Một số phương pháp giải toán Hình học theo chuyên đề: Phần 1

Phần 1 tài liệu Phương pháp giải toán Hình học theo chuyên đề do NXB Đại học Quốc gia Hà Nội ấn hành cung cấp cho người đọc cách giải các bài toán hình học theo phương pháp tọa độ trong mặt phẳng. Mời các bạn cùng tham khảo nội dung chi tiết.

T R U N G T A M L U Y C N T H I D A I H O C V I N H V I £ N S A I G O N

Tdng chu bi§n: PHAM H 6 N G D A N H NGUYEN PHU KHANH - NGUYIN TAT THU NGUYEN TAN SIENG - TRAN VAN TOAN - NGUYEN ANH TRUCfNG

(Nhdm giao vien chuyen luyen thi B^i hpc)

PHUONG PHAP GIAI TOAN

HtNH HOC

Chiu trdch nhiem xuat ban

Gidm doc - Tong bi&n tap : TS P H A M THj T R A M

Bien tap : N G Q C L A M

Che ban : C O N G TY K H A N G V I E T

Trinh bay bia : C O N G TY K H A N G V I E T

Tong phdt hanh va doi tdc lien ket xuat ban:

I Tpa dp trong mat phang

• Cho u ( x p y j ) ; v(x2;y2) va k e R K h i do:

1) u + v = (xi + X 2 ; y i + y 2 ) 2) u - v = ( x i - X 2 ; y i - y 2 )

3) k u = ( k x i ; k y i ) 4) Z=Jx\+y\) u=vc ^r^ "''""''^

6) U V = X]X2 + y ] y 2 = > u l v < ; : > u v = 0<=> \-^\2 + y ] y 2 = 0

• H a i v e c t a u ( x j , y j ) ; v ( x 2 ; y 2 ) ciing phirang v a i nhau <=>

• Goc giija hai vec to u ( x j , y j ) ; v ( x 2 ; y 2 ) :

• Cho tarn giac A B C v o i A{x^;y^), B(xB;yB), C{x^;y^) K h i do trong tarn

G ( x( , ; y g ) ciia tarn giac A B C la :

V _ X A + X B + X C

X G - ^

y G = I

II PhirotTg trinh duong thang ,, ,^,

1 'Phuang trinh duong thdng 1.1 Vec to chi phucmg (VTCP), vec to phdp tuyen (VTPT) cua duong thang:

Cho d u o n g thang d

• n = (a;b) ?t 0 goi la vec to phap tuyen cua d neu gia ciia no v u o n g v o i d

3

• u = ( u j; u 2 ) ^ 0 goi la vec ta chi phuong cua d ne'u gia cua no trung hoac

song song voi duong thang d

Mot duong thang c6 v6 so VTPT va v6 so VTCP ( Cac vec to nay luon cung

phuong voi nhau)

• Moi quan he giua VTPT va VTCP: n.u = 0

• Ne'u n = (a; b) la mpt VTPT cua duong thang d thi u = (b; -a) la mot VTCP

cua duong thang d

• Duong thang AB c6 AB la VTCP

1.2 Phuwig trinh dumig thang

1.2.1 Phuatig trinh tong qudt cua duong thang:

Cho duong thMng d d i qua diem A(xQ;yQ) va c6 n = (a;b) la VTPT, khi do

phuong trinh tong quat ciia d c6 dang: a(x - X Q ) + b(y - yp) = 0

1.2.2 Phuovg trinh tham so cua duong thang:

Cho duong thSng d di qua diem A(xo;yo) va c6 u = (a;b) la VTCP, khi do

X = X Q + at phuong trinh tham so cua duong thang d la: , t G R

[y = y ( , + b t

2 Vi tri tuang doi giua hai duang thdng

Cho hai duong thcing dj : a^x + bjy + c^ = 0; d2 : a2X + b2y + C2 = 0 Khi do vi tri

|a,x + b,y + Cj = 0 tuong doi giua chung phu thuoc vao so nghiem cua h^ : < , (I)

[a2X + b2y+ C2 =0

• Neu (I) v6 nghiem thi d^ / /d2

• Ne'u (I) v6 so nghiem thi d^ = d j

• Ne'u (I) CO nghiem duy nha't thi dj va d2 cat nhau va nghiem ciia he la toa

do giao diem '

3 Goc giua hai dijcang thdng

Cho hai duong thang dj : a j X + b^y+ Cj =0; d2 :a2X + b2y + C2 = 0 Goi a

la goc nhon tao boi hai duong thang dj va d2

Ta CO : cosa = aja2 + bjb2

^/a^Tb^ ^/af+b

4 JChodng each tit mot diem den ducrng thdng

Cho duong th5ng A : ax + by + c = 0 va diem M ( X Q ; y ^ ) Khi do khoang each

tu M den A dugc tinh boi cong thuc:

Cty TNHH MTV DWH Khang Viet

d(M,(A)): axp + byp + c

Va^+b^

5 (phuong trinh duang phdn gidc cua goc tao boi hai duang thdng

Cho hai duong thang d^ : a^x + b^y + c^ = 0 va d2 : ajX + b2y + Cj = 0

Phuong trinh phan giac ciia goc tao boi hai duong thang la: - , v , •

a j X + b^y + Cj a2X + b2y + C2

+ ^ / a ^ ^ b [ • , i c - i ; ;

1 <Phuang trinh duang tron:

Cho duong tron (C) tam I(a; b ) , ban kinh R, khi do phuang trinh ciia (C) la: ( x - a) 2 + ( y - b ) 2 = R 2

Ngoai ra phuong trinh: x^ + y ^ - 2 a x - 2 b y + c = 0 voi a ^ + b ^ - o O cQng

la phuong trinh ciia duong tron c6 tam I(a;b), ban kinh R = Va^ + b^ - c

2 Phuang trinh tiep tuyen:

Cho duong tron ( C ) : ( x - a ) ^ + ( y - b ) ^ = R ^

• Tiep tuyen A ciia (C) tai diem M la duong thang d i qua M va vuong goc vai I M

• Duong thang A : Ax + By + C = 0 la tiep tuyen ciia (C) <=> d(I, A) = R '

• Duong tron ( C ) : (x - a ) ^ + (y - b)^ = R^ c6 hai tiep tuyen cung phuong voi

Oy la x = a ± R Ngoai hai tiep tuyen nay cac tiep tuyen con lai deu c6 dang:

y = kx + m

IV E lip

1 'i)inh nghra.-Trong mat phang cho hai diem co'djnh Fi ,F2 c6 Y^Yj =2c Tap

hop cac diem M cua mat phang sao cho MF^ +MF2 =2a (2a khong doi va

a > c > 0) la mot duong elip

• F,,F2 : la hai tieu diem va 2c la tieu cu ciia elip

• MF|,MF2 : la cac ban kinh qua tieu

2 Phuang trinh chinh tdc cua elip:

4 + 4 = ^ voi b^=a^-c^ K'

a 2 b^

Vay diem M(xo;y(,) e (E) • = 1 va <a Yo < b ,

Phumtg phcip giiii Toan Hhih hoc theo chuyen tie- Nguyen Phu Khdnh, Nguyen Tat Thu

3 Tinh chat v>d hlnh dang cua elip: Cho (E): — + ^ = 1 , a > b

a b

• True doi xung Ox,Oy Tarn dói xiing O j ,

• Dinh: A[(-a;0), A2(a;0), 6^(0;-b) va 62(0; b ) ÂA2 = 2a goi la do dai

true Ion, B]B2 = 2b goi la do dai true bẹ

• Noi tiep trong hinh ehir nhat co so PQRS

C O kích thuoc 2a va 2b voi b^ = â - ệ

1 ^inh nghia: Trong mat phang voi h$ toa do Oxy eho hai diem Fi, F2 eo

FjF2 =2c Tap hop cac diem M ciia mat phSng sao eho MF^ - M F j =2a (2a

khong doi va c > a > 0 ) la mpt Hypebol

• Fp F2 : la 2 tieu diem va F|F2 = 2e la tieu eụ

• 1VIF[,MF2 : la eac ban kinh qua tieụ

2 'Phimng trinh chinh idc cua hypebok x^ y^

â

= 1 voi h^=c^-ậ

3 Tinh chat vd hlnh dang cua hypebol (fi):

• True doi xung Ox (true thuc), Oy (true ao) Tam doi xung O

• Dinh: Aj(-a;0), A2 (a;0) D Q dai true thuc: 2a va do dai true ao: 2b

• Tieu diem Fi(-e; 0), Fj ( c; O)

• Hai tiem can: y = ± —x

a

• Hinh eho nhat co so PQRS c6 kieh thuoe 2a, 2b voi b^ = c^ - ậ

• Tam sai: e = — =

a

• Hai duong chuan: x = ±— = ± —

Cty TNHH MTV DWH Khang Viet

• D O dai cac ban kinh qua tieu cua M ( x o ; y ( , ) e ( H ) : +) MF^ = ex„ + a va MF2 = e X ( , - a khi X Q > 0 +) MFj = -exp - a va MF2 = -exp + a khi X Q < 0

Parabol la tap hop cae diem M cua mat phang each deu mot duong thang

A c o ' d i n h v a m o t diem F co dinh khong thuoe A

A : duong chuan; F : tieu diem va d(F,A) = p > 0 la tham sótieụ

2 'Phuxmg trinh chinh tdc cua ^arabd: = 2px 3.jrinh dang cua Parabol (<P):

• True Ox la true dói xung, dinh Ọ Tieu diem F ( ^ ; 0 )

• Duong chuan A : x =

• M ( x ; y ) e ( P ) : MF = x + ^ voi x > 0

B, CAC BAI THlfONG GAP

§ 1. cAc B A I T O A N C O B A N

1 Xg.p phuang trinh duang thang

De lap phuong trinh duong thang A ta thuong dung cac each sau

• T i m diemM(xo;yo) ma A di qua va mot VTPT n = (a;b) Khi do phuong trinh duong thang can lap la: ăx - X Q ) + b ( y - yp) = 0

• Gia su duong thang can lap A : ax + by + e = 0 Dua vao dieu kien bai toan ta tim dugc a = mb,c = n b Khi do phuong trinh A : m x + y + n = 0 Phuong phap nay ta thuong ap dung doi voi bai toan lien quan den khoang each va goe

• Phuong phap quy tich: M(xQ;yQ)e A:ax + by + e=^Oc:> axy + by^ + e = 0

Vidu 1.1.1.Trong mat phSng voi he toa do Oxy cho duong tron

( C ) : ( x- ] ) 2 + ( y - 2 ) 2 = 2 5

1) Viet phuong trinh tiep tuyen ciia (C) tai diem M(4;6), ' 2) Viet phuong trinh tiep tuyen cua (C) xuát phat t u diem N ( - 6 ; l )

Phucntg phap giai ToAn Ilinh hoc theo chuycn lic- Nguyen Pliii Khanh, Nguyen Tat Thii

3) T u E(-6;3) ve hai tie'p tuye'n EA, EB (A, B la tie'p diem) den (C) Viet

phuong trinh d u o n g thang A B

D u o n g tron (C) c6 tam 1(1; 2), ban kinh R = 5

1) Tie'p tuyen d i qua M va vuong goc v o i I M nen nhan I M = (3;4) lam VTPT

Nen p h u o n g trinh tie'p tuye'n la: 3(x - 4) + 4(y - 6) = 0 <=> 3x + 4y - 36 = 0

2) Gpi A la tie'p tuye'n can t i m

Do A d i qua N nen p h u o n g trinh c6 dang

A : a ( x + 6) + b ( y - l ) = 0<=>ax + by + 6 a - b = 0, a^ + b^ (*)

Ta c6:

7a+ b d(I,A) = R o

l A N A = 0 [(a - l)(a + 6) + (b - 2)(b - 3) = 0

= ^ 7 a - b + 20 = 0

T u do ta suy ra duoc A e A : 7 x - y + 20 = 0

Tuong t u ta cung c6 dug-c B e A = > A B = A = > A B : 7 x - y + 20 = 0

2 Cdch lap phimng trinh dizcrng tron

De lap p h u o n g trinh d u o n g tron (C) ta thuong su dung cac each sau

Cdch 7;Tim tam I(a;b) va ban kinh ciia d u o n g tron K h i do p h u o n g trinh

d u o n g tron co dang: (x -a)^ + ( y - b)^ =

Cdch 2;Gia su p h u o n g trinh d u o n g tron co dang: x^ + y^ - 2ax - 2by + c = 0

8

Cty TNHH MTV DWH Khang Viet

Dua vao gia thie't cua bai toan ta tim dugc a,b,c Cach nay ta t h u o n g ap dung khi yeu cau viet phuong trinh d u o n g tron di qua ba diem

Vi du 1.1.2 Lap p h u o n g trinh d u o n g tron (C), bie't

1) (C) d i qua A(3;4) va cac hinh chie'u ciia A len cac true toa do

2) Goi I(a;b) la tam ciia d u o n g tron (C), v i l € ( C i ) nen: ( a - 2 ) + b = - (1)

Do (C) tie'p xuc voi hai d u o n g t h i n g A ^ A j nen d(I, A j ) = d(I, A2)

a - b a - 7 b

<=>b = -2a,a = 2b

• b = -2a thay vao (1) ta CO dugc:

(a - if- + 4a^ = - <=> 5a^ - 4a + — = 0 phuong trinh nay v6 nghiem

• a = 2b thay v a o ( l ) taco: ( 2 b - 2 r + b ' ' = - < : : > b = - , a = -

o 0 0

Suy ra R = D ( I , A , ) = Vay p h u o n g trinh ( C ) :

3 Cac diem, ctqc biet trong tam gidc

Cho tam giac ABC K h i do:

x — + y

Phumig phdpgidi Todn Hiith hoc theo chiiyen de - Nguyen Phi't Klidnh, Nguyen Tat Thu

• Trong tam G

• True tam H :

3 ' 3 AH.BC = 0

BH.AC = 0 Tam duong tron ngoai tiep I: lA^ = IB^ lA^ = IC^

• Tam duong tron noi tiep K :

Chu y:C6 the tim K theo each sau:

* Ta CO AK = KD tu day ta c6 K

BD ^ Tam duong tron bang tiep (goc A) J: AB.AJ AC.AJ AB AC

BJ.BC AB.BJ

BC AB l?jdui.i.3.Cho tam giac ABC c6 A(1;3),B(-2;0),C 5 3

1) Tim toa do true tam H, tam duong tron ngoai tiep I va trong tam G cua

tam giac ABC Tu dp suy ra I, G, H thang hang;

2) Tim toa do tam duong tron noi tiep va tam duong tron bang tiep goc A

cua tam giac ABC

AH = (x-l;y-3),BH = (x + 2;y),BC = 21 3 ,AC = ( 3 _21 8' 8

in

CUj TNHH MTV DWH Khang Viet

Ma < AH.BC = 0 nen ta eo BH.AC = 0

3 1

7(x-l) + (y-3) = 0 j7x + y-10 = 0 (x + 2) + 7y = 0 [x + 7y + 2 = 0

3

X = —

2

y = -: 1 2' 2

Suy ra H Goi I(x;y), taeo:

8 ^ Vay K(0;1)

Goi J(a;b) la tam duong tron bang tiep goc A eiia tam giac ABC Ta co:

Phuvng phlip gidi Todn Hinh hoc theo chiiyen dc- Nguyen Phu Khdnh, Nguyen Tat Tltu

BC ~ AB

2a - b = - 1 2a + b = -4

4 Cdc duang ddc hiet trong tam gidc

4.1. D u a n g trung tuyen cua tam giac: K h i gap duong trung tuyen cua tam

giac, ta chu yeu khai thac tinh chat d i qua dinh va trung diem cua canh do'i dien

4.2. D u o n g cao cua tam giac: Ta khai thac tinh chat d i qua d i n h va vuong

goc voi canh do'i dien

4.3. D u o n g trung true cua tam giac: Ta khai thac tinh chat d i qua trung

diem va vuong goc voi canh do

4.4. D u o n g phan giac trong: Ta khai thac tinh chat ne'u M thuoc AB, M ' doi

xung voi M qua phan giac trong goc A thi M ' thuoc A C

Vidu 7.i.4.Trong mat ph^ng v o i he tpa do O x y , hay xac d j n h toa do d i n h C

cua tam giac ABC bie't rang hinh chie'u vuong goc cua C tren d u o n g thang

AB la diem H ( - l ; - l ) , d u o n g phan giac trong cua goc A c6 p h u o n g trinh

x - y + 2 = 0 va d u o n g cao ke t u B c6 phuong trinh 4x + 3y - 1 = 0

JCffigidi

K i hi?u d , : X - y + 2 = 0, d2 : 4x + 3y - 1 = 0

Goi H ' la diem doi x u n g voi H qua d j K h i do H ' E A C

Goi A la d u o n g thang d i qua H va vuong goc v o i d j

x + y + 2=::0 Phuong trinh cua A : x + y + 2 = 0 Suy ra A n d j = I :

x - y + 2 = 0 I(-2;0)

Nen A C n d j = A :

Ta CO I la t r u n g diem ciia H H ' nen H ' ( - 3 ; l )

D u o n g thang A C d i qua H ' va vuong goc voi d j nen c6 p h u o n g trinh :

•A(5;7)

12

Cty TNHH MTV DWH Khang Vie

Vi du 1.1.5 Trong mat phang vai he toa do Oxy , cho tam giac ABC biet

A ( 5 ; 2 ) Phuong trinh d u o n g trung true canh BC, d u o n g t r u n g tuyen C C Ian l u ^ t la x + y - 6 = 0 va 2 x - y + 3 = 0 T i m toa do cac d i n h B,C cua tam giac ABC

Xgfi gidi

Goi d : x + y - 6 = 0, C C : 2 x - y + 3 = 0 Ta c6: C(c;2c + 3) Phuong trinh BC : x - y + c + 3 = 0

Goi M la t r u n g diem ciia BC, suy ra M :

3 ' 3 , C

14 37

3 ' 3

5 Mot sobdi todn dung hinh ca ban

5.1. H i n h chie'u v u o n g goc H cua diem A len d u o n g thang A

• Lap d u o n g thang d d i qua A va vuong goc v o i A

• D u n g r doi x u n g v o i I qua d u o n g thang A

• D u o n g tron ( C ) c6 tam I ' , ban kinh R

Chii y: Giao diem ciia (C) va ( C ) chinh la giao diem cua va A

5.4. D u n g d u o n g thang d ' doi xung voi d qua d u o n g thang A

• Lay hai diem M , N thuoc d D u n g M ' , N ' Ian luot d o i x u n g v o i M , N qua A

'if!', r<(.:

• d ' = M ' N '

Phumig phdp gidi Todii Uinh hoc theo chuyen dc - Nguyen Pliii Khdnh, Nguyen Tat Thti

Vidu 1.1.6.Trong mat phang Oxy cho d u o n g thang d : x - 2 y - 3 = 0 va hai

diem A(3;2), B ( - l ; 4 )

1) T i m diem M thuoc d u a n g thang d sao cho M A + M B nho nhat,

2) Viet p h u o n g t r i n h d u a n g thang d ' sao cho d u o n g thang A : 3x + 4y + 1 = 0

la d u o n g phan giac ciia goc tao boi hai d u o n g thang d va d '

JCffigidi

1) Ta tha'y A va B n a m ve m o t phia so v o i d u o n g thang d Goi A ' la diem doi

x u n g v o i A qua d K h i do v a i m o i diem M thuoc d, ta l u o n c6: M A = M A '

V i A la phan giac cua goc h g p bai giiia hai d u a n g thang d va d ' nen d va

d ' do'i x u n g nhau qua A , do do l e d '

'3 _ 1 6 ' 5 ' " 5

Lay E(3;0) G d , ta tim dugc F la d i e m do'i x i i n g v a i E qua A , ta c6

1) T i m toa do true tam, tam d u o n g tron ngoai tiep tam giac A B C 2) Viet p h u o n g t r i n h d u o n g tron ngoai tiep tam giac ABC

Jiuang ddn gidi

1) Goi H ( x ; y ) la true tam tam giac ABC, ta c6: A H B C = 0

BH.AC = 0

'(x - 2)(-7) + (y - 1 ) ( - 4 ) = 0 J7x + 4y - 1 8 = 0 (x - 4)(-5) + (y - 3)(-2) = 0 ^ [Sx + 2y - 26 = 0 ^

X = 34

y = -46 Vay H 34 46

Goi I ( x ; y ) la tam d u a n g tron ngoai tiep tam giac ABC, ta c6:

Bai 1.1.2. Trong mat phang toa do Oxy cho tam giac ABC c6 A(3;2) va

p h u o n g t r i n h hai d u a n g trung tuyen B M : 3x + 4y - 3 = 0 , C N : 3x - lOy - 1 7 = 0 Tinh toa do cac diem B, C

Jiuang dan gidi : ? ; • ;

Goi G la trong tam ciia tam giac, suy ra toa do ciia G la nghiem cua he '3x + 4y - 3 = 0

3 x - 1 0 y - 1 7 = 0

7

^ = 3 [ y = - l

> ; r J J ' I ' i

-Phumig phdpgiiii Toan Hitih hoc theo chuyen de- Nguyen Phi'i Khanh, Nguyen Tat Thu

Goi E la trung diem ciia BC, suy ra EA = - G A => E(2;

Gia sir B(a;b), suy ra C ( 4 - a ; - 5 - b ) T u do ta c6 h^:

Bai 1.1.3. Trong mat phang toa do Oxy cho tam giac A B C c6 A ( - 3 ; 0 ) va

p h u o n g trinh hai d u o n g phan giac trong B D : x - y - 1 = 0,CE : x + 2y +17 = 0

Tinh toa do cac diem B, C

Jiu&ng ddn gidi

Gpi A^ d o i x i i n g v o i A qua BD, suy ra A j e BC va A ^ ( l ; - 4 )

A j do'i x u n g v o i A qua CE, suy ra A 2 e BC va A 2 ( - — ; - — )

5 5 Suy ra p h u o n g trinh BC : 3x - 4y - 1 9 = 0

x - y - l = 0 Toa dp B la nghi^m cua he:

Toa do C la nghiem cua he:

Bai 1.1.4.Trong mat phSng toa do Oxy cho tam giac A B C c6 C(5;-3) va

p h u o n g trinh d u o n g cao A A ' : x - y + 2 = 0 , d u o n g trung tuyen

B M : 2x + 5y - 1 3 = 0 Tinh toa do cac diem A , B

Jiixang ddn gidi

Ta CO p h u o n g trinh BC: x + y - 2 = 0

fx = - l Suy ra toa do ciia B la nghiem cua he: x + y - 2 = 0

2x + 5 y - 1 3 = 0 l y = 3 • B ( - l ; 3 ) Gpi A(a;a + 2), suy ra toa do ciia trung diem A C la M + 5 a - 1 ^

M a M e B M nen 2 ^ y ^ + 5 ^ - 1 3 = 0 « a = 3 =^ A ( 3 ; 5 )

Vay A(3;5),B(-1;3)

Bai 1.1.5. Trong mat phang toa dp Oxy cho tam giac A B C CO B(l; —3) va

p h u o n g trinh d u o n g cao A D : 2 x - y + 1 = 0, d u o n g phan giac C E : x + y - 2=::0

.Tinh toa dp cac diem A , C

Cty TNHH MTV DWH Khang Viet

Jiic&ng ddn gidi

Ta CO p h u o n g trinh B C : x + 2y + 5 = 0

[x + y - 2 = 0 [x = 9 Tpa dp diem C la nghiem '^"^ L ^ 2y + 5 = 0 ^ |y = - 7 Gpi B' la diem d o i x u n g v o i B qua CE, suy ra B'(5;l) va B' e A C

Bai 1.1.6. Trong mat phang v o i h^ tpa dp Oxy, cho tam giac ABC co M (2; 0)

la trung diem cua canh AB D u o n g trung tuyen va d u o n g cao qua d i n h A Ian

lupt CO p h u o n g trinh la 7x - 2y - 3 = 0 va 6x - y - 4 = 0 Viet p h u o n g trinh

duong thang A C

Jiu&ng ddn gidi

| ' 7 x - 2 y - 3 = 0 Toa do A thoa m a n he: <^

• • [ 6 x - y - 4 = 0

V i B do'i xiing v o i A qua M nen suy ra B = (3; - 2 )

D u o n g thSng BC d i qua B va vuong goc v o i d u o n g thSng: 6x - y - 4 = 0 nen suy ra

Phuong trinh B C : x + 6y + 9 = 0

' 7 x - 2 y - 3 = 0 ' x + 6 y + 9 = 0

Phumig phtip giai Toan Hinh hoc theo chuyen dS"- Nguyen Phu Khdnh, Nguyen Tat Thii

1) Gia six A : ax + by + c = 0 la tiep tuyen ciia (C)

Bai toan co ban ciia phuong phap toa do trong mat phang la bai toan xac

dinh toa do ciia mot diem ChSng han, de lap phuong trinh duong thang can

tim mot diem di qua va VTPT, voi phuong trinh duong tron thi ta can xac djnh

tarn va ban kinh Chung ta co the gap bai toan tim toa do ciia diem dugc hoi

true tiep hoac gian tiep

• Ve phuong dien hinh hgc tong hgp thi de xac dinh toa do mot diem, ta

thuong chiing minh diem do thugc hai hinh (H) va (H') Khi do diem can tim

chinh la giao diem ciia (H) va (H')

• Ve phuong di^n dai so, de xac dinh toa do ciia mot diem (gom hai toa do) la

bai toan di tim hai an Do do, chiing ta can xac djnh dugc hai phuong trinh

chiia hai an va giai he phuong trinh nay ta tim dugc toa do diem can tim Khi

thiet lap phuong trinh chiing ta can luu y:

+) Tich v6 huong ciia hai vec to cho ta mgt phuong trinh,

+) Hai doan thang bang nhau cho ta mgt phuong trinh,

+) Hai vec to bang nhau cho ta hai phuong trinh,

18 '

Cty TNHH MTV DWH Khang Viet

+) Neu diem M e A : ax + by + c = 0,a ^ 0 thi M - b m - c - ; m , liic nay toa

do ciia M chi con mgt an va ta chi can tim mgt phuong trinh

Vi da 1.2A. Trong mat phang Oxy cho duong tron (C): (x - 1 ) ^ + (y - 1 ) ^ = 4

va duong thang A : x - 3 y - 6 = 0 Tim tga dg diem M nam tren A , sao cho tvr M ve dugc hai tiep tuyen MA, MB (A,B la tiep diem) thoa AABM la tam giac vuong

Xgigiai

Duong tron (C) co tam 1(1; 1), ban kinh R = 2

Vi AAMB vuong va I M la duong phan giac ciia goc AMB nen A M I = 45°

Trong tam giac vuong l A M , ta co:

IM = 2V2, suy ra M thugc duong tron tam I ban kinh R' = 2 Mat khac M e A nen M la giao diem ciia A va (I,R') Suy ra tga do ciia M la nghiem ciia he

x = 3 y + 6

x - 3 y - 6 = 0 ( x - i ) 2 + ( y - i ) 2 =8 " [(3y + 5)2 + ( y - l ) ' =8 'x = 3y + 6

Vi du 1.2.2. Trong mat phSng voi he tga do Oxy cho cac duong thang

d i : x + y + 3 = 0, d j : x - y - 4 = 0, dg : x - 2 y = 0 Tim tga do diem M nam tren duong thSng sao cho khoang each tu M den duong thang d^ bang hai Ian khoang each tu M den duong thang d2

Xffi gidi

Taco M e d 3 , s u y r a M(2y;y) Suy ra d(M,di) = — ^ ; d ( M , d 2 ) = ^ ^ Theo gia thiet ta co: d(M,di) = 2d(M,d2) <^ 3y + 3 ^ 2 l y - 4

Phuvng plidp gidi Toiin Hiith hoc theo chuyen dc- Nguyen Phii Klidnh, Nguyen Tn't Thu

Vi du 1.2.3 Tron g he toa do O x y , cho die m A(0; 2) va d u o n g th3ng

d : x - 2 y + 2 = 0 T i m tren d u o n g thang d hai diem B, C sao cho tam giac

Vay CO hai bp d i e m thoa yeu cau bai toan la:

Cty TNHH MTV DWH Khang Viet

Vi du 1.2.5 Cho parabol (P): y^ = x va hai diem A(9; 3), B ( l ; -1) thupc (P)

Gpi M la diem thupc cung A B cua (P) (phan ciia (P) bi chan b o i day A B ) Xac djnh tpa dp d i e m M nam tren cung A B sao cho t a m giac M A B c6 dien tich ion nha't

JCgi gidi

P h u o n g t r i n h A B : x - 2y - 3 = 0

Vi M G (P) => M ( t ^ ; t) t u gia thiet suy ra - 1 < t < 3

Tam giac M A B c6 dien tich ion nha't o d ( M , AB) Ion nha't

Vi du 2.6 T r o n g mat phang Oxy cho d u o n g tron (C): (x - 1 ) ^ + y^ = 2 va

hai diem A ( l ; - 1 ) , B(2;2) T i m tpa diem M thupc d u o n g tron (C) sao cho dien tich tam giac M A B bang ^

Xffi gidi

Ta CO A B = Vio va S^^^AB = - d ( M , A B ) A B = d ( M , A B ) =

Lai CO A B = (1;3) nen n = ( 3 ; - l ) la VTPT ciia d u o n g thang A B Suy ra p h u o n g t r i n h A B : 3(x - 1 ) - ( y +1) = 0 hay 3 x - y - 4 = 0 Gpi M ( a ; b) e (C) => (a - i f + b^ = 2

b = 3a - 5 ( a - 1 ) 2 + ( 3 a - 5 ) 2 = 2

b = 3a - 5

3 a - b - 4 = l

( a - l ) 2 + b 2 = : 2 i ) k > , J

3 a - b - 4 = - l ( a - l ) 2 + b2 = 2

b = 3a - 3

hoac ( a - 1 ) ' + ( 3 a - 3 ) 2 = 2 v i , h ; ^ / , „

b = 3 a - 3

21

Phuang phdp gidi Todn Hinh hgc theo chuyen de- Nguyen Phu Khdnh, Nguyen Tat Thu

Bai 1.2.1 Trong mat phang voi he toa do Oxy, cho duong tron (C) c6

phuong trinh: (x - 1 ) ^ + y^ = 1 Ggi I la tarn ciia (C) Xac dinh toa do diem M

thuQC (C) sao cho I M O = 30"

Jiudng ddn gidi

Ggi diem M ( a ; b ) D o M e ( C ) nen {a-lf +h^=l

Mat khac O € (C) =:> l O = I M = 1

Tam giac I M O c6 O I M - 1 2 0 °

nen O M ^ = lO^ + I M ^ - 2IO.IM.cos 120° o a^ + b^ = 3

Toa do diem M la nghiem cua he : ( a - l) ' + b 2 = l

Bai 1.2.2 Trong mat ph^ng voi he toa do Decac vuong goc Oxy, cho

parabol (P) c6 phuong trinh y^ = x va diem 1(0; 2) Tim toa do hai diem M , N

thuQC (P) sao cho I M = 4iN

Jiuong dan gidi

n = l

m = 2 V {

Vay CO hai cap diem thoa yeu cau bai toan la:

1 l ^ M(4;2),N(1;1) hoac M

Bai 1.2.3 Trong mat phang toa dp Oxy cho diem A(3;2), cac duong thang

dj : X + y - 3 = 0 va: d2 : x + y - 9 = 0 Tim toa do diem B G d j , va C e d2 sao

cho tam giac ABC vuong can tai A ,

u ^ + ( u + 2)^ = ( v - l ) 2 + ( v - 5 ) ^

u ( v - l ) + (u + 2 ) ( v - 5 ) = 0 3u + 5

Vay CO hai cap diem thoa yeu cau bai toan la: ,.j *

B(4;-1),C(6;3) hoac B(0;3),C(4;5)

Chti i/.-Ngoai each tren, ta c6 the giai theo each khac nhu sau:

Tjnh tien he true toa dp Oxy ve he tuc XAY theo vec to O A , ta c6 cong thuc 'x = X + 3 ' ;

Trong he true moi, ta c6 phuong trinh cua dj :X + Y + 2 = 0, d2 :X + Y - 4 = 0

Vi tam giac ABC vuong can tai A nen phep quay Q Q : B -> C

( A , ± 9 0 ) |,

, , , '23

doi true:

Phumtg phapgiiii Toan Hinh hqc theo chuyen rfe - Nguyen Pht'i Khanh, Nguyen Tat Thu

Ma B e => C e d 1 = Q^^.^^^jo/cli), do do C ^ d2 n d ,

• Xet phep quay Q^^ ^^^^^, ta c6 phuong trinh d , : X - Y - 2 = 0

Do do toa dp cua C la nghiem cua he: X - Y - 2 = 0

X + Y - 4 = 0 <=> {

X = 3

Y = r

Xet phep quay Q^^ ^^^(y ta c6 phuong trinh d j : X - Y + 2 = 0

Do do tga do cua C la nghiem cua he: • ^ ^ ^ ^ ^ <=>

Bai 1.2.4. Trong he true toa do Oxy cho AABC voi A(2;3), B(2;l), C(6;3),

Goi D la giao diem cua d u o n g phan giac trong goc BAG v o i BG T i m tat ca cac

diem M thuoc d u o n g tron (G): (x - 3)^ + (y -1)^ = 25 sao cho : 5^^^ = 2SADB •

Jiuang ddn gidi

Ta CO A B = (0; 2), A C = (4; 0), BG - (4; 2)

Phuong trinh AB: x - 2 = 0,

nen d(D, AB) = | =^ S ^^BD = {AB.d(D,AB) = 1-2.^ = 1

Phuong trinh DG : x - 2y = 0 Goi M(a; b) => (a - 3)^ + (b -1)^ = 25 (1)

Cty TNHH MTV DWH Khang Vie I

Bai 1.2.5. Trong mat phSng v o i he true toa do Oxy, cho d u o n g thSng

Bai 1.2.6. Trong mat phSng Oxy cho diem A ( l ; 4 ) T i m hai diem M , N Ian lugt

nam tren hai d u o n g tron ( q ) : ( x - 2 ) 2 + ( y - 5 ) 2 =13 va {C^):{x-lf+{y-2f ^75

sao cho tam giac M A N vuong can tai A

x2 + y 2 _ i 0 y + 12 = 0 x2 + y 2- 2 x - 4 y - 2 0 = 0

x 2 + y 2- 1 0 y + 1 2 = : 0 ^ J5y2 - 53y +134 = 0

x = 3 y - 1 6 -1 + 37T29

Phumtg phdp gidi Todn Hinh hgc theo chuyen de- Nguyen Phti Khihih, Nguyen Tat Thu

Q(A-9oO)'*^™P''^""^*""^ (Ci):(x-2)2+(y-3)2 =13

Toa do diem N la nghiem ciia he: (x-2)2+(y-3)2 =13 {x = A \x = 5 <=> < V <

( x - ] ) 2 + ( y - 2 ) 2 =25 ly = 6 [y = 5

Tmonghgp nay c6 hai bo diem: M(-1;7),N(4;6) va M(0;8),N(5;5)

Bai 1.2.7. Trong mat phSng Oxy cho duong tron (C): (x - if + (y - 4)^ = y

va duong thang d : 5x + 2y -11 = 0 Tim diem C tren d sao cho tam giac ABC

CO trong tam G nam tren duong tron (C) biet A(l;2),B(3;-2)

29' 29

Bai 1.2.8. Trong he toa dp Oxy cho duong thang d : x - y + l= 0 va duong

tron (C) CO phuong trinh x'^ + y^ + 2x - 4y = 0 Tim diem M thuoc duong thSng

d sao cho tir M ke dupe hai duong thing tiep xuc vai duong tron tai A va B,

sao cho AMB = 60"

Jlucrng ddn gidi

Duong tron c6 tam I(-l;2) va ban kinh:R = Vs

Tam giac AMB la tam giac deu va MI la phan giac goc AMB nen IMA = 30°

Do do: MI = ^ = iS IM^ = 20

sin30"

Do M e d nen suy ra M ( X Q ; XQ +1)

Khi do ta c6: MI^ = ( X ( , +1)^ + (x„ -1)^ = 20 o x^ = 9 x^ = 3; XQ = -3

Vay CO 2 diem M thoa man dieu kien bai toan: (3; 4); (-3;-2)

Bai 1.2.9. Trong mat phang voi he toa dp Oxy cho diem C(2;-5) va duong

th^ng A : 3x - 4y + 4 = 0 Tim tren A hai diem A va B doi xung nhau qua 1(2; | )

sao cho dien tich tam giac ABC banglS

Vi du 1.3.1 Trong mat phSng Oxy cho hai duong thang di: x - 2y + 1 = 0,

d2: 2x + 3y = 0 Xac djnh tpa dp cac dinh cua hinh vuong ABCD, biet A thupc

j y o n g thang di, C thupc duong thang d2 va hai diem B, D thupc true Ox

27

Phucnig plidp giai Toan Hinh hgc theo chuyen de - Nguyen Phu Kluhili, Nguyen Tat Thu

M a t khac A C 1 BD = Ox nen suy ra 2a - 1 = 3c o c = 1

T u do, ta t i m dugc A(3;2), C(3;-2), 1(3; 0)

V i B e Ox =^ B(b;0), ma IB = IA = 2 =^ |b - 3| = 2 o b = 5,b = 1

Vay toa do cac d i n h ciia hinh vuong ABCD la:

A(3;2), B(1;0), C(3;-2), D(5;0) hoac A(3;2), B(5;0), C(3;-2), D(1;0)

Vidu 7.3.2.Trong mat phang Oxy cho ba diem 1(1; 1), J(-2;2), K ( 2 ; - 2 ) T i m

toa do cac d i n h cua hinh vuong A B C D sao cho I la tam hinh vuong, J thuoc

canh A B va K thuoc canh CD

Goi J' doi x i i n g v a i J qua I , ta c6 J'(4;0) va J' € CD

Vidu 2.3.3.Trong mat phang Oxy cho d u o n g tron (C): (x - 2)^ + (y - 1)^ = 10

T i m toa do cac d i n h cua hinh vuong M N P Q , biet M t r u n g v o i tam cua

d u o n g tron (C); hai d i n h N , Q thuoc d u o n g tron (C); d u o n g thang PQ d i qua

E ( - 3 ; 6 ) va X Q > 0

Ta CO M ( 2 ; l ) va EQ la tiep tuyen cua ( C )

Phuong trinh EQ c6 dang:

ci>(5a-5b)^ =10(a2+b2)

<:i>3a^ -10ab + 3b^ = 0 <=> a = 3b,b = 3a , a = 3b, ta c6 p h u o n g trinh E Q : 3x + y + 3 = 0

K h i do toa do Q la nghiem ciia he ( x - 2 ) 2 + ( y - l ) 2 = 1 0

3x + y + 3 = 0

x = - l

y = 0

Truong hop nay ta loai v i X Q > 0

• b = 3a, ta CO p h u o n g trinh E Q : x + 3y - 1 5 = 0 K h i do toa do Q la nghiem

fx = 3

C i i a he ( x - 2 ) 2 + ( y - l ) 2 = 1 0

3x + y + 3 = 0 <=> < y = 4 •Q(3;4)

Taco P ( 1 5 - 3 x ; x ) va QP = M Q => (12 - 3x)^ + (4 - x f = 10 x = 3,x

X = 3, ta CO P(6;3), suy ra tam cua hinh vuong 1(4;2) nen N(5;0)

X = 5, ta CO P(0;5), suy ra tam cua hinh vuong 1(1;3) nen N ( - l ; 2 ) Vay CO hai bo diem thoa yeu cau bai toan:

M(2;1),N(5;0),P(6;3),Q(3;4) va M(2;1),N(-1;2),P(0;5),Q(3;4)

= 5

Vi du 1.3.4 Trong mat phang voi he toa do Oxy cho hinh chir nhat A B C D

CO diem 1(6; 2) la giao diem cua 2 d u o n g cheo A C va B D D i e m M ( 1 ; 5) thuQc d u o n g th3ng AB va trung diem E ciia canh C D thuoc d u o n g thang

d : x + y - 5 = 0 Viet p h u o n g trinh d u o n g thang AB

gidi

V i E € d = : > E ( a ; 5 - a ) = > i E = ( a - 6 ; 3 - a ) Goi N la t r u n g diem cua A B , suy ra I la trung diem cua E N nen :

Phumtgphdp gidi Todn Hinh hoc theo chuyen de- Nguyen Phu Khdnh, Nguyen Tat Thu

a = 6 => M N = (5;0), suy ra phuang trinh

A B : y - 5 = 0

• a = 7 => M N = (4;1), suy ra phuong trinh AB: x - 4y +19 = 0

Vi du 1.3.5. Trong mat phang voi he true toa do Oxy cho hinh chu nhat

ABCD CO dien tich bang U, tam I la giao diem cua duong thang

d i : x - y - 3 = 0 va d 2 : x + y - 6 = 0 Trung diem cua AB la giao diem aia

dj voi true Ox Tim toa dp cac dinh cua hinh chi> nhat

Taco dj n d j = 1 : x - y - 3 = 0 (9 3^

x + y - 6 = 0 yi 1,

Goi M la giao cua duong thang dj voi Ox, suy ra M(3;0)

Vi AB 1 M I nen suy ra phuong trinh AB:x + y - 3 = 0

Do I la tam ciia hinh chii nhat nen C(7;2), D(5;4)

Vay toa do cac dinh cua hinh chu nhat la: A(2;l), B(4;-l), C(7;2), D(5;4)

Vi da 1.3.6. Trong mat phang Oxy cho ba duong thang d j : 4x + y - 9 = 0,

d 2 : 2 x - y + 6 = 0, d 3 : x - y + 2 = 0 Tim toa dp cac dinh ciia hinh thoi

ABCD, biet hinh thoi ABCD c6 dien tich bang 15, cac dinh A, C thuoc ds, B

A(3;5),B(2;1),C(-2;0),D(-1;4) hoac A(-2;0),B(2;1),C(3;5),D(-1;4)

Vidu 1.3.7. Trong mat phMng he toa do Oxy, cho hinh thoi ABCD c6 tam

I H ^ lA^ I B ^ 4IB2 .IB = I H V S

Mat khac B e AB => B(b;^—^),b > 0 =^ IB^ = (b -2)^ +

3 Vay B(l;-1)

4b + 2

= 5 o b = l

du 1.3.8. Trong mat phang Oxy cho hai duong thang di: c + y - 1 = 0,

d2: 3x - y + 5 = 0 Tim toa dp cac dinh ciia hinh binh hanh ABCD, bie't 1(3; 3)

la giao diem ciia hai duong cheo; hai canh ciia hinh binh hanh nam tren hai

duong thang di, d2 va giao diem ciia hai duong thang do la mpt dinh ciia

Jynh binh hanh

Xffi gidi

Tpa dp giao diem ciia dj va d j la nghi^m ciia h?: <! ^ <=> x + y - l = 0 fx = - l

3 x - y + 5 = o'^|y = 2

Phumig pitdp giai Todn Hinh hoc theo chuyen de- Nguyen Phu Khdnh, Nguyen Tat Thu

D

Ta gia sir A ( - l ; 2) va AB = dp AD = d 2, suy ra C(7; 4) ^

Goi d la dtrong thang di qua I va

song song voi AB, suy ra phuang

A B

la trung diem ciia A D

Vidi^ 7.3.9 Cho hinh binh hanh ABCD c6 B(l;5), duang cao AH:x+2y-2=0,

duong phan giac trong ciia goc ACB c6 phuang trinh x - y - 1 = 0 Tim toa

do cac dinh con lai ciia hinh binh hanh

Ggi B' do'i xung voi B qua d, ta tim dugc B'{6;0) va B' e AC

Suy ra phuang trinh A C : x - 2y - 6 = 0

Toa dp diem A la nghiem ciia h$: x - 2 y - 6 - 0

x + 2 y - 2 = 0

x = 4

A ( 4 ; - l )

Vi A D = B C = > D ( - 1 ; - 1 1 )

Vidu 7.3.iO.Trong mat phang voi he toa do Oxy cho hinh vuong A B C D biet

M ( 2 ; 1 ) , N ( 4 ; - 2 ) ; P(2;0); Q(1;2) Ian lupt thupc canh A B , B C , C D , A D Hay

l^p phuang trinh cac canh cua hinh vuong

Truoc het ta chung minh tinh chat sau day:

"Cho hinh vuong ABCD, cac diem M,N,P,Q Ian luot nam tren cac duang

thSng AB, BC, CD, DA Khi do MP = NQ MP 1 N Q "

Chung minh: Ve ME 1 CD, E € CD; N F 1 AD, F e AD

Cty TNHH MTV DWH Khang Vi$t

Hai tarn giac vuong M E P va N F Q c6 N F = M E

Do do M P = N Q p A M E P = A N F Q

<-> E P M = F Q N <=> Q I M = 90" o M P 1 N Q

Tra lai bai toan:

Taco: MP = (0;-1)=>MP = 1 Gpi d la duong thang di qua N va vuong goc voi MP

Suy ra phuong trinh d: x - 4 = 0

Gpi E la giao diem cua d voi duong thang

AD, ap dung tinh chat tren ta suy ra NE = MP

Ma E(4;m) nen NE = MP o (m -2)^ = 1 <=> m = 3,m = 1

• Voi m = 3, suy ra E(4; 3) QE = (3; 1), suy ra phuang trinh

A D : x - 3y + 5 = 0 Phuong trinh A B : 3x + y - 7 = 0, BC : x - 3y -10 = 0, CD: 3x + y - 6 = 0

• Voi m = 1, suy ra E(4;l) QE = ( 3 ; - l ) , suy ra phuang trinh

A D : x + 3 y - 7 = 0 ^ ^ Phuang trinh A B : 3x - y - 5 = 0, BC: x + 3y + 2 = 0, CD: 3x - y - 6 = 0

Bai 1.3.1 Trong mat phang voi h? toa dp Oxy cho hai duang thSng di: x - y = 0, d2: 2x + y - 1 = 0 Tim tpa dp cac dinh hinh vuong ABCD biet rang dinh A thupc d j , d i n h C thuoc d j va cac dinh B,D thupc tryc hoanh

J^Iu&ng ddn gidi

Vi A e d j => A ( t ; t ) , A va C doi xung nhau qua BD va B,D€Ox=>C(t;-t)

Vi C G d 2 r ^ 2 t - t - l = 0 o t = l V a y A ( l ; l ) , C ( l ; - l )

flA = IB = l Trung diem cua AC la l ( l ; 0 ) Vi I la tarn cua hinh vuong nen

Phuong fihdp gidi Todn Hinh hoc thee chuyen de- Nguyen Phil Khdnh, Nguyen Tat Thti

Bai 1.3.2. Trong mat phang toa do Oxy cho dirong tron

(C):x^+y^-8x + 6y + 21=0 va duong thSng (d):x + y - l = 0

Xac dinh tga do cac dinli cua hinh vuong ABCD ngoai tiep (C) biet A e (C)

Jiuang ddn gidi

Ta CO I(4;-3),R = 2 Ian lugt la tarn va ban kinh cua (C)

Ta CO led , hinh vuong ABCD ngoai tie'p duong tron nen

Vay toa do cac dinli cua hinh vuong la A(2;-l);C(6;-5); B(2;-5),D(6;-1)

va cac hoan vi A cho C, B cho D

Bai 1.3.3. Biet A(1;-1),B(3;0) la hai dinh ciia hinh vuong ABCD Tim toa

* Voi Ci(4;-2)=>Di(2;-3)

* Voi C 2 ( 2 ; 2 ) = ^ D 2 ( 0 ; 1 )

Bai 1.3.4 Viet phuong trinh canh AB( AB c6 he so'goc duong), AD cua

hinh vuong ABCD biet A (2; -1) va duong cheo BD: x + 2y - 5 = 0

Cty TNHH MTV DWH Khang Viet

Hal gia tri ciia b tuong I'mg toa do hai diem B va D

Vi AB CO he so goc duong nen B(5;0), D(l;2) * - i ' :

Goi I la tarn cua hinh chu nhat =^ I la trung diem ciia BD => I d + 4 d + 2

Vi A, I, M thang hang nen ta c6:

Bai 1.3.6. Trong mat phang voi he toa do Oxy cho hinh vuong A B C D biet

M ( 2 ; l ) , N ( 4 ; - 2 ) ; P(2;0); Q(1;2) Ian lupt thupc canh A B , BC, C D , A D Hay lap phuong trinh cac canh ciia hinh vuong

Jiu&ng ddn gidi

Gia su duong thSng A B qua M va c6 vec to phap tuyeh la fi(a; b) '

(a +b 7i 0) suy ra vec to phap tuyeh ciia B C la:nj(-b;a) ^ , ,

Phuong trinh A B c6 dang: ax + by - 2a - b = 0

Phuongphdpgidi Todn Hinh hoc theo chuyen de- Nguyen Phu Khdnh, NguySn Tat Thii

• b = -2a suy ra phirong trinh cac canh can tim la:

A B : x - 2 y = 0 ; CD : x - 2 y - 2 = 0; BC: 2x + y - 6 = 0; A D : 2x + y - 4 = 0

• b = -a Khi do

A B : - x + y + l = 0 ;BC: - x - y + 2 = 0

A D : - x - y + 3 = 0 ; C D : - x + y + 2 = 0

B a i 1 3 7 Trong mat phang voi h§ toa do Oxy cho ba diem I ( 1 ; 1 ) , E ( - 2 ; 2 ) ,

F ( 2 ; - 2 ) T i m tpa do cac dinh ciia hinh vuong A B C D , biet I la tam cua hinh

vuong, AB di qua E va CD di qua F

Jiucmg dan gidi

Duong thang AB c6 phuong trinh dang: a(x + 2) + b(y - 2J = 0

Phuong trinh BC va D A c6 dang x + y + c = 0

B a i 1 3 8 Trong mat phang voi h§ toa do Oxy, cho hinh chii nhat ABCD c6

canh AB: x -2y -1 =0, duong choo BD: x- 7y +14 = 0 va duong cheo AC d i qua

diem M(2;l) Tim tQa do cac dinh cua hinh chu nhat

Cty TNHH MTV DWH Khang Viet

B a i 1.3.9- Cho hinh binh hanh ABCD c6 di?n tich bang 4 Biet A ( l ; 0), B(0; 2)

va giao diem I ciia hai duong cheo nam tren duong thSng y = x T i m toa do

I 6 t - 4 I Ngoai ra: d ( C ; AB) = C H o — ^ = -j=

Bai 1 3 1 0 Trong mat phang voi he toa do Oxy, cho hinh vuong ABCD

Goi M la trung diem canh BC, N la diem nam tren canh CD sao cho CN = 2ND

2 - A N - A M 2

Do do, phuong trinh duong thang A M qua M va tao voi A N mot goc

Phuang phdpgiai Todn Hinh hgc theo chuyen de- Nguyen Phi'i Khdnh, Nguyen Tat Thu

Gia sir duang thang A M c6 phap vector la ri = (a, b) (a^ +b'^ ^ 0) Khi do,

2a - b

ta ti'nh du-oc cos N A M =

VsVa^+b^'

Tu day, do cos N A M = — nen ta c6

A/2 |2a - b| = Vs Va^Tb^ o 3a^ - 8ab - 3b^ = 0 <::^ a = 3b v b = -3a

• Voi a = 3b : Chon b = 1, a = 3, ta c6 A M :

X - -11

1- = 0, tiVc 3x + y - 1 7 = 0

Tu day de dang tim dugc A(4, 5)

• Voi b = -3a : Chon a = 1, b = -3, ta c6 A M : 1 • ( i r ^ 0 X 3

-I 2 J

= 0 , tuc

X - 3y - 4 = 0 Vai ke't qua nay, ta tim duoc A ( l , -1)

Vay CO tat ca hai diem A thoa man yeu cau de bai la : A(4, 5) va A ( l , -1)

Bai 1 3 1 1 Cho hinh thang vuong ABCD, vuong tai A va D Phuang trinh

AD : X - y V 2 = 0 Trung diem M cua BC c6 toa do M ( l , 0) Biet BC = CD = 2AB

Tim toa do cua diem A

Jiu&ng dan gidi

2 / i Goi H la hinh chieu cua M len AD ta c6 H

3' 3 Nell cho AB = X ta c6 BC = CD = 2x de dang ta thay

Cty TNHH MTV DWIl Khnng Viet

Vay CO hai diein can tim la: A 6±V6 3j2±S

g^j 1.3.12 Trong mat phang Oxy cho bon diem M(4;5), N(6;5), P(5;2), Q(2;l) • Viet phuong trinh canh AB cua hinh chir nhat ABCD biet cac duong thSng AB, BC, CD, DA Ian luxit di qua M, N , P, Q va dien tich hinh chir nhat

Jiu&ng dan gidi "

Toa do ciia A la nghiem cua he: x + 2 y - 2 = 0

_ _4

3

y = • ' 3 ' 3 ,

2 x + y + l = 0 Gpi a la canh ciia hinh thoi, ta suy ra:

SABCD = 2SABD = 2(SAMB + S A M D ) = A B ) + d(M, A D ) ] = ~ 8a

Mat khac, S^^Q^ - AB.AD.sinBAD = a sin a

Ma cos a = cos( AB, AD) = - =:> sin a = - => S^J^Q^ = —

3' 3

39

Phuong phapgiai Todn Hinh h(fc thco chuyen de- Nguyen Phu Khdnh, Nguyen Tai Thu

§ 4 C A C B A I T O A N V E Dl/CfNG T R O N V A C O N I C

1 J^fhdm cdc bdi todn lien quan den du&ng tron

Khi giai cac bai toan ve duang tron chiing ta can luu y:

1) Vi tri turnip doi ^iim hai dican^ tron

Cho hai duong tron (C,) c6 tarn I , , ban kinh Rj va duong tron (Cj) c6

tarn , ban kinh Rj Khi do, ta c6 cac ket qua sau:

• (C,) va (C2) khong CO diem chung khi va chi khi

I j l 2 > R , + R 2 hoac I , l 2 < | R , - R 2 •

• (Cj) va (C2) tiep xuc ngoai khi va chi khi I j l 2 = R i + R 2

-• (Cj) va (C2) tiep xiic trong khi va chi khi I,l2 = Rj - R 2 -•

• (C,) va (C2) c^tnhau khi vachi khi | R , R 2 | < I , l 2 < R i + R 2

-2) Vi tri turnip doi ^im ditxtn^ than;^ va dinrnif tron

Cho duong tron (C) c6 tam I , ban kinh R va duong thang A Goi H la hinh

chieu cua I len A va dat d = I H = d(I, A) Khi do:

• (C) va A khong c6 diem chung khi va chi khi d > R

• (C) va A CO diing mot diem chung khi va chi khi d.= R Luc nay A goi la

tiep tuyen cua (C), H la tiep diem

Chu y: Tir mot diem M nSm ngoai duong tron (C) luon ve duoc hai tiep

tuyeh MA, MB (A,B la cac tiep diem) den (C) Khi do M A = MB va I M la

phan giac ciia goc A M B

• (C) va A CO diem A,B chung khi va chi khi d < R Khi do H la trung diem

cua AB va ta c6 cong thuc R^ = d^ + -^^^

Vidu 1.4.1 Trong mat phang voi he true toa do Oxy, cho tam giac ABC c6

A(0;2),B{-2;-2), C(4;-2).Goi H la chan duong cao ke tu B; M , N Ian lugt

la trung diem cua AB, A C Viet phuong trinh duong tron di qua cac diem

Cty TNHH MTV DWli Khang Vic,

Gia su phuong trinh duong tron: x^ + + ax + by + c = 0

Ba diem M , N , H thuoc duang tron nen ta c6 he phuong trinh : a-c = l |'a = - l

• a - 2b + c = -5 o <^ b = 1 ;

a + b + c = -2 [c = -2 Phuong trinh duong tron: x ^ + y ^ - x + y - 2 = 0

'•Yi

Vidu /.4.2.Trong mat phang voi h^ toa do Oxy, cho cho hai diem A(2;0)

va B(6; 4) Viet phuong trinh duong tron ( C ) tiep xuc voi true hoanh tai A

va khoang each tu tam cua ( C ) den diem B bang 5

Goi I(a;b) va R Ian luot la tam ciia va ban kinh cua ( C ) (c ; ^ r ; t >

Vi ( C ) tiep xiic voi Ox tai A nen a = 2 va R = b J ( - <*

Matkhac: IB = 5 o 4 2 + ( b 4 f = 5 ^ ^ b = l,b = 7 • • r

-2 -2 ' '

Voi b - 1 thi phuong trinh duong tron ( C ) : (x - 2) + (y -1) = 1

Voi b = 7 thi phuong trinh duong tron ( C ) : (x - i f + (y - i f = 49

Vi du 1.4.3. Trong mat phing Oxy cho diem M(6;6) va hai duong thang

A j : 4x - 3y - 24 = 0, Aj : 4x + 3y + 8 = 0 Viet phuong trinh duong tron (C) di qua M va tiep xuc voi hai duong thang Aj, A j

JCffigidi

Gpi I(a; b) la tam va R la ban kinh ciia duong tron (C) ' J i h i ;

Vi (C) tiep xiic voi hai duong th^ng Aj va A2 nen ta c6 d(I,Aj) = d(I,A2)

hoac ( C ) : ( x - 2 ) ^ ( y - f ) 2 = ^

4 16

4-Phumtg phdpgiai Todn Hinh hgc theo chuyen de- Nguyen Phu Khdnh, Nguyen Td't Tim

t r i n h d u o n g t r o n ( C ) c6 t a m M tren d, ban k i n h bang 2 Ian ban k i n h d u o n g

t r o n (C) va tiep xuc ngoai v o i d u o n g t r o n (C)

Vidu 1.4.5 T r o n g mat p h a n g O x y cho d u o n g t r o n ( C ^ ) : x ^ + y ^ - 2 x - 2 y - 1 8 = 0

v a d u o n g t r o n ( € 3 ) : (x + if + (y - 2)^ = 8 C h u n g m i n h rang hai d u o n g tron

( C j ) va (C2) cat nhau tai hai diem phan biet A, B Viet p h u o n g trinh d u o n g

cat n h a u tai hai d i e m phan biet A , B

Toa d p giao d i e m ciia ( C j ) v a (C2) la nghiem ciia he:

C/iuiy.-Ngoai each giai tren, ta c6 the sir d u n g c h u m d u o n g t r o n de giai C u the:

V i (C) d i qua cac giao d i e m cua ( C j ) va (C2) nen p h u o n g trinh cua (C)

CO dang: m ( x ^ + y^ - 2x - 2y - 1 8 ) + n(x^ + y^ + 2x - 4y - 3) = 0

Do (C) d i qua M ( 0 ; 6 ) n e n t a c6: 2 m + 3n = 0 , ta chon m = 3 , n = - 2

K h i d o p h u o n g t r i n h (C): x^ + y^ - lOx + 2y - 48 = 0

Vi du 1.4.6 T r o n g he toa d o O x y , cho d u o n g t r o n ( C ) : (x - 6)^ + (y - 2)^ = 4

Viet p h u o n g t r i n h d u o n g t r o n ( C ) tiep xuc v o i hai true tga d o O x , O y d o n g

J h o i tiep xiic ngoai v o i (C)

f

Xffigidi

D u o n g t r o n (C) c6 t a m I (6; 2 ) , ban k i n h R 2 Goi ( C ' ) : ( x - a f + ( y - b f = R'2 thi ( C ) c6 tam I ' ( a ; b ) , ban kinh R '

Phuang phiip gidi Todn Hinh hoc theo chuyen dc- Nguyen Pltii Khduh, Nguyen Tat Thu

( C ; ) : ( x - 2 f + ( y - 2 f =4 va ( q ) : {x - 1 8 ) ^ (y - i s f = 18^

T H 2 : a = -b = R => ( C ) : (x -a)^ + (y + a)^ = a^

Tuong tu nhu truong hop 1, ta CO :

ir = R + R ' » ^(a - 6)^ + (a + 2 ^ = 2 + a o a = 6

Vay truong hgp nay c6 1 duong tron la (C3 j : (x - 6)^ + (y + 6)^ = 36

Tom lai, c6 3 duong tron thoa can tim la :

( x - 2 f + ( y - 2 f =4, (x-18)^+(y-18^=182 va ( x - 6 ) ^ + ( y + 6 f =36

Vi du 1.4.7 Trong mat phSng Oxy cho duong tron (C): (x -1)^ + (y - 2)^ = 9

CO tam I va diem M(5;-3) Chung minh rang tu M, ta c6 the ve den (C) hai

tiep tuyen MA, MB (A,B la tiep diem) Tinh dien tich ciia tu giac MAIB

JCgigidi

Duong tron (C) c6 tam 1(1; 2), ban kinh R = 3

Vi MI = N/41 > R nen M n3m ngoai duong tron (C), do do tu M ta luon ve

duoc hai tiep tuyen toi duong tron (C)

Ta CO SMAIB = 2 3 ^ , 1 = lA.MA = R N / M I ^ - R ^ = 3.V41-9 =12^2 (dvdt)

Vi du 1.4.8 Trong mat phang voi he tga do Oxy, cho duong tron

(C): (x -1)^ + (y + 2)^ = 9 va duong thing d : 3x - 4y + m = 0 Tim m de tren

d CO duy nhat mpt diem P ma tu do c6 the ke dupe hai tiep tuyen PA,PB

toi (C) (A,B la cac tiep diem) sao cho tam giac PAB deu

JCffigidL

Duong tron (C) c6 tam va ban kinh Ian lupt la: 1(1;-2); R = 3

Do tam giac PAB deu nen

44

Cty TNIUI MTV DVVII Kliang Viet

API = 30" ^ IP = 2IA = 2R = 6

Suy ra P thupc vao duang tron (C)

CO tam I va ban kinh R' = 6

Ma P e d nen P chinh la giao diem ciia duong thing d va duong tron (C) Suy ra tren d c6 duy nhat diem P thoa man yeu cau bai toan khi va chi khi duong thang d tiep xuc voi duong tron (C) tai P hay la d(I,d) = 6 m = 19,m =-41

Vi du 1.4.9 Cho duong thang A : x + y + 2 = 0 v a duong tron

(C): x^ + y^ -4x -2y - 0 Gpi I la tam va M thuoc duong thang A Qua M ke

tiep tuyen MA,MB Tim M sao cho di§n tich tu giac MAIB hang 10

(De thi D H Khoi A - 2011)

JCffigidi

Duong tron (C) c6 tam 1(2; 1), ban kinh R = yl5=i'Al = S

Matkhac S^MAI = 2SAIBM =5

V^y M(2;-4) va M(-3;l) la hai diem can tim

Vi du 1.4.10 Trong mat phSng Oxy, cho duong tron (C): (x - 4)^ + y^ = 4 va

diem E(4;1) Tim tpa dp diem M tren tryc tung sao cho tu M ke dupe hai tiep tuyen MA, MB den duong tron (C) voi A,B la hai tiep diem sao cho duong thSng AB di qua diem E

JCgigidi

Duong tron (C) c6 tam 1(4;0), ban kinh R = 2

Gpi M(0; m), gia su T(x; y) la tiep

<Jiem ciia tiep tuyen ve tu M toi (C)

45