hinh giai tich trong mp
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Trn Thnh Minh Phan Lu Bin - Trn Quang Ngha
Phng Php Ta
Trong Mt Phng
www. saosangsong.com.vn
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1. Phng trnh tng qut ca ng thngA. Tm tt gio khoa .1. Vect n
khc 0
vung gc ng thng gi l vectphp tuyn (VTPT)ca .
Phng trnh ca ng thng qua M0( x0 ; y0 ) v c VTPT n = (a ;b) l : a(x x0) + b(y y0)
Phng trnh tng qutca ng thng c dng : ax+ by + c = 0trong n
= (a ; b) l mt VTPT . vung gc Ox : ax + c = 0
vung gc Oy : by + c = 0 qua gc O : ax + by = 0
qua A(a ; 0) v B(0 ; b) :x y
1a b
+ = (Phng
trnh theo an chn )Phng trnh ng thng c h s gc l k : y = kx +
m vi k = tan , l gc hp bi tia Mt ca pha trn Ox v tiaMx
2. Cho hai ng thng 1: a1x + b1y + c1 = 0 v 2 : a2x + b2y + c2 = 0Tnh D = a1
b2 a2b1, Dx = b1
c2 b2 c1, Dy = c 1
a2 c2a1
1 , 2 ct nhau D 0 . Khi ta giao im l :x
y
Dx
DD
yD
=
=
1 // 2 xy
D 0
D 0
D 0
=
1 , 2 trng nhau D = Dx = Dy = 0Ghi ch : Nu a2, b2 , c2 0 th :
1 , 2 ct nhau2
1
2
1
b
b
a
a .
n
a
M
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1 // 22
1
2
1
2
1c
c
b
b
a
a
=
1 , 2 trng nhau2
1
2
1
2
1
c
c
b
b
a
a==
B. Gii tan .Dng tan 1 : Lp phng trnh tng qut ca ng thng : Cn nh:
Phng trnh ng thng qua im M(x0 ; y0 ) v vung gc n = (a;b) l : a(x x0 ) + b(y y0) = 0
Phng trnh ng thng qua im M(x0 ; y0 ) v cng phng)a;a(a 21= l :
2
o
1
o
ayy
axx =
Phng trnh ng thng song song ng thng : ax + by + c = 0 cdng : ax + by + m = 0 vi m c .
Phng trnh ng thng qua M(x0 ; y0 ) :a(x x0 ) + b(y y0) = 0 ( a
2 + b2 0 )
Phng trnh ng thng qua A(a ; 0) v B(0 ; b) l : x y 1a b
+ =
V d 1 : Cho tam gic ABC c A(3 ; 2) , B(1 ; 1) v C(- 1; 4) . Vit phngtrnh tng qut ca :
a) ng cao AH v ng thng BC .b) trung trc ca ABc) ng trung bnh ng vi ACd) ung phn gic trong ca gc A .
Gii a) ng cao AH qua A(3 ; 2) v vung gc BC
= (- 2 ; 3) c phng trnhl : - 2( x 3) + 3(y 2) = 0 - 2x + 3y = 0
ng thng BC l tp h p nhng im M(x ; y) sao cho )1y;1x(BM =
cng phng )3;2(BC = nn c phng trnh l : x 1 y 12 3 =
( iu kin cng
phng ca hai vect) 3(x 1) + 2(y 1) = 0 3x + 2y 5 = 0
b) Trung trc AB qua trung im I( 2 ; 3/2 ) ca AB v vung gc AB
= (- 2 ; -1) nn c phng trnh tng qut l : 2(x 2) + 1.(y 3/2) = 0 4x + 2y 11 =0
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c) ng trung bnh ng vi AB qua trung im K( 0 ; 5/2) v cng phng AB
= (- 2 ; - 1) . ng ny l tp hp nhng im M(x ; y) sao cho
)2
5y;0x(KM = cng phng )1;2(AB = nn c phng trnh l :
x 0 y 5/ 2
2 1
= ( iu kin cng phng ca hai vect)
x 2y + 5 = 0
d) Gi D(x ; y) l ta ca chn ng phn gic trong . Theo tnh cht ca
phn gic :DB AB
ACDC=
M AB = 2 2 2 22 1 5,AC 4 2 2 5+ = = + = , do :
DB 12DC DC
2DC= =
2(1 x) x 1 x 1/ 3
2(1 y) y 4 y 2
= + =
= =
Vy D = (1/3 ; 2) . V yA = yD = 2 nn phng trnh AD l y = 2 .
V d 2 : Cho hnh ch nht ABCD ,phng trnh ca AB : 2x y + 5 = 0 ,ng thng AD qua gc ta O , v tm hnh ch nht l I( 4 ; 5 ) . Vit
phng trnh cc cnh cn li
Gii VAD vung gc vi AB nn VTPT n
= (2 ; - 1) ca AB l VTCP ca AD
Phng trnh AD qua O l :x y
2 1=
x + 2y = 0
Ta A l nghim ca h :2x y 5 0
x 2y 0
+ =
+ =
Gii h ny ta c : x = - 2 ; y = 1 => A(- 2 ; 1)I l trung im ca AC , suy ra :
A C I C
A C I C
x x 2x 8 x 10y y 2y 10 y 9
+ = = = + = = =
: C(10 ; 9)
ng thng CD song song vi AB nn n
= (2 ; - 1)
cng l VTPT ca CD . CD qua C(10 ; 9) , do phng trnh CD l :
2(x 10) - (y 9) = 0 2x y 11 = 0
ng thng BC qua C v song song AD , do phng trnh BC l :
A B
D C
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(x 10) + 2(y 9) = 0 x 2y 28 = 0
V d 3 : Cho ng thng d : 3x 4y 12 = 0 .a) Tnh din tch ca tam gic m d hp vi hai trc ta .
b) Vit phng trnh ng thng d i xng ca d qua trc Ox .c) Vit phng trnh ng thng d i xng ca d qua im I(- 1 ; 1) .
Gii : a) Cho x = 0 : - 4y 12 = 0 y = - 3 => d ct Oy tai A(0 ; - 3)
Cho y = 0 : 3x 12 = 0 x = 4 => d ct Ox tai B(4 ; 0)
Din tch tam gic vung OAB l : .OA.OB = . 3. 4 = 6 vdtb) Gi A(0 ; 3) l i xng ca A
qua Ox . Ta c d qua A v B ,
cng phng )3;4(B'A = c
phng trnh l :3
3y
4
0x
=
3x + 4y 12 = 0
c) Gi B1l i xng ca B qua I=> B1 (- 6 ; 2) . ng thng d
qua B1v song song vi d , c phng trnh :
3(x + 6) 4(y - 2) = 0 3x 4y + 26 = 0
*V d 4 : Vit phng trnh ng thng qua M(3 ; 2) , ct tia Ox ti A, tiaOy ti B sao cho :
a) OA + OB = 12b) hp vi hai trc mt tam gic c din tch l 12
Gii : Gi A(a ; 0) v B(0 ; b) vi a > 0 , b > 0 ,phng trnh ng thng cn tm c dng :x y
1a b
+ = . V ng thng qua M(3 ; 2) nn :
3 21
a b+ = (1)
A
B
x
y
A
B
AB1
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a) OA + OB = 12 a + b = 12 a = 12 b (2)
Th (2) vo (1) : 3 2 112 b b
+ =
3b + 2(12 b) = (12 b)b b2 11b + 24 = 0 b = 3 hay b = 8
b = 3 : a = 9 , phng trnh cn tm : x y 1 x 3y 9 09 3
+ = + =
b = 8 : a = 4 , phng trnh cn tm : x y 1 2x y 8 04 8
+ = + =
b) Din tch tam gic OAB l OA.OB = ab = 12 a = 24/b (3)
Th (3) vo (1) : 3b 2 124 b
+ = b2 + 16 = 8b
(b 4)2 = 0 b = 4
Suy ra : a = 6 , phng trnh cn tm l :x y
16 4
+ = 2x + 3y 12 = 0
Dng 3 : Tm v tr tng i ca hai ng thng .
V d 1 : Tm v tr tng i ca cac ng thng sau :a) 9x 6y 1 = 0 , 6x + 4y 5 = 0
b) 10x 8y + 2/3 =0 ; 25x 20y + 5/3 = 0Gii a) Ta c :
9 6
6 4
nn hai ng thng ct nhau .
b) Ta c :10 8 2 / 3 2
25 20 5 / 3 5
= = =
nn hai ng thng trng nhau .
* V d 2 : Cho d : (m + 1)x 2y + m + 1 = 0d : mx - 3y + 1 = 0
a) nh m hai ng thng ct nhau . Tm ta giao im M.b) Tm m Z ta giao im l s nguyn .
Gii a) Ta giao im M l nghim ca h :(m 1)x 2y m 1 0 (1)
mx 3y 1 0 (2)
+ + + =
+ =
Hai ng thng ct nhau D = 3mm2)1m(33m
21m=++=
+ 0
m - 3
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Ta c : Dx = 13
1m2
+= - 2.1 + 3(m + 1) = 3m +1
Dy = =++
m1
1m1mm(m + 1) 1.(m+1) = m2 - 1
Ta giao im M :
+
+=
+=
3m
1m-
D
D=y
3m
1-3m-.
D
D=x
2y
x
b) Ta c : x =3(m 3) 8
m 3
+ +
+
= - 3 +8
m 3+
y =3m
83m
++
x v y Z th 8 chia ht cho (m + 3) (m + 3) { 1 ; 2 ; 4 ; 8 } m {- 2 ; - 4 ; - 1 ; - 5 ; 1 ; - 7 ; 5 ; - 11 }
V d 3 : Cho ng thng d : 2x + y - 13 = 0 v im A (1 ; 1)a) Vit phng trnh ng thng d qua A v vung gc d .
b) Tm ta hnh chiu ca A ln d v ta im A , i xng ca Aqua A .
Gii a) ng thng d vung gc d nn VTPT n
= (2 ; 1) ca d l VTCP ca d. Suy ra phng trnh ca d l :
x 1 y 1
2 1
= x 2y + 1 = 0
b) Ta giao im H ca d v d tha h :2x y 13 0
x 2y 1 0
+ =
+ =
x 5
y 3
=
=: H(5 ; 3) , l hnh chiu ca
A ln d..
H l trung im ca AA , suy ra :)5;9('A:
5yy2y
9xx2x
AH'A
AH'A
==
==
.
C. Bi tp rn luyn3.1. Cho ng thng d : y = 2x 4
H
A
A
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a) Vng thng d . Xc nh giao im A v B ca d vi Ox v Oy.Suyra din tch tam gic OAB v khong cch t O ti d.
b) Vit phng trnh ng thng d song song vi d , ct Ox ti M , Oyti N sao cho MN = 3 5
3.2. Vit phng trnh tng qut ca ng thng d :a) qua im A(1 ; - 2) v c h s gc l 3 .
b) qua B ( - 5; 2 ) v cng phng a
= ( 2 ; - 5)
c) qua gc O v vung gc vi ng thng : y =2 3
4
x
d) qua I(4 ; 5) v hp vi 2 trc ta mt tam gic cn .e) qua A(3 ; 5) v cch xa im H(1 ; 2) nht.
3.3 . Chng minh cc tp hp sau l cc ng thng :
a) Tp hp nhng im M m khong cch n trc honh gp i khongcch n trc tung .
b) Tp hp nhng im M tha 2 2 2MA MB 2MO+ = vi A(2 ; 1 ) v B(
1 ; - 2)
3. 4 . Cho tam gic ABC c A(4 ; 1) , B(1 ; 7) v C(- 1; 0 ) . Vit phng trnhtng qut ca
a) ng cao AH , ng thng BC .b) Trung tuyn AM v trung trc ca ABc) ng thng qua C v chia tam gic thnh hai phn , phn cha im A
c din tch gp i phn cha im B .
3. 5. Cho tam gic ABC c phng trnh cc ng thng AB, BC v CA l :AB : x 3 = 0BC : 4x 7y + 23 = 0AC : 3x + 7y + 5 = 0a) Tm ta A, B, C v din tch tam gic .
b) Vit phng trnh ng cao v t A v C . Suy ra ta ca trc tm H3. 6.Cho hai ng thng d : mx y + m + 1 = 0 v d : x my + 2 = 0
a) nh m hai ng thng ct nhau . Tm ta giao im M , suy ra M ding trn mt ng thng cnh .
b) nh m d v d v ng thng : x + 2y 2 = 0 ng quy.
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3. 7. Cho hai im A(5 ; - 2) v B(3 ; 4) . Vit phng trnh ca ng thng d
qua im C(1 ; 1) sao cho A v B cch u ng thng d .
3.8. Cho hnh bnh hnh hai cnh c phng trnh 3x y 2 = 0 v x + y 2 = 0 .
Vit phng trnh hai cnh cn li bit tm hnh bnh hnh l I(3 ; 1) .
* 3. 9 . Cho tam gic ABC c trung im ca AB l I(1 ; 3) , trung im AC l
J(- 3; 1) . im A thuc Oy v ng BC qua gc ta O . Tm ta im A ,
phng trnh BC v ng cao v t B .
* 3.10. Cho im M(9 ; 4) . Vit phng trnh ng thng qua M , ct hai tia Ox
v tia Oy ti A v B sao cho tam gic OAB c din tch nh nht .
* 3.11. Cho im M(3 ; 3) . Vit phng trnh ng thng qua M , ct Ox v Oy
ti A v B sao cho tam gic MAB vung ti M v AB qua im I(2 ; 1) .
D. Hng dn hay p s :
3.1. a) A(2 ; 0) , B(0 ; - 4) ; S = 4 vdt .
Ta c : 5
4
OH16
5
16
1
4
1
OB
1
OA
1
OH
1222
==>=+=+=
b) Phng trnh d c dng : y = 2x + m , ct Ox ti M(- m/2 ; 0) , ct Oy
ti N(0 ; m) . Ta c MN =2
5|m|ONOM 22 =+ = 3 5
Suy ra : m = 6 .
3.2 . a) y + 2 = 3(x 1) y = 3x 5
b) 021y2x55
2y
2
5x=++
=
+
c) y = x3
4( hai ng thng vung gc tch hai h s gc l 1)
d) V d hp vi Ox mt gc 450 hay 1350 nn ng thng c h s gc l tan450 = 1 hay tn0 = - 1 , suy ra phng trnh l : y = x + 1 ; y = - x + 9
e) ng thng cn tm qua A v vung gc )3;2(AH = .3.3 . a)Gi (x ; y) l ta ca M : |y| = 2|x| y = 2x hay y = - 2x
b) MO2 = x2 + y2 , MA2 = (x 2)2 +(y 1)2 , MB2 = (x 1)2 + (y + 2)2 .
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Suy ra : 3x y 5 = 0
3. 4 . c) ng thng cn tm qua im D sao cho : DA 2DB=
D = (2 ; 5)
3. 5. a) A(3 ; - 2) ; B(3 ; 5) ; C(- 4 ; 1) , S = .AB . CH = 47/ 2 vdtb) AH : y = 1 , AK : 7x + 4y 13 = 0 , H(9/7 ; 1)
3. 6 . a) D = 1 m2 0 m 1 , ta giao im : 3
x
y
D m 2 1x 1
D m 1 m 1D 1
yD m 1
+= = = + +
= =
+
=> x + y + 1 = 0 => M di ng trn ng
thng : x + y + 1 = 0
b) Th ta ca M vo ng thng x + 2y 2 = 0 , ta c : m = - 2/3
3. 7. d l ng thng qua C :
v qua trung im I(4 ; 1) ca AB hay cng phng )6;2(AB =
3.8. Gi AB : 3x y 2 = 0 v AD : x + y 2 = 0 .
Gii h , ta uc A = (1 ; 1) . Suy ra C = (5 ; 1 ) .
CD : 3x y 14 = 0 ; BC : x + y 6 = 0
* 3. 9 . A = (0 ; a) => B(2 ; 6 a) v C(- 6 ; 2 a)
BC qua gc O nn OB v OC cng phng 2(2 a) = (6 a) ( - 6)
a = 5 .
3. 10. t A(a ; 0) v B(0 ; b),vi a , b > 0 .Phng trnh ng thng cn tm
c dng : 1=+b
y
a
x. ng ny qua I 1
49=+
ba
p dng bt Csi cho hai s : 1 =abbaba
124.
92
49=+
=> 722
112 ==> abSab OAB
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Vy tam gic OAB c din tch nh nht l 72 khi ==== baba ;182
1498
v PT ng thng cn tm l : 072941818
=+=+ yxyx
3.11. t A(a ; 0) , B(0 ; b) , ta c : 0)3)(3()3)(3(. =+= baMBMA
a + b = 6 (1)
Mt khc phng trnh ng thng AB : 1=+b
y
a
x.
(AB) qua I(2 ; 1) 112 =+ba
2b + a = ab (2)
Th (1) vo (2) : 2b + (6 b) = (6 b)b b2 5b + 6 = 0
b = 2 hay b = 3 .
Suy ra : (a = 4 ; b = 2) hay (a = 3 ; b = 3)
2. Phng trnh tham s ca ng thng
A. Tm tt gio khoa
1. a
khc 0
cng phng vi ng thng gi l vectchphng(VTCP)ca .
Phng trnh tham sca ng thng qua M0 (x0 ; y0)v c VTCP a
= (a1 ; a2 ) l :o 1
o 2
x x ta
y y ta
= +
= +
Phng trnh chnh tc ca ng thng qua M0 (x0 ; y0) vc VTCP a
= (a1 ; a2 ) l : o o
1 2
x x y y
a a
= ( a1 0 v a2
0)
2. Nu n
= (a; b) l VTPT ca th a
= (b ; - a) hay ( - b ; a)l mt VTCP ca .
B. Gii ton.Dng ton 1 : Lp PT tham s . . . ca ng thng
n
a
M
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Tm mt im M(x0 ; y0 ) v mt VTCP (a 1 ; a2) : phng trnh tham s l :
+=
+=
tayy
taxx
o
o
2
1
phng trnh chnh tc l : o 01 2
x x y y
a a
= (a1, 2 0)
phng trnh tng qut l : a2(x x0) a1( y y0) = 0 Tm mt im M(x0 ; y0 ) v mt VTPT (a ; b) => VTCP (b ; - a) .
p dng nh trn .
V d : Cho A( 1 ; 2) , B(3 ; - 4) , C(0 ; 6) . Vit PT tham s , chnh tc v tngqut ca :
a) ng thng BC .b) ng cao BHc) ng thng qua trng tm G ca tam gic ABC v song song vi d
: 3x -7y = 0
Gii a) BC qua B(3 ; - 4) v c VTCP )10;3(=BC nn c PTTS l :
+=
=
ty
tx
104
33=> PTCT l :
10
4
3
3 +=
yx
v PTTQ l : 0)4(3)3(10 =++ yx 10x + 3y -18 = 0
b) ng cao BH qua B(3 ; - 4) v vung gc )4;1(AC nn c VTCP l (4 ; 1) .Suy ra PTTS :
+=
+=
ty
tx
4
43
PTCT :1
4
4
3 +=
yx
PTTQ : 1(x 3) 4(y + 4) = 0 x 4y 19 = 0
c) ng thng song song vi d : 3x 7y = 0 nn vung gc VTPT dn (3 ; - 7)
, suy ra VTCP l (7 ; 3) . Ta trng tm G l : (4/3 ; 4/3 ) .
PTTS ca ng thng cn tm :
=
+=
ty
tx
33/4
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PTCT :3
3
4
73
4
= yx
PTTQ : 3(x 4/3) 7(y 4/3) = 0 3x 7y +3
16= 0
Dng ton 2 : Tm im ca ng thng
Ta im M ca ng thng cho bi PTTS . ng vi mi t , ta c mtim ca ng thng.Bi ton thng a v vic gii mt phng trnh hay h phng trnh m t tnhcht ca im y.
V d : Cho ng thng d :
+==
tytx
3123
a) Tm trn d im M cch im A(4 ; 0) mt khong l 5 .b) Bin lun theo m v tr tng i ca d v d: (m + 1)x + my 3m 5 = 0
Gii : a) Ta im M thuc d cho bi phng trnh tham s ca d : M =
(3 2t ; 1 + 3t) . Ta c : AM = (-1 2t ; 1 + 3t ) => AM2 = (1 + 2t)2 + (1 + 3t)2 =13t2 + 10t + 2.Ta c : AM2 = 25 13t2 + 10t + 2 = 25
13t2 + 10t 23 = 0 t = 1 hay t = - 23/13M = (1 ; 4) hay M = ( 85/13; - 56/13)
b) Th phng trnh tham s ca d vo phng trnh ca d , ta c phngtrnh tnh tham s t ca giao im , nu c :
(m + 1)(3 2t) + m(1 + 3t) 3m 5 = 0 (m 2)t + m 2 = 0 (1)
m 2 = 0 m = 2 : (1) tha vi mi m d v d c v simchung d , d trng nhau.
m 2 0 m 2 : (1) c ngh duy nht d v d ct nhau .Ghi ch : C th bin i d v dng tng qut : 3x + 2y 11 = 0 v bin lun
theo h phng trnh 2 n .
C. Bi tp rn luyn .
3.12 : Cho ng thng d c hng trnh tham s : x = 3 +2
3
t; y = 2 -
5
6
t(1)
a) Tm mt VTCP ca d c ta nguyn v mt im ca d . Vit mtphng trnh tham s khc ca d
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b) Tm trn d mt im A c honh gp i tung .c) Tm trn d mt im B cch gc O mt khong l 58 .
3. 13 . Cho tam gic ABC c A(1 ; - 2) , B(0 ; 4) v C(6; 3) . Tm mt VTCP, suyra phng trnh tham s v chnh tc ca cc ng thng sau :
a) ng thng d qua A v c mt VTCP l (3 ; - 2 )b) ng trung trc ca BC .c) ng thng ABd) ng trung bnh ca tam gic ABC ng vi cnh BC .e) ng phn gic ngoi ca ca gc B
3.14 . Cho tam gic ABC vi BC : 2x y 4 = 0 , ng cao BH : x + y - 2 = 0 ,ng cao CK : x + 3 y + 5 = 0 . Vit phng trnh cc cnh tam gic .
3.15. Cho hnh ch nht ABCD c AB : 2x y 1 = 0 , AD qua M(3 ; 1) v tm I
c ta l ( - 1 ; ) . Vit phng trnh cc cnh AD , BC v CD .
*3. 16. Cho tam gic ABC c trung im M ca AB c ta (- ; 0) , ngcao CH vi H(- 1; 1) , ng cao BK vi K(1 ; 3) v bit B c honh dng .
a) Vit phng trnh AB .
b) Tm ta B, A v C
3.17 . Chn cu ng : Phng trnh no di y l phng trnh tham s ca
ng trung trc ca AB vi A(3 ; - 5) v B(5 ; 9) :
4 1) )
2 7 7 7
4 7 4 7) )
2 2
x t x t a b
y t y t
x t x t c d
y t y t
= + = +
= + = +
= + = +
= + =
3.18 . Chn cu ng : Phng trnh no di y l phng trnh tng qut ca
ng thng qua A(4 ; - 5) v vung gc vi ng thng d : 4 31 2
x ty t
= + = +
l :
a) 3x + 2y 2 = 0 b) 3x - 2y 12 = 0
c) 2x 3y 23 = 0 d) 4x + 5y 22 = 0
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Phng php ta tronbg mt phng
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3.19 . Chn cu ng : ng thng d :3 2
5 2
x y+
= xc nh vi hai trc ta
mt tam gic c din tch l :
a) 64/5 b) 128/5 c) 16/ 5 d) p s khc
3.20 . Chn cu ng : Gi d l ng thng qua M(4 ; - 3) v song song vi
ng thng y = 2x 4 .
a) d qua im ( 10 ; 10) b) trn d khng c im no c ta l s nguyn
chn .
c) C (a) v (b) u sai d) C (a) v (b) u ng .
3.21 . Chn cu ng : Cho tam gic ABC cn ti A(1 ; - 2) , trng tm l G(5 ;
6) . Phng trnh ng thng BC l :
a) x + 2y + 27 = 0 b) x + 2y 27 = 0
c) x 2y 27 = 0 d) 2x y 4 = 0
C. Hng dn hay p S.
3.12. a)a
= ( 4 ; - 5) , x = 3 + 4t , y = 2 5t b) Gii
xA = 2yA t = 1/14c) Dng phng trnh tham s ca d : (3 + 4t)2 + (2
5t)2 = 58
3.13. a) x = 1 + 3t , y = - 2 2t b) x = 3 + 8t , y = 7/2 + 3t
c) Trung trc vung gc )1;6( =BC nn cng phng vect(1 ; 6) . Suy ra
phng trnh tham s l :
+=
=
ty
tx
64
3.14 . BC v BH ct nhau ti B(2 ; 0) . BC v CK ct nhau ti C(1 ; - 2) . Phng
trnh AB qua B v vung gc CK l : 3(x 2) 1(y 0) = 0 . . .
3.15. AD qua M v vung gc AB c phng trnh : 1.(x 3) + 2(y 1) = 0
x + 2y 5 = 0 .
Suy ra ta A = AB AD = (7/5 ; 9/5) . Suy ra ta C , i xng ca A qua I
B C
A
G
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Phng php ta tronbg mt phng
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. . .
*3. 16. a) Phng trnh AB qua H v M : 2x + y + 1 = 0b) B thuc AB B = (b ; - 2b 1)
A i xng ca B qua M A = (- 1 b ; 2b + 1) .
Mt khc 0=BKAK 5b2 + 5b 10 = 0 b = 1 .Vy B = (1 ; - 3) , A = (- 2 ; 3) , C = (3 ; 3)
3.17 . (d) 3.18. (a) 3.19. (a) 3.20. (b) 3.21. (b)
3. Khong cch v gcA. Tm tt gio khoa .
I. 1. Khang cch t M (x0 ; y0 ) n ng thng : ax + by + c = 0 l :
d(M, ) =22
0 ||
ba
cbyax o
+
++
*2. Gi M l hnh chiu ca M ln , th th :
22.'
ba
cbyaxnkMM MM
+
++== . Suy ra :
M, N nm cng pha i vi (axM + byM+ c)((axN+ byN+ c) > 0
M, N nm khc pha i vi (axM + byM
+ c)((axN+ byN+ c) < 0
* 3. Phng trnh hai ng phn gic ca gc hp bi hai ng thng :a1x + b1 y + c1 = 0 v a2x + b2 y + c2 = 0 l :
02
22
2
22
21
21
111 =+
++
+
++
ba
cybxa
ba
cybxa
II.Gc ( khng t ) to 1: a1x+ b1y + c1 = 0 v 2 : a2x + b2y + c2 = 0 l :
cos(1 ; 2 ) =2
22
22
12
1
2121 ||
baba
bbaa
++
+
1 2 a1a2 + b1b2 = 0
B. Gii ton .
M
M
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Phng php ta tronbg mt phng
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Dng 1 : Tnh khang cch v lp phng trnh ng thng lin quann khang cch
V d 1 :
a) Tnh khong cch t im A(1 ; 3) n ng thng d : 3x 4y + 4 = 0
b) Tnh bn knhng trn tm O tip xc ng thng d : 2x +y + 8 = 0
c) Tnh khong cch t im P(3 ; 12) n ng thng :2
5 3
x t
y t
= +
=
d) Tnh khong cch gia hai ng thng song song : d : 5x + 3y 5 = 0 vd : 5x + 3y + 8 = 0
Gii a) d(A, d) =2 2
3 4 4 3.1 4.3 4 51
5 53 4
A Ax y + += = =
+
b) Bn knh ng trn l khong cch t O n ng thng
d :R = d(O , d) =2 2
2.0 0 8 8
52 1
+ +=
+
c) Ta vit phng trnh di dng tng qut :
2 53( 2) 5
1 3
x yx y
= =
3x + y - 11 = 0
d(P, ) =2 2
3.3 12 11 1010
103 1
+ = =
+
d) Chn trn d : 5x + 3y - 5 = 0 im M ( 1; 0 ) , th th :
d(d , d ) = d(M, d) =2 2
5.1 .0 8 13 132265 1
+ + = =+
V d 2 :a) Tm trntrc honh im cch ng thng : 2x + y 7 = 0 mt khong l
2 5
d
d'
M
d
O
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Phng php ta tronbg mt phng
18
b) Tm trn ng thng d : x + y + 5 = 0 im cch ng thng d : 3x 4y +
4 = 0 mt khong l 2 .
c) Cho im M ( m 2 ; 2m + 5 ) di ng v im A (2 ; 1) cnh . Tm gi tr
nh nht ca khong cch AM khi m thay i .
Gii a) Gi M(x , 0 ) l im cn tm , ta c :
d(M , d) = 2 2 2 7
2 5 2 7 105
xx
= = =
2x 7 = 10 hay 2x 7 = - 10 x = 17/2 hay x = - 3/2
Vy ta tm c hai im M(17/2 ; 0 ) v M(- 3/2 ; 0 )
b) Gi x l honh ca im M cn tm , tung ca M l : y = - x 5 . Ta cphng trnh : d(M, d ) = 1
+
=3 4 6
25
M Mx y
+ =3 4( 5) 4 10x x
| 7x +24 | = 10 7x + 24 = 10 hay 7x + 24 = -10 x = - 2 hay x = - 34/ 7
Vy ta tm c hai im M(- 2; 0 ) v M(- 34/7 ; 0 )
c) Ta c :2
2 5
x m
y m
=
= +
2 52 9 0
1 2
x yx y
+ = + =
Vy M di ng trn ng thng d : 2x y + 9 = 0 . Suy ra khong cch nh
nht ca AM chnh l : d(A, d) =2.2 1 9 12
5 5
+=
V d 3 :
a)Vit phng trnh ng thng song song v cch u hai ng thng songsong d : x 3y 1 = 0 v d : x 3y + 7 = 0
b)Vit phng trnh ng thng d :song song vi ng thng d : 3x + 2y - 1 =
0 v cch d mt khong l 13 v nm trong na mt phng bd v cha
im gc O.
d M
A
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Phng php ta tronbg mt phng
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c)Vit phng trnh ng thng d qua im A( 6 ; 4) v cch im B( 1 ; 2)
mt khong l 5 .
GII a) ng thng cn tm l tp hp nhng im M(x ; y) sao cho :
d(M, d) = d(M, d)2222 31
|73|
31
|13|
+
+=
+
yxyx
+=
+=
7y3x1y3x
)VN(7y3x1y3x
2x 6y + 6 = 0
x 3y + 3 = 0b) Phng trnh ng thng d song song
vi d c dng : 3x + 2y + m = 0 . Ta nh
m d(d , d ) = 13 .
Chn trn d im A(0 ; ) , ta c : d(d, d)= d(A ,d ) = 13
13.0 2.
213 1 13
13
m
m
+ +
= + =
m + 1 = 13 hay m + 1 = - 13
m = 12 hay m = - 14
d : 3x + 2y + 12 = 0 hay d : 3x + 2y 14 = 0
Xt d : 3x + 2y + 12 = 0 . Chn im M (0 ; - 6) thuc dTh ta M vo d : 0.3 + 2( - 6) 1 = - 13 > 0
Th ta O(0 ; 0) vo d : 0.3 + 0(2) 1 = - 1 < 0
Vy O v M cng mt pha i vi d tc d : 3x + 2y + 12 = 0 l ng thng
cn tm .
Cch khc : Gi M(x ; y) l im bt k , ta c :
M(x ; y) dd(M, d) 13 v O v M nm cng pha i vi d
O
5
d
d
A
d
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Phng php ta tronbg mt phng
20
1313
1y2x3
0)10.20.3)(1y2x3(
1313 |1y2x3| =
>
=
3x 2y + 12 = 0
c) Phng trnh d l ng thng qua A (6 ; 4) c dng :
a(x 6) + b(y 4) = 0 vi a2 + b2 0 .
ax + by 6a 4b = 0 (1)
Ta c : d(B, d) = 5 5|462.1|
22
=
+
+
ba
baba )(25)25( 222 baba +=+
20ab 21b2 = 0 b(20a 21b) = 0
b = 0 hay a =20
21b
* Vi b = 0 : (1) thnh ax 6a = 0 x 6 = 0 (chia hai vchoa a 0 , coi nh
chn a = 1)
* Vi a =20
21b: (1) thnh 0
20
41
20
21=+
bbybx
21x + 20y 41 = 0 ( Chia hai vcho b/20 , coi nhchn b = 20 => a = 21 )
Vy c hai ng thng tha bi l : 21x + 20y 41 = 0 v x = 6 .
Cck khc : C thxt* d : x = 6 ( qua A v vung gc Ox , khng c h sgc ).
* d : y = k(x 6) + 4 kx y 6k + 4 = 0
Gii : d(B , d) = 5 k = - 21/ 20 .
Dng 2 : Vit phng trnh phn gic , phn gic trong , ngoi .
V d 1 : Cho tam gic ABC vi AB : 3x 4y + 6 = 0
AC : 5x + 12y 25 = 0 , BC : y = 0
a)Vit phng trnh cc phn gic ca gc B trong tam gic ABC .b)Vit phng trnh phn gic trong ca gc A trong tam gic ABC.
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Phng php ta tronbg mt phng
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Gii : a) AB ct BC ti B(- 2 ; 0) , AC ct BC ti
C( 5 ; 0)
Phng trnh cc phn gic ca gc B trong tam
gic ABC l phn gic ca gc hp bi AB v BC
, l :
015
643=
+ yyx
3x + y + 6 = 0 hay 3x 9y + 6 = 0b) Phng trnh cc phn gic ca gc A , to bi
AB v AC l :
(t) : 047864013
25125
5
643=+=
++
+yx
yxyx(1)
(t) : 020311214013
25125
5
643=+=
+
+yx
yxyx
Th ta B(- 2 ; 0) vo (1) : 64(-2) 47 < 0
Th ta C(5 ; 0) vo (1) : 64.5 47 > 0Vy B v C nm khc pha i vi (t) , nn (t) l phn gic trong ca gc A .
* V d 4 : Cho d : 3x 4y + 5 = 0 v d : 5x + 12y 1 = 0
a)Vit phng trnh cc phn gic ca gc to bi hai ng thngb)Vit phng trnh ng thng qua gc O v to vi d, d mt tam gic cn
c cnh y l .
Gii a) Phn gic (t) ca gc to bi d , d :
013
1125
5
543=
+
+ yxyx
13(3x 4y + 5) = 5(5x + 12y 1)hay 13(3x 4y + 5) = - 5( 5x + 12y 1)
(t1) : 14x - 112y + 70 = 0 hay(t2) : 64x + 8y + 60 = 0
d
d
t1
t2
1
2O
A
B C
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Phng php ta tronbg mt phng
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l hai ng phn gic cn tm .
b) Nhn xt trong tam gic cn , phn gic trong ca gc ti nh th vung gcvi cnh y . Ta c hai ng thng :
1 qua O v vung gc t1 c phng trnh 112x + 14y = 0 2 qua O v vung gc t2 c phng trnh 8x 64y = 0
Dng 3 : Tnh gc ca hai ng thng v lp phng trnh ng thnglin quan n gc \
V d 1 : Tnh gc hai ng thng sau :
a) 2x + y 3 = 0 ; 3x - y + 7 = 0
b) 3x + 4y - 2 = 0 ,2
5
x t
y t
= +
=
Gii a) cos =2.3 1( 1) 1
5. 10 2
+ = => = 450
b) VTPT ca hai ng thng l : (3;4) , ' (1;1)n n= =
. Suy ra :
cos =2 2 2 2
3.1 4.1 7cos( , ')
5 23 4 1 1n n
+= =
+ +
V d 2 : Tm k bit ng thng y = kx + 1 hp vi ng thng : x y = 0 mt
gc bng 600
Gii : Ta c kx y + 1 = 0 . Ta c phng trnh :
cos 600 = 2 22
.1 1 12( 1) 1
21 2
kk k
k
+= + = +
+
2 4 1 0 2 3k k k+ + = =
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*V du 3 : Cho hnh vung ABCD c ng cho BD : x + 2y 5 = 0 , nh A(2 ;
- 1) . Vit phng trnh cnh AB v AD bit AB c h s gc dng .
Gii : Gi k l h s gc ca AB , AD , phng trnh AB , AD c dng :
y = k(x 2 ) 1 kx y 2k 1 = 0
Ta c AB v AD u hp vi BD mt gc 450
cos 450 = 2 22
2 12( 2) 5( 1)
25 1
kk k
k
= = +
+
3k2 + 8k 3 = 0 k = 1/3 ( ng AB) , k = - 3 ( ng AD ) .
Vy phng trnh AB : - 3x y + 5 = 0 , AD : x 3y 5 = 0 hay ngc li
C. Bi tp rn luyn .
3.22. Chn cu ng : Gi l gc ca hai ng thng : x - y 3 = 0 v 3x + y
8 = 0 , th th cos =
a) 1/ 5 b) 2/ 5
c) 2/ 10 d) p s khc3.23. Chn cu ng : Khong cch t A(1 ; 3) n ng thng 3x 4y + 1 = 0
l :
a) 1 b) 2 c) 3 d) p s khc
3.24. Chn cu ng : C 2 gi tr m ng thng x + my 3 = 0 hp vi
x + y = 0 mt gc 600 . Tng 2 gi tr y l :
a) 1 b) 1 c) 4 d) 4
3.25. Chn cu ng : Cho A(3; 4) , B(1; 1) , C(2 ; - 1) . ng cao tam gic vt A c di l :
a)1
5b)
7
5c)
13
5d) p s khc
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Phng php ta tronbg mt phng
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3.26. Chn cu ng : im A ( a, b) thuc ng thng :3
2
x t
y t
= + = +
cch ng
thng d : 2x y 3 = 0 mt khong 2 5 v a > 0 , th th a + b =
a) 20 b) 21 c) 22 d) 23
3.27 Cho tam gic ABC vi B(1 ; 2) v C(4 ; - 2) .
a) Vit phng trnh ng thng BC v tnh di ng cao AH .
b) Tm ta im A bit din tch tam gic l 10 v A thuc trc tung .
3.28 Cho tam gic ABC c AB : 2x + y 3 = 0 ; AC : 3x - y + 7 = 0 v BC : x
y = 0 .
a) Tnh sinA , BC v bn knh ng trn ngai tip tam gic ABC .
b) Vit phng trnh ng thng i xng ca AB qua BC .
3.29. Cho hnh vung ABCD c tm I ( 2; 3) , phng trnh AB : 3x + 4y 4 =
0 .
a) Tnh cnh hnh vung .
b) Tm phng trnh cc cnh CD , AD v BC .3. 30. Cho hnh vung ABCD c AB : 3x 2y 1 = 0 , CD : 3x 2y + 5 = 0 v
tm I thuc d : x + y 1 = 0
a) Tm ta I .
b) Vit phng trnh AD v BC
* 3.31. Cho tam gic u c A( 3 ; - 5) v trng tm G (1 ; 1) .
a) Vit phng trnh cnh BC .
b) Vit phng trnh cnh AB v AC .*3.32. Trong mt phng Oxy cho tam gic ABC c A(2 ; - 3) , B(3 ; - 2) , din
tch tam gic bng 3/2 v trng tm G thuc ng thng d : 3x y 8 = 0 . Tm
ta nh C .
* 3.33. Cho hnh thoi ABCD c A(- 2; 3) , B(1 ; - 1) v din tch 20 .
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Phng php ta tronbg mt phng
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a) Tnh ng cao hnh thoi v phng trnh cnh AB .
b) Tm ta im D bit n c honh dng .
* 3.34. Cho hnh ch nht ABCD c tm I(2 ; 2) , AB : x 2y 3 = 0 v AB =
2AD v yA > 0 .
a) Tm ta hnh chiu K ca I ln AB.
b) Tm ta A v B.
* 3.35. Cho ng thng d : x + 2y 4 = 0 v A(1 ; 4) , B(6 ; 4)
a) Chng minh A, B nm mt pha i vi d. Tm ta A i xng ca A
qua d .
b) Tm M d sao cho tng MA + MB nh nht .
c) Tm M d sao cho | MA MB| ln nht .
* 3.36. Cho hnh thoi c phng trnh ba cnh l : 5x 12y 5 = 0 , 5x 12y +
21 = 0 v 3x + 4y = 0 . Vit phng trnh cnh cn li .
*3.37. Vit phng trnh 4 cnh hnh vung bit 4 cnh ln lt qua bn im I(0
; 2) , J(5 ; - 3) , K(- 2 ; - 2) v l(2 ; - 4) .
D. Hng dn hay p s
3.22. (a) 3.23. (d) 3.24. (c) 3.25. (b) 3.26. (d)
3.27. a) BC : 4x + 3y 10 = 0 .
Ta c BC = 5 , suy ra AH = =BC
S2 ABC 4 .
b) Gi A( 0 ; a) . Ta c : d(A, BC) = 4
4
5
|10a3|=
a = 10 hay a = - 10/3
3.28. a)Ta c : sinA = sin(AB, AC) = Acos1 2
|cosA| =2
1
10.5
|)1(13.2|=
+=> sinA =
2
1.
B
AC
D
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Phng php ta tronbg mt phng
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Ta B , giao im ca AB v BC , l ( 1 ; 1) .Ta C , giao im ca AC v BC , l (- 7/2 ; - 7/2 ) .
Suy ra : R = =Asin2
BC 2/9
21.4
29=
b) Phng trnh ng thng cn tm BD qua B c dng y = k(x 1) + 1
kx y k + 1 = 0
Ta c : cos (BA, BC) = cos (BD, BC)1k.2
|11.k|
2.5
|)1(11.2|2 +
+=
+
k2 + 1 = 5(k + 1)2 4k2 + 10k + 4 = 0
k = - hay k = - 2 . Ch k = - 2 l ng vi h s gc ca BA nn b lai , tanhn k = - . Phng trnh ng thng BD : x + 2y - 3 = 0
3.29. a) Cnh hnh vung bng 2.d(I, AB) = 4
b) * Phng trnh CD : 3x + 4y + m = 0 vi
5
4)3(4)2(3
5
m)3(4)2(3 +=
++ - 6 + m = 2 m = 8
=> CD : 3x + 4y + 8 = 0* Phng trnh AD v BC : 4x 3y + m = 0
Ta c : d(I, AB) = d(I, AD) 2 =
5
|m17| +
m = - 7 hay m = - 27AD : 4x 3y - 7 = 0 , BC : 4x 3y 27 = 0 hay ngc li .
3.30. a) I d => I = (x ; 1 x) . Ta c : d(I, AB) = d(I, CD) x = 0 => y = 1 :
I(0 ; 1)
b) Nh cu b ( bi 3. 29)
B C
A
I
G
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Phng php ta tronbg mt phng
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3.31. a) Gi I l trung im BC , ta c :
=+
=+=>
=++
=++
GIA
GIA
GCBA
GCBA
y3y2y
x3x2x
y3yyy
x3xxx=> I = (0 ; 4)
Phng trnh BC qua I v vung gc )9;3(AI = : - (x 0 ) + 3(y 4) = 0
- x + 3y 12 = 0
b) Phng trnh AB, AC qua A c dng : kx - y 3k
- 5 = 0
Ta c : cos(AB, BC) = cos60 =
2
1
1k.10
|3k|2
=+
+
3k2 12k 13 = 0 k =3
356. Phng trnh
AB v AC :
03153y3x)356(:AC
03153y3x)356(:AB
=+
=+
3.32 . G d => G = (a ; 3a - 8) .
Ta c ; SGAB = 1/3 . SABC = . M AB = 2 , suy ra : d(G; AB) = 1/ 2
Phng trnh AB : x y - 5 = 0 , suy ra :
1|a23|2
1
2
|58a3a|==
+
. . . . . .
3.33. a) Ta c : h = 4AB
SABCD = . AB : 4x + 3y 1 = 0
b) Gi D = (x ; y) vi d > 0 . Ta c :
==
=
5ABAD
4)AB,D(d
B C
A
I
G
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Phng php ta tronbg mt phng
28
=++
=+
)2(25)3y()2x(
)1(45
|1y3x4|
22
(1) y =3
21x4 +hay y =
3
19x4
Th vo (2) , gii ta c : x = 3 => y = 3 . Vy D = (3 ; 3)
3. 34.a) Phng trnh IK : 2x + y 6 = 0 . Suy ra K(3 ; 0)
c) V AB = 2AD nn KA = 2KI (1) . Ta K(2y + 3 ; y ) AB .Gii (1) , ta c : y = 2 , suy ra A(7 ; 2)
3.35. a) A(- 1; 0 )
b) Ta c : MA + MB = MA + MB AB = 65
Vy GTNN l 65 M = AB d . Vit phng trnh AB , suy ra : M = (4/3
; 4/3)
c) Ta c : |MA MB| AB = 5 .
Vy GTNN l 5M = giao im ca d v AB ko diM = ( - 4 ; 4)
3.36. Ch trong hnh thoi khang cch gia hai cnh bng nhau .
AB : 5x 12y 5 = 0 , CD : 5x 12y + 21 = 0 . Chn M(1 ; 0) AB , ta c :
d(AB, CD) = d(M, CD) = 2
AD : 3x + 4y = 0 , BC : 3x + 4y + m = 0 . Chon O(0 ; 0) AD , ta c :
d(AD, BC) = d(O, BC) = 2 m = 10 .
=> BC : 3x + 4y 10 = 0
3.37. Phng trnh AB qua I : ax + by 2 = 0Phng trnh CD qua K : ax + by + 2a + 2b
= 0
Phng trnh BC qua J : bx ay 5b 3a =
0
A B
D C
I
J
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Phng php ta tronbg mt phng
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Phng trnh AD qua L : bx ay 2b 4a = 0
Ta c : d(I, CD) = d(J, AD) 2222 ab
|ab3|
ba
|a2b4|
+
=
+
+
b = - 3a hay a = - 7b
Chn :
=
=
=
=
1b
7ahay
3b
1a
4. ng trnA. Tm tt gio khoa .
1.Trong mt phng ta Oxy , phng trnh ng trntm I(h ; k) bn knh R l : (x h)2 + (y k)2 = R2 .
Phng trnh ng trn (O, R) l : x2 + y2 = R22. Trong mt phng ta Oxy , mi phng trnh c dng :
x2 + y2 + 2ax + 2by + c = 0 vi a2 + b2 c > 0l phng trnh ng trn :
Tm I(- a ; - b) Bn knh R = 2 2a b c+
3. Tip tuyn vi ng trn(x h)2 + (y k)2 = R2 ti tip im T(x0 ; y0) l :ng thng qua T v vung gc )ky;hx(IT 00 = c
phng trnh : (x0 h)(x x0) + (y0 k)(y y0) = 0 ng thng l tip tuyn ca ng trn (I, R) d(I, ) = R
B . Gii tan ..Dng ton 1 : Xc nh tm v bn knh . iu kin mt phng trnh lng trn .
V d 1 : Xc nh tm v bn knh cc ng trn sau :
a) (x + 1)2 + ( y 4)2 = 1 b) (x 2)2 + y2 = 5c) x2 + y2 + 8x 4y 5 = 0 d) 3x2 + 3y2 + 4x + 1 = 0
Gii :a) ng trn tm I(- 1 ; 4) , bn knh R = 1
x
y
I
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T
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b) ng trn tm I(2 ; 0) , bn knh R = 5c) a = - 4 , b = 2 , c = - 5 => I(- 4 ; 2) , R = 2 2 2 2a b c 4 2 5 5+ = + + = d) Vit li phng trnh ng trn bng cch chia hai v cho 3 :
x2 + y2 +4 1
x 03 3
+ =
Tm I( -2
;0)3
, bn knh R =2
2 1 3
3 9 3
=
V d 2 : Cho phng trnh : x2 + y2 + 2mx 2my + 3m2 4 = 0 (1)a) nh m (1) l phng trnh mt ng trn .
b) Chng minh tm cc ng trn ny di ng trn mt an thng khi mthay i .
c) Vit phng trnh ng trn (1) bit n c bn knh l 1 .
d) Tnh bn knh ng trn (1) bit n tip xc vi : 2x y = 0
Gii :a) Ta c : a = m , b = - m , c = 3m2 4 . (1) l phng trnh ng trn th :
a2 + b2 c > 0 m2 + m2 (3m2 4) > 0 4 m2 > 0 - 2 < m < 2 .
Vi 2 < m < 2 , ng trn c c tm l I
====mby
maxI
I (1) => xI + yI = 0
Li c : - 2 < m < 2 - 2 < xI < 2 (2)T (1) v (2) suy ra tp hp ca I l an AB c phng trnh x + y = 0 ( - 2 < x< 2)
b) Vi 2 < m < 2 , ng trn c bn knh l R = 24 m .
Ta c : R = 1 4 m2 = 1 m 2 = 3 m = 3
m = 3 : phng trnh ng trn l : x2 + y2 2 3 x + 2 3 y + 5 =0
m = - 3 : phng trnh ng trn l : x2 + y2 + 2 3 x - 2 3 y + 5= 0
c) ng trn tip xc d(I, ) = R
2| 2m m |
4 m5
=
9m2 = 5(4 m2 ) ( bng phng hai v)
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14m2 = 20 m = 107
V d 3 : Cho ng trn (C ) : x2 + y2 2x + 4y 4 = 0a) Tm tm v bn knh ca (C).
b) Cho A(3 ; -1) , chng minh A l im trong ng trn .Vitphng trnh ng thng qua A v ct (C) theo mt dy cung c di nh nht .
c) Cho d : 3x 4y = 0 , chng minh d ct (C) . Tnh di dy cung .Gii : a) a = 1 ; b = - 2 , c = - 4 => tm I c ta (1 ; - 2) , bn
knh R = 2 2a b c 3+ = .b) Ta c : IA2 = (3 1)2 + (- 1 + 2)2 = 5 => IA < R
Vy A bn trong ng trn .ng thng qua A ct (C) theo dy cung nh nht khi d cch xatm I nht d vung gc IA
= (2 ; 1) ti A(3 ; - 1) d c phng trnh : 2(x 3) + 1.(y + 1) = 0 2x + y 5 = 0
c) d ct (C) d(I, d) < R .
Ta c : d(I,d) =2 2
| 3.1 4.( 2) | 53
103 1
= d ct (C) theo mt dy cung MN .
K IH vung gc MN , th th : IH = 510
, IM = R = 3 , suy ra :
MH2 = IM2 IH2 = 9 -25 65 13
10 10 2= =
Vy di MN = 2MH = 2.13
262
=
Cn nh: Cho ng trn (I , R) v ng thng : tip xc (I)d(I,) = R ct (I) d(I,) < R ngai (I)d(I,) > R
Dng ton 2 : Thit lp phng trnh ng trn .C 2 cch thit lp phng trnh ng trn :
1.Tm ta (h ; k) ca tm v tnh bn knh R , phng trnh ng trn cntm l : (x h)2 + (y k)2 = R2 .
2.Tm a , b, c , phng trnh ng trn cn tm l : x2 + y2 + 2ax + 2by + c =0
IA
M
NHd
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Cn nh: ng trn (I, R) qua M(x0 ; y0) IM2 = R2
(x0 h)2 + (y0 k)
2 = R2 x02 + y02 + 2ax0 + 2by0 + c = 0
ng trn (I, R) tip xc d(I, ) = R ng trn (I, R) tip xc trc Ox |h| = R ng trn (I, R) tip xc trc Oy |k| = R
V d 1 : Vit phng trnh ng trn :a) ng knh AB vi A(3 ; 1) v B(2 ; - 2) .
b) c tm I(1 ; - 2) v tip xc vi ng thng d : x + y 2 = 0c) c bn knh 5 , tm thuc Ox v qua A(2 ; 4) .d) c tm I (2 ; - 1) v tip xc ngai vi ng trn : (x 5)2 + (y 3)2 = 9e) tip xc hai trc v c tm trn ng thng : 2x y 3 = 0
Gii :a) Tm ng trn l trung im I ca AB, c ta
A B A Bx x y y 5 1; ;2 2 2 2
+ + =
Bn knh R = IA =2 2
1 3 10
2 2 2
+ =
. Phng trnh ng trn l :
(x - 2 252 1 5
) (y )2 2 2
+ + =
b) Bn knh ng trn l R = d(I, d) =2 2
|1 2 2 | 3
21 1
=
+
Phng trnh ng trn l : (x 1)2 + (y + 2)2 =9
2
c) V tm I Ox nn I = (h ; 0) .Ta c : IA = R (h 2)2 + (4 0)2 = 25 (h 2)2 = 9
h 2 = 3 hay h 2 = - 3 h = 5 hay h = - 1 .
Phng trnh ng trn cn tm : (x 5)2 + y2 = 25 hay (x + 1)2 + y2 = 25d) ng trn (x 5)2 + (y 3)2 = 9 c tm K(5 ; 3) , bn knh r = 3ng trn (I, R) cn tm tip xc ngai vi (K) IK = R + r
M IK = 2 2(5 2) (3 1) 5 + + = , suy ra : R = 5 r = 2 .
Vy phng trnh ng trn (I) l : (x 2)2 + (y + 1)2 = 4
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e) Gi (h; k) l tm v R l bn knh ng trn . Ta c :
(I) tip xc Ox , Oy
==
==
R|h|)Oy,O(d
R|k|)Ox,O(d
Suy ra : |h| = |k| h = k (1) hay h = - k ( 2)Mt khc : I 2h k 3 = 0 (3)
Gii (1) v (3) : h = k = 3 => R = 3 Gii (2) v (3) : h = 1 , k = - 1 => R = 1 .
Phng trnh ng trn cn tm :(x 3)2 + (y 3)2 = 9 hay (x 1)2 + (y + 1)2 = 1
V d 2 : Vit phng trnh ng trn :a) qua A(- 2 ; - 1) , B(- 1 ; 4) v C(4 ; 3)b) qua A(0 ; 2) , B(- 1; 1) v c tm trn ng thng 2x + 3y = 0
c) qua A(5 ; 3) v tip xc ng thng d : x + 3y + 2 = 0 ti im T(1 ; - 1)Giia) Phng trnh ng trn c dng (C) : x2 + y2 + 2ax + 2by + c = 0
(C) qua A(- 2 ; - 1) 22 + 12 + 2a(-2) + 2b(-1) + c = 0 4a + 2b - c = 5 (1)
(C) qua B(- 1 ; 4) 2a 8b - c = 17 (2)(C) qua C(4 ; 3) 8a + 6b + c = - 25 (3)
Gii h (1), (2), (3) , ta c : a = b = - 1 , c = - 11 Phng trnh ng trn cn
tm l :x2 + y2 2x 2y 11 = 0
b) Phng trnh ng trn c dng (C) : x2 + y2 + 2ax + 2by + c = 0(C) qua A(0 ; 2) 4b + c = - 4 (1)(C) qua B(- 1 ; 1) - 2a + 2b + c = - 2 (2)Tm I(a ; b) 2a + 3b = 0 (3)
Gii h (1), (2), (3), ta c a = - 3 , b = 2 , c = - 12 . Phng trnh ng trn
cn tm l : x2 + y2 6x + 4y 12 = 0
I K
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I
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c) Phng trnh ng trn c dng (C) : x2 + y2 + 2ax + 2by + c = 0(C) qua A(5 ; 3) 10a + 6b + c = - 34 (1)(C) qua T( 1 ; - 1) 2a - 2b+ c = - 2 (2)
Tm I(a ; b) ng thng vung gc vi d : x + 3y + 2 = 0 tiT(1 ; - 1) c phng trnh l : 3(x 1) (y + 1) = 0 3x y 4 = 0Do : - 3a + b = 4 (3) .Gii h (1), (2), (3), ta c : a = b = - 2 , c = - 2 . Phng trnhng trn cn tm l : x2 + y2 4x 4y 2 = 0
V d 3 : Cho A(2 ; 0) v B(0 ; 1) , chng minh tp h p nhng im M tha MA2
MB2 = MO2 l mt ng trn . Xc nh tm v bn knh ng trn y .
Gii Gi (x ; y) l ta ca M , ta c :MA2 MB2 = MO2
[(x 2)2 + y2 ] [(x2 + (y 1)2 ] = x2 + y2 x2 + y2 + 4x 2y 3 = 0y l phng trnh ng trn tm I(- 2 ; 1) , bn knh R = 2 2 .
Dng 3: Lp phng trnh tip tuyn vi ng trn .Cn nh: Cho ng trn tm I(a ; b) , bn knh R :
Nu bit tip im l T (x0 ; y0) th phng trnh tip tuyn l ngthng qua (x0 ; y0) v vung gc vi IT = (x0 h ; y0 - k)
Nu khng bit tip im th dng iu kin sau gii : l tip tuyn ca ng trn (I, R) d(I, ) = R
V d 1 ( Tip tuyn ti mt im cho trc)a) Vit phng trnh tip tuyn ca ng trn (x 3)2 + (y + 1)2 = 25 ti im
c honh l 1 .b) Vit phng trnh tip tuyn ca ng trn (C) : x2 + y2 + 4x 2y 5 = 0
ti im m ng trn ct trc Ox.
Gii
a) Tm I(3 ; - 1) , bn knh r = 5Th x = - 1 vo phng trnh ng trn , ta c :16 + (y + 1)2 = 25 (y + 1)2 = 9
y + 1 = 3 y = 2 hay y = - 4 . Vy ta tip im l (- 1 ; 2) hay ( - 1 ; - 4)
Vi tip im T (- 1; 2) , tip tuyn vung gc IT = (- 4 ; 3) c phngtrnh l : - 4(x + 1 ) + 3(y 2 ) = 0 - 4x + 3y 10 = 0
I
Td
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Vi tip im (- 1; - 4 ) , tip tuyn vung gc IT = (- 4 ; - 3) c phng
trnh l : 4(x + 1) + 3(y + 4) = 0 4x + 3y + 16 = 0
b) Th y = 0 vo phng trnh ng trn : x2 + 4x 5 = 0 x = 1 hay x = - 5Vy ta tip im l (1 ; 0) hay ( - 5 ; 0) .ng trn c tm l I(- 2 ; 1) .
Tip tuyn ti T(1 ; 0) vung gc vi IT = ( 3 ; - 1) c phng trnh :3(x 1) 1.(y 0) = 0 3x y 3 = 0
Tip tuyn ti T(- 5 ; 0) vung gc vi IT = ( - 3 ; - 1) c phng trnh :3(x + 5) 1.(y 0) = 0 3x y + 15 = 0
V d 2 ( Tip tuyn c phng cho trc )a) Vit phng trnh tip tuyn ca ng trn x2 + y2 = 2 bit tip tuyn
c h s gc l 1 .b) Vit phng trnh tip tuyn ca ng trn (C) : x2 + (y 1) 2 = 25
bit tip tuyn vung gc vi ng thng 3x 4y = 0
Gii :a) ng trn c tm O(0 ; 0) , bn knh 2 . Phng trnh ng thng d
c h s gc l 1 c dng : x y + m = 0 ( m l s cha bit) . Ta c :d tip xc (C) d(I, d) = R
2 2| m | 2 | m | 21 1
= =+
m = 2Vy phng trnh tip tuyn l : x y 2 = 0
b) ng trn c tm I(0 ; 1) , bn knh R = 5 . Phng trnh ng thng vung gc vi 3x 4y = 0 c dng : 4x + 3y + m = 0 .
tip xc (C) d(I, ) = R
2 2
| 4.0 3.1 m |5
4 3
+ +=
+ |3 + m| = 25
m = 22 hay m = - 28 .Vy phng trnh tip tuyn l : 4x + 3y = 22 hay 4x + 3y 28 = 0
V d 3 ( Tip tuyn qua mt im cho trc )Vit phng trnh tip tuyn ca ng trn : x2 + y2 4x 2y 4 = 0 tm I(2 ;
1) , bn knh R = 3 bit tip tuyn qua im A(- 1 ; 2) .
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Giia) ng trn c tm I(2 ; 1) , bn knh R = 3 .Phng trnh ng thng qua A(- 1 ; 2) c dng : y 2 = k(x + 1) kx y + k + 2 = 0 (*) , k l h s gc ca .
tip xc (C) d(I, ) = R
2
| 2.k 1 k 2 |3
k 1
+ +=
+ | 3k + 1 | = 3. 2k 1+
Bnh phng hai v : 9k2 + 6k + 1 = 9(k2 + 1) 6k = 8 k = 4/3Th vo (*) , ta c phng trnh tip tuyn cn tm : ( 4/3) x y + 4/3 + 2 = 0 4x - 3y + 10 = 0 .
Ghi ch : Thng tmtim c thkc 2 tip tuyn vi ng trn , yta chc mt l v ta cha xetn ng thng qua A vung gc vi Ox,ng ny khng c h sgc
* Xt : x 2 = 0 ( qua A v vung gc Ox) :
Ta tnh d(I, ) =|| 1 2 |
31
= , vy d(I, ) = R , do : x 2 = 0 cng l mt
tip tuyn cn tm .Qua A(2 ; 1) c hai tip tuyn l : x 2 = 0 v 4x - 3y + 10 = 0 .
Ghi ch : C thvit phng trnh tip tuyn qua A( - 1 ; 2) di dng tng qut :
a(x + 1) + b(y 2)= 0
ax + by + a 2b = 0 .iu kin tip xc : d(I,) = R 3
ba
|b2a1.ba.2|22
=+
++
(3a b)2 = 9(a2 + b2 ) b(8b + 6a) = 0
b = 0 hay a = - 4b/3
* V d 34 : Cho (C) : x2 + y2 = 1 v (C) : (x 2)2 + (y 3)2 = 4 . Vit phng
trnh tip tuyn chung trong ca hai ng trn .
Gii(C) c tm O , bn knh 1 v (C) c tm I , bn knh 2 .Phng trnh tip tuyn chung d c dng : ax + by + c = 0 ( a2 + b2 0 ) tha :
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=++
=+
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c) Tm tm v bn knh ng trn (C) : x2 + y2 + 6x + 6y + 13 = 0 .
Chng minh (C) v (C) tip xc ngai ti T . Vit phng trnh tip tuyn chung
ti T.
3.42. Cho ng trn (C) : x2 + y2 + 4x 6y + 7 = 0a) im M(- 1; 1) trong hay ngai ng trn . Lp phng trnh dy
cung qua M v c di ngn nht .b) Lp phng trnh ng thng qua O v ct (C) theo mt dy cung c
di l 2 .
3. 43.Lp phng trnh ng trn :
a) c tm I(3 ; - 2) , bn knh 2 b) c tm I(2 ; - 4) v qua gc ta c) c tm I(1 ; - 2) v tip xc ng thng x y 0
* 3. 44. Lp phng trnh ng trn :a) qua A(1 ; 2) v tip xc hai trc ta .
b) tip xc hai ng thng song song : 2x y 3 = 0 , 2x y + 5 = 0 v ctm trn Oy.
c) tip xc ng thng 2x + y 5 = 0 ti im T(2 ; 1) v c bn knh2 5
* d) ti p xc vi hai ng thng .x 2y + 5 = 0 v x + 2y + 1 = 0 v qua
gc O.3.45. Lp phng trnh ng trn :
a) qua A(0 ; 4) , B( - 2; 0) v C(4 ; 3)b) qua A(2 ; - 1), B(4 ; 1) v c tm trn Ox .c) qua A(3 ; 5) v tip xc ng thng x + y 2 = 0 ti im T(1 ; 1) .
3.46. Cho ng trn (C) : (x 2)2 + (y + 1)2 = 4 .a) Tm trn Oy im t kc 2 tip tuyn ca (C) v hai tip tuyn
vung gc nhau .b) Tm trn (C) im gn gc O nht.
3.47. Chng minh ng thng :2x y = 0 v ng trn : x2 + y2 4x + 2y
1 = 0 ct nhau . Tm di dy cung to thnh .
3.48. Cho hai ng trn ( C) : x2 + y2 2x 4y + 1 = 0 v (C) : x2 + y2+ 4x +
4y - 1 = 0 .
a) Chng minh hai ng trn tip xc ngai . Tm ta tip im T.
b) Vit phng trnh tip tuyn chung ti T.
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* 3. 49. Cho ng trn (x 3)2 + (y + 2)2 = 9 v im M(- 3 ; 1)
a) Chng minh M ngai ng trn .
b) Tnh phng tch ca M i vi ng trn v tnh di tip tuyn MT.
* 3.50. Cho hai ng trn (C ) : x2 + y2 2x 2y + 1 = 0 v (C) : x2 + y2 4x +6y + 9 = 0
a) Chng minh hai ng trn c 4 tip tuyn chung .b) Chng minh bn im chia cc an tip tuyn chung theo t s - 2 cng
nm trn mt ng trn .
3.51. a) Vit phng trnh tip tuyn ca ng trn x2 + y2 2x + 4y 5 = 0 ti
im (2 ; 1) .b) Vit phng trnh tip tuyn ca ng trn (x + 1)2 + (y - 3)2 = 5 ti
im m ng trn ct Oy .
*3.52.Vit phng trnh tip tuyn ca ng trn x2 + y2 2x + 8y 1 = 0 :a) bit tip tuyn song song ng thng x y + 3 = 0
b) bit tip tuyn qua im (2 ; 1) .
*3.53.Vit phng trnh tip tuyn ca ng trn x2 + y2 2x - 4y 5 = 0 :a) bit tip tuyn vung gc ng thng 3x + y = 0
b) bit tip tuyn pht xt tim A(3 ; - 2) .
c) Vit phng trnh ng trn ngai tip tam gic AT1T2 v ng thngqua hai tip im T1, T2 .
*3.54.Cho hai ng trn : x2 + y2 2x - 2y 2 = 0 v x2 + y2 8x 4y + 16 = 0a) Chng minh hai ng trn bng nhau v ct nhau .
b) Vit phng trnh ng thng qua giao im ca hai ng trn .b) Tm phng trnh tip tuyn chung ca chng .
*3.55. Cho A(3 ; 0) v B(0 ; 4) . Vit phng trnh ng trn ni ti p tam gic
OAB .
*3.56. Bin lun theo m v tri tng i ca ng thng v ng trn (C )
a) : x + 3y + m = 0 ; (C) : (x 2)2 + y2 = 10
b) : x my + m 4 = 0 ; (C ) : x2 + y2 - 2x 4y + 4 = 0
*3.57. Cho hai ng thng : x + 1 = 0 v : x 1 = 0 , ct Ox ti A v B . .M v N l hai im di ng trn v c tung l m v n sao cho lun c :mn = 4.
a) Vit phng trnh ng thng AN v BM .
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b) Chng minh giao im I ca AN v BM thuc mt ng trn cnh .
3.58.Chn cu ng : Tm tm I v bn knh R ca ng trn (x + 2)2 + (y 1)2 = 4
a) I(2 ; - 1), R = 2 b) I(- 2 ; 1), R = 2c) I(2 ; - 1) , R = 4 d) I(- 2 ; 1) , R = 4
3.59.Chn cu ng : Tm tm I v bn knh R ca ng trn : 2x2 + 2y2 3x +4y - 1 = 0 1
a) I(3/2 ; - 2) , R =29
2b) I(- ; 1) , R =
33
4
c) I(3/4 ; - 1) , R = 334
d) I(3/4 ; - 1) , R = 174
3. 60..Chn cu ng : C bao nhiu s nguyn m : x2 + y2 2(m + 1)x + 2my+ 3m2 + 2m 12 = 0 l phng trnh mt ng trn ?
a) 5 b) 7 c) 9 d) v s
3.61. Chn cu ng : Cho A(1 ; 1) v B(2 ; 3) , tp h p ccim M tha :3MA2 2MB2 = 6 l mt ng trn . Bn knh ca n l :
a) 3 b) 4 c) 5 d) 6
3.62. Chn cu ng : C hai ng trn c tm trn Ox , bn knh 5 v quaim A(1 ; - 38) . Khang cch hai tm ca chng l :
a) 2 b) 4 c) 6 d) 8
3. 63. Chn cu ng :ng trn qua A(1 ; 0), B(2 ; 0) v C(0 ; 3) c bn knhgn nht vi s no di y ?
a) 1, 1 b) 1, 2 c) 1, 3 d) 1, 4
D. Hng dn hay p s :
3. 38.a)I (- 5/.2 ; 3/2 ) , R = 1 b) I(- ; - ) , R =3
2
c) ( - 3/2 ; 0), R =5
2d) I(1 ; - 3/2) , R = 5/4
3.39. a) m , tp h p I lng thng 2x + y 6 = 0b) m < 0 , tp h p l na ng thng x + y = 0 vi x > 0c) 1 < m 0 , mb) Bn knh nh nht khi m = - 1 .c) iu kin tip xc 2m2 + 4m 26 = 0 : phng trnh ny c hai
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nghim .
3.41. a) 4 b) 2 5 c) V khang cch hai tm bng tng hai bn knh . Phng
trnh tip tuyn chung l : 2x + y + 4 = 0
3.42. a) ngai v IM > R . Dy cung qua M v vung gc IM .b) V dy cung c di 2 nn khang cch t I n ng thng l :
2R 1 5 = . Phng trnh ng thng : kx y = 0 . Gii : d(I, ) = 5 , tac k .
3.44. Gi I(h ; k) l tm v R l bn knh :
a) Ta c h :
=+
==
)2(h)2k()1h(
)1(R|k||h|222
Th ln lt k = h v k = - h vo (2) , ta c phng trnh tnh h .
b) I(0 ; k) , ta c h phng trnh :d(I, ) d(I, ')
d(I, ) R
=
=
c) Ta c h :
=
n//IT
52),I(d
=
=+
1
1k
2
2h
525
|5kh2|
=
=+
=+
0k2h
105kh2
105kh2
3.45. Phng trnh ng trn c dng : x2 + y2 + 2a + 2by + c = 0a) Th ta A, B, C , ta c h phng trnh tnh a, b, c .
b) Ta c : b = 0 , th ta A v B , ta c h tnh a v c .c) Phng trnh ng thng qua T v vung gc x + y 2 = 0 l :
x y = 0 . Ta c h :
=+
=+++=+++
0ba
0cb2a220cb10a634
qua A(3 ; 5) v tip xc ng thng x + y 2 = 0 ti im T(1 ; 1) .
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3.46. a) im cn tm cch tm mt khang l R 2 .b) im cn tm l giao im ca OI v ng trn .
3.47. a) ng trn c tm I(2 ; - 1) , bn knh R = 6
Ta c : d(I, ) = 55
5= < R => ct ng trn .
di dy cung : 2 2dR 22 =
3. 48. (C) c tm I(1 ; 2) . (C) c tm I(- 2 ; - 2).
im chung ca hai ng trn tha h :
=+++=++
(2).01-4y4xyx(1)014y2x-yx 22
22
Ly (1) tr (2) : - 6x 8y + 2 = 0 x =3
1y4 +.
Th vo (1) : (5y 2)2 = 0 y = 2/5 => x = - 1/5 . Hai ng trn c mt imchung T nn tip xc nhau ti T(- 1/5 ; 2/5) .Li c xI < xT < xI nn T an II ,chng t hai ng trn tip xc ngai .
Ghi ch :C thchng minh cch khc x (C) c tm I(1 ; 2) , bn knh R = 2 .(C) c tm I(- 2 ; - 2), bn knh R = 3 . V II = R + R = 5 nn hai ngtrn tip xc ngai . Nhng vi cch ny , ta khng tm c tip im .
b) Tip tuyn chung l ng thng vung gc vi )4;3('II = v qua T , cphng trnh : 3x + 4y 1 = 0
3.49. a) Khang cch t tm I n M l IM = 37 > R = 3
b) Phng tch ca M l : IM2 R2 = 28 v di tip tuyn l 7228 =
Ghi ch : Tng qut c th chng minh c rng : Phng tch ca im M(x0 ;
y0) i vi ng trn : x2
+ y2
+ 2ax + 2by + c = 0 l : x2
0+ y2
0+ 2ax0 + 2by0 +
c .
3.50. a) (C) c tm I(1 ; 1 ) , bn knh R = 1 . (C) c tm I(2 ; - 3) , bn knh R= 2 . V II = 17 > R + R = 3 nn hai ng trn ct nhau . Suy ra chng c 4tip tuyn chung .
b) Gi M l im chia an tip tuyn chung TT theo t s - 2 , th th :MT = 2MTMT2 = 4MT 2
IM2 R2 = 4(IM2 R2 )
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x2 + y2 2x 2y + 1 = 4(x2 + y2 4x + 6y + 9 ) 3x2 + 3y2 - 14x + 26y + 35 = 0
y l phng trnh mt ng trn .
3.51. a) x + 3y 5 = 0 b) x + 2y 10 = 0 hay x + 2y 6 = 0
3.52.. a) x y + 1 = 0 , x y 11 = 0b) x + y 3 = 0 , 7x 17y + 3 = 0
3.53. c) ng trn ngoi tip tam gic AT1T2 c ng knh l AI , c phngtrnh : x2 + y2 4x 1 = 0 .* Ta cc im T1 , T2 tha h :
=+
=+
05y4x2yx
01x4yx22
22
nn cng tha :
(x2 + y2 4x 1) (x2 + y2 2x 4y 5) = 0 - 2x + 4y + 4 = 0 x 2y 2 = 0Do phng trnh ng thng T1T2 lx 2y 2 = 0
3.54. a) (C) c tm I(1 ; 1) , R = 2 . (C) c
tm I(4 ; 2) . R = 2 .
V R R < II < R + R nn (C) , (C) ct nhau .
b) Ta gii tng qut : Ta (x ; y) ca cc giao im ca hai ng trntha h :
=++++
=++++
)2(0'cy'b2x'a2yx
)1(0cby2ax2yx22
22
=> chng cng tha phng trnh :
(1) (2) : 2(a a)x + 2(b b)y + c c = 0
c) Tip tuyn chung c VTCP l (3 ; 1) v cch I mt khong l 2 .
3.55. Bn kinh ng trn l r = 1p
S= . Phng trnh phn gic trong gc O l
x y = 0 . Ta I l (1 ; 1) . Phng trnh ng trn ni ti p l :(x 1)2 + (y 1)2 = 1 .
3.56. a) (C) c tm I(2 ; 0) , R = 10 . d = d(I, ) =10
|m2| +
d < R - 12 < m < 8 : d v (C) ct nhaud = R m = 8 hay m = - 12 : d v (C) tip xc
IA
T1
T2
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d > R m < - 12 hay m > 8 : d v (C) ngai nhau .b) (C ) c tm I (1 ; 2) , R = 1 . d = d(I, ) =
1m
|3m|2 +
+
d < R 3/4m08m611m
|3m|2
- 4/3 : d v (C) ngai nhau
3. 57. a) Phng trnh chnh tc AN
qua A(- 1; 0) v N(1 ; n) : ny2 1x =+ (1)
Phng trnh chnh tc BM qua B(1 ; 0)
v M(- 1 ; m) :m
y
2
1x
=
(2)
b) Ta (x ; y) ca I tha (1) v (2)
=> (x ; y) tha :m
y.
n
y
2
1x.
2
1x
=+
4y
mny
41x 222 == x2 + y2 = 1
Vy I thuc ng trn (O ; 1)
3. 58 (b) 3.59.(c) 3.60. (b) 3.61. (d) 3.62 (d) 3.63 (d)
&5 .lipA. Tm tt gio khoa
1.nh ngha . Cho hai im cnh F1 , F2 vi 1 2 2F F c=
v mt dikhng i 2a ( a > c) Elip l tp hp nhng im M sao cho :
1 22F M F M a+ =
F1 , F2 : tiu im , F1F2 : tiu c , F1M, F2M : bn knh qua tiu .2. Phng trnh chnh tc .
Vi F1( - c ; 0) , F2(c ; 0) :
M
x
O
y
A1 A2
B1
B2
F1 F2
A B
N
MI
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M(x ; y) (E)
2 2
2 2 1x ya b+ = vib
2 = a2 - c 2 . ( 1)
(1) : phng trnh chnh tc ca (E)3. Hnh dng ca elip .-
* A1 ( - a ; 0 ) , A2 ( a ; 0 ) ,B1(0 ; - b) , B2 ( 0 ; b) : nh .
* on A1A2 = 2a : trc ln , B1B2 = 2b : trc nh .* Hnh ch nht gii hn bi cc ng x = a, y = b gi l hnh chnht csca elip.
* e = 1
a
c< : tm sai lip .
* F1M = a +a
cx M = a + exM ; F2M =a
cxa M = a - exM
B. Gii tan .Dng ton 1 : Xc nh cc yu t ca lip
V d : Hy xc nh nh , di cc trc , tiu c , tiu im tm sai v v elipc phng trnh sau :
a) (E) : +2 2
4 1
x y=1 b) (E) : 2 29 16 144x y+ =
Gii : a)Ta c : a2 = 4 , b2 = 1 => a = 2 v b = 1Suy ra A1 (- 2; 0 ) , A2 (2 ; 0 ) , B1(0 ; - 1 ) , B2 ( 0 ; 1) di trc ln 2a = 4 , trc nh 2b = 2 .
Ta c : c = =2 2 3a b . Tiu c 2c = 2 3 , tiu im F1( - 3 ; 0 ) , F2( 3 ; 0 ) . Tm sai : e = c/a = 3 /2 .
c) Vit li phng trnh (E) :2 2
116 9
x y+ = => a2 = 16 ; b2 = 9 => a = 4 , b = 3 v c
=2 2
7a b = Suy ra A1 (- 4; 0 ) , A2 (4 ; 0 ) , B1(0 ; - 3 ) , B2 ( 0 ; 3) di trc ln 2a = 8 , trc nh 2b = 6 .Tiu c 2c = 2 7 , tiu im F1( - 7 ; 0 ) , F2( 7 ; 0 ) .
Tm sai e = c/a =4
7
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Dng ton 2 : Lp phng trnh chnh tc ca lip :T gi thit , lp h phng trnh theo a v b . Gii h , tm c a , b . Suy ra
phng trnh (E) . Cn nh: M(x0 ; y0) (E)2 2
o o2 2
x y1
a b+ =
V d 1 : Lp phng trnh ca elip (E) bit :a) C di hai trc l 6 , 4 .
b) (E) c mt nh l ( 5 ; 0 ) v tiu c l 6 .c) (E) c mt nh l (0 ; 3 ) v (E) qua im M( 4 ; 1) .d) (E) qua hai im ( 1 ; 3
2
) v (- 2 ;2
2
) .
e) (E) c tiu im F2 ( 2 ; 0 ) v qua im (2, 5/3)Gii a) 2 a = 6 = > a = 3 , 2b = 4 = > b = 2 . Phng trnh elip l :
2 2
19 4
x y+ =
b) Phng trnh (E) :2 2
2 21
x y
a b+ =
nh (5 ; 0 ) Ox do n l nh A2 (a ; 0 ) . Suy ra : a = 5Tiu c = 2c = 6 c = 3 . Suy ra : b2 = a2 - c2 = 25 9 = 16
Vy phng trnh (E) l :2 2
125 16x y+ =
c) Phng trnh (E) :2 2
2 21
x y
a b+ =
nh (0 ; 3 ) Oy do n l nh B2 ( 0 ; b ) . Suy ra : b = 3 v :
(E) :2 2
2 9
x y
a+ = 1
y
O
x
O
y
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M(4; 1) (E)
2
22 24 1 16 81 189 9
aa a+ = = =
Vy phng trnh (E) :2 2
118 9
x y+ =
d) Phng trnh (E) :2 2
2 21
x y
a b+ =
( 1 ;3
2) (E)
2 2
1 31
4a b+ = (1)
N(- 2 ;2
2 ) (E) 2 22 2
14a b+ = (2)
Gii h (1) v (2) vi hai n l : u =2 2
1 1, v
a b= , ta c : u = , v = 1 .
Vy phng trnh (E) :2 2
14 1
x y+ =
e) F2( 2 ; 0 ) => c = 2 . Suy ra : F1 ( - 2 ; 0 ) .
Ta c : F2M =2
2 5 5(2 2)
3 3
+ =
, F1M =
2
2 5 13(2 2)
3 3
+ + =
Theo nh ngha elip : 2a = F1M + F2M =13 5
63 3+ = => a = 3 .
Suy ra : b2 = a2 c2 = 5 v phng trnh eip l :2 2
9 5
x y+
Cch khc : c= 2 = > a2 = b2 + 4 . Phng trnh elip :2 2
2 21
x y
a b+ =
Thta ca M , ta c :
2 2 4 2
2 2
4 251 36 25 100 9 36
4 9b b b b
b b+ = + + = +
+
9b4
25b2
100 = 0 .
Gii phng trnh trng phng ny , ta c : b2 =5 . Suy ra a2 = 9 .
V d 2 : Cho on AB c di khng i bng 3 . u A( 0 ; a) di ng trntruc honh , u B (b ; 0) di ng trn trc tung . M l im chia onAB theo t s 2. Tm ta ca M , suy ra M di ng trn mt elip .
Gii Gi (x; y) l ta ca M , ta c :
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2. MA MB=
2 2
3 3
2
3 3
A B
A B
x xbx
y y ay
+= =
+ = =
V a2 + b2 = AB2 = 3 , suy ra : (3y)2 +2
3
2
x
= 92 2
14 1
x y+ =
Vy M di ng trn elip c phng trnh2 2
14 1
x y+ =
Dng ton 3 : Tm
im thu
c (E)
Cn nh: * M(x0 ; y0) (E)2 2
o o2 2
x y1
a b+ = F1M + F2M = 2a .
* F1M = a +a
cx M ; F2M =a
cxa M
V d 1 : Cho elip (E) :2 2
16 2
x y+ =
a) Tm trn (E ) im M c honh l 2 .b) Tm ta giao im ca (E) v ng thng y = x 3 - 2 .
c) Tm trn (E) im M sao cho gc F1MF2 = 900
.d) Tm trn (E) im M tha F1M F2M = 6
GII a) Th x = 2 vo phng trnh ca (E) :2 2
2( 2) 4 21
6 2 3 3
yy y+ = = =
Ta tm c 2 im M c ta (2 ;2
3) , ( 2 ; -
2
3) .
b) Ta giao im l nghim ca h :
+ =
=
2 2
1 (1)
6 23 2 (2)
x y
y x
Th (2) vo (1) : x2 + 3(x 3 -2)2 = 6 x2 + 3(3x2 4x 3 + 4) = 6
5x2 - 6x 3 + 3 = 0
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Phng trnh ny c 2 nghim : = =1 2 33 ; 5x x
Th vo (2) : = = = = 1 1 2 2
73 2 1; 3 2
5 y x y x
Ta c 2 im c ta (x1 ; y1) , (x2 ; y2 ) .
c) Gi (x; y) l ta ca M . Ta c : F1MF2 = 900 OM = OF1 = OF2
2 2 2 2 4 x y c x y+ = + = ( c2 = a2 b2
= 6 2 = 4 )
Mt khc v M (E) nn ta E tha :
2x2 + 6y2 = 12
Ta c h :2 2
2 2
2 6 12
4
x y
x y
+ =
+ =
2
2
3 3
11
x x
yy
= =
= =
Ta tm c 4 im c ta ( 3 ; 1) , ( 3 ; - 1) , (- 3 ; 1) , ( - 3 ; - 1)
d) Theo nh ngha : F1M + F2M = 2a = 2 6 m F1M F2M = 6
Suy ra : F1M =3 6
2, F2M =
6
2
T :3 6
2= a +
a
cxM
3 6
2= 6 +
6
x2 M xM =
2
3
Th li vo phng trnh (E) , ta c :
2
29 15 5 51
24 2 48 16 4
yy y+ = = = =
Vy ta im cn tm (3 5
; )2 4
v (3 5
; )2 4
M
F1 F2
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V d 4 : Cho elip (E) :
2 2
2 2 1x ya b+ = c tiu im F
1 , F2. M l im bt k
trn (E) .a) Tm trn (E) : x2 + 4y2 = 4 im M sao cho F1M = 2F2M
b) Chng minh F1M . F2M + OM2 = a2 + b2 .
Gii a) Vit li phng trnh (E) :2 2
14 1
x y+ = => a2 = 4 ; b2 = 1 => c2 = 3
Theo chng minh trn : F1M = 2F2M a +c
xa
= 2( a -c
xa
)
2
33
cx aa xa c
= =
Th a2 = 4 , c = 3 : x =4
3 3. Th vo phng trnh (E) , ta c :
2
2 24 234 4
273 3y y
+ = =
y =
23
27
b) Ta c : F1M . F2M = (a + )( )c c
x a xa a
=2
2 2
2
ca x
a ( 1)
OM2 = x2 + y2 (2)
Cng (1) v (2) : F1M . F2M + OM2 = a2 + (1 -
2
2
c
a) x2 + y2
= a2 +2 2 2 2 2 2
2 2
2 2
b x b x a yy a
a a
++ = +
V M (E) nn b2 x2 + a2 y2 = a2 b2 , suy ra : F1M . F2M + OM2 = a2 + b2 : gi
tr khng i .
C. Bi tp rn luyn.
3.64 . Xc nh di cc trc , ta nh , tiu im v v cc elip sau :
a)2 2
112 9
x y+ = b)
2 2
15 1
x y+ = c) 4x2 + 9y2 = 36
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3. 65 . Cho elip (E) :
2
2 14
x y+ = . Tm trn (E) :
a) im M c tung . b) im N c tung gp i honh .
c) im P sao cho gc F1PF2 = 900 .
d) ta cc nh ca hnh vung ni tip (E) bit hnh vung c cc cnh
song song vi cc trc ta .
3.66. Cho elip (E) c di trc ln l 6 v qua im M(3 2
; 22
a) Lp phng trnh (E) .b) Tnh di dy cung ca (E) vung gc vi trc ln ti tiu im .
c) Tm trn (E) im M cch tm O mt khong l11
2. .
3.67. Lp phng trnh (E) bit :a) tiu c 4 v khong cch t mt nh n tiu im l 5 .
b) di trc nh l 4 v mt tiu im l ( 2 ; 0 )c) mt tiu im l F2 ( 5 ; 0 ) v khong cch gia hai nh l 9.
3.68. Lp phng trnh (E) bit :a) di trc ln l 8 v qua im ( 3 ; 2) .
b) qua hai im P2 2 1 5
; , 2;3 3 3
Q
.
c) c tiu c l 4 v qua im ( 1 ;2
5)
d) qua im M3 4
;5 5
v F1MF2 = 900 .
3.69 . Cho (E) : 4x2
+ 9y2
= 36a) Xc nh tiu im , di cc trc .b) Mt ng thng thay i d : y = x + m . nh m d ct (E) ti hai im
P, Q .c) Tm ta trung im I ca PQ . Chng t I di ng trn mt on cnh
khi d thay i .d) Gi P v Q ln lt l i xng ca P v Q qua gc O . T gic PQPQ l
hnh g ? nh m n l hnh thoi .
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3.70. Cho hai lip : x2 + 8y2 = 16 v 4x2 + 9y2 = 36 . Vit phng trnh ngtrn qua cc giao im ca hai lip .
3.71. Cho ng trn tm F1 ( - 2; 0) v bn knh 6 v im F2 (2 ; 0) . M l tmng trn di ng qua F2 v tip xc trong vi (F1) .Chng minh M thuc mt lip (E) . Vit phng trnh (E).
* 3.72.a) Vit phng trnh ca (E) bit n c mt tiu im l F(- 2 ; 0) vkhong cch t F n nh trn trc nh l 3 .
b) Hai ng thng d : mx y = 0 v d : x + my = 0 ln lt ct (E) ti M , Pv N, Q . T gic MNPQ l hnh g? Tnh din tch ca n theo m .
c) nh m MNPQ l hnh vung .
*3.73. Cho lip : 5x2 + 9y2 = 45 c tiu im F1 , F2 . M l im bt k trn (E) .
a) Chng minh chu vi tam gic F1MF2 khng i . Tm m din tch tam gicF1MF2 l 2 vdt.
b) Tim M sao cho : T =MF
1
MF
1MFMF
2121 +++ ln nht .
*3.74. Cho ng trn tm O , bn knh 2 . AB l ng knh trn Ox. Gi M, Nl hai im di ng trn tip tuyn ca (C) ti A v B , c tung l m, n luntha mn = 4.
a) Chng minh MN l tip tuyn ca ng trn (O).b) AN v BM ct nhau ti I. Chng minh I di ng trn mt elip (E).c) Gi H, K ln lt l trung im ca AM v BN .Chng minh ng trn
ng knh HK qua hai tiu im ca (E).
*3.75. Cho im M di ng trn lip : 9x2 + 16y2 = 144 . H, v K l hnh chiuca M ln hai trc . Tm M din tch OHMK ln nht .
*3.76. Cho M, N l hai im bt k trn lip : 4x2 + 9y2 = 36 v khng trng vicc nh .Gi I l trung im ca MN.
a) Chng minh tch h s gc ca ng thng MN v ng thng OI c gitr khng i .b) Vit phng trnh ng thng MN bit trung im I c ta (1 ; 1)
* 3. 77. Cho ng trn (O; a) v elip (E) : bx2 + ay2 = a2b2 .
a) Chng minh php co v trc hanh theo h s k =a
bbin (O) thnh (E).
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b) Gi T, M l hai im trn (O) ( MT ct Ox ) , php co trn bin ngthng MT thnh ng thng no . Chng minh hai ng thng ng qui .Khi M tin v T ( T cnh ) th MT , MT tin n v tr no . Suy ra cch vtip tuyn ca (E) ti mt im cho trc . Tm phng trnh tip tuyn bit tipim T c ta (x0 ; y0) .
c) Php co trn bin mt hnh vung n v c cc canh song song vi cc trchay nm trn hai trc thnh hnh g , c din tch bao nhiu . T hy suy ancng thc tnh din tch hnh lip.
3.78. Chn cu ng : Cho (E) : 6x2 + 9y2 = 54 . Khong cch t tiu im nnh trn trc nh l :
a) 6 b) 3 c) 15 d) 63.79 . Chn cu ng : Cho (E) : 4x2 + 5y2 = 20 . Khong cch gia hai tiuim l :
a) 1 b) 2 c) 3 d) 2 5
3.80. Chn cu ng : Cho (E) : 3x2 + 4y2 = 12. im M c honh l 1 thuc(E) . Th th F1M = ( F1 l tiu im bn tri )
a) 3/2 b)13
2c) 5/2 d)
3 5
2
3.81. Chn cu ng : Cho (E) : 4x2 + 9y2 = 36 . Tnh di dy cung vung gcvi Ox v qua tiu im F .
a) 3 b) 4/3 c) 5 d) 8/3
3.82. Chn cu ng : Tung giao im ca (E) :2
21
4
xy+ = vi ng trn
x2 + (y 1)2 = 1 gn nht vi s no di y ?
a) 0 , 86 b) 0 , 88 c) 0, 9 d) 0, 92
3.83. Chn cu ng : Elip c hnh
bn c tiu c l :
a) 4 b) 6
c) 2 11 d) 2 14
64
OA1
B1
F1
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3.84. Chn cu ng : Elip c hnh di bn tri c di trc nh gn ngvi s no di y ?
a) 48
3b) 8
8
3c) 2 96 d) p s khc
3.85. Chn cu ng : Elip c hnh trn bn phi c di trc ln l :a) 5/ 3 b) 8/3 b) 3 d) 10/3
D. Hng dn gii hay p s
3.65. a) Th y = vo phng trnh (E) b) Th y = 2x vo phng trnh (E) .
c) Ta (x ; y) ca P tha phng trnh (E) v OM2 = c2 x2 + y2 = 3
d) Gi(x ; y) l ta mt nh bt k ca hnh vung , ta c h :
: x2 + 4y2 = 4 v x2 = y2 .
3.66. a) a = 3 v2 2
9 21
2a b+ = => (E) :
2 2
19 4
x y+ = => c = 5
b) Th x = 5 : y = 4/ 3 => di dy cung l 8/ 3.
c) im (x ; y) cn tm tha h :
2 2
2 2
4 9 36
11
4
x y
x y
+ =
+ =
3.67. a) c = 2 . Phn bit cac trng hp :
O
M(2;2)
N(-1 ; - 3)
O
M(- 2;4)
B(0; - 5)
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55
(1) B2F2 =2 2
5b c a+ = = .(2) A2F2 = a c = 5 => a = 7
(3) A2F1 = a + c = 5 => a = 3
b) b = 2 , c = 2 .
c) c = 5 . Phn bit 2 trng hp :
(1) B1B2 = 2b = 9 b = 9/ 2
(2) A1A2 = 2a = 9 a = 9/2 < c : loi .
d) A1B1 =2 2 2 2
9 81a b a b+ = + = v a2
b2
= c2
= 253. 68. a) a = 4 v
2 2
9 41
a b+ =
b)2 2
2 2
8 11
9 9
4 51
9
a b
a b
+ =
+ =
c) c = 2 v2 2
1 41
5a b+ = . Th a2 = b2 + 4
d) OM2 = c2 =9 16
55 5
+ = . Gii nh bi .
3.69 . b) Th y = x + m : 4x2 + 9(x + m)2 = 36 13x2 + 18mx + 9m2 36 = 0 (1)
YCBT 0 m2 13 - 13 13m (*)
c )
1 2 9
2 13
4
13
x x mx
Im
y x m
+ = =
= + =
=> y = -9
4x vi -
9 9
13 13x do (*)
=> I di ng trn on thgng c phng trnh y = -9
4x vi
9 9
13 13x
d) Do i xng PQPQ l hnh bnh hnh . Gi (x1 ; y1) v (x2 ; y2) ln lt l ta ca P v Q , trong x1, 2 l nghim ca phng trnh (1) v y1,2 = x1, 2 + m .
YCBT
1 2 1 2 1 2 1 2. 0 ( )( ) 0OP OQ x x y y x x x m x m + = + + + =
2x1x2 + m(x1 + x2 ) + m2 = 0
Th x1 + x2 = - 18m/ 13 , x1x2 = (9m2 36) /13 ( nh l Viet ca phng trnh (1)
) , ta c phng trnh tnh m .
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3.70. Ta giao im l nghim ca h :
=+
=+
36y9x416y8x 22
22
2=
=
23
8y
23
144x
2
2
=> x2 + y2 =23
172: y l
phng trnh cn tm .
3.71. Gi r = MF2 l bn knh ng trn (M)
.Ta c : MF1 + MF2 = MF2 + r = 6 . Do Mthuc lip c 2a = 6 v 2c = 4 . Suy ra : b2 = a2
c2 = 9 4 = 5 Phng trnh (E) l : 15
y
9
x 22=+
3.72. a) c = 2 , a = 3 :2 2
19 5
x y+ =
b) Ta M, P :
2 2 2
2
53
5 9 45 9 5
53
9 5
xx y m
y mxy m
m
=
+ = +
= = +
Tng t , ta N, Q :
2 2 2
2
53
5 9 45 5 9
53
5 9
yx y m
x myx m
m
=
+ = +
= = +
T gic l hnh thoi v d v d vung gc .
Din tch hnh thoi MNPQ : 4. SOMN = 2 . OM. ON = 2 .
2 2 2 2. M M N N x y x y+ +
= 18(m2 + 1)2
2 2 2 2
5 5 90( 1).
9 5 5 9 (9 5)(5 9)
m
m m m m
+=
+ + + +
r
r
F1 F2
M
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c) YCBT OM = ON 9m2 + 5 = 5m2 + 9 m = 1
3.73.a)Chu vi l : 2a + 2c = 6 + 4 = 10 . Din tch tam gic l : .|yM| . 2c = 2 |yM| = 1 . Suy ra xM.
b) T = 2a +MF.MF
a2
21
m F1M.F2M = a2 -
2
22
a
xc= 9 - 2x
9
4( - 3 x 3)
Vy T ln nht F1M.F2M nh nht x2 = 3
3.74. a) Phng trnh MN : (n m)x + 4y + 2(m + n) = 0
Ta c : d(O; MN) = 2
mn2nm
|nm|2
16mn2nm
)nm(2|2222
=
++
+=
++
+( v mn = 4)
=> MN tip xc ng trn (O; 2) .b)Xem bi tp 3.57 .
c) Ta chng minh : 0KF.HF 2,12,1 =
3.75. Dng bt ng thc C si cho hai s
3.76. a) Ta c : 4xM2 + 9yM
2 = 36 (1) v 4xN2 + 9yN
2 = 36 (2) .Ly (1) (2) : 4(xM
2 xN2
) = - 9(yM2 yN
2 ) 4(xM xN) (xM + xN) = - 9(yM yN) (yM + yN)
9
4
xx
yy.
x
y
NM
NM
I
I =
kOI . kMN = - 4/9
b) H s gc ca OI l 1 , do kMN = - 4/9 . Vy phng trnh MN l :. . . . .3.77. b) Cc ng thng qua T ,M v vung gc vi Ox ct (E) lnlt ti T v M . ng thngTM co li thnh ng thngTM. Hai ng thng ny ngqui ti K Ox .
Khi M tin v T , ng thngTM bin thnh tip tuyn ca (O)
ti T , khi ng thng TMbin thnh tip tuyn ca (E) ti T. Hai tip tuyn ny ng qui ti Ivi IT vung gc bn knh OT.
T
MT M
IO K
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Nu (x0 ; y0) l ta ca T th (x0 ; oyb
a) l ta ca T. Phng trnh tip
tuyn ca ng trn ti T vung gc )yb
a;x(OT oo= l :
x0 (x x0) + 0)yb
ay(y
b
aoo =
b2 x0 x + yabyo = b2 x0
2 + a2 yo2 = a2 b2 (TI)
Thay y bng yb
av gi nguyn x , ta c phng trnh tip tuyn IT ca lip
ti T : b
2
x0 x + yabyo = a
2
b
2
1b
yy
a
xx2
o
2
o =+
c) Php co v Ox h s k , bin hnh vung n v c cnh song song hay nmtrn hai trc thanh hnh ch nht c cnh song song hay nm trn hai trc c dintch l kvdt .
Din tch hnh trn l a2 . Vi s chn on v di nh tng ng vivic lm trn s , hnh trn coi nh cha a2 hnh vung n v . Suy ra qua
php co , hnh lip coi nh cha a2 hnh ch nht c din tcha
bvdt . Do
hnh lip c din tch l : a2 . =a
bab .
3.78 (b) . FB = 2 2b c a+ = = 3
3.79 (b)F1F2 = 2c = 2
3.80 (b) . yM = 3 /2 => F1M =2
2 3(1 1)
2
+ + =
5/2
3.81 (d) . Th x = 5 = c : 9y2 = 36 20 = 16 y = 4/3Vy di dy cung l 8/ 3 .
3.82 (a). Th x2 = 1 (y 1)2 vo phng trnh (E) : 1 (y 1)2 + 4y2 = 4 3y2 + 2y 4 = 0
Phng trnh ny c 2 nghim :1 2
1 13 1 13;
3 3y y
+= =
V x2 = 1 - (y 1)2 0 (y 1)2 1 - 1 ( y 1)2 1 0 y 2
nn ch nhn y =13 1
3
0,868
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3.83 (d). BF =2 2
c b a+ = =5 , 2 2 6a b+ = b2 = 36
25 = 11.
Suy ra : c= 25 11 14 =
Vy tiu c l 2 14
3.84 (b). Ta c h :
2 2
2 2
4 41
1 91
a b
a b
+ =
+ =
Nhn phng trnh sau cho 4 ri tr vi phng trnh u , ta c :
2
32 323
3b
b= = .
di trc nh l 232
3= 8
8
3
3.85 (d) .Ta c h :
2
5
1 0 / 34 161
25
b
a
a
=
=> =+ =
di trc ln l : 20 / 3 .
64
OA1
B1
F1
O
M(- 2;4)
B(0; - 5)
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* 6. HypebolA. Tm tt gio khoa1.nh ngha .Cho hai im cnh F1 , F2 vi 1 2 2F F c= v mt di khngi 2a ( a > c) . Hypebol l tp hp nhng im M sao cho :
1 22F M F M a =
F1 , F2 : tiu im , F1 F2 : tiu c .2. Phng trnh chnh tc :
Vi F1( - c ; 0) , F2(c ; 0) :
M(x ; y) (H)2 2
2 21
x y
a b
= vib2 = c2 - a 2 ( 1)
(1) : phng trnh chnh tc ca hypebol .
3. Hnh dng ca hypebol .-* A1 ( - a ; 0 ) , A2 ( a ; 0 ) : nh .* Ox : trc thc , di 2a . Oy : trc o , di 2b .* Hypebol gm 2 nhnh : nhnh tri gm nhng im c x - a, nhnh
phi gm nhng im c x a .* Hnh ch nht gii hn bi cc ng x = a , y = b gi l hnh ch
nht csca hypebol.
* ng thng y = b xa
gi l hai
tim cn .
* Tm sai : e = 1a
c>
* F1M =
+
=+
trinhnhM,axa
c
phainhnhM,axa
c
aex
M
M
M
F2M =
+
=
trinhnhM,axa
c
phainhnhM,axa
c
aex
M
M
M
F1 F2A1 A2
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B. Gii ton .Dng ton 1 : Xc nh cc yu t ca hypebol
V d : Hy xc nh nh , di cc trc , tiu c , tiu im , tim cn , tmsai v v hypebol c phng trnh sau :
a) (H) :2 2
14 2
x y = . b) (H): 16x2 9y2 = 144
Gii : a)Ta c : a2 = 4 , b2 = 2 => a = 2 v b = 2 Suy ra nh A1 (- 2; 0 ) , A2 (2 ; 0 ) . di trc thc 2a = 4 , trc o 2b = 2 2 .
Ta c : c = 2 2 6a b+ = . Tiu c 2c = 2 6 , tiu im F1( - 6 ; 0 ) ,
F2( 6 ; 0 ) .
Tim cn : y =2
2
bx x
a = . Tm sai e = c/a = 6 /2
b)Vit li phng trnh (H) :2 2
19 16
x y = => a2 = 9 ; b2 = 16
=> a = 3 , b = 4 v c = 2 2 5a b+ = Suy ra A1 (- 3; 0 ) , A2 (3 ; 0 ) .
di trc thc 2a = 6 , trc o 2b = 8 .Tiu c 2c = 10 , tiu im F1( - 5 ; 0 ) , F2(5; 0 ) .
Tim cn : y = 4
3
bx x
a= . Tm sai e = c/a = 5/3
Dng ton 2 : Lp phng trnh chnh tc ca hypebol
F1 A1 A2 F2 F1 A1 A2 F2
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T gi thit , lp h phng trnh theo a v b . Gii h , tm c a , b . Suy raphng trnh (H) .
Cn nh: M(x0 ; y0) (H) 1b
y
a
x2
2o
2
2o =
V d 1 : Lp phng trnh ca hypebol (H) bit :a) (H) c di trc thc l 6 , tiu im l ( 4; 0 )
b) (H) c mt nh l ( 5 ; 0 ) v tim cn l y = 2x .c) (H) c mt tim cn l y = - 2 x v qua im M( 4 ; 2 ) .
d) (H) qua hai im ( 1 ; 3 ) v (- 2 ; 2 2 ) .
e) (H) c tiu im F2 ( 3 ; 0 ) v qua im ( 3;
4
5 )
Gii a) 2 a = 6 = > a = 3 , c = 4 = > b 2 = c2 a2 = 16 9 = 7 .Phng trnh hypebol l :
2 2
19 7
x y =
b) Phng trnh (H) :2 2
2 21
x y
a b =
nh (5 ; 0 ) do a = 5.
Tiu cn y = 2x => ba = 2 b =10 .
Vy phng trnh (H) l :2 2
125 100
x y =
c) Phng trnh (H) :2 2
2 21
x y
a b =
Tim cn y = - 2 x => 2b
a= b2 = 2a2 (1)
M(4 ; 2 ) thuc (H)2 2
16 21
a b = (2)
Th (1) vo (2) : 22
151 15a
a= = . Suy ra b2 = 30 .
Vy phng trnh (H) :2 2
115 30
x y = = 1
d) Phng trnh (H) :2 2
2 21
x y
a b =
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( 1 ; 3 ) (H) 2 21 3
1a b = (1)
N(- 2 ; 2 2 ) (H)2 2
2 81(2)
a b = )
Gii h (1) v (2) vi hai n l : u =2 2
1 1, v
a b= , ta c : u = 5/2 , v = 1/ 2 .
Vy phng trnh (H) :2 2
15 / 2 2
x y =
e) F2( 3 ; 0 ) => c = 3 . Suy ra : F1 ( - 3 ; 0 ) .
c = 3 = > a2 = 9 b2 . Phng trnh hypebol :2 2
2 2
1x y
a b =
Th ta ca M , ta c :2 2 2 2
2 2
9 161 45 16(9 ) (9 )5
9 5b b b b
b b = =
45b2 144 + 16b2 = 45b2 5b4 5b4 + 16b2 144 = 0
Gii phng trnh trng phng ny , ta c : b2 = 4. Suy ra a2 = 5 .
Vy phng trnh (H) :2 2
15 4
x y =
V d 2 : Cho ng trn (M) di ng lun chn trnhai trc ta hai dy cung c di l 6 v 4 .Chng minh tm ng trn di ng trn mthypebol cnh .
GiiGi M(x ; y) l tm cc ng trn (M) . K MH ,
MK vung gc Ox v Oy , ta c : HA = HB = 3 , KC= KD = 2Suy ra : MB2 = MD2= r
2MH2 + HB2 = MK2 + KD2 y2 + 9 = x2 + 4
x2 y2 = 5 =5
y
5
x 221
Chng t M (H) : =5
y
5
x 221 .
Dng ton 3 : Tm im trn hypebol
rr
x
y
M
D
C
A BH
K
O
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Cn nh: * M(x0 ; y0) (H) 1byax 2
2
o2
2
o =
| F1M + F2M| = 2a .
* F1M = |a
cxM + a | ; F2M = | aa
cx M |
V d 1 : Cho hypebol (H) :2 2
19 3
x y =
a) Tm trn (E ) im M c tung l 3 .b) Tm trn (H) im M sao cho gc F1MF2 = 900 .c) Tm trn (H) im M sao cho F1M = 2F2M
Gii a) Th y = 3 vo phng trnh ca (H) :2 2
2( 3) 41 9. 2 3
9 3 3
xx x = = =
Ta tm c 2 im M c ta (2 3 ; 3 ) , ( - 2 3 ; 3 ) .
b) Gi (x; y) l ta ca M . Ta c : F1MF2 = 900 OM = OF1 = OF2
2 2 2 2 12 x y c x y+ = + = ( c2 = a2 + b2 = 9 + 3 = 12 )
Mt khc v M (H) nn ta E tha : 3x2 - 9y2 = 27
Ta c h :
22 2
2 22
45
3 9 27 4
312
4
xx y
x yy
= =
+ = =
3 5
2
3
2
x
y
=
=
Ta tm c 4 im c ta (3 5
2;
3
2) , (
3 5
2; -
3
2), (-
3 5
2;
3
2) ,
( -3 5
2
; -3
2
)
c) V F1M = 2F2M => F1M > F2M => M thuc nhnh phi vF1M F2M = 2a = 6
Suy ra F2M = 6 v F1M = 12 .
M F1M = =+ axa
cM 123x3
32M =+ x =
9 3
2
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Th vo phng trnh (H) , ta suy ra : y = 692 . Ta im cn tm :
(9 3 69
; )2 2
.
V d 2 : a) Cho hypebol (H) :2 2
2 21
x y
a b = c tiu im F1 , F2.
M l im bt k trn (H) .a) Chng minh tch khong cch t M n hai tim cn c gi tr khng i
b) Cho hypebol (H) :
2 2
11 2
x y
= . Mt ng thng d bt k : y = x + m ct(H) ti M, N v hai tim cn ti P v Q . Chng minh MP = NQ .
Giia) Phng trnh hai tim cn : 1 : bx + ay = 0 v 2 : bx ay = 0 . Gi (x; y) lta ca M , ta c :
d(M; 1) =2 2
bx ay
a b
+
+, d(M, 2) =
2 2
bx ay
a b
+
d(M,1).d(M,2) =2 2 2 2
2 22 2 2 2
.b x a ybx ay bx ay
a ba b a b
+ =
++ +
V M(x; y) thuc (H) :2 2
2 2 2 2 2 2
2 21
x yb x a y a b
a b = = suy ra :
d(M,1).d(M,2) =2 2 2 2
2 2 2
a b a b
a b c=
+: gi tr khng i .
M
M
P
N
Q
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b) (H) : 2x2 y2 = 2 .Phng trnh honh giao im M, N : 2x2 (x + m)2 = 2 ( th y = x + m vo
phng trnh ca (H) ) x2 2mx m2 2 = 0 (1)Phng trnh hai tim cn : ( 2 x + y)( 2x y) = 0 2x2 y2 = 0Phng trnh honh giao im P, Q : 2x2 (x + m)2 = 0 ( th y = x + m vo
phng trnh hai tim cn ) x2 2mx m2 = 0 (2)
Nu (1) c hai nghim x1, x2 , th th honh trung im ca MN l : (xM + xN ) = . 2m = m ( nh l Viet ca (1))
Nu (2) c hai nghim x3, x4 , th th honh trung im ca PQ l :
(xP + xQ ) = . 2m ( nh l Viet ca (2) )Chng t MN v PQ c cng trung im hay MP = NQ.
Ghi ch : Tnh cht ny ng vi mi hypebol
C. Bi tp rn luyn .
3.86 . Xc nh di cc trc , ta nh , tiu im , tim cn v v cc
hypebol sau :
a)
2 2
14 5
x y
= b)
2 2
14 4
x y
= c) 4x2 - 9y2 = 36
3.87 . Cho hypebol (H) :2
21
4
yx = .
Tm trn (H) :
a) im M c honh 2 . b) im N cch u hai trc ta .
c) im P sao cho gc F1PF2 = 900 .
d) ta cc nh ca hnh ch nht ni tip (H) bit hnh ch nht c cc
cnh song song vi cc trc ta v c din tch l 8 2 vdt.
e) im Q sao cho F2Q = 2F1Q .
3.88. Cho hypebol (H) c di trc thc l 4 v qua im M ( )5 ; 2
a) Lp phng trnh (H) .
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b) Tnh di dy cung ca (H) vung gc vi trc thc ti tiu im .
c) Tm giao im ca (H) v ng trn ng knh F1F2 , F1 , F2 l cc
tiu im ca (H) .
3.89. Lp phng trnh (H) bit :
a) tiu c 8 v khong cch t nh trn trc thc n tiu im l 1 .
b) di trc o l 4 v mt tiu im l ( 3 ; 0 )
c) mt tiu im l F2 ( 5 ; 0 ) v mt tim cn l y = 2x .d) mt tim cn l y = 3 x v qua im ( 3 ; 15 )
e) mt tiu im l ( 2 ; 0) v qua im (3 ; 2 ) .3.90. Lp phng trnh chnh tc ca hypebol (H) bit :
a) di trc thc l 6 v qua im ( 10 ; 2) .
b) qua hai im P ( ) 510 ; 2 , ;12
Q
.
c) c tiu c l 4 2 v qua im ( 3 ; 5 )
3.91. Lp phng trnh chnh tc ca hypebol (H) bit :
a) qua im M ( )3 ; 1 v F1MF2 = 900 b) mt tiu im (2 ; 0 ) v khong cch t n n tim cn l 1.
c) tiu im l( 3 ; 0) v dy cung qua tiu im v vung gc Ox c di l 5 .d) mt tim cn c h s gc 2/ 5 v khang cch t tiu im n timcn l 2 .
3.92 Cho ng trn tm I( - 6; 0) , bn knh 4 v im J(6 ; 0 ) .(M) l ng trn di ng lun qua J v tip xc vi (I) . Chng minh tphptm M cc ng trn M l mt hypebol . Vit phng trnh hypebol .
3.93 . Cho (H) : 9x2 - 4y2 = 36a) Xc nh tiu im , di cc trc v tim cn . V (H) .
b) M ty ca (H) , chng minh rng : (F1M+ F2M)
2 4OM2 l mt hng s. c) Mt ng thng thay i d : x + y + m = 0 . Chng minh d lun ct (H)ti hai im phn bit P, Q . Tnh di on PQ theo m .
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3. 94. a) Vit phng trnh ca (H) bit n c mt nh l (1 ; 0) v mt tiuim l ( 5,0) .
b) nh m hai ng thng d : mx y = 0 v d : x + my = 0 u ct (H). c) Gi M , P v N, Q ln lt l giao im ca d v d vi (H) . T gicMNPQ l hnh g ? Tnh din tch ca n khi m = 2 .
3.95. Cho (H) : 5x2 4y-2 = 20 v ng thng d : 2x y + m = 0a) nh m d ct (H) ti 2 im M, N phn bit .
b) Tm tp h p trungim ca MNc) Gi P, Q ln lt l i xng ca M, N qua O . nh m MNPQ l hnh
thoi.
3.96. Cho (H) : x2 3y2 = 12
a) Tm cc nh, tiu im , tim cn .
b) Tm trn (H) im M sao cho gc F1MF2 = 1200 .
c) Tm M (H) sao cho : T = F1M F2M +MF
1
MF
1
12
ln nht
d) Cho M bt k (H) , tnh tch cc khang cch t M n hai tim cn .
3.97. Cho lip (E) v hypebol (H) bit chng c cng tiu im F(2 ; 0) , tim cnca (H) cha ng cho ca hnh ch nht csca (E) v hp vi Ox mt gc300 .
a) Vitphng trnh chnh tc ca (E) v (H) .
b) Vit phng trnh ng trn qua cc giao im ca (E) v (H) .
3. 98 .Cho hai im A1 ( 2; 0) v A2( 2 ; 0 ) . Gi (I) l ng trn di ng qua
A1 , A2 v MM l ng knh ca (I) cng phng vi Ox . Chng minh tp hp
nhng im M, M l mt hypebol .
3.99. Cho ng trn tm O , bn knh 1 . Gi A v A l hai im trn ngtrn c honh l 1, 1 . ng thng di ng x = m ( 0, 1m ) ct ng
trn ti M v M ( M c tung dng) .a) Tm ta M v M .
b) Vit phng trnh ng thng AM v AM . Chng minh giao imca AM v AM di ng trn mt hypebol cnh.
3. 100. Chn cu ng :Cho (H) : 6x2 - 9y2 = 54 . Phng trnh mt tim cn l :
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a) y = 63
x b) y = 36
x c) y = 69
x d) y = 96
x
3.101 . Chn cu ng :Cho (H) : 4x2 - 5y2 = 20 . Khong cch gia hai tiu im l :
a) 1 b) 2 c) 3 d) 6
3. 102. Chn cu ng :Cho (H) : 3x2 - y2 = 3. im M c tung l 3 thuc (H) . Th th
F1M = ( F1 l tiu im bn tri )a) 3 b) 4 c) 5 d) p s khc
3.103. Chn cu ng :Cho (H) : 4x2
- 9y2
= 36 . Tnh khong cch t tiu imn mt tim cn l :
a) 2 b