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Chuyen e 14: HNH HOC GIAI TCH TRONGMAT PHANG

A. KIEN THC C BAN:

PHNG PHAP TOA O TRONG MAT PHANGTOA O IEM - TOA O VEC T

91

I. He truc toa o E-CAC trong mat phang:

x'Ox : truc hoanh y'Oy : truc tung O : goc toa o : vec t n v (1 2,e e

1 2 11 vae e e e= =

2 )

x

y

1e

2e

O' x

' y Quy c :Mat phang ma tren o co chon he truc toa o e-Cac vuong goc Ox

Oxy va ky hieu la : mp(Oxy)II. Toa o cua mot iem va cua mot vec t :1. nh ngha 1:Cho ( ) M mp Oxy . Khi o vec tOM c bieu dien mot cach duy nhat t

e e bi he thc co dang :OM 1 2,

xe ye1 2 vi x,y= +

.Cap so (x;y) trong he thc tren c goi la toa o cua iem M.

Ky hieu: M(x;y) ( x: hoanh o cua iem M; y: tung o cua i' x

y

2

' /

1 2( ; ) n

M x y OM xe ye = +

Y ngha hnh hoc:

va y=OQ x OP=

2. nh ngha 2:Cho a m ( ) p Oxy . Khi o vec ta c bieu dien mot cach duy nhat theoe e bi he thc co dang :1 2,

1 1 2 2 1 2vi a ,aa a e a e= +

.

Cap so (a1;a2) trong he thc tren c goi la toa o cua vec t .a Ky hieu: 1 2( ; )a a a=

/ 1 2 1 1 2 2=(a ;a )

na a a = +

e a e

Y ngha hnh hoc:

1 1 1 2 2 2va a =Aa A B B=

x1e

e

O

M Q

P

y

y

xO

x'

' y

M Q

P x

y

x

y

1e

2e

O' x

' y

P

a

y

xO

' x

' y

1 A 1 B

2 A

2 B BK

A H

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BAI TAP AP DUNG:Trong mat phang Oxy hay ve cac iem sau: A(2;3), B(-1;4), C(-3;-3), D(4;-2), E(2;

III. Cac cong thc va nh ly ve toa o iem va toa o vec t :nh ly 1:Neu B( ; ) va B(x ; ) A A B A x y y th

92

( ; ) B A B A AB x x y y=

nh ly 2:Neua a th1 2 1 2( ; ) va ( ; )a b b b= =

* a b 1 12 2

a b

a b

== =

* a b 1 1 2 2( ; )a b a b+ = + +

)a b a b =

)ka ka=* a b 1 1 2 2( ;* k a ( )1 2. ( ; k

BAI TAP AP DUNG:Bai 1: Cho A(1;3), B(-2;-1), C(3;-4). Tm toa o iem D sao cho t giac ABCDla hnh bnh hanh.

Bai 2: Cho A(1;2), B(2;3), C(-1;-2). Tm iem M thoa man 022 =+ CB MB MA IV. S cung phng cua hai vec t:

Nhac lai Hai vec t cung phng la hai vec t nam tren cung mot ng thang hoa

thang song song . nh ly ve s cung phng cua hai vec t:

nh ly 3 : Cho hai vec t va vi 0a b b

a k b

a b cung phng !k sao cho . =

Neu 0a

th so k trong trng hp nay c xac nh nh

k > 0 khia cung hngb k < 0 khia ngc hngb

ak

b=

nh ly 4 : , , thang hang cung phng A B C AB AC

(ieu kien 3 iem thang hang ) nh ly 5:Cho hai vec t 1 2 1 2( ; ) va ( ; )a a a b b b= =

ta co :

a b 1 2 2 1cung phng a . . 0b a b =

(ieu kien cung phng cua

);( A A y x A

);( B B y x B

a

b

a

b

A B

C

a b

2 5a b , b - a

5 2= =

a

b

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)4;2(

)2;1(

=

=

b

a:VD

);(

);(

21

21

bbb

aaa

=

=

BAI TAP AP DUNG:

93

Bai 1: Cho 1(0; 1); (2;3); ( ;0)2

A B C . Chng minh A, B, C thang hang

Bai 2: Cho A(1;1), )4

31;23(

+ B , )4

31;32(

C . Chng minh A, B, C thang hang

V. Tch vo hng cua hai vec t:Nhac lai:

x

y

. . .cos( , )a b a b a b=

22

a a=

a b . 0a b =

nh ly 6:Cho hai vec t 1 2 1 2( ; ) va ( ; )a a a b b b= =

ta co :

a b (Cong thc tnh tch vo hng theo to1 1 2 2. a b a b= +

nh ly 7:Cho hai vec t 1 2( ; )a a a= ta co :

2 21 2a a a= + (Cong thc tnh o dai vec t )

nh ly 8:Neu B( ; ) va B(x ; ) A A B A x y y th2 2( ) ( ) B A B A AB x x y y= + (Cong thc tnh khoang cac


ta co :

a b (ieu kien vuong goc cua1 1 2 2a 0b a b + =


ta co

1 1 2 22 2 2 2

1 2 1 2

.cos( , )

. .

a b a b a ba b

a b a a b b

+= =+ +

(Cong thc tnh goc cua 2 vec

b

BAI TAP AP DUNG:Bai 1:Chng minh rang tam giac vi cac nh A(-3;-3), B(-1;3), C(11;-1) la tam giaBai 2:Cho )7;342(),336;8(),3;2( ++ C B A . Tnh goc BAC.

O' x

' y

aa

b b

aO

B

A

);( B B y x B);( A A y x A

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2. Cac bat ang thc vec t c ban :nh ly 13:Vi hai vec tu,v

bat ky ta luon co :

u

v

vu +

u v u v+ +

. .u v u v

Dau bang xay ra khi va ch khi ,u v

la hai vec t cung phng cung chieuhoac laco motrong hai vec t la vec t khong .

BAI TAP AP DUNG:Bai 1: Tm dien tch tam giac co cac nh A(-2;-4), B(2;8), C(10;2)Bai 2: Cho tam giac ABC co dien tch bang 3 vi A(3;1), B(1;-3)

1. Tm C biet C tren Oy2. Tm C biet trong tam G cua tam giac tren Oy

Bai 3: Cho A(1;1), B(-3;-2), C(0;1)1. Tm toa o trong tam G, trc tam H va tam ng tron ngoai tiep I cua

2. Chng minh rang G, H, I thang hang vaGI GH 2= 3. Ve ng cao AA' cua tam giac ABC. Tm toa o iem A'

Bai 4: Cho tam giac ABC biet A(6;4), B(-4;-1), C(2;-4).Tm toa o tam va ban knh ng tron ngoai tiep tam giac ABC

Bai 5: Tm toa o trc tam cua tam giac ABC, biet toa o cac nh( 1; 2), (5;7), (4; 3) A B C Bai 6: Cho ba iem A(1;6), B(-4;-4), C(4;0)

1. Ve phan giac trong AD va phan giac ngoai AE. Tm toa o D va E2. Tm toa o tam ng tron noi tiep tam giac ABC

Bai 7: Cho hai iem A(0;2), )1;3( B . Tm toa o trc tam va toa o tam ng tron

cua tam giac OAB (TS A 2004)Bai 8:Cho tam giac ABC co cac nh A(-1;0), B(4;0), C(0;m) vi0m . Tm toa o trong tam cua tam giac ABC theo m. Xac nh m e tam giac GAB vuong tai G. (TS D 2

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NG THANG TRONG MAT PHANG TOA

A.KIEN THC C BAN

I. Cac nh ngha ve VTCP va PVT cua ng thang:1. VTCP cua ng thang :

a la VTCP cua ng thang ( )n

0a co gia song song hoac trung vi ( )a

n la VTPT cua ng thang ( )n

0n co gia vuong goc vi ( )n

96

* Chu y: Neu ng thang ( ) co VTCP 1 2( ; )a a a= th co VTPT la 2 1( ;n a a= )

a

a )(n

(

Neu ng thang ( ) co VTPT ( ; )n A B= th co VTCP la ( ; )a B A=

an

)(

BAI TAP AP DUNG:Cho ng thang i qua hai iem A(1;-2), B(-1;3). Tm mot VTCP va mot VTPT c( ) ( ) II. Phng trnh ng thang :1. Phng trnh tham so va phng trnh chnh tac cua ng thang :

a. nh ly :Trong mat phang (Oxy). ng thang ( ) qua M0(x0;y0) va nhan 1 2( ; )a a a= lamVTCP se co :

Phng trnh tham so la : 0 10 2

.( ) : (

. x x t a

t y y t a

= + = +

Phng trnh chnh tac la : 0 01 2

( ) : x x y ya a

=

y

BAI TAP AP DUNG:Bai 1:Cho hai iem A(-1;3), B(1;2). Viet phng trnh tham so va chnh tac cua nBai 2: Cac iem P(2;3); Q(4;-1); R(-3;5) la cac trung iem cua cac canh cua mot ta

phng trnh chnh tac cua cac ng thang cha cac canh cua tam giac o

);( 000 y x M

a );( y x M

xO

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2. Phng trnh tong quat cua ng thang :a. Phng trnh ng thang i qua mot iem M0(x0;y0) va co VTPT( ; )n A B= la:

97

0 0( ) : ( ) ( ) 0 A x x B y y + =

BAI TAP AP DUNG:Bai 1: Cho tam giac ABC biet( 1; 2), (5;7), (4; 3) A B C

1. Viet phng trnh cac ng cao cua tam giac2. Viet phng trnh cac ng trung trc cua tam giac

Bai 2: Cho tamgiac ABC vi A(1;-1) ; B(-2;1); C(3;5).a) Viet phng trnh ng vuong goc ke t A en trung tuyen BK cua tam gb) Tnh dien tch tam giac ABK.

b. Phng trnh tong quat cua ng thang :nh ly :Trong mat phang (Oxy). Phng trnh ng thang ( ) co dang :

Ax + By + C = 0 vi2 2 0 A B+

Chu y:T phng trnh ( ):Ax + By + C = 0 ta luon suy ra c :

1. VTPT cua ( ) la ( ; )n A B= 2. VTCP cua ( ) la ( ; ) hay a ( ; )a B A B A= =

3. ( ;0 0 0 0 0) ( ) 0 M x y Ax By C + + = Menh e (3) c hieu la :

ieu kien can va u e mot iem nam tren ng thang langhiem ung phng trnh cua ng thang .

BAI TAP AP DUNG:Bai 1: Viet phng trnh tham so cua ng thang biet phng trnh tong quat cua n5 2 3 x y + =Bai 2: Viet phng trnh tong quat cua ng thang qua M(-1;2) va song song( ) : 2 3 4 0 x y + =

Bai 3: Viet phng trnh tong quat cua ng thang qua N(-1;2) va vuong goc( ) : 2 3 4 0 x y + =Bai 4: Cho hai iem A(-1;2) va B3;4) . Tm iem C tren ng thang x-2y+1=0 sao chABC vuong C.

Bai 5: Cho A(1;1) ; B(-1;3) va ng thang d:x+y+4=0.a) Tm tren d iem C cach eu hai iem A, B.b) Vi C tm c . Tm D sao cho ABCD la hnh bnh hanh .Tnh dien tch hnh

) y M ;( 000 x

);( y x M n y

xO

);( y M 000 x

); An ( B=

x

y

);( A Ba =O

);( A Ba =

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3. Cac dang khac cua phng trnh ng thang :a. Phng trnh ng thang i qua hai iem A(xA;yA) va B(xB;yB) :

( ) : A A B A B A

x x y y AB

x x y y

=

( ) : A AB x x= ( ) : A AB y y=

98

BAI TAP AP DUNG:Cho tam giac ABC biet A(1;-1), B(-2;1), C1;5). Viet phng trnh ba canh cua tam giab. Phng trnh ng thang i qua mot iem M0(x0;y0) va co he so goc k:

nh ngha:Trong mp(Oxy) cho ng thang . Goi ( , )Ox = k tgth = c goi la he scuang thang

nh ly 1:Phng trnh ng thang qua 0 0 0( ; ) M x y co he so goc k la :

(1)0 0y-y = k(x-x )

Chu y 1:Phng trnh (1) khong co cha phng trnh cua ng thang i qua M0 va vuong goOx nen khi s dung ta can e y xet them ng thang i qua M0 va vuong goc Ox lx = x0

Chu y 2: Neu ng thang co phng trnh y ax b= + th he so goc cua ng thang k a= nh ly 2:Goi k1, k2 lan lt la he so goc cua hai ng thang1 2, ta co :

1 2 1 // k k = 2 1 2 1 2k . 1k =

BAI TAP AP DUNG:Viet phng trnh ng thang qua A(-1;2) va vuong goc vi ng thang3 4 x y + =0c. Phng trnh t i qua mot iem va song song hoac vuong goc vi m

i. 1 1Phng trnh ng thang ( ) //( ): Ax+By+C=0 co dang: Ax+By+m =0

ii. 1 2Phng trnh ng thang ( ) ( ): Ax+By+C=0 co dang: Bx-Ay+m =0

x

y

O

);( y x M

x

y y);( A A y x A );( B B y x B y);( A A y x A

);( B B y x B

A x B x A y

B y);( A A y x A

);( B B y x B A y B y

x xO

)

y

O

;( y M x

0 x

0 y

x

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Chu y : c xac nh bi mot iem co toa o a biet nam tren1 2;m m 1 2; 0: 11 =++ m By Ax

x

y

O 0 x

0: 1 =++ C By Ax

1 M

0: 21 =+ m Ay Bx

x

y

O 0 x1 M

0: 1 =++ C By Ax

BAI TAP AP DUNG:Bai 1: Viet phng trnh tong quat cua ng thang qua M(-1;2) va song song( ) : 2 3 4 0 x y + =Bai 2: Viet phng trnh tong quat cua ng thang qua N(-1;2) va vuong goc( ) : 2 3 4 0 x y + = III. V tr tng oi cua hai ng thang :

99

Trong mp(Oxy) cho hai ng thang :1 1 1 12 2 2 2

( ) : 0( ) : 0

A x B y C

A x B y C

+ + = + + =

V tr tng oi cua phu thuoc vao so nghiem cua he phng tr1( ) va ( ) 2

hay1 1 12 2 2

0

0

A x B y C

A x B y C

+ + =

+ + =

1 1 1

2 2 2(1) A x B y C

A x B y C

+ =

+ = Chu y: Nghiem duy nhat (x;y) cua he (1) chnh la toa o giao iem M cua1 2( ) va ( )

nh ly 1:1 2

1 2

1 2

. He (1) vo nghiem ( )//( ). He (1) co nghiem duy nhat ( ) cat ( ). He (1) co vo so nghiem ( ) ( )

i

ii

iii

nh ly 2:Neu 2 2 2; ; A B C khac 0 th

=

= =

1 11 2

2 2

1 1 11 2

2 2 2

1 11 2

2 2

A. ( ) cat ( )AA. ( ) // ( )AA. ( ) ( )A

1

2

Bi

B

B C ii

B C

B C iii

B C

1

x

y

O

2

21 //

1

x

y

O

2

y

O

1

x

2

21 21 cat

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BAI TAP AP DUNG:

Bai 1:Cho tam giac ABC co phng trnh ba canh la( ) :8 3 17 0

( ) : 3 5 13

( ) : 5 2 1 0

AB x y

AC x y

BC x y

0

+ = =+ =

Tm toa o ba nh A, B, CBai 2:Cho tamgiac ABC co nh A(2;2) .Lap phng trnh cac canh cua tam giac AB

cac ng thang 9x-3y-4=0 va x+y-2=0 lan lt la cac ng cao cua tam gB va C.

Bai 3:Tuy theo m, hay bien luan v tr tng oi cua hai ng thang sau:1

2

: 1

: 2 0

d mx y m

d x my

0+ =+ =

IV. Goc gia hai ng thangnh ly :Trong mp(Oxy) cho hai ng thang :1 1 1 1

2 2 2 2

( ) : 0( ) : 0

A x B y C

A x B y C

+ + = + + =

Goi (0 ) la goc gia0 090 21( ) va ( ) ta co :

1

x

y

O

2

1 2 1 22 2 2 21 1 2 2

cos.

A A B B

A B A B

+=

+ +

100

He qua:

( 1 2 1 2 1 2) ( ) A 0 A B B + = BAI TAP AP DUNG:Bai 1: Viet phng trnh ng thang i qua iem A(0;1) va tao vi ng thang : x

mot goc bang 450

Bai 2: Lap phng trnh cac canh cua hnh vuong co nh la (-4;5) va mot ng trnh 7x-y+8=0.V. Khoang cach t mot iem en mot ng thang :nh ly 1:Trong mp(Oxy) cho hai ng thang( ) : 0 Ax By C + + = va iem0 0 0( ; ) M x y

Khoang cach t M0 en ng thang( ) c tnh bi cong thc:

0 00 2 2

( ; ) Ax By C d M A B

+ + =

+

nh ly 2:Trong mp(Oxy) cho hai ng thang :1 1 1 12 2 2 2

( ) : 0( ) : 0

A x B y C

A x B y C

+ + = + + =

va ( )

Phng trnh phan giac cua goc tao bi ( )1 2 la :

1 1 1 2 22 2 2 21 1 2 2

2 A x B y C A x B y C

A B A B

+ + + += + +

0 M y

O

H

)(

y

O

1

2

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nh ly 3:Cho ng thang 0:)( 1 =++ C By Ax va hai iem M(xM;yM), N(xN;yN) khong nam

tren ( ). Khi o: M

N

M

N

Hai iem M , N nam cung pha oi vi ( ) khi va ch khi0))(( >++++ C By AxC By Ax N N M M

Hai iem M , N nam khac pha oi vi ( ) khi va ch khi0))((

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Chu y:

102

= =

1

2

0 va 0 th0 va 0 th

ac biet :

+ + + + + =+ + + + + =

1 1

1 1 1 2 2 2

1 1 1 2 2 2

Neu 0 va 0 th va trong trng hp nayphng trnh co the viet di dang sau:

1. m(A ) (A ) 0hoac 2. (A ) (A ) 0

x B y C x B y C

x B y C n x B y C

M

21

I

BAI TAP AP DUNG:Viet phng trnh ng thang i qua giao iem cua hai ng thang3 5 2 0 & 5 2 4 0 x y x y + = + =

va vuong goc vi ng thang( ) : 2 4 0d x y + =

.

BAI TAP REN LUYEN

Bai 1:Phng trnh hai canh cua tam giac trong mat phang toa o la 5x-2y+6=0 va 4Viet phng trnh canh th ba cua tam giac biet trc tam cua tam giac trung

Bai 2: Cho tam giac ABC , canh BC co trung iem M(0;4) con hai canh kia co phng2x+y-11=0 va x+4y-2=0.a) Xac nh nh A.b) Goi C la iem tren ng thang x+4y-2=0, N la trung iem AC . Tm iem

toa o B, C.Bai 3: Cho tam giac ABC co M(-2;2) la trung iem cua BC , canh AB co phng trnh

canh AC co phng trnh : 2x+5y+3=0.Xac nh toa o cua cac nh cua tamBai 4: Cho tam giac ABC co nh B(3;5) ng cao ke t A co phng trnh 2x-5y+3

trung tuyen ke t C co phng trnh x+y-5=0 .a) Tnh toa o iem A.b) Viet phng trnh cua cac canh cua tam giac ABC.

Bai 5:Cho tam giac ABC co trong tam G(-2;-1) va co cac canh AB:4x+y+15=0 vaAa) Tm toa o nh A va toa o trung iem M cua BC .

b) Tm toa o iem B va viet phng trnh ng thang BC.Bai 6: Cho tam giac ABC co nh A(-1;-3).a) Biet ng cao BH: 5x+3y-25=0, ng cao CK: 3x+8y-12=0. Tm toa o nb) Biet ng trung trc cua AB la 3x+2y-4=0 va trong tam G(4;-2). Tm B, C

Bai 7: Lap phng trnh cac canh cua tam giac ABC biet nh C(4;-1) ng cao vake t mot nh co phng trnh 2x-3y+12=0 va 2x+3y=0.

Bai 8: Lap phng trnh cac canh cua tam giac ABC neu biet A(1;3) va hai ng tphng trnh la x-2y+1=0 va y-1=0.

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Bai 9: Cho tam giac ABC biet C(4;3) phan giac trong (AD):x+2y-5=0, trung tuyen (AE4x+13y-10=0.Lap phng trnh ba canh.

Bai 10: Cho tam giac ABC biet A(2;-1) va phng trnh hai ng phan giac trong culan lt la d: x-2y+1=0 va x+y+3=0 .Tm phng trnh cua ng thang cha

Bai 11: Cho iem M(-2;3) . Tm phng trnh ng thang qua M va cach eu hai ieva B(2;1).

Bai 12: Cho A(2;-3) , B(3;-2) .Trong tam G cua tam giac nam tren ng thang d: 3xtch tam giac ABC bang 3/2 . Tm C.

Bai 13: Viet phng trnh ng thang song song vi d: 3x-4y+1=0 va co khoang cathang d bang 1.

Bai 14: Cho tam giac can ABC biet phng trnh canh ay AB:2x-3y+5=0 canh ben ATm phng trnh canh ben BC biet rang no i qua iem D(1;1).

Bai 15: Cho tam giac ABC co nh A(-1;3) , ng cao BH nam tren ng thang y=trong goc C nam tren ng thang x+3y+2=0 . Viet phng trnh canh BC .

Bai 16: Cho ng thang d: 2x+y-4=0va hai iem M(3;3) , N(-5;19).Ha MKd va goi P la ieoi xng cua M qua d:a) Tm toa o cua K va P.

b) Tm iem A tren d sao cho AM + AN co gia tr nho nhat va tnh gia tr Bai 17: Cho tam giac ABC vuong A , phng trnh BC la3x y 3 0 = , cac nh A va Bthuoc truc hoanh va ban knh ng tron noi tiep bang 2. Tm toa o trontam giac ABC.

Bai 18: Cho hnh ch nhat ABC co tam I(1/2;0) , phng trnh ng thang AB la x-2AB=2AD . Tm toa o cac nh A, B, C, D biet rang nh A co hoanh o a

Bai 19: Trong mp(Oxy) cho hai ng thang1 : 0d x y = va 2 : 2 1 0d x y+ = . Tm toa o cac hnh vuong ABCD biet rang nh A thuoc d1, nh C thuoc d2 va cac nh B,D thuoc tr

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NG TRON TRONG MAT PHANG TOA O

A.KIEN THC C BANI. Phng trnh ng tron:

1. Phng trnh chnh tac:nh ly :Trong mp(Oxy). Phng trnh cua ng tron (C) tam I(a;b), ban knh

104

( ) (1)2 2 2: ( ) ( )C x a y b R + =

Phng trnh (1) c goi la phng trnh chnh tac cua ng tronac biet:Khi I O th (hay:2 2( ) :C x y R+ = 2 2 2 y R x= )

BAI TAP NG DUNG:Bai 1: Viet phng trnh ng tron ng knh AB biet A(1;3), B(3:-5)Bai 2: Viet phng trnh ng tron co tam I(-1;2) va tiep xuc ng thang( ) : 3 4 2 0 x y + =2. Phng trnh tong quat:

nh ly :Trong mp(Oxy). Phng trnh : 2 2 2 2 0 x y ax by c+ + = via b2 2 0c+ >la phng trnh cua ng tron (C) co tam I(a;b), ban knh2 2 R a b= + c

BAI TAP NG DUNG:Bai 1: Xac nh tam va ban knh cua ng tron( ) 2 2: 2 4 20 0C x y x y+ + = Bai 2: Viet phng trnh ng tron (C) i qua ba iem A(3;3), B(1;1),C(5;1)Bai 3: Cho phng trnh : (1)2 2 4 2 2 3 x y mx my m+ + + + =0

nh m e phng trnh (1) la phng trnh cua ng tron (Cm)II. Phng trnh tiep tuyen cua ng tron:

nh ly :Trong mp(Oxy). Phng trnh tiep tuyen vi ng tron( ) tai iem2 2: 2 2 0C x y ax by c+ + = 0 0( ; ) ( ) M x y C la :

( ) 0 0 0 0: ( ) ( ) 0 x x y y a x x b y y c + + + + =

x

y

O

);( ba I

Rb

a

);( y x M

BAI TAP NG DUNG:Xet ng tron (C) qua ba iem A(-1;2), B(2;0), C(-3;1). Viet phng trnh tiep tuyenIV. Phng tch cua mot iem oi vi mot ng tron:

Nhac lai :nh ngha: Cho ng tron (O;R) va mot iem M co nh .

Phng tch cua iem M oi vi ng tron (O) c ky hieu la M/(O) la mot

(C)

I(a;b))(

);( 000 y x M

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105

2c xac nh nh sau: M/(O) = 2d R ( vi d = MO )Chu y:

M/(O) > 0 ngoai ng tron (O) M M/(O) < 0 trong ng tron (O) M M/(O) = 0 tren ng tron (O) M

nh ly:Trong mp(Oxy) cho iem 0 0( ; ) M x y va ng tron2 2 2 2 x y ax by c 0+ + = via b co tam I(a;b) va ban knh2 2 0c+ > 2 2 R a b c= + . Phng tch cua iem M oing tron (C) la

M/(O) = 2 20 0 0 02 2 x y ax by c+ +

(C)

MI

BAI TAP NG DUNG:Cho ng tron (C): va iem A(3;5). Xet v tr cua iem A o(C)

2 2 2 4 4 0 x y x y+ + =

IV. Truc ang phng cua hai ng tron:Nhac lai:nh ly :Tap hp cac iem co cung phng tch oi vi hai ng tron khac ta

ng thang vuong goc vi ng noi hai tam.ng thang nay c goi la truc ang phng cua hai ng tron o

Cach xac nh truc ang phng

)( 1C )( 2C

2 I 1 I

)( 1C

)( 2C

1 I 2 I M

)1C

)(2

C

1 I 2 I

(

M

)( 2C

)( 1C )( 3C

I 1 I 2 I

3 I

2 1

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nh ly :Cho hai ng tron (C1) va (C2) khong cung tam co phng trnh:

2 21 1 1 1

2 22 2 2

( ) : 2 2 0( ) : 2 2 0C x y a x b y c

C x y a x b y c

+ + =

2+ + =

Phng trnh truc ang phng cua (C1) va (C2) la :

( ) 1 2 1 2 2 1: 2( ) 2( ) 0a a x b b y c c + + =

106

Cach nh: 2 2 2 21 1 1 2 22 2 2 2 2 x y a x b y c x y a x b y c+ + = + +

BAI TAP NG DUNG:Xac nh phng trnh truc ang phng cua hai ng tron sau:

2 21

2 22

( ) : 4 5 0

( ) : 6 8 16 0

C x y y

C x y x y

+ =

+ + + =

VI. Cac van e co lien quan:1. V tr tng oi cua ng thang va ng tron:

)(C )(C )(C

I

I I R R H

R M M H H M nh ly:

( ) ( ) d(I; ) > RC = ( ) tiep xuc (C) d(I; ) = R

( ) cat (C) d(I; ) < R

BAI TAP NG DUNG:Bai 1:Cho ng tron (C): . Viet phng trnh tiep tuyen vi (C) bie2 2( 3) ( 1) x y + =4

tuyen nay i qua iem M(6;3)Bai 2:Cho ng tron (C): . Viet phng trnh tiep tuyen vi 2 2 6 2 5 x y x y+ + + =0

tiep tuyen song song vi ng thang( ) : 2 10 0d x y+ + = Bai 3:Cho ng tron va iem M(-3;1). Goi T0662:)( 22 =++ y x y xC 1, T2 la cac tiep ie

cac tiep tuyen ke t M en (C). Viet phng trnh ng thang T1T2.2. V tr tng oi cua hai ng tron :

1 I 1 R

1C

2 I 2 R

2C

1 I 1 R

1C

2C 2 R

2 I

1C

1 I 1 R

2C

2 R2 I

1C

2C

1 I 2 I

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1 2 1 2 1 2

1 2 1 2 1 2 1 2

1 2 1 2 1 2

1 2

( ) va (C ) khong cat nhau I I > R( ) va (C ) cat nhau R < I I < R( ) va (C ) tiep xuc ngoai nhau I I = R( ) va (C ) tiep xuc trong

C R

C R

C R

C

R

+ +

+

1 2 1 2nhau I I = R R

BAI TAP NG DUNG:Xac nh v tr tng oi cua hai ng tron sau:

2 21

2 22

( ) : 4 5 0

( ) : 6 8 16 0

C x y y

C x y x y

+ =

+ + + =

VII: Chum ng tron:

nh ly: Cho hai ng tron cat nhau :

( )

2 21 1 1

2 22 2 2

( ) : 2 2 01

2: 2 2 0

C x y a x b y c

C x y a x b y c

+ + =

+ = +

Phng trnh ng tron (C) i qua giao iem cua (C1) va (C2) co dang :

2 2 2 2 2 21 1 1 2 2 2( 2 2 ) ( 2 2 ) 0 ( + 0) x y a x b y c x y a x b y c + + + + + =

BAI TAP NG DUNG:Viet phng trnh ng tron i qua cac giao iem cua hai ng tron

( ) 2 2 2 21 2: 10 0; ( ) : 4 2 20 0C x y x C x y x y+ = + + =va i qua iem A(1;-1)

BAI TAP REN LUYEN Bai 1: Lap phng trnh ng tron ngoai tiep tam giac co ba nh la A(1;1); B(-1;2Bai 2: Lap phng trnh ng tron ngoai tiep tam giac co ba canh nam tren ba

1 2 3x 2( .d ) : y ;(d ) : y x 2;(d ): y 8 x5 5

= = + =

Bai 3: Lap phng trnh ng tron noi tiep tam giac co ba nh la A(-1;7); B(4;-3)Bai 4: Lap phng trnh ng tron i qua cac iem A(-1;1) va B(1;-3) co tam nam

thang (d):2x - y + 1 = 0.Bai 5: Lap phng trnh ng tron i qua iem A(-1;-2) va tiep xuc vi ng tha

(d): 7x-y-5=0 tai iem M(1;2).Bai 6: Lap phng trnh ng tron co tam nam tren ng thang 2x+y=0 va tiep

thang x-7y+10=0 tai iem A(4;2).Bai 7: Viet phng trnh ng tron co tam nam tren ng thang 4x +3y - 2 = 0 va

xuc vi hai ng thang : x + y + 4 = 0 va 7x - y + 4 = 0.Bai 8: Viet phng trnh ng tron i qua iem A(2;-1) va tiep xuc vi hai truc toBai 9: Cho ng tron (C):(x-1)2 +(y-2)2=4 va ng thang (d):x-y-1=0. Viet phng trnh

ng tron (C') oi xng vi ng tron (C) qua ng thang (d). Tm toa o cua (C) va (C').

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Bai 10:Cho hai ng tron: (C1): 2 2 10 0 x y x+ = va (C2): 2 2 4 2 20 x y x y 0+ + = 1. Viet phng trnh ng tron i qua cac giao iem cua (C1) va (C2) va co tam nam tr

ng thang (d): x + 6y - 6 = 0.2. Viet phng trnh tiep tuyen chung cua cac ng tron (C1) va (C2) .

Bai 11: Cho hai ng tron: (C1): 2 2 4 5 x y y 0+ = va (C2): 2 2 6 8 16 x y x y 0+ + + = Viet phng trnh tiep tuyen chung cua cac ng tron (C1) va (C2) .

Bai 12: Cho hai ng tron :2 2

12 2

2(C ) : x y 4x 2y 4 0(C ): x y 10x 6y 30 0

+ + =

+ + =

co tam lan lt la I va J.1) Chng minh (C1) tiep tiep xuc ngoai vi (C2) va tm toa o tiep iem H.2) Goi (D) la mot tiep tuyen chung khong i qua H cua (C1) va (C2) . Tm toa o giao

iem K cua (D) va ng thang IJ.Viet phng trnh ng tron (C) i quaxuc vi hai ng tron (C1) va (C2) tai H.

Bai 13: Cho iem M(6;2) va ng tron (C):2 2 2 4 x y x y 0+ = . Lap phng trnh ng than(d) qua M cat (C) tai hai iem phan biet A, B sao cho 10 AB =

Bai 14: Cho ng tron (C): va iem A(1;2). Hay lap phng trnh cua 2 2

9 x y+ =

cha day cung cua (C) i qua A sao cho o dai day cung o ngan nhat.Bai 15: Cho ng tron (C): va iem M(2;4)2 2 2 6 6 x y x y+ + =9

1. Chng to rang iem M nam trongng tron.2. Viet phng trnh ng thang i qua iem M, cat ng tron tai hai iem

cho M la trung iem cua AB .3. Viet phng trnh ng tron oi xng vi ng tron a cho qua n

Bai 16: Trong mp(Oxy) cho ho ng tron (Cm) co phng trnh :02 2x y (2m 5)x (4m 1)y 2m 4+ + + + =

1) Chng to rang (Cm) qua hai iem co nh khi m thay oi.2) Tm m e (Cm) tiep xuc truc tung.

Bai 17: Cho ho ng tron (Cm) co phng trnh :2 2x y (m 2)x 2my 1 0+ + = 1) Tm tap hp tam cac ng tron (Cm) .2) Cho m = -2 va iem A(0;-1). Viet phng trnh cac tiep tuyen cua ng tr-2)

ve t A.Bai 18: Viet phng trnh cac tiep tuyen cua ng tron (C):2 2 2 6 9 x y x y 0+ + =

1. Tiep tuyen song song vi ng thang x-y=02. Tiep tuyen vuong goc vi ng thang 3x-4y=0

Bai 19: Cho tam giac ABC eu noi tiep trong ng tron (C):2 2( 1) ( 2) 9 x y + = . Xac nh toao cac iem B, C biet iem A(-2;2).

Bai 20: Trong mp(Oxy) cho ho ng tron (Cm) co phng trnh :2 2x 2mx y 2(m 1)y 12 0 + + + =

1) Tm tap hp tam cac ng tron (Cm) .2) Vi gia tr nao cua m th ban knh cua ho ng tron a cho la nho nh

Bai 21: Cho hai ho ng tron :

108

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'

2 2m

2 2m

(C ) : x y 2mx 2(m 1)y 1 0(C ) : x y x (m 1)y 3 0

+ + + =

+ + + =

Tm truc ang phng cua hai ho ng tron tren. Chng to rang khi m thayang phng o luon luon i qua mot iem co nh.


12 2

2

(C ) : x y 2x 9y 2 0(C ) : x y 8x 9y 16 0

+ =

+ + =

1) Chng minh rang hai ng tron (C1) va (C2) tiep xuc nhau.2) Viet phng trnh cac tiep tuyen chung cua hai ng tron (C1) va (C2).


12 2

2

(C ) : x y 10x 0(C ) : x y 4x 2y 20 0

+ =

+ + =

Viet phng trnh cac tiep tuyen chung cua hai ng tron (C1) va (C2).Bai 24: Cho hai ng tron :

2 21

2 22

(C ) : x y 4x 5 0(C ) : x y 6x 8y 16 0

+ =

+ + + =

Viet phng trnh cac tiep tuyen chung cua hai ng tron (C1) va (C2).Bai 25: Cho hai iem A(2;0), B(6;4). Viet phng trnh ng tron (C) tiep xuc vi tr

A va khoang cach t tam cua (C) en iem B bang 5 (TS.K.B2005)ng dung phng trnh ng tron e giai cac he co cha

Bai 1: Cho he phng trnh :2 2x y

x y a + =

=1

Xac nh cac gia tr cua a e he phng trnh co nghiem duy nhat.

Bai 2: Cho he phng trnh :2 2 0

0 x y x

x ay a

+ = + =

Xac nh cac gia tr cua a e he phng trnh co 2 nghiem phan biet

Bai 3:Tm m e he sau co nghiem duy nhat2 2

2 2

(x 2) y mx (y 2) m + =

+ =

109

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NG ELP TRONG MAT PHANG TOA O

A.KIEN THC C BANI.nh ngha:

Elp (E) la tap hp cac iem M co tong khoang cach en hai iem co n1; F2 bang han* Hai iem co nh F1; F2 c goi la cac tieu iem* F1F2 = 2c ( c > 0 ) c goi la tieu c

110

{ }1 2(E) M/ MF MF 2a= + = ( a>0 : hang so va a>c

(E)

II. Phng trnh chnh tac cua Elp va cac yeu to:1. Phng trnh chnh tac:

2 2

2 2x y(E) : 1a b

+ = vi 2 2 2b a c= ( a > b) (1)

2. Cac yeu to cua Elp:* Elp xac nh bi phng trnh (1) co cac ac iem:- Tam oi xng O, truc oi xng Ox; Oy- Tieu iem F1(-c;0); F2(c;0)- Tieu c F1F2 = 2c- Truc ln nam tren Ox; o dai truc ln 2a ( = A1A2 )- Truc nho nam tren Oy; o dai truc ln 2b ( = B1BB2 )- nh tren truc ln : A1(-a;0); A2(a;0)- nh tren truc nho :B1(0;-b); B2(0;b)- Ban knh qua tieu iem:

2c

M

1 2F F

- a a

(E)

c-c

y

x

R S

PQ

O

M

1r 2r

1 A 2 A

1 B

2 B

1F 2F

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Vi M(x;y) (E) th1 1

2 2

cr MF a x a exacr MF a x a ea

= = + = +

x= = =

- Tam sai : ce (0 ea

= < 1)<

- ng chuan : axe

=

III. Phng trnh tham so cua Elp: (E) x acos t:y bsin t

= =

t [0;2 )

IV. Tiep tuyen cua Elp:nh ly:Phng trnh tiep tuyen vi (E) :

2 2

2 2x y 1a b

+ = tai M0(x0;y0) (E) la :

111

) : 0 02 2x x y y 1a b

+ = (

V. ieu kien e ng thang tiep xuc vi Elp:nh ly:Cho Elp (E) :

2 2

2 2x y 1a b

va ng thang( ) : Ax By C 0 + + = ( A2 + B2 > 0 )+ =

A a2 2 2 2 2( ) tiep xuc (E) B b C+ =

BAI TAP REN LUYEN

y

O

)( E

x

y

O

);( 000 y x M

)( E

Bai 1:Cho (E) co hai tieu iem la1 2( 3;0); ( 3;0F F ) va mot ng chuan co phng trn43

x =

1. Viet phng trnh chnh tac cua (E).2. M la iem thuoc (E). Tnh gia tr cua bieu thc:

P F 2 2 21 2 1 23 . M F M OM F M F M = + 3. Viet phng trnh ng thang (d) song song vi truc hoanh va cat (E) tai h

choOA OBBai 2: 1. Lap phng trnh chnh tac cua (E) co tieu iem1( 15;0)F , tiep xuc vi (d):4 10 x y+ =2. Viet phng trnh tiep tuyen vi (E) vuong goc vi (d):6 0 x y+ + = .

Bai 3: Cho Elp (E) :2 2

19 4

x y+ = va ng thang (d):mx 1 0 = y

1. Chng minh rang vi moi gia tr cua m, ng thang (d) luon cat (E) tai2. Viet phng trnh tiep tuyen cua (E), biet rang tiep tuyen o i qua iem

Bai 4: 1. Lap phng trnh chnh tac cua (E) co tieu iem1 2( 10,0); ( 10;0)F F , o dai truc l

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2 18 .2. ng thang (d) tiep xuc (E) tai M cat hai truc toa o tai A va B. Tm M

nho nhat.OAB


18 4

x y+ = va ng thang (d): 2 2 0 x y + =

1. CMR (d) luon cat (E) tai hai iem phan biet A,B . Tnh o dai AB.2. Tm toa o iem C thuoc (E) sao cho ABC co dien tch ln nhat.

Bai 6: Cho hai Elp :2 2 2 2

1 2( ) : 1 va (E ) : 116 9 9 16 x y x y

E + = + = . Viet phng trnh tiep tuyen chung elp tren.


124 12 x y+ = . Xet hnh vuong ngoai tiep (E) ( tc la cac canh hnh v

vi (E) . Viet phng trnh cac ng thang cha cac canh hnh vuong o.


19 4

x y+ = . Cho A(-3;0),M(-3;a),B(3;0),N(3;b) trong o a,b la hai so t

1. Xac nh toa o giao iem I cua ng thang AN va BM.2. Chng minh rang ieu kien can va u e ng thang MN tiep xuc v

3. Vi a,b thay oi , nhng luon tiep xuc vi (E) . Tm quy tch iem I.

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NG HYPEBOL TRONG MAT PHANG TOA O

A.KIEN THC C BANI. nh ngha:

113

{ }1 2(H) M / MF MF 2a= = ( a > 0 : hang so va a < c )

II. Phng trnh chnh tac cua Hypebol va cac yeu to:1. Phng trnh chnh tac:

2 2

2 2x y(H) : 1a b

= vi 2 2 2b c a= (1)

M

1F 2F c2

xab

y = xab

y =

1F 2F

M x

y

1 B

2 B

1 A 2 Aa

cca

O

2. Cac yeu to cua Hypebol:* Hypebol xac nh bi phng trnh (1) co cac ac iem:

- Tam oi xng O, truc oi xng Ox; Oy- Tieu iem F1(-c;0); F2(c;0)- Tieu c F1F2 = 2c- Truc thc nam tren Ox; o dai truc thc 2a ( = A1A2 )- Truc ao nam tren Oy; o dai truc ao 2b ( = B1BB2 )- nh: A1(-a;0); A2(a;0)

- Phng trnh tiem can : by xa=

- Ban knh qua tieu iem:Vi M(x;y) (H) th :

Vi x > 0 1 12 2

r MF a exr MF a ex

= = + = = +

Vi x < 0 1 1

2 2

r MF (a ex)r MF ( a ex

= = +

)= = +

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- Tam sai : ce (ea

= >1)

- ng chuan : axe

=

IV. Tiep tuyen cua Hypebol:nh ly:Phng trnh tiep tuyen vi (H) :

2 2

2 2x y 1a b

= tai M0(x0;y0) (H) la :

114

0 02 2

x x y y 1a b

= ( ) :

V. ieu kien e ng thang tiep xuc vi Hypebol:nh ly:Cho Hypebol (H) :

2 2

2 2x y 1a b

= va ng thang( ) : Ax By C 0 + + = ( A2 + B2 > 0

2 2 2 2 2A a B b C = ( ) tiep xuc (H)

y

0 M

O

BAI TAP REN LUYENBai 1:Cho Hypebol (H):

2 21

16 9 x y =

1. Tm o dai truc ao, truc thc , tam sai , tieu iem F1,F2 cua (H)2. Tm tren (H) nhng iem sao cho1 2 MF MF

Bai 2:Cho Hypebol (H):

2 2

2 2 1 x y

a b =

.CMR tch cac khoang cach t mot iem M0 bat ky tren (H) en hai tiem can la mo

Bai 3: Cho Hypebol (H): .2 24 4 x y =1. Viet phng trnh tiep tuyen vi (H) tai10 4( ;

3 3 A )

0

2. Viet phng trnh tiep tuyen vi (H) biet no vuong goc vi ng thang: 2 x y = 3. Viet phng trnh tiep tuyen vi (H) ke t M(2;-1)

Bai 4: Cho Hypebol (H):2 2

2 2 1 x y

a b = trong mat phang Oxy

Tm a,b e (H) tiep xuc vi hai ng thang (1 2) : 5 6 16 0 va (D ) :13 10 48 D x y x y = =

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NG PARABOL TRONG MAT PHANG TOA O

115

A.KIEN THC C BANI. nh ngha :

{ }(P) M/ MF d(M,= =

* F la iem co nh goi la tieu iem* ( ) la ng thang co nh goi la ng chuan* HF = p > 0 goi la tham so tieu

II. Phng trnh chnh tac cua parabol:

1) Dang 1:Ptct:y2 = 2px 2) Dang 2:Ptct:y2 = -2px

p

K

HF

M

3) Dang 3:Ptct:x2

= 2py 4) Dang 4:Ptct :x2

= -2py

III.Tiep tuyen cua parabol:nh ly:Trong mp(Oxy). Phng trnh tiep tuyen vi (P): y2 = 2px tai M0(x0;y0) (P) la

( ) : y0y = p.(x + x0 )

x

y

P

y

xp/2F(-p/2;0)

M 2 / :)( p x =

y

x

-p/2 :y = -p/2

F(0;p/2)

O

M F(0;-p/2)

x

( ) : y = p/2p/2

y

O

M

( ) : x=-p/2

y

O -p/2

F(p/2;0)

M

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IV. ieu kien e ng thang tiep xuc vi parabol:nh ly:Trong mp(Oxy) cho (P) : y2 = 2px va ng thang( ) : Ax By C 0 + + = (A2 + B2 > 0)

x

y

o

( ) tiep xuc (P) 2B p 2AC=

BAI TAP REN LUYEN

Bai 1: Cho (P): y2

= 16x1. Lap phng trnh tiep tuyen cua (P), biet tiep tuyen o vuong goc vi (d) : 3x-2y+6=0

2. Lap phng trnh cac tiep tuyen vi (P) ke t M(-1;0) en (P)

Bai 2: Lap phng trnh cac tiep tuyen chung cua elp :2 2

18 6

x y+ = va parabol: .2 12 y x=

Bai 3: Cho A(3;0) va (P): y=x2 1. Cho ( ) M P va M x a= . Tnh AM . Tm a e AM ngan nhat2. Chng minh neu AM ngan nhat th AM vuong goc tiep tuyen tai M cua (

Bai 4: Cho (P):y2= 2x va cho A(2;-2); B(8;4). Gia s M la iem di ong tren cung nhonh toa o cua M sao cho tam diac AMB co dien tch ln nhat.

Bai 5: Cho (P): 2 y x= va iem I(0;2). Tm toa o hai iem M, N thuoc (P) sao cho4 IM IN =

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