Bai Tap Hinh Hoc Giai Tich

8/10/2019 Bai Tap Hinh Hoc Giai Tich

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Trng HSP Tp.H Ch Minh

1

MC LC

NG BC HAICh 1

I MC TIU......................................................................................... 3Ch 2LP PHNG TRNH NG BC HAI............................................. 5Ch 3PHNG TIM CN - NG TIM CN....................................... 13Ch 4GIAOTUYN CA NG BC HAI VI NG THNG......... 19Ch 5PHNG TRNH TIP TUYN CA NG BC HAI................. 21Ch 6NG KNH LIN HP VI MT PHNG................................ 26

Ch 7A PHNG TRNH BC HAI V DNG CHNH TC............... 29

MT BC HAICh 1PARABOLOIT HYPERBOLIC.............................................................. 35Ch 2ELIPXOIT................................................................................................ 39Ch 3HYPEBOLOIT......................................................................................... 43

Ch 4NG SINH THNG........................................................................... 48Ch 5MT S BI TP TNG HP.............................................................. 51

Chuyn QU TCH.............................................................................................. 55

Chuyn CHNG MINH CC NG THC HNH HC............ 62

Chuyn TH..................................................................................................... 70


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2

NG BC HAI


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3

Ch 1I MC TIU

Phng php

S dng cc cng thc i mc tiu hc:

1. Php tnh tin theo vc t OI

: ' 'TOIOxy Ix y

1 1

2 2

' '

' '

o

o

x x a x b y

y y a x b y

2. Php quay tm O mt gc : ( ; ) ' 'Q OOxy Ox y

'cos 'sin

'sin 'cos

x x y

y x y

Bi mu 1: Cho hnh bnh hnh ABCDHy vit cng thc i mc tiu

(A; AB, AC )sang mc tiu

(C;CB,CD).

Gii

Ta c : C(1;1)AC AD AB

0. ( 1;0)CB DA AD AB CB

0. (0; 1)CD BA AD AB CD

Vy ta c cng thc i trc l:1 ' 0. ' 1 '

1 0 ' ' 1 '

x x y x

y x y y

Nhn xt : - i mc tiu trong Afin hay trong trc chun khng kh,nhng trnh sai xt chng ta cn nhn nh ng yu cuca bi.

- bi ny ta vn dung tnh cht bng nhau ca cc cp cnhi ca hnh bnh hnh gii quyt bi ton.

Bi mu 2 : Cho hai h to trc chun xOy v xOy. i vi h xOy,ng thng Ox v Oy ln lt c phng 2x + y - 1 = 0 vx - 2y +4 = 0. Vit cng thc i to t mc tiu xOy sangmc tiu xOy.

Gii

i vi h to Oxy, im Ol nghim ca h phng trnh :


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4

2 1 0 2 9 ' ( ; )

5 52 4 0

x yO

x y

ng thng Ox c vc t ch phng l ( 1; 2)u

5u

Gi 'i

l vc t n v cng phng vi u

th ta c :1 2' ( ; )

5 5i

hoc1 2

' ( ; )5 5

i

ng thng Oy c vc t ch phng l ' (2;1)u

' 5u

Gi 'j

l vc t n v cng phng vi 'u

th ta c :2 1

' ( ; )5 5

j

hoc2 1

' ( ; )5 5

i

Vy ta c cng thc i trc l :2 1 2

' '

5 5 59 2 1' '

5 5 5

x x y

y x y

hoc

2 1 2' '

5 5 59 2 1' '

5 5 5

x x y

y x y

hoc

2 1 2' '

5 5 59 2 1

' '5 5 5

x x y

y x y

hoc

2 1 2' '

5 5 59 2 1

' '5 5 5

x x y

y x y

Nhn xt : - Vic suy ra c 'i

v 'j

l do ta p dng tnh cht ca vc t

u = u . 'i v 'u = 'u . 'j .

- Bi ton ny c th c 4 cng thc i trc.

BBiittppttnnggtt

Bi 1 :Trong h trc chun Oxy cho O= (-4; 2); A = (2; 0) v B = (0; 8)Hy vit cng thc i trc to t mc tiu (O';A,B) sang mc tiu

(O';OA,OB )

p s:

x = -4 + 2x'

y = 2 + 8y'

Bi 2: Trong h trc Oxy, cho tam gic ABCHy vit cng thc i mc tiu

(A; AB, AC )sang mc tiu

(B;BC,BA) .

p s:

x = 1 - x' - y'

y = x'


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Ch 2:LP PHNG TRNH NG BC HAI

Dng 1: Vit phng trnh ng bc hai bit trc hai tim cn

Phng php:

(C) nhn 1 1 1 1

2 2 2 2

(d ) : a x b y c 0(d ) : a x b y c 0

lm hai ng tim cn

Nn (C) c dng: 1 1 1 2 2 2(a x b y c )(a x b y c ) k 0 (*)Da vo iu kin bi tm k (C)

Bi 1: Mt ng cong bc hai i qua im ( 1; 1) v tha nhn cc ng

2x 3y 5 0 v 5x 3y 8 0 lm tim cn. Lp phng trnh ng cong .

Li gii:

(C) nhn cc ng 2x 3y 5 0 v 5x 3y 8 0 lm tim cnnn (C) c dng:

(2x 3y 5)(5x 3y 8) k 0 ( 1; 1) (C) (2 3 5)(5 3 8) k 0 k 36

Vy 2 2(C) :10x 21xy 9y 41x 39y 4 0

Nhn xt: - gii quyt bi ton ny chng ta ch cn vn dng phng trnh(*), sau cho qua im (1, -1) l xong.

Bi mu 2: Lp phng trnh ng cong tip xc vi ng thng4x y 5 0 v tha nhn cc ng thng x 1 0 v 2x y 1 0 lm timcn.

Li Gii:

(C) nhn cc ng x 1 0 v 2x y 1 0 lm tim cn nn (C) c dng:

(x 1)(2x y 1) k 0 Gii h phng trnh:y 4x 54x y 5 0 y 4x 5

2(x 1)(2x y 1) k 0 (x 1)(6x 6) k 0 6x 6 k 0 (1)

(C) tip xc vi ng thng 4x y 5 0 (1) c nghim kp: x 0 k 6

Vy 2(C) : 2x xy x y 5 0


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Bi mu1: ng cong bc hai i qua ccim (0;0); (0; 2); ( 2; 4) v ch ct mi

ng thng sau: 3x 2y 1 0 v 2x y 5 0 ti mt im. Lp phng trnh

ng cong .

Li Gii:2 2(C) :ax 2bxy cy 2dx 2ey f 0,(a,b,c) (0,0,0)

(C) qua im (0;0);(0; 2) nn f 0 v 4c 4e 0 Vy (C) c dng: 2 2ax 2bxy cy 2dx 2cy 0

Thay1

y (3x 1)2

vo (C) ta c:

2 2 21 9 3 3ax bx(3x 1) c(3x 1) 2dx c(3x 1) 0 (a 3b c)x (b c 2d)x c 04 4 2 4

Thay y 5 2x vo (C) ta c:2 2 2ax 2bx(5 2x) c(5 2x) 2dx 2c(5 2x) 0 (a 4b 4c)x (10b 16c 2d)x 15c 0

(C) qua im ( 2; 4) v ct 3x 2y 1 0 v 2x y 5 0 ti mt im nn ta c h

phng trnh:a 4b 2c d 04a 12b 9c 0a 4b 4c 0

(1)

v3

b c 2d 02

10b 16c 2d 0(2)

Lp ma trn cc h s m rng:

3 3 22 2 1

3 3 1

1 4 2 1 0 1 4 2 1 0 1 4 2 1 0d d 2dd d 4d

A 4 12 9 0 0 0 4 1 4 0 0 4 1 4 0d d d

1 4 4 0 0 0 8 2 1 0 0 0 0 7 0

a 3c

1b c

4d 0

(1)

.Chn1

c 2 a 6,b ,d 02

tha (2)

Vy 2 2(C) : 6x xy 2y 4 y 0

Nhn xt: - Bi ton ny chng ta c tt c 5 d kin, do chng ta s thit lpcc phng trnh theo mt tham s khc khng do bi ton cgii quyt.

Bi mu 2:Mt ng cong bc hai ch ct mi trc ta ti gc O. Ngoi rabit n i qua hai im ( 2; 1);( 2; 2) . Lp phng trnh ng cong .

Li Gii:

2 2(C) : ax 2bxy cy 2dx 2ey f 0,(a,b,c) (0,0,0)


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a 1 6db 1 24dc 1 4d

(1)

.Chn

1d 12 a 2,b ,c 3

2 .Vy

2 2(C):2x xy 3y x 6y 15 0

Nhn xt: - Vi iu kin bi ton c tm th bi ton ch cn 4 n s. Do vi 3d kin cn li ta c th tm ba tham s theo mt tham s ( li gii trn ltm theo tham s d). Khi bi ton c gii quyt.

Bi mu 2:Mt ng cong bc hai i qua cc im (0;0);(0; 1);( 1;0) .Ngoi ra bit tm ca n l ( 2; 3) . Lp phng trnh ca ng cong .

Li Gii:(C) c tm ( 2; 3) c dng: 2 2a(x 2) 2b(x 2)(y 3) c(y 3) d 0,(a,b,c) (0,0,0)

2 2(C) :ax 2bxy cy (4a 6b)x (4b 6c)y 4a 12b 9c d 0

(C) qua (0;0);(0; 1);( 1;0) nn ta c h phng trnh:a 6b 9c d 0

4a 12b 9c d 04a 8b 4c d 0

(1)

2 2 13 3 2

3 3 1

11 6 9 1 0 1 6 9 1 0 1 6 9 1 0d (d 4d ) d d 4d3A 4 12 9 1 0 0 4 9 1 0 0 4 9 1 0d d 4d

4 8 4 1 0 0 16 32 3 0 0 0 4 1 0

a 5 8d

b 5 16dc 1 4d(1)

.Chn5

d 8 a 5,b ,c 22

.Vy 2 2(C) : 5x 5xy 2y 5x 2y 0

Nhn xt: - Bi ton ny tng t bi ton trn l (C) c tm v ngoi racn c thm 3 d kin.

Dng 4: ng bc hai qua cc im cho trc

Phng php:2 2(C) :ax 2bxy cy 2dx 2ey f 0,(a,b,c) (0,0,0)

Da vo iu kin bi thit lp mi lin h gia cc h s , t tm c phngtrnh ng bc hai (C).


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Bi mu1:Lp phng trnh ng cong bc hai i qua 5 im:(0;0); (0; 2); ( 1;0); ( 2; 1); ( 1; 3).

Li Gii:2 2(C) :ax 2bxy cy 2dx 2ey f 0,(a,b,c) (0,0,0)

(C) qua im (0;0);(0; 2);( 1;0) nn f 0;4c 4e 0 v a 2d 0 Vy (C) c dng: 2 2ax 2bxy cy ax 2cy 0

(C) qua im ( 2; 1);( 1; 3) nn ta c h phng trnh:2a 4b c 0 a 3b

6b 3c 0 c 2b

2 2(C) :3bx 2bxy 2by 3bx 4by 0

Vy 2 2(C) : 3x 2xy 2y 3x 4y 0

Nhn xt: - Bi ton ny c 5 gi thit, do ta thit lp 5 n theo n cn li lbi ton c gii quyt.

- Li gii trn trnh by mt cch gii l rt gn dn cc h s lmcho cch gii n gin hn.

Bi mu 2:Lp phng trnh Parabol i qua 4 im:(0; 15); ( 3; 0); ( 5; 0); ( 2; 3).

Li Gii:(P) c dng: 2 2(C) :ax 2bxy cy 2dx 2ey f 0,(a,b,c) (0,0,0)

(P) qua ( 3;0); ( 5;0) nn ta c h phng trnh:9a 6d f 0 9a 6d f 0 f 15a

25a 10d f 0 16a 4d 0 d 4a

2 2(P) : ax 2bxy cy 8ax 2ey 15a 0 (P) qua (0; 15);( 2; 3) nn ta c h phng trnh:

3a 45c 6e 0 3a 45c 6e 0 3a 45c e 03a 12b 9c 6e 0 12b 36c 0 b 3c

(1)

(P) khng c tm nn h pt:ax by 4a 0bx cy e 0

v nghima b 4a

(2)b c e

(1) (2) a 9cb 3c

e 72c

2 2(P) :9cx 6cxy cy 72cx 144cy 135c 0

Vy (P): 2 29x 6xy y 72x 144y 135 0

Nhn xt: - Bi ton ny yu cu l tm Parabol, do chng ta phi ch ti mtiu kin cho n, l (P) khng c tm. Bi ton c gii quytnh trn.


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Dng 5: ng bc hai tip xc vi cc ng thng ti cc im

Bi mu 1:Lp phng trnh Parabol tip xc vi trc Ox ti ( 3;0) vtrc Oy ti (0; 5).

Li Gii:(P) c dng: 2 2ax 2bxy cy 2dx 2ey f 0,(a,b,c) (0,0,0)

Thay y 0 vo (P) ta c: 2ax 2dx f 0 (1)

Thay x 0 vo (P) ta c: 2cy 2ey f 0 (2) (P) tip xc vi Ox ti( 3;0) (1) c nghim kp:

2 d 3adx 3 v d af 0f 9aa

(P) tip xc vi Oy ti(0; 5) (2) c nghim kp:

2e 5c

ey 5 v e cf 0 f 25cc

a 1 9f c 1 25f

d 1 3f e 1 5f

.Chn f 225 a 25,c 9,d 75,e 45

2 2(P) : 25x 2bxy 9y 150x 90y 225 0 (P) khng c tm nn h phng trnh:

25x by 75 0

bx 9y 45 0

v nghim

25 b 75b 15

b 9 45

Vy 2 2(P) : 25x 30xy 9y 150x 90y 225 0

Nhn xt: - Cc d kin ca bi ton r: khng tm (v l Parabol), Qua 2 imv tip xc vi hai ng thng.- Vic tnh ton hi phc tp do phi cn thn v cc bi ton dngny thng l s hi ln.

Bi mu 2:Lp phng trnh ng cong bc hai qua gc ta tip xc ving thng 1) :( 4x 3y 2 0 ti ( 1; 2) v vi ng thng 2 ) :( x y 1 0 ti (0; 1) .

Li Gii:(C) i qua gc ta c dng: 2 2(C) :ax 2bxy cy 2dx 2ey 0,(a,b,c) (0,0,0)

Phng trnh tip tuyn ca (C) ti( 1; 2) l:

1 1(d ): ax b(y 2x) 2cy d(1 x) e(y 2) 0 (d ) : (a 2b d)x (b 2c e)y d 2e 0 Phng trnh tip tuyn ca (C) ti(0; 1) l:

2 2(d ) : bx cy dx e(y 1) 0 (d ) : (d b)x (e c)y e 0


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Theo bi ta c:a b e b 3e b e

4 3 2a 2b d b 2c e d 2e(d ) ( )1 1 c 2e4 3 2(d ) ( )2 2 d b c e e d b e

2 2

a 12e

b 3e

c 2e

d 2e

(C) : 12ex 6exy 2ey 4ex 2ey 0

Vy 2 2(C) :6x 3xy y 2x y 0 Nhn xt: - Cc d kin ca bi ton ny d dng thy r, tuy nhin vic tnh ton

hi phc tp. Chng ta s dng k thut nh trn s n gin ho biton i rt nhiu.

BBiittppttnnggtt

Bi 1 :Cho ng cong 2 2(C) : 2 6xy 5y 2x 2y 10 0x . Tnh tin h trc titm. Vit phng trnh ng cong trong h mi.

p s: 2 22X 6XY 5Y 11 0 Bi 2: Tm a,b 2 2(C) :2x 6xy ay 3x by 4 0 biu din:

a. Mt ng cong c tm.b. Mt ng cong thuc loi Parabol.

c.

Mt ng cong c v s tm.

p s: a.9a b2

b. 9 9a b2 2

c. 9 9a b2 2

Bi 3:Vit phng trnh Parabol i qua 2 im (0;0);(0; 1) bit rng trc ca n songsong vi ng thng x y 0

p s: 2 2(P) : x 2xy y kx y 0, k 1


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Ch 3PHNG TIM CN - NG TIM CN

A-I VI NG BC HAI: (C):ax2+2bxy+cy2+2dx+2ey+f=0

Dng 1: Tm - Phng tim cn - ng tim cn

Phng php:

ng bc hai (C) c dng: F(x,y)= ax2+2bxy+cy2+2dx+2ey+f =0

Tm I(x0; y0) l nghim ca h phng trnh:o o

o o

F' (x , y ) = 0xF' (x , y ) = 0y

Phng tim cn v= ),( (0,0) tho h phng trnh:02 22 cba

=> Ta tm c , Nu tm khng thuc (C) th ng tim cn l ng thng qua tm v c vc t chphng l phng tim cn.

Bi mu 1:Tm tm ca cc ng cong:2 2

2 2

2 2

a)(C): x - 4xy+4y +10x- 20y+25=0

b)(C): 9x - 6xy+y +2x- 7=0

c)(C): x +6xy+9y +4x+12y- 5=0

Li Gii .

2 2a) ( C): x - 4xy+4y +10x- 20y+25=0

Tm I(x,y) l nghim ca h phng trnh:2 5 0 2 4 10 0

2 4 10 0 2 4 10 0

x y x y

x y x y

H c v s nghim suy ra ( C) c v s tm.2 2b) ( C): 9x - 6xy+y +2x- 7 =0.

Tm I(x,y) l nghim ca h phng trnh:9 3 1 0 9 3 1 0

3 0 9 3 0

x y x y

x y x y

H v nghim suy ra (C ) khng g tm:2 2c) ( C): x +6xy+9y +4x+12y- 5 =0.

Tm I(x,y) l nghim ca h phng trnh:3 2 0

3 9 6 0

x y

x y


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H c v nghim suy ra (C ) c v s tm.

Bi mu 2: Tm phng tim cn ca cc ng bc hai sau:

a) 2 23x + 2xy - y + 8x + 10y + 14 = 0

b) (C ):2 2

3x + 10xy +7y + 4x + 2y + 1 = 0

a) 2 23x +2xy- y +8x+10y+14 =0Gi ( , )v

l phng tim cn . Ta gii phg trnh:

2 2

1

2

3 2 0

(1,3)3

( 1,1)

v

v

Vy c 2 phng tim cn l:

1

2

v =( 1,3)

v =( -1,1)

b) (C ): 2 23x +10xy+7y +4x+2y+1=0 Gi ( , )v

l phng tim cn . Ta gii phg trnh:

2 2

1

2

3 10 7 0

7( 7, 3)

3( 1,1)

v

v

Vy (C ) c hai phng tim cn l :

1

2

v =( -7,3)

v =( -1,1)

Bi mu 3: Tm tm - Phng tim cn - ng tim cn ca ccng bc hai sau:

a)9x2-2xy+6y2-16x-8y-2=0b)8x2+6xy-26x-12y+11=0c)x2-2xy+y2-10x-6y+25=0

Li Gii

a/ Tm I(x; y) l nghim ca h phng trnh: )53

28,

53

44(

046

089

I

yx

yx

Phng tim cn v

= ( , ) . Ta gii pt:2 2

2 2 2

9 2 6 0

( ) 8 5 0

0

Vy khng c phng tim cn.

Vy khng c ng tim cn.

b) Tm I(x,y) l nghim ca h pt:


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8 3 13 0(2, 1) ( )

2 0

x yI C

x

Phng tim cn v

= ( , ) . Xt pt:

2

1

2

8 6 0

2 (4 3 ) 0

0(0,1)

3( 3, 4)

4

v

v

Vy ta s c hai ng tim cn c pt l :

1

2

2( )

1

2 3( )

1 4

xd t R

y t

x td t R

y t

c) Tm I(x; y) l nghim ca h phng trnh:5 0

3 0

x y

x y

H v nghim .suy ra ( C) khng c tm.

Phng tim cn v

=( , ) . ta xt pt:

2 2

2

2 0

( ) 0

Vy ch c mt phng tim cn: v

=(1,1)

V ( C) khng c tm nn khng c ng tim cn.

Nhn xt: -ng bc hai ( C) c th c duy nht mt tm hoc v s tm, cngc th khng c tm .-Khi ng bc hai khng c tm hoc khng c phng tim cn thng bc hai khng c ng tim cn.


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Dng 2:Lp phng trnh ng cong ( C) vi cc iu kin c lin quan ti tm,phng tim cn , ng tim cn.

Phng php:

- Vit pt tng qut ca ng cong ( C).- Da vo cc iu kin cho ta tm ra cc h s ca phng trnh.- Sau th cc h s vo phng trnh tng qut ta c phng trnh ng bc 2 (C) cn tm.

Bi mu 1:Lp phng trnh Hypebol i qua cc im (1,2),(-1,-1),1 -1

( , )2 4

vi iu kin mt tim cn ca n trng vi Ox.

Li Gii:

Phng trnh tng qut ca Hypebol: (H):ax2

+2bxy+cy2

+2dx+2ey+e=0.V (H) c 1 tim cn v

= ( , ) trng vi Ox nn (0,1) l nghim ca pt:

2 22 0 0a b c a

(H) tr thnh :2bxy+cy2+2dx+2ey+f=0

V c, bkhc khng nn ta c th vit (H) nh sau:2xy+y2+2dx+2ey+f=0

V (H) i qua 3 im (2,1),(-1,-1),1 1

( , )2 4

nn ta c h pt:

21

44 2 567

2 2 38

1 337

2 164

dd e f

d e e e

d e ff

Vy phng trnh ca (H) l:16xy+8y2+42x-67y-74=0.

Nhn xt: - Nhng bi ton lp phng trnh vi d kin c lin quan n tm,phng tim cn v ng tim cn th khng kh. Tuy nhin i hingi hc phi bin i tt da vo cc d kin ca bi.

Bi mu 2:Lp pt ng cong bc hai c tm ti I(0,-1), qua (3,0) v ch ctmi ng thng d1, d2ti 1 im:

d : 2x - 3y + 1 = 01d x + y -5 = 02


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Li Gii:

ng cong ( C) c tm I(0,1) nn c pt:

a(x-0)2+2b(x-0)(y+1)+c(y+1)2+f=02 2 2 2 2 0ax bxy cy bx cy c f (*)

Qua (3,0) nn ( C) c dng:9a+6b+c+f=0 (1)

1 3: ( )1 1 2

2: ( )2 3

x td t R

y t

x td t R

y t

V d1 ct (C ) ti mt im nn pt sau ch c 1 nghim:a(1+3t)2+2b(1+3t)(1+2t)+c(1+2t)2+2b(1+3t)+2c(1+2t)+c+f=0

2(9 12 4 ) (6 16 8 ) 4 4 0a b c t a b c t a b c f (*)

(*) ch c 1 nghim9 12 4 0

(2)6 16 8 0

a b c

a b c

V d2ct ( C) ti mt im tng t nh trn ta c pt:

a(2+t)2+2b(2+t)(3-t) +c(3-t)2+2b(2+t)+2c(3-t)+c+f=0

(a-2b+c)t2+(4a+4b-8c)t+4a+16b+16c+f=0 (*)

(*)ch c mt nghim 2 0

(3)

4 4 8 0

a b c

a b c

T (1),(2),(3) chn f=12 a=2,b=1

2

,c=-3

Vy pt ca (C ) l:

2x2-xy-3y2-x-6y+9=0.

Nhn xt: -Phng trnh ng bc 2 ( C) c tm I(xo,yo) c dng:

a(x-xo) +2b(x-xo)(y-y0)+c(y-y0)2+f=0.

-Phng trnh :Ax2+2Bx+C=0 ch c 1 nghim khi v ch khi :

00

AB


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BBiittppttnnggtt

Bi 1 :Tm phng tim cn ca cc ng bc hai sau:1)10xy-2y2+6x+4y-21=0

p s:

1

2

v = (1,0)v = (-1,1)

2)2x2-3xy-x+3y+4=0

p s:

1

2

v = (0,1)

v = (3, 2)

Bi 2 : Tm tm - Phng tim cn - ng tim cn ca cc

ng bc hai sau:a) (C) 9x2-2xy+6y2-16x-8y-2=0.b) (C): 8x2+6xy-26x-12y+11=0 (1)

p s: a.I(-44 28,53 53

), Khng c phng tim cn

b. I(2,-1) ,

v = (0,1)1v = (-3,4)2

;

(d1):

x = 2

(t R)y = -1+ t

(d2):

x = 2 - 3t

(t R)y = -1+ 4t


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Ch 4

GIAO TUYN CA NG BC HAI VI NG

THNGPhng php:

Cho ng cong (C): F(x;y) = ax2+2bxy + cy2 +2dx +2ey + f = 0,(a,b,c) (0,0,0)(1)

V ng thng (d) : o

o

x x t

y y t

(2)

Gi M(x0;y0) l giao im ca (C) v (d). khi ta ca M l nghim ca h(1) v (2).(2) th vo (1)

a( tx 0 )2+ 2b( tx 0 )( tx 0 ) + c( tx 0 )2+2d( tx 0 ) +2e( tx 0 )+f=0Pt2+2Qt + R = 0 (3)Trong :P = 2 2a 2b c 2Q = (2ax0+ 2by0+ 2d) + (2bx0+ 2cy0+ 2e) = Fx(x0;y0) + Fy(x0;y0)R = F(x0;y0)* P = 0: (3)..Qt+R = 0 (4)

- Q = 0: (4) R = 0 :M(x0;y0)(C) (d) nm trn (C)M(x0;y0) (C) (d) khng nm trn (C)

- Q 0 : (4) t = Q

R

:(d) giao vi (C) ti 1 im* P 0: ' = Q2- PR

- >0: (3)c 2 nghim phn bit (d) ct (C) ti 2 im phn bit- = 0: (3)c nghim kp (d) ct (C) ti 2 im trng nhau-


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Li Gii:

a. Ta giao im ca 1(d )v (C) l nghim ca h phng trnh:2 2 2 2

2

x 2xy 3y 4x 6y 3 0 x 2x(5x 5) 3(5x 5) 4x 6(5x 5) 3 05x y 5 0 y 5x 5

1x 1 5x y284x 126x 42 0

2 2x 1y 5x 5 x 1 y 0y 5x 5

Vy (d1) v (C) c 2 giao im ln lt c ta l1 5

( ; ) v ( 1; 0)2 2

b. Ta giao im ca 2(d ) v (C) l nghim ca h phng trnh:2 2

2 2

x 2y 2x 2xy 3y 4x 6y 3 0x 2y 2 0 (2y 2) 2(2y 2)y 3y 4(2y 2) 6y 3 0

2

x 2y 2

5y 14x 15 0

H phng trnh v nghim

Vy (C) v 2(d ) khng c im chung.c. Ta giao im ca 3(d ) v (C) l nghim ca h phng trnh:

2 2

2 2

2

x 1 4yx 2xy 3y 4x 6y 3 0

x 4y 1 0 (1 4y) 2(1 4y)y 3y 4(1 4y) 6y 3 0

x 1 4y x 1y 021y 0

Vy giao im ca (C) v 3(d ) c ta l ( 1; 0)

d. Ta giao im ca 4(d ) v (C) l nghim ca h phng trnh:2 2

2 2

x 3yx 2xy 3y 4x 6y 3 0x 3y 0 (3y) 2(3y)y 3y 4(3y) 6y 3 0

1xx 3y 2

118y 3 0y

6

Vy giao im ca (C) v 4(d ) c ta l1 12 6

( ; )

Nhn xt: - Vic tm giao im thc cht l quy v vic gii phng trnhbc hai. Vic ny tht d dng.


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21

Ch 5PHNG TRNH TIP TUYN CA NG

BC HAI

Dng 1: Tip tuyn tho iu kin cho trc

Phng php:

Trong (xOy) cho ng bc hai2 2(C) :ax 2bxy cy 2dx 2ey f 0,(a,b,c) (0,0,0)

(d) l tip tuyn ca (C) khi n nm trn (C) hoc ct (C) ti hai im trng nhau ,t ta gii h gm hai phng trnh (C) v (d), s dng iu kin tip xc (d)

Bi mu 1:Trong (xOy) cho ng bc hai2 2(C) : 2x 4xy 5y 6x 8y 1 0 .

Vit phng trnh tip tuyn (d) ca (C) trong cc trng hp sau y:a. (d) song song vi ng thng x y 0 b.(d) i qua im ( 5; 0) c. (d) i qua im ( 1; 1)

Li Gii:2 2(C) : 2x 4xy 5y 6x 8y 1 0

a)

Tip tuyn (d) ca (C) song song vi ng thng x y 0 nn (d) c dng: x y m 0 (m 0)

Thay y x m vo (C) ta c:2 2 2 22x 4x(x m) 5(x m) 6x 8(x m) 1 0 3x 2(3m 1)x 5m 8m 1 0 (1)

2 2 2: (3m 1) 3(5m 8m 1) 6m 18m 4(1) (d) l tip tuyn ca (C) (1) c nghim kp

2 2 9 1056m 18m 4 0 3m 9m 2 0 m6

Vy c 2 phng trnh tip tuyn

9 105

(d) : x y 06

b) (d) i qua im ( 5;0) nn (d) c dng:x 5 at

, (a, b) (0, 0)y bt

thay vo (C)

ta c: 2 2 2

2 2

2 2 2 2 2

2(5 at) 4bt(5 at) 5b t 6(5 at) 8bt 1 0

(2a 4ab 5b )t 2(7a 6b)t 19 0

(7a 6b) 19(2a 4ab 5b ) v 2a 4ab 5b 0, a,b

(2)


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22

(d) l tip tuyn ca (C) (2) c nghim kp2 2 2 2 2(7a 6b) 19(2a 4ab 5b ) 0 11a 8ab 59b 0

Chn b 1 ta c: 2 4 665a11

11a 8a 59 0

Vy c 2 phng trnh tip tuyn 4 665x 5 t(d): 11y t

c) (d) i qua im ( 1; 1) nn (d) c dng :x 1 at

, (a, b) (0, 0)y 1 bt

thay vo (C) ta

c: 2 2

2 2

2 2 2 2 2

2(1 at) 4(1 at)(1 bt) 5(1 bt) 6(1 at) 8(1 bt) 1 0

(2a 4ab 5b )t 2(a 3b)t 4 0

(a 3b) 4(2a 4ab 5b ) v 2a 4ab 5b 0, a,b

(3)

(d) l tip tuyn ca (C) (3) c nghim kp2 2 2 2 2(a 3b) 4(2a 4ab 5b ) 0 9a 22ab 29b 0 v nghim

Vy khng c tip tuyn no ca (C) i qua im ( 1; 1)

Nhn xt: - ng thng (d) l tip tuyn ca (C) khi v ch khi (d) ct (C)ti hai im trng nhau hoc (d) nm hon ton trn (C).- C nhiu cch gii dng ton ny. Tuy nhin, tu theo biton m ta s dng phng php cho ph hp. Chng ta c ththam kho cch gii trn.

Bi mu 2:Vit phng trnh tip tuyn vi 2 2(C) : x xy y 2x 3y 3 0 , bit

tip tuyn song song vi trc Ox

Li Gii:2 2(C) : x xy y 2x 3y 3 0

Tip tuyn (d) ca (C) song song vi Ox nn (d) c dng: y m 0 (m 0) th (d) vo

(C) ta c: 2 2 2 2x mx m 2x 3m 3 0 x (2 m)x m 3m 3 0 (1) 2 2 2(2 m) 4(m 3m 3) 3m 8m 16(1):

(d) l tip tuyn ca (C) (1) c nghim kpVy c 2 phng trnh tip tuyn : 1 2(d ) : y 4 0 v (d ) :3y 4 0

Nhn xt: - y l mt bi ton d, li gii trn ch c mt tham s. Nu gii didng tham s ca phng trnh (d) th chng ta s gii quyt vi phngtrnh hai tham s. Tuy nhin chng ta s khng bao gi thiu nghim, cngha l bi ton lun gii quyt c.

Dng 2: Phng trnh tip tuyn ti tip im


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Phng php:

2 2(C) :ax 2bxy cy 2dx 2ey f 0,(a,b,c) (0,0,0) ng thng (d) l tip tuyn ca (C) ti tip im o oM(x ;y )

nn (d) c dng: o o o o o oax x b(x y y x) cy y d(x x) e(y y) f 0 (*)

Bi mu 1:Vit phng trnh tip tuyn ca ng bc hai (C):2 23x 2xy 2y 3x 4y 0

ti tip im M( 2; y)

Li Gii:

Ta c 2 1 2y 1

y 3

M( 2; y) (C) 2y 8y 6 0 M ( 2; 1),M ( 2; 3)

Tip tuyn ca (C) ti o oM(x ; y ) : o o o o o o3

3x x (x y y x) 2y y (x x) 2(y y) 02

Vy Phng trnh tip tuyn ti 1M ( 2; 1) : 7x 4y 10 0 Phng trnh tip tuyn ti 2M ( 2; 3) : 3x 4y 18 0

Nhn xt: Bi ton ny to im M cho khuyt. p dng phng trnh (*) tac hai tip tuyn nh trn.

Bi mu 2:Vit phng trnh tip tuyn ca ng bc hai (C):2 2

2x 4xy y 2x 6y 3 0 bit tip tuyn i qua im A( 3; 4)

Li Gii:Gi (d) l tip tuyn ca (C) ti tip im o oM(x ;y ) nn o o o o o o

o o o o o o

(d) : 2x x 2(x y y x) y y (x x) 3(y y) 3 0

(d) : (2x 2y 1)x ( 2x y 3)y x 3y 3 0

Ta c : o o2 2

o o o o o o o o

o o2 2o o o o o o

3x y 6 0A( 3; 4) (d)M(x ; y ) (C) 2x 4x y y 2x 6y 3 0

y 3(x 2)2x 12x (x 2) 9(x 2) 2x 18(x 2) 3 0

o

2oo o

o o o

o

1

2

x 4x 3 0

y 3(x 2)

x 1M ( 1; 3)

y 3

x 3M ( 3; 3)

y 3

Vy Phng trnh tip tuyn ti 1M ( 1; 3) : 7x 2y 13 0


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24

Phng trnh tip tuyn ti 2M ( 3; 3) : x 3 0

Nhn xt: - y l dng ton vit phng trnh tip tuyn qua mt im khngthuc (C). Bi ton trn s dng phng php tm tip im trc sau tm c hai tip tuyn nh trn. Ta cng c th vit phng trnh

ng thng (d) qua A dng:y = k(x - 3) + 4 hoc A(x-3) + B(y-4) = 0 sau s dng iu kin tipxc tm tham s.

Bi mu 3:Ti cc giao im ca ng thng (d) :3x y 6 0 vi ng cong2 2(C) : x 2xy y 2x 6y 0 k cc tip tuyn vi ng cong. Tm giao im

ca cc tip tuyn

Li Gii:Ta giao im ca (d) vi (C) l nghim ca h phng trnh:

2 2 2 2

3x y 6 0 y 3x 6

x 2xy y 2x 6y 0 x 2x(3x 6) (3x 6) 2x 6(3x 6) 0

2

x 0M( 0; 6)y 3x 6

y 6y 3x 6x 0

4x 8x 0 x 2x 2 N( 2;0)

y 0

Phng trnh tip tuyn ca (C) ti tip im o oM(x ;y ) l (d) c dng:

o o o o o o

o o o o o o

x x (x y y x) y y (x x) 3(y y) 0

(d) : (x y 1)x ( x y 3)y x 3y 0

Phng trnh tip tuyn ti M(0; 6) c dng: 5x 3y 18 0 Phng trnh tip tuyn ti N( 2; 0) c dng: x y 2 0

Ta giao im ca 2 tip tuyn l nghim ca h :

3x

27

y2

5x 3y 18

x y 2 0

Nhn xt: - y l mt bi ton n gin, ch cn tm to giao im sau vit phng trnh tip tuyn theo phng trnh (*).

Bi mu 4:Qua im M( 3; 1) k c 2 tip tuyn vi ng bc hai (C):2 23x 2xy 3y 4x 4y 4 0 .

Vit phng trnh ng thng i qua cc tip im.

Li Gii:


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Phng trnh tip tuyn ca (C) ti tip im o oM(x ;y ) l (d) c dng:

o o o o o o

o o o o o o 4

3x x (x y y x) 3y y 2(x x) 2(y y) 4 0

(d) : (3x y 2)x ( x 3y 2)y 2x 2y 0

(d) qua im o oM( 3; 1) 10x 2y 4 0

Vy phng trnh ng thng i qua cc tip im l: 5x y 2 0 Nhn xt: - Bi ton ny ch yu cu vit phng trnh ng thng qua cc tip

im. Do nu vit phng trnh tip tuyn ri tm tip im th biton tr nn phc tp. Do chng ta s dng th thut trn th yu cubi ton tr nn d dng hn rt nhiu.

BBiittppttnnggtt

Bi 1 :Cho ng cong :x2+ xy + y2+2x + 3y -3 = 0

Lp tip tuyn song song vi ng thng 3x + 3y -5 = 0. Xc nh to cctip im.p s : x + y -1 =0 v 7x + 7y -17 =0 ; M1=(1 ; 0) v M2(-5 ; -6)

Bi 2 : Trong tt c nhng ng thng tip xc vi ng cong :x2+ xy + y2+2x + 3y -3 = 0

Hy tm nhng ng song song vi trc honh.p s : 3y - 4=0 v y + 4 =0


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26

Ch 6NG KNH LIN HP VI MT PHNG.

Phng php:

+Trc tin cn nm r khi nim ng knh v ng knh lin hp vi 1 phngcho trc.+ng knh v ng knh lin hp vi 1 phng u l 2 ng thng qua tm vc phng khc phng tim cn.+ng knh lin hp vi 1 phng l ng thng i qua trung dim cc dy cungca ng bc hai (C ).+ng knh lin hp vi phng v

=(a,b) (khc phng tim cn )c dng:

aF'x(x,y)+bF'y(x,y)=0 (*)

Bi mu 1:Tm 2 ng knh lin hp vi ng cong :(C ):x2-2xy+2y2-4x-6y+3=0

Bit rng 1 ng knh i qua gc ta .

Li Gii

Tm I(x,y) l nghim ca h pt:

2 0

7,52 3 0

x yI

x y

Gi ng knh qua gc ta c pt: ax+by=0(1)ng knh qua I(7,5) th vo (1) a=5,b=7.

Vy pt ng knh l:7

( )5

x tt R

y t

c v=(7,5)

ng knh lin hp vi phng v c pt:7F'x+5F'y=0

7(2x-2y-4)+5(-2x+4y-6)=0

2x+3y-29=0.

Vy pt ng knh lin hp vi phng v l :2x+3y-29=0.

Nhn xt: - Vic tm ng knh v ng knh lin hp th trc ht phi tm tmca (C).- Nu bit trc phng v th p dng phng trnh (*) tm ngknh lin hp.


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Bi mu 2:cho ng cong:2x2+5xy-3y2+3x+16y=0Lp phng trnh ng knh song song vi trc honh v ng knh lin hpvi n.

Li Gii

Tm I(x,y) l nghim ca h pt:

016650354

yxyx I(-2,1)

Phng trnh ng knh song song vi trc Ox c dng:y = m

ng knh qua (-2,1) ng knh (d):y=1 y-1=0 c v

=(1,0)

Phng trnh ng knh lin hp vi (d) c dng:1.F'x(x,y)+0.F'y(x,y)=0

4x+5y+3=0

Vy phng trnh ng knh lin hp vi (d)l:4x+5y+3=0

Nhn xt: - ng knh lin hp vi ng knh song song vi trc honh tc l

lin hp vi phng v

=(1,0). S dng cng thc (*) gii quyt.

Bi mu 3:Lp phng trnh ng knh ca 2x2+4xy+5y2-8x+6=0 song song ving thng (d):2x-y+5=0

Li Gii

Tm I (x,y) l nghim ca h pt:2 2 4 0

2 5 0

x y

x y

I(10 4,

3 3

)

ng knh song song vi ng thng(d) c dng: 2x-5y+m=0(1)

V ng knh qua tm I nn m= 403

Vy phng trnh ng knh l:2x-5y-40

3=0 6x-15y-40=0.

Nhn xt: - Bi ton ny tht n gin, s dng cch gii trn rt hiu qu. Hyth gii bi ton ny vi iu kin l lp phng trnh ng knh linhp vi ng knh vung gc vi ng thng (d) u bi.

Bi mu 4:Cho ng cong :3x2+7xy+5y2+4x+5y+1=0.Tm qu tch trung imnhng dy:

a)Song song vi trc x (y=0).b)Song song vi trc y (x=0).c)Song song vi ng thngx+y+1=0. c vtcp: v

=(1,-1)

Li Gii

a)Song song vi trc Ox: (y=0)


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Yu cu bi ton tng ng vi yu cu:vit pt ng knh ca (C) lin hp viphng vi t y=0

Ta c:aF'x(x,y)+0F'y(x,y)=0 a(6x+7y+4)=0. V 0a nn ta c th vit pt ng knhnh sau:6x+7y+4=0,

b)Song song vi trc Oy : (x=0)Tng t nh cu a:ta c phng trnh ng knh nh sau:7x+10y+5=0.

c)Song song vi ng thngx+y+1=0. c vec t ch phng: v=(1,-1)

Phng trnh ng knh c dng:1F'x(x,y)-1F'y(x,y)=0

(6x=7y+4)-(7x+10y+5)=0

x+3y+1=0

Nhn xt: - c th yu cu tm qu tch trung im cc dy cung ca (C )tng ng vi yu cu vit pt ng knh lin hp vi 1 phng.

BBiittppttnnggtt

Bi 1 : Qua im A(1; -2) dng ng knh ca (C):3x2 - 2xy + 3y2 + 4x + 4y - 4 = 0

v ng knh lin hp vi n.p s :(d) : x + 2y + 3 =0 v (d) : 7x - 5y +2 =0

Bi 2 : Cho ng cong (C) :3x2 + 2xy + 2y2 + 3x - 4y = 0

V mt ng knh ca n l (d) : x + 2y -2 =0. Lp phng trnh ng knh

lin hp vi ng knh trn.p s: x + 1 =0


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Ch 7A PHNG TRNH BC HAI V DNG

CHNH TC

Dng 1: Phng trnh chnh tc ca ng bc hai trong AFINPhng php

Dng php i mc tiumt cch thch hp ta s a ng bc hai (C) v mt trong9 dng sau :

1. Elip thc X2+ Y2= 12. Elip o X2+ Y2= -13. Hypebol X2- Y2= 14. Parabol X2+ 2Y=05. Hai ng thng thc ct nhau X2- Y2= 06. Hai ng thng thc song song X2- 1 = 07. Hai ng thng thc trng nhau X2= 08. Hai ng thng o ct nhau X2+ Y2= 09. Hai ng thng o song song. X2+ 1 = 0

Bi mu: Trong h to Afin, hy a cc ng bc hai sau v dngchnh tc:

1. x2- 2xy + 2y2- 4x - 6y + 3 = 02. x2- 2xy - 2y2- 4x - 6y + 3 = 03. x2- 2xy + y2- 4x - 6y + 3 = 0

4. x2- 2xy - 2y2- 4x - 6y - 133 = 0

Li Gii

1. x2- 2xy + 2y2- 4x - 6y + 3 = 02 2( ) 4 6 3 0x y y x y (1)

t'

'

x x y

y y

2 2(1) ' ' 4 ' 10 ' 3 02

' 2 ' 5 1

26 26

x y x y

x y

t

' 2

26' 5

26

xX

yY

2 2(2) 1X Y


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Vy ng bc hai cho thuc loiElip

2. x2- 2xy - 2y2- 4x - 6y + 3 = 02 2( ) 3 4 6 3 0x y y x y (1)

t

'

'

x x y

y y

225 223( ' ) ( ' 2)

3 3

3(

2 2(1) ' 3 ' 4 ' 10 ' 3 0

223 ' 5 ' 2)

122 22

y x

x y x y

y x

t

3 ' 5

223( ' 2)

22

yX

xY

(2)

2 2(2) X 1Y Vy ng bc hai cho thuc loiHypebol.3. x2- 2xy + y2- 4x - 6y + 3 = 0

2( ) 4 6 3 0x y x y (1)

t'

'

x x y

y y

2 10 ' 1 (2)

2(1) ' 4( ' ') 6 ' 3 0

( ' 2) y

x x y y

x

t

' 2

10 ' 1

2

x

y

X

Y

2(2) X 2Y Vy ng bc hai cho thuc loiParabol.

4. x2- 2xy - 2y2- 4x - 6y -133

= 0

2 13 33

2( ) 4 6 0yx y x y (1)

t'

'

x x y

y y


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2

22

13

3

53( ' ) =0 (2)

3

2(1) ' 3 ' 4 ' 10 ' 0

( ' 2) y

x y x y

x

t

' 2

53( ' )

3

x

y

X

Y

2 2(2) X 0Y Vy ng bc hai cho l hai ng thng thc ct nhau.

Nhn xt : - Bi ton dng ny tuy khng kh nhng n i hi ngi lmphi bit nhm cc n s mt cch thch hp a phngtrnh v dng n gin, sau dng php i mc tiu an v dng chnh tc.

- Cc bi ton 2, 3, 4 l m rng ca bi ton 1, tc l ch cnthay i h s t do v h s ca y2. V vy chng ta thy rngt 1 bi ton chng ta c th m rng ra nhiu bi ton khc nhn c tt c 9 loi ng nu.

Dng 2: Phng trnh chnh tc ca ng bc hai trong TRC CHUN

Phng php

Dng php Quay quanh gc to O mtgcthch hp ta s a ng bc hai (C)v mt trong 9 dng sau :

1. Elip thc X

2

+ Y

2

= 12. Elip o X2+ Y2= -13. Hypebol X2- Y2= 14. Parabol X2+ 2Y=05. Hai ng thng thc ct nhau X2- Y2= 06. Hai ng thng thc song song X2- 1 = 07. Hai ng thng thc trng nhau X2= 08. Hai ng thng o ct nhau X2+ Y2= 09. Hai ng thng o song song. X2+ 1 = 0

Bi mu: Trong h to Trc chun, hy a cc ng bc hai sau vdng chnh tc:1. x2+ 6xy + y2+ 6x + 2y - 1 = 02. 3x2- 2xy +3y2+ 4x + 4y - 4 = 0

Li Gii

1. x2+ 6xy + y2+ 6x + 2y - 1 = 0 (1)

Thc hin php quay tm O mt gc thch hp : ( ; ) ' 'Q OOxy Ox y


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Khi ta c :1 1

cot 2 02 6

a cg

b

Chn 22 4

Ta dng php i mc tiu : 'cos 'sin'sin 'cos

x x yy x y

1

2

1

2

( ' ')

( ' ')

x x y

y x y

2

2 2

2 2 1

1 12( ' ) ( ' ) 1 (2)

2 2

2(1) 4 ' 2 ' 4 ' 2 ' 0

x y

x y x y

t

1'2

1'

2

x

y

X

Y

1

2

2X 2(2) 1Y

Vy ng bc hai cho thuc loiHypebol.2. 3x2- 2xy +3y2+ 4x + 4y - 4 = 0 (1)

Thc hin php quay tm O mt gc thch hp : ( ; ) ' 'Q OOxy Ox y Khi ta c :

3 3cot 2 0

2 2

a cg

b

Chn 22 4

Ta dng php i mc tiu :'cos 'sin

'sin 'cos

x x y

y x y

1

21

2

( ' ')

( ' ')

x x y

y x y

2

2 2

2 4

( ' 2) '1 (2)

4 2

2(1) 2 ' 4 ' 4 ' 0

x y

x y x

t' 2

'

x

y

X

Y


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2

4 2

2X(2) 1

Y

Vy ng bc hai cho thuc loiElip

BBiittppttnnggttBi 1 :Trong h to Afin, hy a cc ng bc hai sau v dng

chnh tc:1. 5x2+ 4xy + 8y2 - 32x - 56y + 80 = 02. 5x2+12xy -22x -12y -19 = 03. x2- 4xy + 4y2+ x + 2y -2 = 04. x2 - 5xy + 4y2+ x + 2y - 2 = 0

p s: 1.X2 - Y2= 1.y l ng hypebol2. X2+ Y2= 1.y l ng elip.3. X2 - 2Y = 0.y l ng parabol.4. XY = 0. y l cp ng thng ct nhau

Bi 2 : Trong h to Trc chun, hy a cc ng bc hai sau vdng chnh tc:1. x2+ 2xy + y2+ 2x +2y -4 = 02. -5x2- 4xy + y2 + 12x - 6y + 9 = 0

p s : 1. X2- 1 = 0. Hai ng thng thc song song.2. X2= 0 . Hai ng thng thc trng nhau.


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MT BC HAI


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Ch 1

PARABOLOIT HYPERBOLIC

Dng 1: Tm giao tuyn ca mt bc hai (trong c mt yn nga)

Phng php:Lp h phng trnh gm hai phng trnh ca hai mt mt bc hai sau dng cc

php bin i nhn dng giao tuyn l ng g.

Bi mu 1: Tm giao tuyn ca hai mt bc hai:2 2 2 2

2 2

x y z a

x y 2az

Li Gii:Giao tuyn ca mt bc hai l nghim ca h phng trnh:

222 2 2 2

2 2 22

x 2 z a

2x z a x 2 z ax y z a

x y 2az y 2 z a2y z a

y 2 a z

Vy hai mt bc hai giao nhau ti bn ng thng c phng trnh:

x 2 z a 0 x 2 z a 0 x 2 z a 0 x 2 z a 0; ; ;

y 2 z a 0 y 2 z a 0 y 2 z a 0 y 2 z a 0

Nhn xt: - Vic tm giao tuyn ca mt bc hay bi ton ny i hi ngi hcphi kho lo bin i thy r giao tuyn.

Dng 2: Xc nh phng trnh ca mt yn nga

Phng php:

Paraboloit hyperbolic2 2

2 2

x y(S) : 2kz 0 (k 0)

a b

Da vo iu kin bi thit lp mi lin h gia cc gi tr a,b,k (S)

Bi mu 1: Vit phng trnh ca paraboloit hyperbolic qua hai ng thngy x v z 0 v qua im ( 1; 2; 3) bit rng Oz l trc i xng.

Li Gii:


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36

Paraboloit hyperbolic2 2

2 2

x y(S) : 2kz 0 (k 0)

a b (Oz l trc i xng)

(S) qua hai ng thng y x v z 0 nn2 2

2 22 2

x x0 a b

a b(1)

Mt khc (S) qua ( 1; 2; 3) nn 2 2 21 4 1

6k 0 k a b 2b (2) 2 2

2 2 2

x y 2(S) : z 0

b b 2b Vy 2 2(S) : x y z 0

Nhn xt Dng bi ny ty vo yu cu ca ma ta c phng php gii thch hp.

Dng 3: ng sinh thng ca mt yn nga

Phng php: gii bi ton dng ny ta thc hin cc bc:B1: Vit phng trnh tham s ca (d) da vo iu kin cho trc (nu c)B2: (d) l ng sinh ca mt yn nga (S) th (d) (S) B3 :Tm vect ch phng ca (d) v vit phng trnh ng sinh (d).

Bi mu 1: Tm ng sinh thng ca2 2x y

(S) : z16 4

;bit n song song vi mt

phng (P) :3x 2y 4z 0

Gii:

Gi (d) qua o o oM( x ; y ; z )v c VTCP u ( ; ; ) ( 0; 0; 0)

o

o

o

x x t

(d) : y y tz z t

Thay (d) vo (S) ta c: 2 2o o o2 2 2 2 2

o o o o o4y 16z

(x t) 4(y t) 16(z t)

( 4 )t 2( x 4 y 8 )t x 0

(d) l ng sinh thng ca (S)

o o2 2

2 2o o o

o o2 2o o o

o o

2 2o o o

2x 4 y 8 0

4 0 x 4y 16z 0(d) (S) x 4 y 8 0

2x 4y 16z 0x 4 y 8 0

x 4y 16z 0

Vio o

2x 4 y 8 0

.Chn o ox y

1 2,4 2

Vio o

2x 4 y 8 0

.Chn o ox y

1 2,4 2


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37

H hai ng sinh ca (S) c VTCP l: o o o o1 2x y x y

u ( 2; 1; ) ; u ( 2; 1; )4 2 4 2

Mt khc (d) // (P) c VTPT n ( 3, 2, 4)

nn ta c:

o o o1o 1 2

o o o2

8 x 2y 0 x 6u n 0z 2 u ( 2; 1; 2) ; u ( 2; 1; 1)4 x 2y 0 y 1u n 0

..

Vy c 2 ng sinh thng tha yu cu bi l:

1 2

x 6 2t x 6 2t(d ) : y 1 2t ; (d ): y 1 2t

z 2 2t z 2 t

Nhn xt: - y thay v bi cho cc ng sinh song song vi mtphng th c th cho ng sinh vung gc vi mt ng thng hochp vi mt phng mt gc .

Bi mu 2: Tm cc ng sinh thng ca mt yn nga2 2x y

(S) : 2z16 9

i qua

im ( 8; 3 2; 1)

Gii:

Gi (d) qua ( 8; 3 2; 1) v c VTCP

x 8 at

u (a; b; c) (0; 0; 0) (d) : y 3 2 bt

z 1 ct

Thay (d) v (S):2 2 2 2 2(8 at) (3 2 bt) a b 2 22(1 ct) ( )t (a b 2c)t 0

16 9 16 9 3

(d) l ng sinh ca (S)

2 2 3a 4ba b3a 4b16 9(d) (S)

2 2 1 2 2a b 2c 0 c (a b)3 2 3

Chn a 4 b 3, c 2 2

Chn a 4 b 3, c 2 2 Vy c hai ng sinh thng ca (S) l

1 2) )

x 8 4t x 8 4t

(d : y 3 2 3t ; (d : y 3 2 3t

z 1 (2 2)t z 1 (2 2)t


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38

Bi mu 3: Cho mt yn nga 2 2(S) : x y 2z .Tm qu tch nhng giao im ca

nhng ng sinh thng vung gc vi nhau.

Li Gii:

Gi (d) qua o o oM(x ; y ; z ) (S) v c VTCP u (a;b;c) (0; 0; 0)

o

o

o

x x at(d) : y y bt

z z ct

Thay (d) vo (S): 2 2 2 2 2o o o o o(x at) (y bt) 2(z ct) (a b )t 2(ax by c)t 0

(d) l ng sinh thng ca (S) (d) (S)

1 o o2 2o o o o

o o2 o o

o o o o

b a b au ( 1; 1;x y )

a(x y ) c 0 c a(x y )a b 0

ax by c 0 b a b au ( 1; 1;x y )

a(x y ) c 0 c a(x y )

Ta c 2 2 2 21 2 o o o ou .u 0 1 1 (x y ) 0 x y 0

o o o oM(x ; y ; z ) (S) z 0 Vy qu tch giao im ca nhng cp ng sinh thng vung gc vi nhau l

hai ng thng ct nhau:x y 0 x y 0

vz 0 z 0


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39

Ch 2

ELIPXOIT

Dng 1: Vit phng trnh Elipxoit tho iukin cho trc

Phng php:

Elipxoit:2 2 2

2 2 2(S):

x y z1

a b c

S dng cc d kin bi cho trc tm cc h s a, b, c.Vit phng trnh mt elipxoit tm c.

Bi mu 1: Vit phng trnh elipxoit c trc trng vi trc to , nu

bit rng n i qua im I( 3; 1; 1) v i qua ng trn c pt:2 2 2x y z 9

z x

Li Gii:

Phng trnh Elipxoit c dng2 2 2

2 2 2(S):

x y z1

a b c(1)

(S) qua ng trn2 2 2x y z 9

z x

nn (S) qua ( 2; 1; 2);( 2; 5; 2)

(S) qua ( 3; 1; 1);( 2; 1; 2);( 2; 5; 2) nn ta c h pt:

2 2 2

2 2 2

2 2 2

9 1 11

a b c4 1 4

1a b c2 5 2

1a b c

2 2 2 36a 12;b 9;c5

.Vy2 2 2

(S):x y z

13612 95

Nhn xt: - Phng trnh Elipsoid (1) nhn cc trc to lm trc ixng. Nhn gc to lm tm i xng.- Li gii trn kho lo tm thm hai im m (E) i qua dn ti h 3 phng trnh 3 n.- Ta cng c th c hng khc vn i n kt qu trn nh sau :


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40

Giao ca (E) vi mt phng z =x l :2

22 2 2

1 1 yx 1

a c b

z x

(*)

Phng trnh ng trn c vit li:2

22 yx 19 9

z x

(**)

So snh (*) v (**) v cho (E) qua I ta s tm li c kt qu nh li gii trn.

2 2 2

(S):x y z

13612 95

Dng 2 : Giao tuyn ca Elipxoit vi ng v mt

Bi mu 1 : Cho Elipxoit2 2 2

2 2 2

x y z(S) : 1

a b c

a)Nu a b c th elipxoit tr thnh mt g?b) CMR: Nu 0 c b a th M (S) ta u c : c OM a c) Chng minh rng nu a b c th giao tuyn ca Elipxoit vi cc mt phng:

2 2 2 2

1 1 1 1x z 0

b a c b

l nhng ng trn.

Li Gii:a) Khi a b c Elipxoit tr thnh mt cu tm O bn knh R a

b) Gi s M (x ; y; z )nm trn Elipxoit ,tc l :2 2 2

2 2 2

x y z1

a b c

Khi 0 c b a .Ta c:2 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2

2 2

2 2

x y z x y z x y za a a a b c c c c

OM OM1 c OM a

a c

c)

Phng trnh ca cp mt phng cho c th vit di dng:2 2

2 2 2 2

1 1 1 1x z 0

b a c b

Giao tuyn ca Elipxoit vi cc mt phng cho c phng trnh :2 2 2 2

2 22 2 2 2 2 2 2 2

2 2 22 2 2

2 2 22 2 2 b b

1 1 1 1 x z x zx z 0b a c b b b a c

x y zx y z 11ba b c


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2 22 2 2 2

2 2 2 2

1 1 1 1x z

b a c b

x y z b

(1)

(2)

Phng trnh (1)cho ta mt cp mt phng ct nhau theo trc Oy, cn phng trnh (2)

cho ta mt mt cu tm O bn knh b. Vy h phng trnh trn cho ta mt cp ngtrn tm Obn knh l b, ln lt nm trn cp mt phng (1).

Nhn xt: Bi ton ny tng i n gin do li gii trn l hiu qu. cu c/i hi chng ta phi linh hot bin i thnh giao tuyn ca mt phng vi mt cuth xem nh yu cu bi ton c thc hin v giao tuyn lun l ng trn.

Dng 3: Cho phng trnh elipxoit, tim tip tuyn, tm iu kin 1 ng thng ltip tuyn , tm iu kin 1 mt phng l thit din ca elipxoit

Phng php gii:

Nm vng l thuyt v iu kin ca mt ng thng l tip tuyn ca elipxoitKt hp vi d kin bi choTm cc h s (i vi bi i tm tip tuyn),cc iu kin (vi bi bin lun), hoc tmthit din tho yu cu tng bi.

Bi mu: cho elipxoit (S) co pt22 2

2 2 21

x y z

a b c

a) gi M0(x0, y0, z0 l im nm trn (S).Mt ng thng (d) i qua M0gi l tiptuyn ca (S) ti M0nu n ch ct (S) ti im duy nht M0. hy tm k ca vectch phng ( ; ; )u

ca (d) (d) l tip tuyn ca (S) ti M0. .

b) chng minh rng mi tip tuyn ca (S) ti M0 nm trn 1 mt phng . Mtphng ny gi l tip din ca (S) ti M0.

Li Gii:a) ng thng (d) i qua M0vi vcto ch phng ( ; ; )u

c

pt: 0 0 0; ;x x t y y t z z t Giao im ca (d) v (S) ng vi gi tr ca t lnghim ca h gm 3 pt vi pt ca (S).

Vic gii h phng trnh tng ng vi vic gii phng trnh : 2 2 2

2 2 21

x t y t z t

a b c

hay2 2 22 2 2

2 0 0 0 0 0 02 2 2 2 2 2 2 2 2

2 1x y z x y z

t ta b c a b c a b c

Vi ch rng M0(x0, y0, z0) nm trn (S) ta suy ra :2 2 2

2 0 0 02 2 2 2 2 2

2 1x y z

t ta b c a b c


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k (d) ct (S) ch ti im duy nht M0l pt (*) ch c nghim duy nht t = 0 . Vy

k\k l : 0 0 02 2 2 0x y z

a b c

(**)

b)ta xt vect 0 0 00 2 2 2; ;x y z

na b c

, vect ny khc 0 v x0;y0;z0khng ng thi bng 0 .

iu kin (**) chng t rng vect u ca d vung gc vi vect 0n . Vy cc tip tuyn

nm trn d phi i qua M0v c vect php tuyn 0n

.mt phng c pt:

0 0 0 0 0 02 2 2

( ) ( ) ( )0

x x x y y y z z z

a b c

hay : 0 0 02 2 2 1

x x y y z z

a b c

chnh l pt tip din ca (S) ti im M0.


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43

Ch 3HYPEBOLOIT

Dng 1 : Bin lun - Tm giao ca Hypeboloit vi ng thng v mt phng

Phng php:Bng cch gii h phng trnh ta xc nh c dng ca phng trnh giao tuyn cntm

Bi mu 1: Mt phng x = a ct hyperboloit 1 tng2 2 2

2 2 2(H):

x y z1

a b c theo

ng g?.

Li Gii:

Giao tuyn ca (H) vi mt phng x a l nghim ca h phng trnh:

2 2 2 2 2

2 2 2 2 2

x ax a x acy bz 0x y z y z

1 0cy bz 0a b c b c

Vy giao tuyn cn tm l hai ng thng ct nhau:x a x a

vcy bz 0 cy bz 0

Bi mu 2:Tm giao tuyn ca mt phng x 9 v mt hypeboloit :2 2 2

(H): x y z 19 8 2

Li Gii:Giao tuyn c xc nh bi:2 2 2

2 2 2 2x y z y z y z18 19 8 2 8 2 64 16

x 9

y l dng Hyperbol trong mt phng x 9

Nhn xt: - y l (H) nm trong mt phng x = 9.- Chng ta c th tm giao tuyn ca (H) vi cc mt phng i xng vxt xem th giao tuyn l ng g.

Dng 2: ng sinh thng ca mt Hypeboloit

Bi mu 1: Tm ng thng i qua im I( 1; 1; 1) v nm trn Hyperboloit 1

tng 2 2 2(H) : x y z 1


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Li Gii:

(d) qua I( 1; 1; 1) c phng trnh :x 1 aty 1 bt ,(a; b; c) (0; 0; 0)z 1 ct

Thay (d) vo (H) ta c:

2 2 2

2 2 2 2(1 at) (1 bt) (1 ct) 1(a b c )t 2(a b c)t 0

(d) nm trn (H)2 2 2 2 2 2a b c 0 a b (a b) 0

a b c 0 c a b

a 0c bab 0

c a b b 0c a

Vy c hai ng thng nm trn (H) l 1 2x 1 x 1 t

(d ) : y 1 t v (d ) : y 1z 1 t z 1 t

Nhn xt: y thc cht l i tm ng sinh thng ca Hypeboloit 1 tng.Chng ta s lm r vn ny trong ch ng sinh thng.

Bi mu 2: Cho Hyperboloit 1 tng2 2 2

2 2 2

x y z(H) : 1

a b c .Vit phng trnh

ng thng i qua o o oM(x ; y ; z ) (H) v nm trn (H)

Li Gii:

(d) qua o o oM(x ; y ; z ) (H) vi VTCP u ( ; ; ) (0; 0; 0) o

o

o

x x t

(d) : y y t

z z t

Thay (d) vo (H) ta c:2 2 2 2 2 2

2o o o o o o2 2 2 2 2 2 2 2 2

(x t) (y t) (z t) x y z1 ( )t 2( )t 0

a b c a b c a b c

(d) nm trn (H) (d) (H)

2 2 2o o o

2 2 2 2 2 2

x y z

va b c a b c (1)

2 2 2 2 2 22 2 2 2 2

o o o o o o o o o2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

2 2o o o o

2

x y z x y x y y x1 2

a b c c a b a b a b a b a b a b

y x y xab ab ab ab cc

(1)


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45

Xt h phng trnh:o o

o o o2 2 2

y xab ab c

x y za b c

Gii h pt trn, xem v l n s, ta c:2 2 2 2o o o o o o o o o o2 2 2 2

x y y x z x y x y z;a b ac c b a bc ca b a b

Nh vy nu ta chn2 2 2o o o2 2 2

z

c

x yc c 1

a b

th o o o o o oy x z x y z

a ; bb ac a bc

Tm li c hai phng trnh ng thng tha yu cu bi l:

o o o

2o o o o o o o2

x x y y z z

y x z x y z za b c 1b ac a bc c

Dng 3: Cc bi ton lin quan n Hypeboloit

Bi mu 1: Chng minh Hyperboloit 1 tng:2 2 2

2 2 2

x y z1 (a b 0)

a b c ct mt

cu 2 2 2 2x y z a theo 2 ng trn c bn knh R a.

Li Gii:

Giao tuyn ca Hyperboloit v mt cu c phng trnh:2 2 22 2 2 2 2 2 2 2 2 2 2

2 2 22 2 2 2 2 2 2

2 22 2 22 2 2 2 2 2 2 22 2 2 2

2 2 2

x y zx y z x y z a1 x y z a1 a b ca b c 1 1 1 1y z y z y ( ) z ( ) 0x y zx y z a 1 b a a cb c a a

a a a

t

2 2 2 21 22 2 2 2 2 2 2 2

1 1 1 1 1 1 1 1(S): x y z a ; ( ) : y z 0 ; ( ): y z 0

b a a c b a a c

(S) c tm O(0; 0; 0) v bn knh R a Khong cch

1 2(O,( )) (O, ( ))0d d

Vy hai mt bc hai ct nhau theo hai ng trn c bn knh R a. Bi mu 2:Vit phng trnh ng thng (d) i qua o o oM(x ; y ; z ) v ch ct

2 2 2

2 2 2

x y z(H) : 1

a b c ti 1 im duy nht.

Li Gii:


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47

v n ct cc mt (Oxz) v (Oyz) theo cc ng :2 2 2 2

1 1 v25 16 9 16

0 0

x z y z

y x

p s:

2 2 2

125 9 16x y z

Bi 2:Lp phng trnh Elipsoid qua M (1, 2, 23 ) v ct mt phng (Oxy) theong (C):

2 2

19 16

0

x y

z

p s:

2 2 2

19 16 36

x y z

Bi 3: Xc nh mt bc hai do mt ng thng chuyn ng to nn bit n ta lnba ng thng:

1 2 3

1 2 1: ; : ; :

2 0 1 0 1 1 2 0 1

x y z x y z x y zd d d

p s:

22 2 1

4

xy z


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48

Ch 4NG SINH THNG

Bi 1:Tm ng sinh thng ca mt:

a/2(S) : y 3xy 2yz zx 3x 2y 0 qua O(0; 0; 0)

b/2 2 2x y 5z 6xy 2yz 2zx 9 0

v song songx 1 y 3 z

(D):2 1 1

Gii:a.

(d) qua O(0; 0; 0) v c VTCP u (a; b; c) (0; 0; 0)

nn

x at

(d) : y btz ct

Thay (d) vo (S) ta c: 2

2 2

(bt) 3(at)(bt) 2(bt)(ct) (ct)(at) 3(at) 2(bt) 0

(b ab bc ac)t (3a 2b)t 0

(d) l ng sinh ca2

3a 2b 0(S) (d) (S)

b 3ab 2bc ac 0

2 2

3a 3ab b

2 29 3a 3a 9

a 3a( ) 2( )c ac 0 a 4ac 04 2 2 4

1

2

a 0u (0; 0; c) // u (0; 0; 1)3a

b2

3ab

3a 9a2 u (a; ; ) // u ( 16; 24; 9)9a 2 16

c16

Vy c 2 ng sinh thng ca (S) l 1 2

x 0 x 16t

(d ) : y 0 v (d ) : y 24tz t z 9t


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b/(d) //(D) nn (d) qua o o oM(x ; y ;z ) (S) v c VTCP u ( 2; 1; 1)

nno

oo

x x 2t

(d) : y y t

z z t

Thay (d) vo (S) ta c:2 2 2

o o o o o o o o o

2 2 2o o o o o o o o o o o

(x 2t) (y t) 5(z t) 6(x 2t)(y t) 2(y t)(z t) 2(z t)(x 2t) 9 0

( 12y 12z )t x y 5z 6x y 2y z 2z x 9 0

(d) l ng sinh cao o

2 2 2o o o o o o o

z y(S) (d) (S)

x 6y 6x y 2y 2x y 9 0

o o o o o o

2 2 2o oo o o o o o

z y z y z y

x 2y 3x 4x y 4y 9 0 (x 2y ) 9 0

Vy cc ng sinh thng ca (S) c dng

x 2k 3 2t(d) : y k t , k

z k t

R

Bi 2:Mt mt phng song song vi mt x y z 0 ct mt 2 2 2(H) : x y z 1 theo hai ng sinh thng. Tm giao im ca hai ng sinh thng ny v gcto bi chng.

Gii:

(P) song song vi mt phng x y z 0 nn (P) : x y z m 0 (m 0) c VTPT n ( 1; 1; 1)

(H) x z x z 1 y 1 y Phng trnh hai h ng sinh thng ca (H):

:

p x z q 1 y p x z q 1 y; :

q x z p 1 y q x z p 1 y(d) (d )

vi (p; q) (0; 0) v (p ;q ) (0;0)

px qy pz q 0 p x q y p z q 0: ; :

qx py qz p 0 q x p y q z p 0(d) (d )

(d) c cp VTPT 1

2

n (p; q;p)n (q;p; q)

nn c VTCP 2 2 2 2u (q p ;2pq; p q )

(d ) c cp VTPT 1

2

n (p ;q ;p )

n (q ; p ; q )

nn c VTCP 2 2 2 2u (p q ;2p q ; p q )


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2 212

2 22

1 2

; 0; q ) // ( 1; 0; 1)

; 2q ) // (0; 1; 1)

p 0 u (q u(d) (P) u.n 0 2pq 2p 0

p q u (0; 2q u

x y z 1 0y 1 0(d ) : ; (d ) :

x y z 1 0x z 0

2 212

22

1 2

; 0; q ) // ( 1; 0; 1)

// (0; 1; 1)

p 0 u ( q u(d ) (P) u .n 0 2p q 2p 0

p q u (0; 2p q ; 2q ) u

x y z 1 0y 1 0(d ): ; (d ):

x y z 1 0x z 0

Ta c: 1 1 2 2(d ) (d ) v (d ) (d )// //

Giao im ca 2 1(d ) v (d ) l im ( 1; 1; 1) trong mt phng 1(P ) : x y z 1 0 Giao im ca 1 2(d ) v (d ) l im ( 1; 1; 1) trong mt phng 2(P ) : x y z 1 0

1 2

2 1 1 21 2

u .u 1cos(d ,d ) cos(d ,d ) cos

2 3u . u

BBiittppttnnggtt

Bi 1: Tm ng sinh thng ca (S): zyx

416

22

bit n song song vi mt phng (P)

: 3x+2y-4z=0

p s : 1=

tz

tytx

21

14/325

2:

tz

tytx

1

24

Bi 2:Tm ng thng i qua im I( 1; 1; 1) v nm trn Hyperboloit 1 tng2 2 2(H) : x y z 1

p s: 1 2

x 1 x 1 t(d ) : y 1 t v (d ) : y 1

z 1 t z 1 t


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52

Li Gii:Giao ca mt trn xoay vi mt phng bt kz = t l mt ng trn c bn knh rphthuc vo tv c tm thuc trc Oz.Do , vi mi im M(x, y, z) bt k thuc ng trn v x2+ y2= r2V rph thuc vo tnn t phng trnh ng thng (d) ta c :

2 2 2 2(1 2 ) ( 3 3 )x y z z 2 2 2x + y -13z +14z -10 = 0

Bi 3 : Cho mt yn nga zyx 222 .Tm qu tch nhng giao im ca nhngng sinh thng vung gc vi nhau.

Li GiiGi (d) l ng thng i qua im M( );; 000 zyx (S)v VTCP )0;0;0();;( cbau

ctzz

btyy

atxx

d

0

0

0

Xt h phng trnh hp bi(d) v(S): )(2)()( 02

02

0 ctzbtyatx

0)(2)( 00222 tcbyaxtba (*)

(d) l ng sinh thng ca (S) khi v ch khi h (*) c nghim ng vi mi t

)(

)(

0)(

0)(

0

0

00

00

00

00

00

22

yxbc

ba

yxbc

ba

cyxa

ba

cyxa

ba

byax

ba

00)(110

;1;1();;1;1(

20

20

20

2021

00200

yxyxuu

yxuyxu

Vy qu tch giao im ca nhng cp ng sinh vung gc vi nhau l hai ng

thng ct nhau:

0

0

z

yxv

0

0

z

yx

Bi 4:Tm tm ,bn knh ca mt cu :2 2 2

( 4) ( 7) ( 1) 36( )3 0( )x y z Sx y z P

Li Gii.(S) c tm I(4,7,-1) v c bn knh = 6.Tm ca (C) l hnh chiu vung gc ca I(4,7,-1) ln(P)Phng trnh ng thng qua I v vung gc vi (P) l :


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53

(d)

4 3

7 ( )

1

x t

y t t R

z t

Giao im M(x,y,z) ca (d) v (P) l nghim ca h:3 9 0

4 3( )

7

1

x y z

x tt R

y t

z t

Ta c :3(4+3t)+7+t+1+t-9=0 t=-1. Vy ta ca M l (1,6,0) y l tm cang trn (C)Bn knh (C) :p dung nh l Pitago trong tam gic cho ta :r2=R2-d2(M,(P))[Trong:r:bn knh ng trn,R:bn knh mt cu (S), d2(M,(P)):khong cch t M n (P).]Vy ta c bn knh (C)l:r= 36 11 =5.

Bi 5:Qua im M(1,-2) dng ng knh ca ( C):2 23 2 3 4 4 4 0x xy y x y V ng knh lin hp vi n.Vit phng trnh cc ng knh :Li Gii .V ng knh l ng thng qua tm I nn trong trng hp ny ng knh cn tml ng thng MI.

Tm I(x,y) l nghim ca h:3 2 0

( 1, 1)3 2 0

x yI

x y

Phng trnh IM:1 2

2 3 01 1 1 2

x yx y

Phng trnh ng knh lin hp vi ng knh trn c dng:1F'x(x,y)+2F'y(x,y)=0 6x-2y+4+2(-2x+6y+4)=0 x+5y+6=0

Bi 6 : Xc nh mt do 1 ng thng chuyn ng to nn bit rng n ta trn 3 ngthng

1 3

2

x y 1 z x y 1 zd : ; d :

2 0 1 2 0 1x 2 y z

d : 0 1 1

Gii

Ly 0 0 0 0M x , y , z (d) v v a,b,c 0,0,0,

l vtcp ca (d)

d ct d1

0 0 0

a b c

2 0 1 0

x y 1 z

0 0 0 0a y 1 x 2z b 2y 2 c 0 1


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54

d ct d2

0 0 0

a b c

0 1 1 0

x 2 y z

0 0 0 0a z y b 2 x c 2 x 0 2

d ct d3

0 0 0

a b c2 0 1 0

x y 1 z

0 0 0 0a y 1 b 2z x c 2y 2 0 3

H (1), (2), (3) l h thun nht c nghim (a, b, c) (0, 0, 0)

0 0 0 0

0 0 0 0

0 0 0 0

y 1 x 2z 2y 2

z y x 2 2 x 0

y 1 2z x 2y 2

0 0 0 0

0 0 0 0

0 0

y 1 x 2z 2y 2

z y x 2 2 x 0

2 4z 4y

2 2 20 0 0x 4y 4z 4 0

Vy phng trnh ca mt phng cn tm l : 2 2 20 0 0x 4y 4z 4 0

BBiittppttnnggttBi 1:Cho ng cong : 2 23 2 2 3 4 0.x xy y x y v 1 ng knh ca n : x+2y-2=0.Lp phng trnh ng knh lin hp vi ng knh trn.

p s:2x+2y-1=0Bi 2: Mt mt phng song song vi mt x + y z = 0 ct mt x2+ y2 z2= 1 theo haing sinh thng. Tm giao im ca hai ng sinh thng ny v gc to bi chng.


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55

Chuyn QU TCH

Dng 1: Qu tch tm ca ng bc hai (C)

Phng php:ng bc 2 (C) c dng: 0),,( myxF trong m l tham s.

To tm I (x,y) ca (C) tho h phng trnh:

0),,('

0),,('

myxyF

myxxF (1)

Kh tham s m trong h (1) Ta c qu tch cn tm.

Bi mu 1:Tm qu tch tm ca tt c nhng ng cong c phng trnh:

(C): 014my2mx2y2xy2xm)y,F(x, , trong m l tham s.

Gii :

Trc ht, ta i tm to tm I(x;y) ca(C) l nghim ca h phng trnh :

(2)02'

(1)0'

myxyF

myxxF

H phng trnh cho c nghim duy nht vi mi m,do ng cong (C) lun ctm .By gi ta s kh mtrong h trn bng cch rt m= x+y t (1) thay vo (2) ta c qu

tch tm I ca h ng cong (C) l ng thng :3x+y = 0

Nhn xt : - Qu tch va tm c l mt ng thng i qua gc to O.- C th kh m h trn bng cch nhn phng trnh (1) vi 2ri cng vo (2) ta vn c qu tch cn tm.

Bi mu 2 : Tm qu tch tm ca tt c nhng ng cong bc hai(C) i qua bnim O(0 ;0) ; A(2 ;0) ; B(0 ;1) ; C(1 ;2).

Gii :

Trc ht, ta vit phng trnh (C) dng tng qut :F(x,y) = ax2+ 2bxy + cy2 + 2dx + 2ey + f = 0

V (C) qua O(0;0) nn: f = 0(C) qua A(2 ;0) nn : d = -a(C) qua B(0 ;1) nn : c = - 2e

Khi (C) c vit li l : ax2+ 2bxy - 2ey2- 2ax +2ey = 0Mt khc, v C(1 ;2) (C) nn ta c : a = 4(b-e) (1)To tm I (x ;y) ca (C) tho h phng trnh :


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56

0

0

eeybx

abyax

(2)

0)1(

0)1(

yebx

byxa

Thay (1) vo (2) ta c :

0)1(

0)1)((4

yebx

byxeb

)1(

)1(4)44(

yebx

xeyxb

Chia v theo v ta c :

1)1(444

yx

xyx

4x2- 4xy - y2+ 5y - 4 = 0 (C)

Vy qu tch cc tm ca ng cong (C) chnh l ng cong (C) v tho

iu kin : 0)(4

eb

beb

Nhn xt: - Qu tch bi ny cha hai tham s trong h phng trnh nn tala chn phng php lp t s kh n.

- y l bi ton qu tch c iu kin nn phi ch n iu kin

Dng 2: Qu tch trung im cc dy cung c phng v

ca ng bc hai (C).

Phng php:

Trung im ca cc dy cung c phng );(v

l ng knh lin hp vi phng:- Trc ht ta vit phng trnh tng qut (d) c phng );(v

-Tm iu kin (d) v (C) c giao im A,B.Ti y ta c 2 cch gii:

Cch 1: To trung im tho:

)(2

1

)(21

By

Ay

Iy

BxAxIx (I)

- Kh tham s trong h (I) ta qu tch l mt ng thng vi iu kin bc 2Cch 2:

Qu tch c dng:.Fx(x;y) + .Fy(x;y) = 0

V c iu kin nh bc 2.


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57

Bi mu 1: Cho ng cong (C) c phng trnh:F(x;y) = 3x2+ 7xy + 5y2+ 4x + 5y + 1 = 0 (1)

Tm qu tch trung im ca nhng dy song song vi trc Ox.

Gii:

ng thng song song vi trc Ox c dng: y = m. Thay vo (1) ta c:3x2+ (7m+4)x + 5m2+ 5m + 1 = 0 (*)

phng trnh (*) c nghim th = (7m + 4)2- 12(5m2+ 5m + 1) 0 hay:-11m2- 4m + 4 0

11

23411

234

m (**)

Vi iu kin (**) th phng trnh (*) c 2 nghim x1v x2l honh ca cc giaoim A,B ca (C) v ng thng y = m. Trung im I ca AB c to :

my

mxxx )47(

6

1)

21

(

2

1

(***)

Kh m trong (***) ta c qu tch trung im I ca cc dy song song vi trc Ox lng thng:

6x + 7y +4 = 0

vi iu kin11

234y

11234

Nhn xt : - y s dng cch 1 gii quyt bi ton. Tuy nhin ta cc th s dng cch 2 cng i n kt qu trn.

- Trong iu kin(**) khng nht thit phi c 0, v khi = 0th phng trnh (*) c 2 nghim trng nhau, nn xem nh vnc trung im ca nhng dy ny.

Bi mu 2:Cho ng cong (C) c phng trnh:F(x;y) = 3x2+ 7xy + 5y2+ 4x + 5y + 1 = 0 (1)

Tm qu tch trung im ca nhng dy vung gc vi ng thng(d): x - y + 1 = 0.

Gii:

Phng trnh ng thng (D) vung gc vi (d) c dng: x + y + m = 0.Giao im ca (D) v (C) l nghim ca phng trnh:

x2+ (3m-1)x + 5m2- 5m + 1= 0 (*) phng trnh (*) c nghim th:

= (3m-1)2- 4(5m2- 5m +1) 0 -11m2+ 14m - 3 0

1113

m (**)

Honh trung im I ca AB l: )13(21

)(21

mBxAxIx (***)


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58

(xA, xBl nghim ca (*))

T (**) v (***) ta c :111

1 Ix (****)

Vi iu kin (****) th qu tch trung im I ca AB l ng knh lin hp viphng )1;1( v

l vect ch phng ca (D) c phng trnh:

1.Fx(x;y) - 1.Fy(x;y) = 0 x + 3y + 1 = 0

Vy qu tch trung im I ca AB l:

11

1x1-

013yx

Nhn xt: - y s dng cch 2 gii quyt bi ton, nu s dng cch1 ta vn i n kt qu trn, nhng vic s dng cch no giiquyt dng ton ny l tu vo k nngca mi ngi.

- y gii hn qu tch (****) c c l do chng ta linh hot

da vo iu kin (**) v (***)

Dng 3: Qu tch giao im ca nhng cp ng sinh thng ca mt k bchai (S) vung gc nhau.(Mt Hypeboloit 1 tng v Paraboloit hypebolic)

Phng php:- Vit phng trnh hai h ng sinh thng d v d ca mt bc hai (S).- Tm vecto ch phng ca d v d.- d d th 0'. dudu

=> Qu tch cn tm.

Bi mu 1: Cho mt Paraboloit Hypebolic : x2- y2= 2z (S).Tm qu tch nhng cp ng sinh thng vung gc vi nhau.

Gii:

Hai h ng sinh thng ca (S) l :

0)(q'

2p'y)(xq'

zq'y)-(xp':d' v0)(p

)(

2)(:

pzyxq

qyxpd

0)(q'

0'2''0zq'-yp'-xp':d'v0)(p

002:

pyqxqpzqyqxqpypxd

ng thng d c cp vec t php tuyn );;(2

)0;;(1pqqn

ppn

nn c vec t ch phng

)2;;//()2;2;2(1 qpppqppu


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59

ng thng d c cp vec t php tuyn)0;';'(2'

)';';'(1'

qqn

qppn

nn c vec t ch phng

)'2;';'//()''2;2';2'(2 pqqqpqqu

d d th 0'

. d

ud

u

4pq = 0

Nu p= 0 th qu tch cc giao im cn tm l ng thng:

0

0

z

yx

Nu q = 0 th qu tch cc giao im cn tm l ng thng:

0

0

z

yx

Vy qu tch cc giao im ca hai ng sinh thng vung gc nhau l hai ngthng va tm c trn.

Nhn xt: - y phng trnh mt bc 2 c cho dng chnh tc. Nu

(S) khng dng chnh tc th bi ton tr nn phc tp hnnhng c th gii quyt nh trn khi chuyn n v dng chnhtc.

- bi gii ny s dng mt tnh cht c bit ca mtParaboloit hypebolic l: hai ng sinh thng khc h luncng nm trong mt mt phng v ct nhau.

Bi mu 2: Cho mt yn nga: b)(a2z2b

2y2a

2x . Tm qu tch giao

im ca nhng cp ng sinh thng vung gc vi nhau.

Gii:

H hai ng sinh thng ca (S) l:

0)(q')(2''2)('

)(1'')(':d'v0)(p

(2))(

(1)2)(:

pby

ax

q

zqby

ax

p

pzby

ax

q

qby

ax

pd

)0(q'

0'2''

0'''

:d'v0)(p

0

02:

pybq

xaq

zqyb

px

ap

pzybq

xaq

qybp

xap

d

ng thng d c vec t ch phng )2

;a

p;

b

p(//)

2;

2;

2(

ab

q

ab

pq

a

p

b

pdu

ng thng d c vc t ch phng )'2

;'

;b

q'(//)

''2;

2';

2'(' ab

p

a

q

ab

qp

a

q

b

qdu

V d v d l hai h ng sinh thng ca mt Paraboloit hypebolic nn chng lun ctnhau.


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60

d d th 0'. dudu

0'

42

'2

'

abqp

a

pq

b

pq

')22('4 pqabqp (3)

Ly (1) nhn vi (2) ri so snh vi (3) ta c: 2222

22 ab

b

y

a

x

Ly (2) nhn vi (1) ri so snh vi (3) ta c:4

2)22(2 abz

Vy qu tch cn tm l ng cong:

2

22

222

2

2

2

abz

abb

y

a

x

Nhn xt: - y l mt dng tng qut ca mt Paraboloit hypebolic. biny phng trnh (3) c 4 n p,p,q,q nn ta chn phngphp so snh kh n.

- Trong phng trnh4

2)22(2 abz

ta ch nhn2

22 abz

v2

22 abz

khng tho phng trnh mt yn nga.


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61

BBiittppttnnggtt

Bi 1: Cho ng cong bc 2 (C):F(x;y) = x2- 2xy + 2y2- 4mx - 6my + 3 = 0

Tm qu tch tm I ca ng cong (C).p s:x - 7y = 0

Bi 2: Cho ng cong bc 2 (C):F(x;y) = 5x2+ 4xy + 8y2- 32x - 56y + 80 = 0

Mt ng thng thay i song song vi trc Oy ct ng bc hai (C)ti cc im M v N. Tm qu tch trung im ca on thng MN.

p s:

222x222

0144yx

Bi 3: Cho mt yn nga: 2z9

2y

4

2x Tm qu tch giao im ca nhng cp

ng sinh thng vung gc vi nhau.

p s:

2

5z

59

2y

4

2x


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62

Chuyn CHNG MINH CC NG THC

HNH HC

Dng 1: Chng minh cc tnh cht ca ng bc hai

Phng php- Nm vng cc kin thc v ng trn, Elip, Hypebol, Parabol.- Nm vng cc kin thc v tm, phng tim cn, ng tim cn.- C k nng tnh ton v bin i.

Bi mu 1: Mt dy cung AB tu i qua tiu im F ca Elip, Hypebol,

Parabol. Chng minh rng : FB

1

FA

1

lun khng i.

CM:

- Xt Parabol : y2= 2px. ng thng (d) qua F c phng trnh )2

(p

xky

Giao ca (d) v (P) l : pxp

xk 22)2

(2

04

22)22(22

kpxkpxk (*)

Gi xA ; xBln lt l nghim ca phng trnh (*) th ta c :

4

2.

pBxAx

0v4

2 Ax

Axp

Bx

Ta c:2

2)2

(22)2

(p

Axp

AxAyp

AxFA

Tng t ta c : )2

2(

2)1

2(

224

2

2 AxAxpp

Axppp

Axpp

BxFB

Khi :

)2

2(

2

1

2

111

AxAxppp

AxFBFA

)2

(

2

2

1p

Axp

Ax

pAx


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63

=p2

(pcm)

- Tng t vi Elip 12

2

2

2

b

y

a

x ta c

211 2

bFBFAa

- Vi Hypebol 122

22

by

ax ta c

211 2

bFBFAc

Nhn xt: - Dng bi tp chng minh i hi ngi lm phi c kin thc vng vc k nng tnh ton,v vy chng ta phi bin i cn thn v vn dngcc gi thit ca bi ton.- Cch chng minh cho Elip v Hypebol hon ton tng t nh (P).

Bi mu 2: Chng minh rng:Nu mt ng bc hai c tm nm trn Ox v tip xc vi trc Oy ti O th phngtrnh ca n c dng: ax2+ cy2+ 2dx = 0.

CM:

Gi s ng bc hai (C) c phng trnh:

ax2+ 2bxy + cy2+ 2dx + 2ey + f = 0

Tm I ca (C)l nghim ca h phng trnh:

0

0

ecybx

dbyax

Do tm I Ox nn ta c:

0

0

ebx

dax

Mt khc, v (C) tip xc vi Oy ti O nn phng trnh cy2+ 2ey + f = 0 phi c

nghim kp y = 0, do ta c:

0

0

fe

c

V e = 0 nn t iu kin bx + e =0 ta suy ra b= 0 v t phng trnh ax + d = 0 ta suyra a 0.Vy ng bc hai (C) c dng: ax2+ cy2+ 2dx = 0.

Nhn xt: - y l mt bi ton d, nhng i hi ngi hc phi pht hinra y cc d kin th mi c th gii quyt c.

- Ch trng hp nghim kp y = 0 trong bi ton ta ch cn e= f= 0 v c 0.


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64

Bi mu 3: Trong h to cc vung gc Oxy cho Hypebol:

12b

2y2a

2x

a. CMR: tip tuyn ti M ca Hypebol l phn gic trong ca gc

21 FMF (F1, F2) l hai tiu im..b. CMR :din tch ca tam gic to bi hai ng tim cn caHypebol vi tip tuyn tu ca n l khng i.

CM:

Gi A,B l 2 giao im ca hai ng tim cn ca Hypebol v mt tip tuyn bt kca n.Phng trnh ca tip tuyn AB ti im M(x0, y0) bt k l :

12020 ybyx

ax

Phng trnh hai tim cn OA v OB l :

xab

xab

y yv

To ca A,B ln lt l :

o o o o o o o o

a b a bA ; B ;

x y x y x y x ya b a b a b a b

Tnh di on thng AB.Tnh khong cch t O n ng thng AB.Khi din tch tam gic OAB l :

abABOdABOABS ),(.21

(pcm)

Nhn xt: - V M(H) nn2 2o o2 2

x y1

a b

- Khi tnh ton cn ht sc cn thn trnh sai st dn n sai.

Dng 2: Chng minh cc tnh cht ca mt bc hai

Phng phpy l mt dng bi tp kh, v vy gii tt ta phi c mt s k nng sau :- Nm vng php chiu v tm giao im ca mt bc hai vi mt mt phng.- Bit vit v bin i trn h ng sinh thng.- C k nng tnh ton cn thn.


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65

Bi mu 1 :CMR : mt phng z = Ax + By + C ct Paraboloit x2+ y2= 2pz

(vi p>0) theo mt Elip th hnh chiu vung gc ca Elip xung mt phng Oxy l mt ng trn.

CM:

Mt phng ct Paraboloit nn h phng trnh sau c nghim :2 2x y 2pz

z Ax By C

(*)

Do tp hp cc im M(x,y,z) tho mn phng trnh sau khng rng :2 2

2 2 2 2

x y 2p(Ax By C)

(x-pA) (y pB) p(2pC A B )

Phng trnh cui l mt mt tr trn xoay c ng chun l mt ng trn trong

mt phng Oxy c tm (pA, pB) v bn knh 2 2R p(2pC A B )

Nhn xt : -Bt k mt bc hai no c dng F(x, y) hoc F(x, z) ; F(y, z)l mt mt tr trn xoay.

- Bi ny tm hnh chiu ca ng cong xung mt phng OxyNn ta kh z trong phng trnh (*).

Bi mu 2 : CMR:Khng c bt k ng thng no nm hon ton trn mt

Elipxoit : 2 2 2

2 2 2

x y z1

a b c

CM

Gi M0(x0,y0, z0) l mt im bt k thuc (E).Phng trnh ng thng (d) qua M0c phng trnh tham s l :

o

o

o

x x mt

y y nt

z z pt

Tham s t ng vi giao im ca(d) v (E) l nghim ca phng trnh :

12

2)0(

2

2)0(

2

2)0(

cptz

bnty

amtx

(m2b2c2+ n2a2c2+ p2a2b2).t2+ 2(mb2c2x0+ na2c2y0+ pa2b2z0).t= 0 (*)V phng trnh (*) c m2b2c2+ n2a2c2+ p2a2b2> 0 nn phng trnh ny c nhiunht l hai nghim do (d) ct (E) nhiu nht ti hai im. Do (d) khng nm honton trn Ellipxoit.

Nhn xt: -Ngoi Elipxoit cn c nhng mt sau cng c kt qu chngminh tng t : Hypeboloit hai tng ; Paraboloit Eliptic.


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66

- ng thng (d) nm hon ton trn mt bc hai th phngtrnh (*) phi nghim ng vi mi t.

Bi mu 3 : Cho Hypeboloit 1 tng : 2 2 2

2 2 2

x y z1

a b c (H)

Xt cc h nm hon ton trn (H). CMR : Qua mi im ca mtc ng mt ng thng ca mi h.

CM

Hai h ng sinh thng ca (H) l :

(0,0))q',(p'1''

1'':d'v(0,0)q)(p,

1

1:

by

pcz

ax

q

by

qcz

ax

p

by

pcz

ax

q

by

qcz

ax

pd

Gi s Mo(x0, y0, z0) (H) th Mophi thuc d v d. T ta c :

+ Nu Mod :

o o o

o o o

x z yp q 1

a c b

x z yq p 1

a c b

V p v q c chn sai khc mt tha s khc 0 nn d l ng thng qua M oduynht.Tng t, v p v q c chn sai khc mt tha s khc 0 nn:

+ Nu Mod :

o o o

o o o

x z yp ' q ' 1

a c b

x z yq ' p ' 1

a c b

th d cng l ng thng qua Moduy

nht.Tm li, qua mi im thuc (H) c ng mt ng sinh thng ca mi h i qua.

Nhn xt: - Cch chng minh cho Paraboloit hypebolic hon ton tng t.- p,q sai khc mt tha s khc 0 ngha l ta lun c th chn

c (p,q) (0,0).


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67

Bi mu 4 : Cho Hypeboloit 1 tng : 2 2 2

2 2 2

x y z1

a b c (H)

CMR:

+ Hai ng thng phn bit ca cng mt h ng sinh thnglun cho nhau.

+ Hai ng sinh thng khc h lun cng nm trong mt mtphng.

CM

+ Gi s hai ng sinh thng ca cng mt h c phng trnh:

1 : ( , ) (0,0)

p q px y z q

a b cd p qq p q

x y z pa b c

v 2

' ' ''

: ( ', ') (0,0)' ' '

'

p q px y z q

a b cd p qq p q

x y z pa b c

Giao im ca d1v d2 l nghim ca h phng trnh:

' ' ''

' ' ''

p q px y z q

a b cq p q

x y z pa b c

p q px y z q

a b cq p q

x y z pa b c

(*)

Gii h phng trnh trn c rank(A)= 3


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68

Khi :' ' ' '

: d':' ' ' '

pX qY pZ q p X q Y p Z qd

qX pY qZ p q X p Y q Z p

T phng trnh ca d ta c:

2 2

2 2

2 ( )

2

X pq q p Z

Y p q pqZ

T phng trnh ca d ta c:2 2

2 2

2 ' ' ( ' ' )

' ' 2 ' '

X p q q p Z

Y q p p q Z

d v d ct nhau khi v ch khi h phng trnh sau c nghim :2 2 2 2

2 2 2 2

2 ( ) 2 ' ' ( ' ' )

2 ' ' 2 ' '

pq q p Z p q q p Z

p q pqZ q p p q Z

Hay2 2 2 2

2 2 2 2

( ' ' ) 2( ' ' )

2( ' ') ' '

q p p q Z p q pq

pq p q Z q p q p

(**)

V p,q v p,q khng ng thi bng 0 v xc nh sai khc mt tha s khc 0 nn tac th chn sao cho : p2+ q2= p2+ q2= 1Khi t (**) chng t d v d ct nhau.

Nhn xt: - Vi Paraboloit hypebolic ta chng minh hon ton tng t.- Trong (*) A l ma trn cc h s,A l ma trn cc h s m

rng v theo nh l Kronecker Capelly h (*) v nghim.

- Vic t ; ;x y z

X Y Za b c

l thun tin cho vic tnh ton.


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BBiittppttnnggtt

Bi 1: Cho Hypeboloit 1 tng : 12c

2z

2b

2y

2a

2x (H)

CMR : hnh chiu vung gc ca cc ng sinh thng ca (H)xung cc mt phng to l tip tuyn ca giao tuyn ca (H) vimt phng to .

Bi 2 : Cho Paraboloit hypebolic : 2 2x y

2pz2 2a b

(P)

CMR :Qua mi im ca mt (P) c ng mt ng thng ca mih ng sinh thng i qua.

Bi 3 : Cho Paraboloit hypebolic : 2 2x y

2pz

2 2a b

(P)

CMR : Hai ng sinh thng khc h lun ct nhau.

Hng dn:

Bi 1 :- Vit mt h ng sinh thng d ca (H)- Tm hnh chiu ca h xung mi mt phng to , gi s l d.- giao im ca d vi mi mt phng thuc (H)

- T hai phng trnh ca dta rt p, q ca (1) thay vo (2) v rt p, qt (2) thay vo (1) ri cng hai phng trnh vi nhau ta ciu cn chng minh.

Bi 2 :Lm tng t nh bi mu 3.Bi 3 :Lm tng t nh bi mu 4.


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Chuyn TH

Dng 1 : th ng bc hai

Phng php

S dng cc php i mc tiu hc bin i phng trnh ca ng bc hai (C) vdng n gin.V th ca ng bc hai .Nm vng hai php i mc tiu quan trng :

1. Php tnh tin theo vc t OI

: ' 'TOIOxy Ix y

' '0 1 1

' '0 2 2

x x a x b y

y y a x b y

2. Php quay tm O mt gc : ( ; ) ' 'Q OOxy Ox y

'cos 'sin

'sin 'cos

x x y

y x y

Bi mu 1 : Cho ng bc hai (C) trong h to cc :F(x, y) = 5x2+ 6xy + 5y2+ 22x - 6y + 21 = 0 (*)

a. ng bc hai (C) thuc loi no ?

b. Hy v (C)

Gii :a.

Bng php quay tm O mt gc ta c :5 5

cot 2 02 6

a cg

b

=> = 450

Khi ta c ( ; ) ' 'Q OOxy Ox y v

1( ' ')

21

( ' ')2

x x y

y x y

( 1)

Thay (1) vo (*) ta c :2 28 ' 2 ' 8 2 ' 14 2 ' 21 0x y x y

1 72 28( ' ) 2( ' ) 32 02 2

x y (2)


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Bng php i mc tiu : ' 'TOIOx y IXY

ta c :

1'

27

'2

X x

Y y

Vy ta tnh tin h trc Oxy sang IXY, trong 1 7( ; )2 2I

Khi (2) c vit li :2 2

14 16

X Y

Vy (C) l mt Elip c trc ln l trc IY v di trc ln l 2a = 8, trc nh l trcIX v di trc nh l 2b = 4.

b.Da vo cu a. ta v c th nh sau :

Nhn xt: - y l mt dng ton bin i khng phc tp nhng n i hi

ngi lm phi ht sc cn thn v chn cch bin i sao chophng trnh cui cng l n gin nht.- bi ton ny v th chng ta thc hin ln lt cc thao

tc : Quay quanh O mt gc 450, ri tnh tin h trc n htrc to IXY v v th trong h trc mi ny.

x

y

xy

O

I

X

Y

-3

450


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Bi mu 2 : Cho ng bc hai (C) trong h to cc :F(x, y) = 3x2+ 10xy + 3y2+ 2x - 2y -9 = 0 (*)

a. ng bc hai (C) thuc loi no ?b. Hy v (C)

Giia.

Bng php quay tm O mt gc ta c :3 3

cot 2 02 10

a cg

b

=> = 450

Khi ta c ( ; ) ' 'Q OOxy Ox y v

1( ' ')

21

( ' ')2

x x y

y x y

( 1)

Thay (1) vo (*) ta c : 2 28 ' 2 ' 2 2 ' 9 0x y y

12 28 ' 2( ' ) 8 02

x y (2)

Bng php i mc tiu : ' 'TOIOx y IXY

ta c :'

1'

2

X x

Y y

Vy ta tnh tin h trc Oxy sang IXY, trong 01

( ; )2

I

Khi (2) c vit li : 2 2 14

X Y

Vy (C) l mt Hypebol c trc thc l trc IX v di trc thc l 2a = 2 ; trc o ltrc IY v di trc o l 2b = 4.Hypebol (C) c hai tim cn l Y = 2X v Y = - 2X

b.

Da vo kt qu cu a. ta v th ca (C) nh sau :

x

yx

XY

y

OI

Y= 2X

Y=- 2X


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Nhn xt: - bi ton ny v th chng ta thc hin ln lt cc thaotc : Quay quanh O mt gc 450, ri tnh tin h trc n htrc to IXY v v th ca Hypebol trong h trc mi ny.

- Vic v th trong h to mi khng cn quan tm n cch to c m ch quan tm n h ang v m thi.

Dng 2 : th mt bc hai

Phng php

v c th ca mt bc hai ta phi ch ti cc tnh cht c bit ca mt nh l trc i xng, cc mt phng i xng, giao ca n vi cc mt to .Ta cn nm vng tnh cht ca hai mt c bit sau:

1. Mt Hypeboloit mt tng: 2 2 2

2 2 2

x y z1

a b c

- Hypeboloit 1 tng i xng vi cc mt phng to , cc trc to v gc to .- Giao ca Hypeboloit 1 tng vi mt phng song song vi trc Oz l Hypebol hoc lnhng cp ng thng.

2. Mt Paraboloit hypebolic : 2 2

2 2

x y2pz

a b

- i xng vi cc mt Oxz ; Oyz v trc Oz.- Giao vi mt phng z = h l 1 cp ng thng ( nu h = 0) hoc l nhng Hypebol(nu h 0).

Bi mu 1 : Trong khng gian cho mt (S) :2 2 2x y z+ - = 1

4 9 16

a. Mt (S) thuc loi no?b. Hy v (S).

Gii

a.Mt bc hai (S) l Hypeboloit 1 tng i xng vi cc trc to ,cc mt to v gcto .

- Giao ca (S) vi mt Oxy l mt Elip c phng trnh:2 2

x y+ = 14 9.y l Elip c

trc ln l Oy, trc nh l Ox.

- Giao ca (S) vi cc mt z = 4 v z = - 4 l Elip c phng trnh l:2 2x y

+ = 18 18

y cng l cc Elip c trc ln song song vi trc Oy.- Nu ct (S) theo cc mt phng song song vi trc Oz ta s c giao tuyn l nhngHypebol c trc thc nm trong mt v song song vi trc Oz.


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x

y

z

O

4

- 4

b.Da vo kt qu cu a. ta v th ca mt S nh sau:

Nhn xt: - v mt (S) ta cn tm cc giao tuyn c bit, tc l giaotuyn vi cc mt phng to .

- v cn xng th ta cn s dng cc ng ging songsong vi cc trc to .

- Vic chn z = 4 v z = -4 l cho kt qu p hn m thi.

Bi mu 2: Trong khng gian cho mt (S) :2 2x y

- = 2z 16 9 a. Mt (S) thuc loi no?b. Hy v (S).

Gii:

a.Mt (S) l mt bc hai Paraboloit hypebolic- Mt (S) nhn trc Oz lm trc i xng, mt Oxz, Oyz lm mt i xng.- Giao ca (S) vi mt phng Oxy l hai ng thng ct nhau ti O :

3 3 v4 4y x y x

- Giao ca (S) vi mt phng z = 2 l hypebol :2 2x y

- = 164 36

(1)

Hypebol ny c trc thc song song vi Ox.

- Giao ca (S) vi mt phng z = -2 l hypebol :2 2y x

- = 136 64

(2)

Hypebol ny c trc thc song song vi Oy.


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b.Da vo phn tch cu a. ta v th ca mt (S) nh sau :

Nhn xt: - Vic v th ca Paraboloit hypebolic rt phc tp nn i hichng ta phi nhn din cc giao tuyn cho tht chnh xc.

- Vic chn z = 2 v z = -2 ch c mc ch gip vic tnh tontin hn m thi.

- Mc ch s dng cc ng ging song song vi cc trc to l cho th c cn i hn. y l mt vic rt quantrng.

BBiittppttnnggtt

Bi 1 : Cho ng bc hai (C) trong h to cc :F(x, y) = x2- 2xy - 2y2- 4x - 6y + 3 = 0

a. ng bc hai (C) thuc loi no ?b. Hy v (C)Hng dn: th l mt Hypebol.

Bi 2 : Cho ng bc hai (C) trong h to cc :F(x, y) = x2- 2xy + 2y2- 4x - 6y + 3 = 0 (*)

a. ng bc hai (C) thuc loi no ?b. Hy v (C)Hng dn: th l mt Elip.

Bi 3 : Trong khng gian cho mt (S) :2 2x y

- = 2z 8 18

a. Mt (S) thuc loi no?b. Hy v (S).

Hng dn: y l th mt Paraboloit hypebolic.

y x

z

O

2

-2


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TI LIU THAM KHO

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