Bai 2. Viet phiTdng trinh canh AB cija hlnh chOr nhat ABCD biet canh AB, Be

Một phần của tài liệu luyện giải đề thi đại học 3 miền môn tóan theo từng chuyên đề (Trang 161 - 164)

^ Hi^dng din giai ' - , , •

Nh§n xet: Ta can xac djnh kich thirdc cua hlnh chiy nhat. Kich thirdc hint, ,/ chff nhat b^ng khoang each tff mot diem thuoc mot canh cua hlnh chO nhai

den canh do'i di$n.

(h.7.5) Gpi phiTdng trinh canh AB (di qua diem M(4; 5)) a(x - 4) + b(y - 5) = 0(a^ + b^ ^ 0).

Khi d6 phiTdng trlnh diTdng th^ng BC (di qua N(6;

5) vuong g6c vdi AB) la b(x - 6) - a(y - 5) = 0.

Di?n tich hlnh chff nhat la ABCD ^

a(5 -4) + b(2 - 5)| [b(2 -6) - a(l -5)| ^ 4[(a -3b)(a - h) Theo bai ra ta c6

d(P; AB).d(Q; BC) =

4(a-3b)(a-b) = 1 6 o (a-3b)(a-b) =4 (a^+b^): b = l,a = - l b = l;a = - j Vay PTcanh AB m -x + y - 1 = 0hoSc -x + 3y - II = 0

B&i 3. Cho hlnh thoi ABCD c6 phiTdng trlnh hai canh AB. AD theo thuT i\i 1^' X + 2y - 2 = 0, 2x + y + 1 = 0. Canh BD chffa diem M( 1; 2). Tim tpa dp cdc dinh

HiTdng din giai

Nh§n xet: Tim tpa dp diem A de dang. Theo tinh chS't hlnh thoi, diTdng cht cung 1^ diTdng phan giic, nen ta c6 the vie't phiTdng trlnh duTdng ch6o kc dinh A. Mat khdc, dffdng ch6o BD qua M vuong g6c vdi AC nen cung >>

dinh diTdc phffdng trlnh.

Tpa dO diem A nghi^m h$ PT [x + 2 y - 2 = 0 1 2x + y +1 = 0 '

4 5

•3*3

Gpi N(x; y) thuOc tia phSn giic AC (cua BAD).

f^hi d6 N cdch d e u hai diTcJng t h i n g AB. AD. Mat k h a c N. M ciing phia doi

y,ji dff^ng t h i n g AB va cilng phia doi vdi diTdng t h i n g AD. Ttfc la

x + 2 y - 2 2x + y + l (x + 2y-2)(l + 4 - 2 ) > 0 (2x + y + l)(2 + 2 - 2 ) > 0

\. ^

x - y + 3 = 0. '

vay PT diTdng t h i n g chffa c a n h AC la x - y + 3 = 0.

PT diTdng c h e o BD la l(x - 1) + l(y - 2) = 0 o x + y - 3 = 0

TiTd6 B (4 ;-l). D ( ^ ; 7).

TSm c u a hlnh thoi la d i e m 1(0; 3). tiT tpa dO

r4

dilm A, suy ra tpa dp d i e m C - ; — . w 3 y

Chu y : C6 the viet phiTdng trlnh canh BD b^ng cdeh: ^ Wi^t PT dffdng t h i n g qua M tao vdi hai dffdng t h i n g AB. AD hai gdc b^ng nhau.

Ta cung c6 the tlm tpa dp d i e m B, D b l n g cdch suf dung cong thffc tinh dien U'ch hlnh thoi. Gpi c a n h hlnh thoi la a. Khi dd dipn tich hlnh thoi la

8a

SABCD = 2SABD = 2(SAMB + SAMD) = a(d(M; AB) + d(M; AD)) =

1.2+2.1 4 3

r- r- =—=>sina=-

S.S 5 5

2 6 475

= a . - => a =

M4t khac SABCD =2SABD =2AB.ADsinBAD = 2a^sina , trong d6 a \h gdc

giffa hai diTdng thing AB, AD diTdc xic dinh bdi cosa=

Vly ta c6

75 5 3

TiT dd ta cung tlm diTdc tpa dp cAc diem B. D (thupc hai diTdng thing da cho each A mot khoang bkng a).

4. Cho tam gidc ABC can tai B. phffdng trlnh canh AB cd dang

>'^x-y-2>/3-0. 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/<1c ky Ihi />// < iniin Bdc, Trun^. Sam IK::: < r , Nguyln VOn TMng

^ Tpa dp B(2; 0). Goi A(&;S&-2y[3)

vdia?t2. Ta c6: IA = IB

•y I r /- \ Fa ==2 (loai) o a ^ + ( N / 3 a - 2 V 3 - l ) =8 o ^

^ ' [a = l + 73

V$y tpa do diem A(I + >/3;3-N/S).

' Do g6c giCfa IB vdi true hoinh \k 45", g6c giffa AB vdi true ho^nh b^ng 60", suy ra iBC = 30"

hay gdc giffa BC vdi true ho^nh \l 30". Vay hp

so g6e eua difcJng thAng B C k = tan30" = -|=. ;( , . Vay phffdng trinh dffdng thing chffa eanh BC y = -U(x - 2).

v3

\ o

^ c

; i-

Gpi tpa dp diem C ( c-2) vdi c ?t 2.

Ta c6 IC = IB o + 1 ( c- 2 ) - 2 \ = 8: e = 2 (loai) . c = ^/3-I

Vay c( V 3 - l ; I - > / 3 ) .

Bai 5. Cho tarn giic ABC c6 phiTdng trlnh canh BC la y = 2. dinh A thupc di/tlnJ thing x + y - 2 = 0va dipn tich tam giac la . Tim tpa dp edc dinh cu|

tam gidc ABC, biet A c6 ho^nh dp dffdng. ' Htfdng din 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.

khdc, AH bSng khoang cdch tff dinh A den dffdng thing BC, tff dd ta tfn^

dffdc tpa dp dinh A. Do BC song song vdi true ho^nh va AH vuong go*- true ho^nh nen dffa v^o dp dii canh da biet ta de d^ng xac dinh dffdc c u a B , C .

(h.7.7) Gpi a la dp d^i canh tam gidc deu ABC.

2 41

Ta CO dipn tich tam giac 1^ S^BC ^ * "4"

AH = a. : ^ = x/2.

2 8.

=>a = , - '3 \

po dinh A thupc dffdng thing x + y - 2 = 0 nen A(t; 2 - t).

jChi <16 AH = d(A; BC) = |(2 -1) - 2| = |t| = >^ I = po A cd ho^nh dp dffdng nen A( > ^ ; 2 - > ^ ) .

pifa v^o hinh ve, ta de suy ra tpa dp dinh:

B = B

\

= c

V 4i

3

>/6

;2

t i v i i i w : ; Hinh7.7 '

pal 6. Cho lam giac ABC vdi dffdng cao AH cd phffdng trlnh x = 3V3 , phffdng trinh hai dffdng phan giac Irong gdc ABC v^ ACB Ian Iffdt 1^ y = — x ;

\/3 y = —|=x + 6. Bdn kinh dffdng trdn npi tiep tam gidc b^ng 3. Viet phffdng

v3

trinh cdc canh cua tam gidc, bie't dinh A cd lung dp dffdng.

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 nhau nen tam gidc n^y can tai A.

(h.7.8) Do dffdng cao AH cd

phffdng trinh \ 3\f3 nen 1 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 nhau

b k g 30" k = ±

y = -

, nen

Hinh 7.8 tam giac A B C deu.

Tam dffdng iron npi tiep la l(3>/3;3). Khoang each tff I den B C bing 3, nen phffdng trinh dffdng thing B C la y = 0 hoSic y = 6. 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 = 0.

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'= - 7 3 , vay phffdng trlnh cua chiing Ian lff(?tm y = V3x; y = -V3x + i8.

Luy?n gi&i 3S IfUOt Jg> IM UH J mten-BiB^TTimgrnam lomnpc- lyguym van i nong

%huyht M IS:

B A D t f d N G C 6N 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)

1. T 6 M T A T Li T H U Y E T ^ '

A.EUP ' M / ' ^ -f^:' V. . ^a..,!:,.;

/. Dfnh nghta Elip: . v

- Trong mat p h i n g cho hai diem co dinh F i , F 2 vdi F 1 F 2 = 2 c > 0. Tap hpp diem M cua m5t p h i n g sao cho ta luon c6 M F i + M F 2 = 2a (a > c) (a la h^n so) goi Ik mpt elip.

- Hai diem F,, F 2 gpi Ik tieu diem cua elip. ' '^^^ " ' ' ^ ' - 2c gpi 1^ tieu c\i cua elip.

- Neu M n^m tren elip thi M F i , M F 2 gpi 1^ cdc bdn kinh qua tieu d i l m cua M.

2. Phuang trinh chtnh tdc cua elip.

- Chpn he tpa dp Oxy sao cho Fi vh F 2 c6 c^c tpa dp F i ( - c ; 0), F2(c; 0). Khi d(

2 ..2

elip c6 phi/dng trinh ^ + ^ = 1 (b^ = a^ - c^) (1) a^ b^

( 1) gpi \k phiTcfng trinh chinh t^c cua elip. \ Elip c6 bon dinh A , ( - a ; 0). A 2( a ; 0), B,(0; - b ) , 6 2 ( 0 ; b).

2a gpi Ik dp d^i ciia true Idn A 1 A 2 ; 2b gpi Ik dp d^i cfla true nh6 B 1 B 2 cua elip F|(-c; 0), F2(c; 0) la hai tieu diem cua elip.

Ta CO cong thufc sau de tim cic bdn kinh qua tieu: Ne'u M ( X o ; yo) n^m tren elip thi MF, = a + ^ ; M F j = a - ^0

a a

- Dai liTdng e = — goi la tam sai cua elip. NhiT vay 0 < e < 1.

a

- Hinh chff nhat gidi han bdi ckc difdng t h i n g x = ±a, y = ±b gpi 1^ hinh chff nhat cd sd cua elip.

- Elip c6 hai di/dng chuan:

+ DiTcJng chuan x = - la difcfng chuan tfng vdi tieu diem F2(c; 0) e

+ DiTcfng chuan x = - - la di/dng chuan uTng vdi tieu diem F i ( - c ; 0) a e

3. Dinh It

Neu M(x,); y„) thupc elip. K i hi?u M H : , M H 2 tiTdng tfng Ik ckc khodng <- tir M den hai diTdng chuan x = - - . x = - . K h i d6: i!lS_^J^, vdi ni'''

^ e e M H 2 M H ,

diem M .

piSu ki$n di dudng thdng tigp xiic vdi elip.

pi/^jng t h i n g A x + By + C = 0 la ti^p tiiyen da elip ndi trdn (1) khi vk chi

' ^ h i: A V + B V = C^ w ,

^ f)inhnghia hypebol

Tr6n mat p h i n g cho hai diem co dinh F, va F 2 , vdi F 1 F 2 = 2 c > 0. Tap hdp c^c diem M cua mat p h i n g sao cho |M F , - M F J] = 2a (trong do a la mot so difdng khong ddi nho hdn c) gpi la mpt hypebol.

F I va F 2 gpi la cac tieu diem cua hypebol. ^ Khoang each F 1 F 2 = 2c gpi la tieu ciT cua hypebol.

N^u M n^m tren hypebol, thi MF|, M F 2 gpi la ckc ban kinh qua tieu diem c u a M . . ^ . j

2. Phuang tiinh chtnh tdc vdcdcyiutdcim hypebol.

Chpn he tpa dp Oxy sao cho F|(c; 0) khi do phi/dng trinh cua hypebol la - ^ - ^ = 1 (2) v d i b ' = c ' - a '

a^ b^

phi/dng trinh tren gpi la phiTdng trinh chinh t i c cua hypebol. i j uv, - Cic diem A i ( - a ; 0) va A 2( a ; 0) gpi la cac dinh ciia hypebol t ^

Ox gpi la true thiTc, Oy gpi la true ao cua hypebol (2) (do no khong c i t true Oy);

Fi(-c; 0) va F2(c; 0) gpi la hai tieu diem cua hypebol, con 2c gpi la tieu ciT hypebol;

•2a gpi la dp dai true thiTc, c6n 2b la dp dai cua true a o ; ' y = — x , y = - — X la hai dirdng tipm can cua hypebol;

a a '^nf'' •"' ' ' Hinh chi? nhat tao bdi cac diTcfng t h i n g x = ±a, y = ±b gpi la hinh chff nhat cd

sdciia hypebol. cjii, .vc , Cac cong thtfc ban kinh qua tieu

Neu x„ > 0 thi MF, = a + ; MF2 = - a + ^

a a , Neu x„ < 0 thi MF, = - a ; MF2 = a - - ^

a a • ••

sai e ciia hypebol diTdc djnh nghia e = - . NhiT vay vdi hypebol ta c6 e > I . a

^u'clng chuan cua hypebol:

EJiTcJng chua'n x gpi la diTdng chuan cua hypebol tiTdng tfng vdi tieu diem

P 2( c ; 0 ) ^ .

Luyen gfM di truOc thi DH 3 miin Bdc, Trung, Nam Todn hoc - Nyuvl^n Van ThOng

+ Di/dng chuan x = — g o i 1^ diTdng chuan c u a hypebol ttfdng uTng vdi ij^^

diemF |(-c ;0)

+ Goi H,, H2 tiTdng lirng 1^ hlnh c h i c u tCf M d e n hai diftlng c h u a n x = , x - ^ c

MH, MH,

K h i d o t a c o : llll-L^ L=e. , rfn!' • • - - t : "nhfin

3. Diiu ki^n di dudng thdng tiSp xOc vdi hypebol

- Dc dirdng thdng Ax + By + C = 0 (A^ + > 0) la ticp tuycn vdi hypcrbol (i) dieu kiOn can va dij la AV - B V =

C.PARABOL im^^-^U'ifi'^i^^^

1. DinhnghJa. ^'

- Parabol la tap hcfp cdc diem cua mat phdng each deu mot dirc:>ng thang A (.6 djnh va mot diem F co djnh khong thupc A .

+ Diem F di/dc goi la tieu diem cua parabol.

+ DiTclng thdng A goi li di/dng chuan cua parabol.

2. Phuang trlnh chtnh tdc cm parabol

Chpn he true toa do: True Ox 1^ difcJng thdng di qua tieu diem F va vuong goc vdi drftfng chuan A, hi/dng difdng tif P den F (P \h giao diem cua Ox vdi

A). True Oy la trung trifc cua PF.

- Trong hp true niy F , P

\2 ) \ )

- Difdng chuan A c6 phifOng trlnh * = " ' ^

- Phirpng trinh c u a parabol \k: y^ =2px v^ goi Ik phiTdng trinh chinh tac cua

parabol

- p > 0 goi la tham s6 tieu (chu y ring p b ^ng khoang cdch tif tieu diem F den

diTiJng chuan A).

- Neu M(x; y) nkm tren parabol y^ = 2px. thi MF goi Ik hkn kinh qua tieu CD^

diem M. Ta co cong thiJc de tinh ban kinh qua tieu: MF = x +

3. Vdi dQng phuang trinh khdc cua parabol

• y^ = -2px

Tieu diem F ; dirCJng chuan " = ^

• x^ = 2py

Tieu diem F 0;il

, 2)

; diftlng chuan y = —j

:-2py

jgu diem F ; dirdng chuan y = —

^ piSu ki$n tiip xuc ciia dudng thdng vdi parabol

' Ax + By + C = 0, (A^ + B^ > 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. 1

PHl/CfNG TRiNH C A C D l / d N G C d N I C

Một phần của tài liệu luyện giải đề thi đại học 3 miền môn tóan theo từng chuyên đề (Trang 161 - 164)

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