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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
nh ly 9:Cho hai vec t 1 2 1 2( ; ) va ( ; )a a a b b b= =
ta co :
a b (ieu kien vuong goc cua1 1 2 2a 0b a b + =
nh ly 10:Cho hai vec t 1 2 1 2( ; ) va ( ; )a a a b b 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 :
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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.
Bai 22: Cho hai ng tron :2 2
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).
Bai 23: Cho hai ng tron :2 2
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 + =
+ =
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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
Bai 5: Cho Elp (E) :2 2
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.
Bai 7: Cho Elp (E) :2 2
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.
Bai 8: Cho Elp (E) :2 2
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
x
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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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