) + b(y - 3) = 0 V i dv 1.1.12. Trong mat phing tpa dp Oxy cho duong tron (C) c6 phuang 3a + 4b trinh: (x-4)^+y^=25 va diem M(l;-1). Tim phuong trinh duong Taco: d(I, AB) = I H = 2%/5 = 2V5 » ( 3 a + 4b)2 = 20(3^ + b^) thing A di qua diem M va cat duong tron (C) tai 2 diem A, B sao cho a = 2b MA = 3MB • o l l a ^ _24ab + 4b^ = 0 o Loi gidi. 11 Duong tron (C) c6 tarn 1(4; 0) va c6 ban kinh R = 5 ; M(l;-1) • a = 2b ta CO phuong trinh A B : 2x + y -11 = 0 MI = VlO < 5 = R nen M nam ben trong duong tron (C) • a = — b ta CO phuong trinh A B : 2x + l l y - 41 = 0 . ''A=4xM-3xB=4-3xg MA = 3MB MA = -3MB: Vi du 1.1.11. Trong mat phSng tpa do Oxy cho duong thing (d) c6 phuong yA=4yM-3yB=-4-3yB trinh : x - y = 0 va diem M(2;l). Tim phuong trinh duong thang A cat A , B e ( C ) nen true hoanh tai A cat duong thing (d) tai B sao cho tarn giac AMB vuong .(xB-4)^+yB'=25 | ( X B - 4 ) 2 + yg2=25 can tai M . YB =-3^^6-3 xg =0;yB - - 3 Lai gidi. ^B +^^6 =0 .XB = - i ; y B = o , Vi A nkm tren Ox nenA(a;0), B nam tren duong thing x - y = 0 nen Duong thang can tim di qua B, M vay c6 hai duong thing thoa man bai B(b;b) toan: Aj : 2x - y + 3 = 0 va A2 : x + 2y +1 = 0. Do M(2;l) =>MA = ( a - 2 ; - l ) , M B = ( b - 2 ; b - l ) Vi dvi 1.1.13. Trong mat phing voi h# tpa dp Oxy, cho duong tron (C): Tarn giac ABM vuong can tai M nen: (x -1)2 + (y + if = 16 tam I va diem A(l + Chung minh rang mpi (a-2)(b-2)-(b-l) = 0 MA.MB = 0 duong thing di qua A deu cat duong tron (C) tai hai diem phan bi?t. Viet \MA = MB ^y(a-2)2+l=V(b-2)^+(b-l)^' phuong trinh duong thing d di qua A va cat (C) tai hai 3x + y - 4V1O -1 = 0 Do b = 2 khong thoa man, v$y diem B, C sao cho tam giac IBC nhpn va c6 di^n tich bJng A\l3 . a-2 = - ^ , b ^ 2 Lai gidi. b-1 a-2 = ,b^2 b-2' Ta c6: Duong tron (C) tam 1(1; -1), ban kinh R = 2 b-2 i N2 b-1 + l-(b-2)2+(b-l)2 (a-2)2+l = (b-2)2+(b-l)2 Ta c6: IA = V3 + 9 = 2V3 < 4 , suy ra diem A nam trong ( C ) . , b-2 a=2 ^lAB = |lA.IB.sin BIC = 4N/3 o i.4.4.sin BIC = 4^3 sin BIG = a-2 = ^ , b ^ 2 b-2 b=l =>Bic = 60°=>d(I;BC) = 2>/3. ' a=4 (b-2)2+(b-l)^ --1 = 0 Duong thing d di qua A, nhan n(a;b) (a^+b2 ^^0) c6 phuong trinh (b-2)^ b=3 a ( x - l - V 3 ) + b ( y - 2 ) = 0 =>d(I;BC) = 273c:>(V3a-b)2 = 0 ^ V 3 a - b = 0 a=2 + Voi \g thing A qua AB c6 phuong trinh x + y - 2 = 0 b=1 Chpn a = 1, b = V s . a=4 Tu do phuong trinh duong thing d: >/3x + 3 y - > / 3 - 9 = 0. - + Voi
Camnang V i d y 1.1.14. T r o n g m a t p h i n g v a i tQa d p O x y , cho d u a n g t r 6 n L o t gidi. D u o n g t r o n (C) c6 t a m 1(1;-1), ban k i n h R = 3 . ( C ) : ( x - 1 ) ^ + ( y - l ) ^ = 1 0 . D u o n g tron ( C ) t a m r ( - 2 ; - 5 ) c i t (C) tai h a i Gpi H la t r u n g d i e m cua A B d i e m A, B sao cho AB = 2Vs . Viet p h u o n g t r i n h d u a n g t h i n g AB. Suy ra IH 1 A B Lai gidi. 4>/2 A SAAIB = T H L A B = 2^2 => A B = D u o n g t r o n (C) c6 t a m 1(1; 1), ban k i n h R = VlO . A D 9 d a i i r = 3V5 M | t khac: G(?i H la giao d i e m cua IF va AB, suy ra H la t r u n g d i e m AB nen A H ^ + Hl2 = IA2 =^ — + Hl2 = 9 + Hl2 = 9 A H = V5 HF Do i n AB nen ta c6: IH = VIA^ - AH^ = Vs HI = 1 =:> A B = 4>/2 A B = 2 ( l o a i ) ' T H I : H t h u p c d o a n IF ^ r H = 2N/5 r : > i H - i i r V i A d i qua M nen p h u o n g t r i n h A c6 d^ng: ax + by + 6a - 3b = 0 M a IH = ( x H - l ; y H - l ) ' i ? = (-3;-6) M a t khac: nen ta c6: 7a-4b HI = lr:>d(I,A) = l < » = l c > 1 5 b 2 - 5 6 a b + 48a^=0b = - a , b = — a X H - 1 =-1 3 5 " " " " ^ 3>H(0;-1) yH-i = -2^1yH=-i Vay A : 3x + 4 y + 6 = 0 hoac A : 5x + 12y - 6 = 0 la d u o n g t h i n g can lap. V i AB d i q u a H va nhan V i d\f 1.1.16. T r o n g m a t p h i n g v o i h ^ tpa d p O x y cho h a i d u o n g tron ( C i ) : x^ + y 2 = 13 va ( C j ) : (x - 6)^ + y^ = 2 5 . G p i A la giao d i e m ciia (Ci) n = - i i F = (l;2) lam VTPT va (C2) v a i y ^ < 0 • Viet p h u o n g t r i n h d u o n g t h i n g d i qu a A va cat (C^), nen p h u o n g t r i n h AB la: x + 2y + 2 = 0. (C ) theo 2 day cung c6 d p d a i bang nhau. T H 2: H k h o n g nam t r o n g doan I F , ' Suy ra I ' H = 4 V 5 r ^ i H = i i F x 2 + y 2 =13 Xet h f : X u -1= — (x-6)^ + y 2=25 Suy ra i _i 4' 2 Gpi A la d u a n g t h i n g can lap. yH-i = -2 YH = A = A B thoa y eu cau bai toan A * A B gia s u A cat hai d u o n g P h u o n g t r i n h A B : x + 2y + - = 0 . tron (Ci), (C2) Ian l u p t tai M , N V i dM 1.1.15. T r o n g mat phSng O x y cho d u o n g tron ( C ) : (x - 1 ) ^ + (y +1)^ = 9 Phep d o i x u n g t a m A bieh M t h a n h N va ( C j ) t h a n h ( C 3 ) CO t a m I. Viet p h u o n g t r i n h d u o n g t h i n g A d i qua M ( - 6 ; 3 ) v a cSt d u a n g Vi M6(Ci)=^N6(C3)=>N€(C2)n(C3) t r o n (C) tai h a i d i e m phan bi?t A, B sao cho t a m giac l A B c6 di?n rich P h u o n g t r i n h (C3): (x - 4)^ + (y + 6f = 13 . bang 2N/2 va A B > 2 .