Hinh Giai Tich Trong MP

8/14/2019 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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Phng php ta trong mt phng

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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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3

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

I


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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

I


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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

73/4


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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

15

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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16

. . .

*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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17

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


18/101


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


19/101


19

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


20/101


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.


21/101


21

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


22/101


22

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+ + = =


23/101


23

*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


24/101


24

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 .


25/101


25

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


26/101


26

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


27/101


27

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


28/101


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


29/101


29

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

O

I

T

R


30/101


30

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

O

I


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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

O

I

a

b


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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 :


37/101


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=++

=+


38/101


38

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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39

* 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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44

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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45

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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46

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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47

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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48

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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49

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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50

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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51

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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52

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).


53/101


53

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


54/101


54

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)


55/101


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 .


56/101


56

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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57

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


58/101


58

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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60

* 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 .


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 .


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

Hinh Giai Tich Trong MP

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