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Chuyen e 1: PHNG TRNH AI SO& BAT PHNG TRNH AI SO

TOM TAT GIAO KHOACAC HANG ANG THC C BAN

1. ( ) + = + +2 2 22a b a ab b22a b a ab b

)a b a b33 3a b a a b ab b33 3a b a a b ab b

)b a b a ab b

)a b a ab b

abbaba 22)(22 +=+

2. ( ) = +2 2 abbaba 22)(22 +=+

3. a b = + 2 2 ( )(

4. ( ) + = + + +3 3 2 2 )(33)(33 baabbaba ++=+

5. ( ) = + 3 3 2 2

6. a + = + +3 3 2 2( )(

7. a b = + +3 3 2 2( )(

Ap dung:Biet va . Hay tnh cac bieu thc sau theo S va PSyx =+

+= 4xD

Pxy =

d 2) ya += 2xA 2y)-(xB =)b 3) yc += 3xC 4) y

A. PHNG TRNH AI SO

I. Giai va bien luan phng trnh bac nhat:

1. Dang : ax + b = 0 (1)

sotham:ba,

soan:x

2. Giai va bien luan:

1

Ta co : (1) ax = -b (2)

Bien luan:

Neu a 0 th (2) a

bx =

Neu a = 0 th (2) tr thanh 0.x = -b* Neu b 0 th phng trnh (1) vo nghiem* Neu b = 0 th phng trnh (1) nghiem ung vi moi x

Tom lai :

a 0 : phng trnh (1) co nghiem duy nhata

bx =

a = 0 va b 0 : phng trnh (1) vo nghiem a = 0 va b = 0 : phng trnh (1) nghiem ung vi moi x

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Ap dung:V du : Giai va bien luan cac phng trnh sau:

mxmx 222 +=+

3. ieu kien ve nghiem so cua phng trnh:

nh ly: Xet phng trnh ax + b = 0 (1) ta co:

(1) co nghiem duy nhat a 0 (1) vo nghiem

=

0

0

b

a

(1) nghiem ung vi moi x

=

=

0

0

b

a

Ap dung:V du : Vi gia tr nao cua a, b th phng trnh sau nghiem ung vi moi x

0)1(24

=++ bxaxaII.Giai va bien luan phng trnh bac hai:

1. Dang: 2 0ax bx c+ + = (1)

sotham:c,ba,

soan:x

2. Giai va bien luan phng trnh :

Xet hai trng hpTrng hp 1

2

: Neu a th (1) la phng trnh bac nhat : bx + c = 00=

b 0 : phng trnh (1) co nghiem duy nhat bcx = b = 0 va c 0 : phng trnh (1) vo nghiem b = 0 va c = 0 : phng trnh (1) nghiem ung vi moi x

Trng hp 2: Neu a 0 th (1) la phng trnh bac hai co

Biet so ( hoac2 4b a = c ' 2 '' vi b2

bb ac = = )

Bien luan:) Neu th pt (1) vo nghiem0 0

V du 2: Tm GTNN cua ham so : xxy += 42

b) Phng phap 2: S dung ieu kien co nghiem cua pt hoac he phng trnh

V du: Tm GTLN va GTNN cua ham so:2232++

+=xx

xy

b) Phng phap 2: S dung ao ham, lap BBT cua ham so f tren D roi suy ra ket quaV du 1: Tm GTLN cua ham so : 4334 xxy =

V du 2: Tm GTNN cua ham so :x

xy22 += vi x > 0

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V du 3: Tm GTLN va GTNN cua ham so : xxy += 42

V du 4: Tm GTLN va GTNN cua ham so : x-2xsin=y tren

2;2

V du 5: Tm GTLN va GTNN cua ham so :cosx2

sinx

+=y tren ;0

V du 6: Tm GTLN va GTNN cua ham so : 22 xxy +=

V du 7: Tm GTLN va GTNN cua ham so :2

12cossin += xxy

V du 8: Tm GTLN va GTNN cua ham so :

)8cos4(cos2

1)4cos.2sin1(2 xxxxy +=

BAI TAP REN LUYENBai 1: Tm GTLN va GTNN cua ham so:

vixxxxy 922334 += ]2;2[x Bai 2: Tm GTLN va GTNN cua ham so :

xxy = 2sin tren

2;2

Bai 3: Tm GTLN va GTNN cua ham so :

exy .2= tren ]2;3[

Bai 4: Tm GTLN va GTNN cua ham so : = y 5cosx cos5x tren [ ;4 4]

Bai 5: Tm GTLN va GTNN cua ham so:2232++

+=xx

xy

Bai 6: Tm GTLN va GTNN cua ham so: 2312 xxy +=

Bai 7: Tm GTLN va GTNN cua ham so: 24)2( xxy += Bai 8: Tm GTLN va GTNN cua ham so:

12)3( += xxy vi ]2;0[x

Bai 9: Tm gia tr ln nhat va gia tr nho nhat cua ham so :+ +

= +

22cos cos 1cos 1

x x

y x Bai 10: Tm gia tr ln nhat va gia tr nho nhat cua ham so

= 4 32sin sin tren oan 0;3y x x

Bai 11: Tm GTNN cua ham so :3 322 xxy = tren oan

3;21

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Bai 12: Cho phng trnh vi . Tm a e nghiem ln013)62(2 =++ axax 1acua phng trnh at gia tr ln nhat.

Bai 13: Cho ham so1

242)1(2

++

=x

mmxmxy (1)

Xac nh cac gia tr cua m e ham so co cc tr. Tm m e tch cac gia tr

cc ai va cc tieu at gia tr nho nhatBai 14: Tm GTLN va GTNN cua ham so :

xxxxxf 2sin32)cos(sin222cos)( ++=Bai 15: Tm gia tr ln nhat va be nhat cua ham so sau :

= + +2 2y 4cos x 3 3sinx 7sin x

Bai 16: Tm GTLN va GTNN cua ham so :1sin2sin

1sin

++

+=xx

xy

Bai 17: Tm GTLN va GTNN cua ham so:

= + 12(1 sin2 cos4 ) (cos4 cos8 )2y x x x x Bai 18: Tm GTLN va GTNN cua ham so: = + +3 32(sin cos ) 8sin .cosy x x x x

Bai 19: Chng minh cac bat ang thc sau : 174sin4)sin1(81 + xx Rx

--------------------------------Het----------------------------------

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Chuyen e12: CAC BAI TOAN LIEN QUAN EN THAM SOCac phng phap giai thng s dung

Phng phap 1: Phng phap ai so. S dung cac phep bien oi tng ng thch hp e tm so nghiem

V du: Tm m e he phng trnh sau co nghiem:

=+

=+

myyxx

yx

31

1

Phng phap 2: Phng phap giai tch S dung cong cu ao ham xet tnh n ieu, cc tr, GTLN & GTNN e tm so

nghiem

V du: Tm m e vi moi034cossin82cos2 + mxxx

4;0

x

Phng phap 3: Phng phap o th cua giai tch Da vao v tr tng oi cua cac o th e e tm so nghiem

V du: Bien luan theo m so nghiem cua phng trnh: mxxxx +=+ 5452 22

Phng phap 4: Phng phap o th cua hnh hoc giai tch Da vao cac o th cua hnh hoc giai tch e tm so nghiem

V du: Bien luan theo m so nghiem cua phng trnh: mxx = 2312 Phng phap 5: Phng phap ieu kien can va u

V du: Cho he phng trnh:

=+

=+222 6 myx

myx

Tm m e he phng trnh a cho co nghiem duy nhat, xac nh nghiem o

Chu y: Khi co s dung an phu th phai tm ieu kien ung cho an phu

80

Phng phap ai so Phng phap giai tch Phng phap o th cua

giai tch

Pt,bpt,hpt, hbptco cha tham so

Phng phap o th cuahnh hoc giai tch

Phng phap ieu kiencan va u

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BAI TAP REN LUYEN:

Bai 1: Tm m e cos vi moi034cossin822 + mxxx

4;0

x

Bai 2: Tm m e phng trnh sau co nghiem:

02sin4

12coscossin 244 =+++ mxxxx

Bai 3: nh m e phng trnh : mxxgxtgxxx =++++++ )cos1sin1cot(211cossin

co nghiem

2;0

x

Bai 4: Cho bat phng trnh : (1)0324 + mm xx

Tm m e bat phng trnh (1) co nghiem.

Bai 5: Cho phng trnh : ( ) 0loglog42

1

2

2 =+ mxx (1)

Tm m e phng trnh (1) co nghiem thuoc khoang (0;1).

Bai 6: Cho ham so: 1)coscos2

()coscos

4

(2

2

2=++

xxmxx

Tm m e phng trnh co nghiem thuoc ).2

;0(

Bai 7: Tm tat ca cac gia tr cua m sao cho ta co:Rxmxxxx ++ ,cos.sincossin 66

Bai 8: Tm m e bat phng trnh sau ung vi moi x [ 4;6] 2(4 x)(6 x) x 2x m+ +

Bai 9: Cho phng trnh : 01)cot(3sin

3 22

=+++ gxtgxmxtgx

Tm tat ca cac gia tr cua m e phng trnh co nghiem.Bai 10: Xac nh m e phng trnh :

4 42(sin x cos x) cos4x 2sin2x m 0+ + + =

co t nhat mot nghiem thuoc oan [0; ]2

Bai 11: Cho phng trnh : mxxx = )sin(cos42sin (1)Tm tat ca cac gia tr cua m e phng trnh (1) co nghiem.

Bai 12: Cho bat phng trnh : 2m. 2x 7 x m+ < + (1)Tm m e bat phng trnh nghiem ung vi moi x .

Bai 13: Tm m e phng trnh : 4 4 6 6 24(sin x cos x) 4(sin x cos x) sin 4x m+ + = co nghiem.Bai 14: Tm tat ca cac gia tr cua tham so m e phng trnh: mm xxxx 2)22)(1(44 2211 ++=+ ++

co nghiem thuoc oan [0;1].

Bai 15:Cho phng trnh : 032)2(2 22 =+ mxxxx Vi gia tr nao cua m th phng trnh co nghiem.

Bai 16: Cho phng trnh cos4 6sin cos 0x x x m+ =

81

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nh m e phng trnh co nghiem 0;4

x

.

Bai 17: Cho ham so 2 3f(x) sin 2x 2(sin x cosx) 3sin 2x m= + + +

Tm m e f(x) 1 vi moi x [0; ]2

Bai 18: Tm m e phng trnh : 0)cos)(sincos.(sin2cos2 =++ xxmxxx co nghiem tren oan

2;0

Bai 19: Cho phng trnh : 0123).2(9211211 =+++ ++ mm xx

Tm m e phng trnh co nghiem.

Bai 20: Cho bat phng trnh: 42)1( 222 ++++ xxmx (1)Tm m e co nghiem x ]1;0[

Bai 21: Tm m e phng trnh sau co hai nghiem trai dau: 013)52(9)3( =+++ mmm xx

Bai 22: Tm m e phng trnh sau co nghiem:mxxxx =+++ )6)(3(63

Bai 23: Tm m e phng trnh : 2 2 22 1 42

(log x) log x 3 m(log x 3)+ = co nghiem thuoc [32; + )

Bai 24: Cho bat phng trnh : mxxx =+++2sin

22cos1

22cos2

2

Xac nh m e bat phng trnh thoa man vi moi x

Bai 25: Cho phng trnh: 022

12

1

22 =++

m

x

x

Tm m sao cho phng trnh co nghiem duy nhat trong oan [0;1]

-------------------------------Het------------------------------

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Chuyen e 13: TCH PHAN VA NG DUNGTOM TAT GIAO KHOA

I. Bang tnh nguyen ham c ban:Bang 1 Bang 2

Ham so f(x) Ho nguyen ham F(x)+C Ham so f(x) Ho nguyen ham F(x)+Ca ( hang so) ax + C

x

1

1

xC

+

++

( )ax b +

a

11( )

1

ax bC

+++

+

1

x ln x C+

1

ax b+

1ln ax b C

a+ +

xa

ln

xa

Ca

+

xe xe C+ ax be + 1 ax be C

a

+ +

sinx -cosx + C sin(ax+b) 1 cos( )ax b C a

+ +

cosx Sinx + C cos(ax+b) 1 sin( )ax b C a

+ +

2

1

cos x tgx + C

2

1

cos ( )ax b+

1( )tg ax b C

a+ +

2

1

sin x -cotgx + C 2

1

sin ( )ax b+

1cot ( )g ax b C

a + +

'( )

( )

u x

u x

ln ( )u x C+ 2 2

1

x a

1ln

2

x aC

a x a

+

+

tgx ln cosx C + 2 2

1

x a+

2 2ln x x a C + + +

cotgx ln sinx C+

Phng phap 1: Phan tch tch phan a cho thanh nhng tch phan n gian co cong thc trong bang nguyen

ham c ban Cach phan tch : Dung bien oi ai so nh mu, luy tha, cac hang ang thc ... va bien oi

lng giac bang cac cong thc lng giac c ban.

V du : Tm ho nguyen ham cua cac ham so sau:

1. 31

( ) cos1

f x xx x

= ++

2.2

2x 5f(x)

x 4x 3

=

+

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Phng phap 2: S dung cach viet vi phan hoa trong tch phan

V du: Tnh cac tch phan: 1. 5cos sinx xdx 2. costgx

dxx 3.

1 lnxdx

x

+

I. TNH TCH PHAN BANG CACH S DUNG N VA CAC TNH CHAT TCH PHAN1. nh ngha: Cho ham so y=f(x) lien tuc tren [ ];a b . Gia s F(x) la mot nguyen ham cua ham so f(x)

th:

[ ]( ) ( ) ( ) ( )b

b

aa

f x dx F x F b F a= = ( Cong thc NewTon - Leiptnitz)

2. Cac tnh chat cua tch phan:

Tnh chat 1: Neu ham so y=f(x) xac nh tai a th : ( ) 0ba

f x dx =

Tnh chat 2: ( ) ( )b aa b

f x dx f x dx=

Tnh chat 3: Neu f(x) = c khong oi tren [ ];a b th: ( )b

a

cdx c b a=

Tnh chat 4: Neu f(x) lien tuc tren [ ];a b va ( ) 0f x th ( ) 0ba

f x dx

Tnh chat 5: Neu hai ham so f(x) va g(x) lien tuc tren [ ];a b va [ ]( ) ( ) x a;bf x g x th( ) ( )

b b

a a

x dx g x dx

Tnh chat 6: Neu f(x) lien tuc tren [ ];a b va ( ) ( m,M la hai hang so)m f x M th( ) ( ) ( )

b

a

m b a f x dx M b a

Tnh chat 7: Neu hai ham so f(x) va g(x) lien tuc tren [ ];a b th[ ]( ) ( ) ( ) ( )

b b

a a

b

a

x g x dx f x dx g x dx =

Tnh chat 8: Neu ham so f(x) lien tuc tren [ ];a b va k la mot hang so th. ( ) . ( )

b b

a a

k f x dx k f x dx=

Tnh chat 9: Neu ham so f(x) lien tuc tren [ ];a b va c la mot hang so th( ) ( ) ( )

b c b

a a c

f x dx f x dx f x dx= +

Tnh chat 10: Tch phan cua ham so tren [ ];a b cho trc khong phu thuoc vao bien so , nghla : ( ) ( ) ( ) ...

b b b

a a a

f x dx f t dt f u du= = =

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Bai 1: Tnh cac tch phan sau:

85

1)1

30

xdx

(2x 1)+ 2)1

0

xdx

2x 1+3)

1

0

x 1 xdx 4)1

20

4x 11dx

x 5x 6

++ +

5)1

20

2x 5dx

x 4x 4

+ 6)

3 3

20

xdx

x 2x 1+ + 7)6

6 6

0

(sin x cos x)dx

+ 8)32

0

4 sin xdx

1 cos x

+

9)4

20

1 sin 2xdx

cos x

+ 10)

24

0

cos 2xdx

11)2

6

1 sin 2x cos2xdx

sin x cos x

+ ++ 12)

1

x0

1dx

e 1+ .

13) dxxx )sin(cos4

0

44

14) +

4

0 2sin21

2cos

dxx

x15)

+

2

0 13cos2

3sin

dxx

x16)

2

0 sin25

cos

dxx

x

17) +

0

22

32

4dx

xx 18)

++

1

12

52xx

dx

Bai 2:

1)

32

3 x 1dx 2)4

2

1 x 3x 2dx + 3)5

3 ( x 2 x 2 )dx + 4)2

2

21

2

1

x 2x+ dx

5)3

x

0

2 4dx 6)0

1 cos 2xdx

+ 7)2

0

1 sin xdx

+ 8) dxxx 2

0

2

Bai 3:1) Tm cac hang so A,B e ham so f(x) A sin x B= + thoa man ong thi cac ieu kien

va'f (1) 2=2

0

f(x)dx 4=

2) Tm cac gia tr cua hang so a e co ang thc :

22 3

0 [a (4 4a)x 4x ]dx 12+ + = II. TNH TCH PHAN BANG PHNG PHAP OI BIEN SO :

1) DANG 1:Tnh I = bang cach at t = u(x)b

'

a

f[u(x)].u (x)dx

Cong thc oi bien so dang 1: [ ] =)(

)(

)()('.)(bu

au

b

a

dttfdxxuxuf

Cach thc hien:

Bc 1: at t dxxudtxu )()( '==

Bc 2: oi can :)(

)(

aut

but

ax

bx

=

=

=

=

Bc 3: Chuyen tch phan a cho sang tch phan theo bien t ta c

[ ]=b

fI (tiep tuc tnh tch phan mi)=)(

)(

)()('.)(bu

aua

dttfdxxuxu

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Tnh cac tch phan sau:

1)2

3 2

0

cos x sin xdx

2)2

5

0

cos xdx

3)4

20

sin4xdx

1 cos x

+ 4)1

3 2

0

x 1 x dx

5)2

2 3

0

sin 2x(1 sin x) dx

+ 6)4

40

1dx

cos x

7)e

1

1 ln xdx

x

+ 8)

4

0

1dx

cosx

9)e 2

1

1 ln xdx

x

+ 10) 11)

15 3 6

0

x (1 x ) dx6

20

cosxdx

6 5sin x sin x

+ 12)3 4

0

tg xdx

cos2x

13)4

0

cos sin

3 sin 2

x xdx

x

+

+ 14) +2

022

sin4cos

2sin

dxxx

x15)

+ 5ln

3ln 32xx

ee

dx16)

+

2

02

)sin2(

2sin

dxx

x

17) 3

4

2sin

)ln(

dxx

tgx18)

4

0

8)1(

dxxtg 19) +

2

4

2sin1

cossin

dxx

xx20)

+

+2

0 cos31

sin2sin

dx

xx

21) +

2

0 cos1

cos2sin

dxx

xx 22) +2

0

sincos)cos(

xdxxe x 23) +

2

1 11dx

x

x 24) +e

dxx

xx

1

lnln31

25) +

4

0

2

2sin1

sin21

dxx

x

2) DANG 2: Tnh I = bang cach at x =b

a

f(x)dx (t)

Cong thc oi bien so dang 2: [ ]==

dtttfdxxfI

b

a)(')()(

Cach thc hien:

Bc 1: at dttdxtx )()( ' ==

Bc 2: oi can :

=

=

=

=

t

t

ax

bx

Bc 3: Chuyen tch phan a cho sang tch phan theo bien t ta c

(tiep tuc tnh tch phan mi)[ ]==

dtttfdxxfI

b

a)(')()(


1)1

2

0

1 x dx 2)1

20

1dx

1 x+ 3)1

20

1dx

4 x 4)

1

20

1dx

x x 1 +

5)1

4 20

xdx

x x 1+ + 6)2

0

1

1 cos sindx

x x

+ + 7)2

22

20

xdx

1 x 8)

22 2

1

x 4 x dx

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9)

2

3

22

1dx

x x 1 10)

3 2

21

9 3xdx

x

+ 11)

1

50

1

(1 )

xdx

x

+ 12)

2

22

3

1

1dx

x x

13)2

0

cos

7 cos2

xdx

x

+ 14)1 4

60

1

1

xdx

x

++ 15) 20

cos

1 cos

xdx

x

+ 16) ++

0

12

22xx

dx

17) ++

1

0 311 x

dx 18)

2

1 5

1 dxx

xx

II. TNH TCH PHAN BANG PHNG PHAP VI PHAN:Tnh cac tch phan sau:

1)8

23

1

1dx

x x + 2)

7 3

3 20 1

xdx

x+ 3)

35 2

0

1x x dx+ 4)ln2

x0

1dx

e 2+

5)

7

3

30

1

3 1

xdx

x

+

+ 6)

22 3

0

1x x d+ x 7) +

32

52

4xx

dx

III. TNH TCH PHAN BANG PHNG PHAP TCH PHAN TNG PHAN:Cong thc tch phan tng phan:

[ ] =b

a

b

a

b

adxxuxvxvxudxxvxu )(').()().()(').(

Hay: [ ] =b

a

b

a

b

a vduvuudv .

Cach thc hien:

Bc 1: at)(

)('

)('

)(

xvv

dxxudu

dxxvdv

xuu

=

=

=

=

Bc 2: Thay vao cong thc tch phan tng tng phan : [ ] =b

a

b

a

b

a vduvuudv .

Bc 3: Tnh [ va]bavu. b

a

vdu


1)2

51

ln xdx

x 2)2

2

0

xcos xdx

3)1

x

0

e sinxdx

4)

2

0

sin xdx

5) 6)e

2

1

x ln xdx3

20

x sin xdx

cos x

+

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7) 8)2

0

xsin xcos xdx

4

2

0

x(2 cos x 1)dx

9)2

21

ln(1 x)dx

x

+

10) 11) 12)1

2 2x

0

(x 1) e dx+e

2

1

(xlnx) dx2

0

cosx.ln(1 cosx)dx

+

13) 21

ln

( 1)

e

e

x

dxx + 14)1

2

0xtg xdx 15)

1

0

2

)2( dxexx

16) 17) +1

0

2)1ln( dxxx

e

dxx

x

1

ln18) +

2

0

3sin)cos(

xdxxx

19) 20) ++2

0

)1ln()72( dxxx 3

2

2)ln( dxxx

MOT SO BAI TOAN TCH PHAN QUAN TRONG VA NG DUNG

Bai 1: 1) CMR neu f(x) le va lien tuc tren [-a;a] (a>0) th :

a

a f(x)dx 0 =

2) CMR neu f(x) chan va lien tuc tren [-a;a] (a>0) th :a a

a 0

f(x)dx 2 f (x)dx

= Bai 2: 1) CMR neu f(t) la mot ham so lien tuc tren oan [0,1] th:

a)2 2

0 0

f (sinx)dx f(cosx)dx

=

b)0 0

xf(sin x)dx f(sin x)dx2

=

AP DUNG: Tnh cac tch phan sau:

88

1)n2

+

n n0

cos xdx vi n Z

cos x sin x

+ 2)

42

4 40

cos xdx

cos x sin x

+ 3)62

6 60

sin xdx

sin x cos x

+

4) 5)5

0

xsin xdx

2

2

2

4 sin

x cosxdx

x

+ 6)

1 4

21

sin

1

x xdx

x

++

7) 20

xsinx

dx4 cos x

8)4 3

0 cos sinx x xd

x

Bai 3:CMR neu f(x) lien tuc va chan tren R th +0

( )( ) vi R va a > 0

1xf x

dx f x dxa

= + ; a 1

AP DUNG : Tnh cac tch phan sau:

2)1 2

1

1

1 2xx

dx

+ 3)

2sin

3 1xx

dx

+ 1)

1 4

1 2 1x

xdx

+

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IV .NG DUNG TCH PHAN TNH DIEN TCH HNH PHANG: Cong thc:

89

1Cy

2Cy

2Cx

1Cx

]dxxgxfS )()(

[ =b

a

[ ] =b

a

dyygyfS )()(

Tnh dien tch cua cac hnh phang sau:

1) (H1):

2

2

xy 4

4

xy

4 2

=

=

2) (H2) :2y x 4x 3

y x 3

= +

= +3) (H3):

3x 1y

x 1y 0

x 0

=

= =

4) (H4): 5) (H2

2

y x

x y

=

= 5):

2

y x

y 2 x

=

= 6) (H6):

2y x 5 0

x y 3 0

+ =

+ =

7) (H7):

ln xy

2 x

y 0

x e

x 1

= = =

=

8) (H8) :2

2

y x 2x

y x 4

=

x= +9) (H9):

2 3 3y x x2

y x

2

= + =

10) (H10): 11)

2

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

=

=

)(

2:)(

:)(

Ox

xyd

xyC

12)

==

=

1:)(

2:)(

:)(

x

yd

eyCx

V. NG DUNG TCH PHAN TNH THE TCH VAT THE TRON XOAY.Cong thc:

=

==

=

bx

ax

xgyC

xfyC

H

:

:

)(:)(

)(:)(

:)(

2

1

2

1

==

=

=

by

ay

ygxC

yfxC

H

:

:

)(:)(

)(:)(

:)(

2

1

2

1

x

y

)(Ha

b

)(:)( 1 yfxC =

)(:)( 2 ygxC =

ay =

by =

O

yx

x

)(H

a b

)(:)(1

xfyCa=

=

)(:)(2

xgyC

bx =

O

=

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b

ax

y

0=x

O

)(:)( yfxC =by =

ay =

a b0=y

)(:)( xfyC =

b

ax =bx =

x

y

O

[ ] dxxfVb

a

2

)(= [ ] dyyfVb

a

2

)(=

Bai 1: Cho mien D gii han bi hai ng : x2 + x - 5 = 0 ; x + y - 3 = 0Tnh the tch khoi tron xoay c tao nen do D quay quanh truc Ox

Bai 2: Cho mien D gii han bi cac ng : y x;y 2 x;y 0= = = Tnh the tch khoi tron xoay c tao nen do D quay quanh truc Oy

Bai 3: Cho mien D gii han bi hai ng : va y = 42y (x 2)= Tnh the tch khoi tron xoay c tao nen do D quay quanh:

a) Truc Oxb) Truc Oy

Bai 4: Cho mien D gii han bi hai ng : 2 24 ;y x y x 2= = + .

Tnh the tch khoi tron xoay c tao nen do D quay quanh truc Ox

Bai 5: Cho mien D gii han bi cac ng :2

2

1;

1 2

xy y

x= =

+

Tnh the tch khoi tron xoay c tao nen do D quay quanh truc Ox

------------------------------Het-------------------------------

90

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Chuyen e 14: HNH HOC GIAI TCH TRONGMAT PHANG

A. KIEN THC C BAN:

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

91

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

x'Ox : truc hoanh y'Oy : truc tung O : goc toa o

: vec t n v (1 2,e e

1 2 11 vae e e e= =

2

)

x

y

1e

2e

O'x

'yQuy c : Mat phang ma tren o co chon he truc toa o e-Cac vuong goc Oxy c goi la mat pha

Oxy va ky hieu la : mp(Oxy)II. Toa o cua mot iem va cua mot vec t:1. nh ngha 1: Cho ( )M mp Oxy . Khi o vec t OM

c bieu dien mot cach duy nhat theo

e e bi he thc co dang : OM1 2,

xe ye1 2 vi x,y

= +

.

Cap so (x;y) trong he thc tren c goi la toa o cua iem M.Ky hieu: M(x;y) ( x: hoanh o cua iem M; y: tung o cua iem M )'x

y

2

'/

1 2( ; ) n

M x y OM xe ye = +

Y ngha hnh hoc:

va y=OQx OP=

2. nh ngha 2: Cho a m ( )p Oxy

. Khi o vec t a


e e bi he thc co dang :1 2,

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

.

Cap so (a1;a2) trong he thc tren c goi la toa o cua vec t .a

Ky hieu: 1 2( ; )a a a=

/

1 2 1 1 2 2=(a ;a ) n

a a a = +

e a e

Y ngha hnh hoc:

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

x1e

e

O

MQ

P

y

y

xO

x'

'y

MQ

Px

y

x

y

1e

2e

O'x

'y

P

a

y

xO

'x

'y

1A 1B

2A

2B BK

A H

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BAI TAP AP DUNG:Trong mat phang Oxy hay ve cac iem sau: A(2;3), B(-1;4), C(-3;-3), D(4;-2), E(2;0), F(0;-4)

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

92

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

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

* a b 1 1

2 2

a

b

a b

==

=

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

)a b a b =

)ka ka=

* a b 1 1 2 2( ;

* k a ( )1 2. ( ; k

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

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

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

thang song song . nh ly ve s cung phng cua hai vec t: nh ly 3 : Cho hai vec t va vi 0a b b

a k b

a b cung phng !k sao cho . =

Neu 0a

th so k trong trng hp nay c xac nh nh sau:

k > 0 khi a

cung hng b

k < 0 khi a

ngc hng b

ak

b=

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

(ieu kien 3 iem thang hang )

nh ly 5: Cho hai vec t 1 2 1 2( ; ) va ( ; )a a a b b b= =

ta co :

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

(ieu kien cung phng cua 2 vec

);( AA yxA

);( BB yxB

a

b

a

b

A

B

C

a

b

2 5a b , b - a

5 2= =

a

b

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)4;2(

)2;1(

=

=

b

a

:VD);(

);(

21

21

bbb

aaa

=

=

BAI TAP AP DUNG:

93

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

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

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

31;23(

+B , )

4

31;32(

C . Chng minh A, B, C thang hang

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

x

y

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

22

a a=

a b . 0a b =


ta co :

a b (Cong thc tnh tch vo hng theo toa o)1 1 2 2. a b a b= +

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

ta co :

2 21 2a a a= +

(Cong thc tnh o dai vec t )

nh ly 8: Neu B( ; ) va B(x ; )A A BA x y y th

2 2( ) ( )B A B AAB x x y y= + (Cong thc tnh khoang cach 2 ie


ta co :

a b (ieu kien vuong goc cua 2 vec t1 1 2 2a 0b a b + =


ta co

1 1 2 2

2 2 2 21 2 1 2

.cos( , )

. .

a b a b a ba b

a b a a b b

+= =

+ +

(Cong thc tnh goc cua 2 vec t)

b

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

O'x

'y

a

a

b

b

a

O

B

A

);( BB yxB);( AA yxA

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VI.iem chia oan thang theo ty so k:nh ngha: iem M c goi la chia oan AB theo ty so k ( k 1 ) neu nh : .MA k M=

A M B

nh ly 11 : Neu B( ; ) , B(x ; )A BA x y y va .MA k MB=

( k 1 ) th

.1.

1

A BM

A BM

x k xxk

y k yy

k

=

=

94

ac biet : M la trung iem cua AB 2

2

BM

BM

x xx

y yy

+=

+ =

VII. Mot so ieu kien xac nh iem trong tam giac :

++=

++=

=++

3

30.1

CBA

G

CBA

yyyy

xxx

GCGBGx

GAABCgiactamtamtronglaG

2.. 0

H la trc tam tam giac ABC. 0

AH BC AH BC

BH AC BH AC

=

=

3.'

''

la chan ng cao ke t Acung phng

AA BCA

BA BC

4.IA=IB

I la tam ng tron ngoai tiep tam giac ABCIA=IC

5. =

D la chan ng phan giac trong cua goc A cua ABC .AB

DB DCAC

6. =

' ' 'D la chan ng phan giac ngoai cua goc A cua ABC .AB

D B DAC

C

7. J la tam ng tron noi tiep ABC .AB

JA J

BD

= D

VIII. Mot so kien thc c ban thng s dung khac:1. Cong thc tnh dien tch tam giac theo toa o ba nh : nh ly 12: Cho tam giac ABC . at 1 2 1 2( ; ) va ( ; )AB a a AC= b b=

ta co :

1 2 2 11

.2ABC

S a b = a b

G

A

BC

H

A

BC

A

C

I

A

B C

B '

A

D

AB

J

C

DB

A

CB

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

bat ky ta luon co :

u

v

vu

+

u v u v+ +

. .u v u v

Dau bang xay ra khi va ch khi ,u v

la hai vec t cung phng cung chieu hoac la co mottrong hai vec t la vec t khong .

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

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

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

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

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

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

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

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

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

-------------------Het-------------------

95

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NG THANG TRONG MAT PHANG TOA O

A.KIEN THC C BAN

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

a

la VTCP cua ng thang ( )n

0

a co gia song song hoac trung vi ( )

a

n

la VTPT cua ng thang ( )n

0

n co gia vuong goc vi ( )

n

96

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

th co VTPT la 2 1( ;n a a= )

a

a )(

n

(

Neu ng thang ( ) co VTPT ( ; )n A B=

th co VTCP la ( ; )a B A=

a

n

)(

BAI TAP AP DUNG:Cho ng thang i qua hai iem A(1;-2), B(-1;3). Tm mot VTCP va mot VTPT cua( ) ( )

II. Phng trnh ng thang :1. Phng trnh tham so va phng trnh chnh tac cua ng thang :

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

lam

VTCP se co :

Phng trnh tham so la : 0 10 2

.( ) : (

.

x x t at

y y t a

= +

= +

Phng trnh chnh tac la : 0 01 2

( ) :x x y y

a a

=

y

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

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

);( 000 yxM

a

);( yxM

xO

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

la:

97

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

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

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

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

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

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

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

1. VTPT cua ( ) la ( ; )n A B=

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

3. ( ;0 0 0 0 0) ( ) 0M x y Ax By C + + =

Menh e (3) c hieu la :ieu kien can va u e mot iem nam tren ng thang la toa o iem onghiem ung phng trnh cua ng thang .

BAI TAP AP DUNG:Bai 1: Viet phng trnh tham so cua ng thang biet phng trnh tong quat cua no la 5 2 3x y + =Bai 2: Viet phng trnh tong quat cua ng thang qua M(-1;2) va song song ( ) : 2 3 4 0x y + =

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

ABC vuong C.Bai 5: Cho A(1;1) ; B(-1;3) va ng thang d:x+y+4=0.

a) Tm tren d iem C cach eu hai iem A, B.b) Vi C tm c . Tm D sao cho ABCD la hnh bnh hanh .Tnh dien tch hnh bnh hanh.

)yM ;( 000 x

);( yxM

ny

xO

);( yM 000 x

);An

( B=

x

y

);( ABa =

O

);( ABa =

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

( ) : A A

B A B A

x x y yAB

x x y y

=

( ) :

AAB x x= ( ) : AAB y y=

98

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

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

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

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

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

Chu y 2: Neu ng thang co phng trnh y ax b= + th he so goc cua ng thang la k a= nh ly 2: Goi k1, k2 lan lt la he so goc cua hai ng thang 1 2, ta co :

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

BAI TAP AP DUNG:Viet phng trnh ng thang qua A(-1;2) va vuong goc vi ng thang 3 4x y + = 0

c. Phng trnh t i qua mot iem va song song hoac vuong goc vi mot t cho trc:

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

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

x

y

O

);( yxM

x

y y

);( AA yxA );( BB yxBy

);( AA yxA

);(BB yxB

Ax BxAy

By);( AA yxA

);( BB yxB

Ay Byx xO

)

y

O

;( yM x

0x

0y

x

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

x

y

O 0x

0: 1 =++ CByAx

1M

0: 21 =+ mAyBx

x

y

O 0x1M

0:1 =++ CByAx

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

99

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

2 2 2 2

( ) : 0

( ) : 0

A x B y C

A x B y C

+ + =

+ + =

V tr tng oi cua phu thuoc vao so nghiem cua he phng trnh :1( ) va ( ) 2

hay1 1 1

2 2 2

0

0

A x B y C

A x B y C

+ + =

+ + =

1 1 1

2 2 2

(1)A x B y C

A x B y C

+ =

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

nh ly 1:1 2

1 2

1 2

. He (1) vo nghiem ( ) //( )

. He (1) co nghiem duy nhat ( ) cat ( )

. He (1) co vo so nghiem ( ) ( )

i

ii

iii

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

=

= =

1 11 2

2 2

1 1 11 2

2 2 2

1 11 2

2 2

A. ( ) cat ( )

A

A. ( ) // ( )

A

A. ( ) ( )

A1

2

Bi

B

B Cii

B C

B Ciii

B C

1

x

y

O

2

21//

1

x

y

O

2

y

O

1

x

2

21 21 cat

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BAI TAP AP DUNG:

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

( ) : 3 5 13

( ) : 5 2 1 0

AB x y

AC x y

BC x y

0

+ =

=

+ =

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

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

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

2

: 1

: 2 0

d mx y m

d x my

0+ =

+ =

IV. Goc gia hai ng thang

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

( ) : 0

( ) : 0

A x B y C

A x B y C

+ + =

+ + =

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

1

x

y

O

2

1 2 1 2

2 2 2 21 1 2 2

cos.

A B B

A B A B

+=

+ +

100

He qua:

( 1 2 1 2 1 2) ( ) A 0A B B + =

BAI TAP AP DUNG:Bai 1: Viet phng trnh ng thang i qua iem A(0;1) va tao vi ng thang : x+2y+3=0

mot goc bang 450

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

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

0 00 2 2

( ; )Ax By C

d MA B

+ + =

+

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

( ) : 0

( ) : 0

A x B y C

A x B y C

+ + =

+ + =

va ( )

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

1 1 1 2 2

2 2 2 21 1 2 2

2A x B y C A x B y C

A B A B

+ + + +=

+ +

0M

y

O

H

)(

y

O

1

2

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

tren ( ). Khi o:M

N

M

N

Hai iem M , N nam cung pha oi vi ( ) khi va ch khi0))(( >++++ CByAxCByAx NNMM

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

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Chu y:

102

=

=

1

2

0 va 0 th

0 va 0 th

ac biet :

+ + + + + =

+ + + + + =

1 1

1 1 1 2 2 2

1 1 1 2 2 2

Neu 0 va 0 th va trong trng hp nay

phng trnh co the viet di dang sau:

1. m(A ) (A ) 0

hoac 2. (A ) (A ) 0

x B y C x B y C

x B y C n x B y C

M

21

I

BAI TAP AP DUNG:Viet phng trnh ng thang i qua giao iem cua hai ng thang 3 5 2 0 & 5 2 4 0x y x y + = + =

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

.

BAI TAP REN LUYEN

Bai 1: Phng trnh hai canh cua tam giac trong mat phang toa o la 5x-2y+6=0 va 4x+7y-21=0Viet phng trnh canh th ba cua tam giac biet trc tam cua tam giac trung vi goc toa o.

Bai 2: Cho tam giac ABC , canh BC co trung iem M(0;4) con hai canh kia co phng trnh2x+y-11=0 va x+4y-2=0.a) Xac nh nh A.b) Goi C la iem tren ng thang x+4y-2=0, N la trung iem AC . Tm iem N roi tnh

toa o B, C.Bai 3: Cho tam giac ABC co M(-2;2) la trung iem cua BC , canh AB co phng trnh x-2y-2=0,

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

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

Bai 5: Cho tam giac ABC co trong tam G(-2;-1) va co cac canh AB:4x+y+15=0 vaAC:2x+5y+3=0a) Tm toa o nh A va toa o trung iem M cua BC .

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

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

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

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

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

Bai 11: Cho iem M(-2;3) . Tm phng trnh ng thang qua M va cach eu hai iem A(-1;0)va B(2;1).

Bai 12: Cho A(2;-3) , B(3;-2) .Trong tam G cua tam giac nam tren ng thang d: 3x-y-8=0, dientch tam giac ABC bang 3/2 . Tm C.

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

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

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

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

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

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

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

---------------------------Het--------------------------

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NG TRON TRONG MAT PHANG TOA O

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

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

104

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

Phng trnh (1) c goi la phng trnh chnh tac cua ng tron

ac biet: Khi I O th (hay:2 2( ) :C x y R+ = 2 2 2y R x= )

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

nh ly : Trong mp(Oxy). Phng trnh : 2 2 2 2 0x y ax by c+ + = vi a b2 2 0c+ >

la phng trnh cua ng tron (C) co tam I(a;b), ban knh 2 2R a b= + c

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

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

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

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

x

y

O

);( baI

Rb

a

);( yxM

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

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

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

(C)

I(a;b))(

);(000 yxM

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105

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

Chu y :M/(O) > 0 ngoai ng tron (O)M

M/(O) < 0 trong ng tron (O)M

M/(O) = 0 tren ng tron (O)M

nh ly:Trong mp(Oxy) cho iem 0 0( ; )M x y va ng tron2 2 2 2x y ax by c 0+ + = vi

a b co tam I(a;b) va ban knh2 2 0c+ > 2 2R a b c= + . Phng tch cua iem M oi ving tron (C) la

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

(C)

MI

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

(C)

2 22 4 4 0x y x y+ + =

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

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

Cach xac nh truc ang phng

)( 1C)( 2C

2I1I

)(1

C

)( 2C

1I 2I M

)1C

)(2

C

1I 2I

(

M

)(2C

)( 1C

)(3

C

I1I 2I

3I

2

1

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

2 21 1 1 1

2 22 2 2

( ) : 2 2 0

( ) : 2 2 0

C x y a x b y c

C x y a x b y c

+ + =

2+ + =

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

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

106

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

BAI TAP NG DUNG:

Xac nh phng trnh truc ang phng cua hai ng tron sau:2 2

1

2 2

2

( ) : 4 5 0

( ) : 6 8 16 0

C x y y

C x y x y

+ =

+ + + =

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

)(C

)(C )(CI

I IRR H

RMM HH Mnh ly:

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

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

BAI TAP NG DUNG:Bai 1: Cho ng tron (C): . Viet phng trnh tiep tuyen vi (C) biet rang tiep2 2( 3) ( 1)x y + = 4

tuyen nay i qua iem M(6;3)Bai 2: Cho ng tron (C): . Viet phng trnh tiep tuyen vi ng tron biet2 2 6 2 5x y x y+ + + = 0

tiep tuyen song song vi ng thang ( ) : 2 10 0d x y+ + =

Bai 3: Cho ng tron va iem M(-3;1). Goi T0662:)( 22 =++ yxyxC 1, T2 la cac tiep iem cucac tiep tuyen ke t M en (C). Viet phng trnh ng thang T 1T2.

2. V tr tng oi cua hai ng tron :

1I 1R

1C

2I2

R

2C

1I 1R

1C

2C

2R

2I

1C

1I 1R

2C

2R

2I

1C

2C

1I

2I

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1 2 1 2 1 2

1 2 1 2 1 2 1 2

1 2 1 2 1 2

1 2

( ) va (C ) khong cat nhau I I > R

( ) va (C ) cat nhau R < I I < R

( ) va (C ) tiep xuc ngoai nhau I I = R

( ) va (C ) tiep xuc trong

C R

C R

C R

C

R

+

+

+

1 2 1 2nhau I I = R R

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

2 2

1

2 2

2

( ) : 4 5 0

( ) : 6 8 16 0

C x y y

C x y x y

+ =

+ + + =

VII: Chum ng tron:

nh ly: Cho hai ng tron cat nhau :

( )

2 2

1 1 1

2 2

2 2 2

( ) : 2 2 01

2: 2 2 0

C x y a x b y c

C x y a x b y c

+ + =

+ = +

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

2 2 2 2 2 2

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

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

( ) 2 2 2 21 2

: 10 0;( ) : 4 2 20 0C x y x C x y x y+ = + + =

va i qua iem A(1;-1)

BAI TAP REN LUYENBai 1: Lap phng trnh ng tron ngoai tiep tam giac co ba nh la A(1;1); B(-1;2); C(0;-1).Bai 2: Lap phng trnh ng tron ngoai tiep tam giac co ba canh nam tren ba ng thang

1 2 3

x 2( .d ) : y ;(d ) : y x 2;(d ): y 8 x

5 5= = + =

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

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

(d): 7x-y-5=0 tai iem M(1;2).

Bai 6: Lap phng trnh ng tron co tam nam tren ng thang 2x+y=0 va tiep xuc vi ngthang x-7y+10=0 tai iem A(4;2).

Bai 7: Viet phng trnh ng tron co tam nam tren ng thang 4x +3y - 2 = 0 va tiepxuc vi hai ng thang : x + y + 4 = 0 va 7x - y + 4 = 0.

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

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

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

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

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

Bai 12: Cho hai ng tron :2 2

12 2

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

+ + =

+ + =

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

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

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

Bai 14: Cho ng tron (C): va iem A(1;2). Hay lap phng trnh cua ng thang2 2

9x y+ =

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

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

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

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

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

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

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

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

Bai 19: Cho tam giac ABC eu noi tiep trong ng tron (C): 2 2( 1) ( 2) 9x y + = . Xac nh toa

o cac iem B, C biet iem A(-2;2).Bai 20: Trong mp(Oxy) cho ho ng tron (Cm) co phng trnh :

2 2x 2mx y 2(m 1)y 12 0 + + + =

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

Bai 21: Cho hai ho ng tron :

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'

2 2m

2 2

m

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

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

+ + + =

+ + + =

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


1

2 22

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

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

+ =

+ + =

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


1

2 22

(C ) : x y 10x 0

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

+ =

+ + =

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

2 21

2 22

(C ) : x y 4x 5 0

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

+ =

+ + + =

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

A va khoang cach t tam cua (C) en iem B bang 5 (TS.K.B2005)

ng dung phng trnh ng tron e giai cac he co cha tham soBai 1: Cho he phng trnh :

2 2x y

x y a

+ =

=

1

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

Bai 2: Cho he phng trnh :

2 2 0

0

x y x

x ay a

+ =

+ =

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

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

2 2

(x 2) y m

x (y 2) m

+ =

+ =

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NG ELP TRONG MAT PHANG TOA O

A.KIEN THC C BANI.nh ngha:

Elp (E) la tap hp cac iem M co tong khoang cach en hai iem co nh F 1; F2 bang hang s* Hai iem co nh F1; F2 c goi la cac tieu iem

* F1F2 = 2c ( c > 0 ) c goi la tieu c

110

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

(E)

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

2 2

2 2x y

(E) : 1a b

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

2. Cac yeu to cua Elp:* Elp xac nh bi phng trnh (1) co cac ac iem:- Tam oi xng O, truc oi xng Ox; Oy- Tieu iem F1(-c;0); F2(c;0)

- Tieu c F1F2 = 2c- Truc ln nam tren Ox; o dai truc ln 2a ( = A1A2 )- Truc nho nam tren Oy; o dai truc ln 2b ( = B1BB2 )- nh tren truc ln : A1(-a;0); A2(a;0)- nh tren truc nho :B1(0;-b); B2(0;b)- Ban knh qua tieu iem:

2c

M

1 2F F

-a a

(E)

c-c

y

x

R S

PQ

O

M

1r

2r

1A

2A

1B

2B

1F

2F

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

2 2

cr MF a x a ex

ac

r MF a x a ea

= = + = +

x= = =

- Tam sai :c

e (0 ea

= < 1)<

- ng chuan :a

xe

=

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

=

=t [0;2 )

IV. Tiep tuyen cua Elp:

nh ly: Phng trnh tiep tuyen vi (E) :2 2

2 2x y

1a b

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

111

) : 0 02 2

x x y y1

a b+ = (

V. ieu kien e ng thang tiep xuc vi Elp:

nh ly: Cho Elp (E) :2 2

2 2x y

1a b

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

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

BAI TAP REN LUYEN

y

O

)(E

x

y

O

);( 000 yxM

)(E

Bai 1: Cho (E) co hai tieu iem la 1 2( 3;0); ( 3;0F F ) va mot ng chuan co phng trnh4

3x =

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

P F 2 2 21 2 1 23 .M F M OM F M F M= +

3. Viet phng trnh ng thang (d) song song vi truc hoanh va cat (E) tai hai iem A, B sao

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

2. Viet phng trnh tiep tuyen vi (E) vuong goc vi (d): 6 0x y+ + = .

Bai 3: Cho Elp (E) :2 2

19 4

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

1. Chng minh rang vi moi gia tr cua m, ng thang (d) luon cat (E) tai hai iem phan biet2. Viet phng trnh tiep tuyen cua (E), biet rang tiep tuyen o i qua iem A(1;-3).

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

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2 18 .

2. ng thang (d) tiep xuc (E) tai M cat hai truc toa o tai A va B. Tm M sao cho dien tchnho nhat.OAB


18 4

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

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

Bai 6: Cho hai Elp :2 2 2 2

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

x y x yE + = + = . Viet phng trnh tiep tuyen chung cua ha

elp tren.


124 12

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

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


19 4

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

1. Xac nh toa o giao iem I cua ng thang AN va BM.2. Chng minh rang ieu kien can va u e ng thang MN tiep xuc vi (E) la ab=4

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

112

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NG HYPEBOL TRONG MAT PHANG TOA O

A.KIEN THC C BANI. nh ngha:

113

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

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

2 2

2 2x y

(H) : 1a b

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

M

1F

2F

c2

xa

by = x

a

by =

1F 2F

M

x

y

1B

2B

1A 2A

acc

a

O

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

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

- Phng trnh tiem can : by xa

=

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

Vi x > 0 1 1

2 2

r MF a ex

r MF a ex

= = +

= = +

Vi x < 0

1 1

2 2

r MF (a ex)

r MF ( a ex

= = +

)= = +

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- Tam sai :c

e (ea

= > 1)

- ng chuan :a

xe

=

IV. Tiep tuyen cua Hypebol:

nh ly: Phng trnh tiep tuyen vi (H) :2 2

2 2x y

1a b

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

114

0 02 2

x x y y1

a b = ( ) :

V. ieu kien e ng thang tiep xuc vi Hypebol:

nh ly: Cho Hypebol (H) :2 2

2 2x y

1a b

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

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

y

0M

O

BAI TAP REN LUYEN

Bai 1: Cho Hypebol (H):2 2

116 9

x y =

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

Bai 2: Cho Hypebol (H):

2 2

2 2 1

x y

a b =

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

Bai 3: Cho Hypebol (H): .2 24 4x y =

1. Viet phng trnh tiep tuyen vi (H) tai10 4

( ;3 3

A )

0

2. Viet phng trnh tiep tuyen vi (H) biet no vuong goc vi ng thang : : 2x y =

3. Viet phng trnh tiep tuyen vi (H) ke t M(2;-1)

Bai 4: Cho Hypebol (H):2 2

2 21

x y

a b = trong mat phang Oxy

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

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NG PARABOL TRONG MAT PHANG TOA O

115

A.KIEN THC C BANI. nh ngha :

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

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

II. Phng trnh chnh tac cua parabol:

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

p

K

HF

M

3) Dang 3: Ptct: x2

= 2py 4) Dang 4: Ptct : x2

= -2py

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

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

x

y

P

y

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

M2/:)( px =

y

x

-p/2 :y = -p/2

F(0;p/2)

O

MF(0;-p/2)

x

( ) : y = p/2p/2

y

O

M

( ): x=-p/2

y

O

-p/2

F(p/2;0)

M

x

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

x

y

o

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

BAI TAP REN LUYEN

Bai 1: Cho (P): y2

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

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

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

18 6

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

Bai 3: Cho A(3;0) va (P): y=x21. Cho ( )M P va Mx a= . Tnh AM . Tm a e AM ngan nhat

2. Chng minh neu AM ngan nhat th AM vuong goc tiep tuyen tai M cua (P)Bai 4: Cho (P):y2= 2x va cho A(2;-2); B(8;4). Gia s M la iem di ong tren cung nho AB cua (P). Xa

nh toa o cua M sao cho tam diac AMB co dien tch ln nhat.

Bai 5: Cho (P): 2y x= va iem I(0;2). Tm toa o hai iem M, N thuoc (P) sao cho 4IM IN=

----------------------------------Het-------------------------------

116

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Chuyen e 15: HNH HOC GIAI TCH TRONGKHONG GIAN

A. KIEN THC C BAN:

PHNG PHAP TOA O TRONG KHONG GIANTOA O IEM - TOA O VEC T

117

I. He truc toa o E-CAC trong khong gian x'Ox : truc hoanh

O

z

'x

y

x

'y 3eK

1

eK 2e

K

'z

y'Oy : truc tung z'Oz : truc cao O : goc toa o

: vec t n v1 2 3, ,e e e

JG JJG JJG

Quy c : Khong gian ma trong o co chon he truc toa o e-Cac vuong goc Oxyz c goi lakhong gian Oxyz va ky hieu la : kg(Oxyz)

II. Toa o cua mot iem va cua mot vec t:1. nh ngha 1: Cho ( )M kg Oxyz . Khi o vec t OM

JJJJGc bieu dien mot cach duy nhat theo

bi he thc co dang :1 2 3, ,e e eJG JJG JJG

1 2 3+ y vi x,y,zOM xe ye e= + JJJJG JG JJG JJG

\ .

Bo so (x;y;z) trong he thc tren c goi la toa o cua iem M.Ky hieu: M(x;y;z)

( x: hoanh o cua iem M; y: tung o cua iem M, z: cao o cua iem M )

z

/

1 2 3( ; ; ) n

M x y z OM xe ye ze = + +JJJJG JG JJG JJG

Y ngha hnh hoc:

; y= OQ ; z = ORx OP=

O

My

x

z

y

x

z

y

x

p1M

M

Q

3M

2MR

O

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2. nh ngha 2: Cho (a kg Oxyz )G

. Khi o vec t aG


bi he thc co dang :1 2 3, ,e e eJG JJG JJG

1 1 2 2 3 3 1 2+ a vi a ,aa a e a e e= + G JG JJG JJG

\ .

Bo so (a1;a2;a3) trong he thc tren c goi la toa o cua vec t .aG

Ky hieu: 1 2( ; )a a a=G

/

1 2 3 1 1 2 2 3 3=(a ;a ;a ) n

a a a e a e = + +G G JG JG JJG

a eJ

118

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

( ; ; )B A B A B AAB x x y y z z= JJJG

nh ly 2: Neu a a th1 2 3 1 2 3( ; ; ) va ( ; ; )a a b b b b= =G G

* a b

1 1

2 2

3 3

a

b

a b

a b

=

= = =

G G

* a b 1 1 2 2 3 3( ; ; )a b a b a b+ = + + +G G

)a b a b a b = G G

)a ka ka ka=G

* a b 1 1 2 2 3 3( ; ;

* k ( )1 2 3. ( ; ; k \

III. S cung phng cua hai vec t:Nhac lai

Hai vec t cung phng la hai vec t nam tren cung mot ng thang hoac nam tren hai nthang song song . nh ly ve s cung phng cua hai vec t: nh ly 3 : Cho hai vec t va vi 0a b b

G G G G

a k bG G

a b cung phng !k sao cho . =G G

\

Neu th so k trong trng hp nay c xac nh nh sau:0a G G

k > 0 khi aG

cung hng bG

k < 0 khi aG

ngc hng bG

ak

b=

G

G

, , thang hang cung phngA B C AB ACJJJG JJJG

nh ly 4 :

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nh ly 11 : Neu B( ; ; ) , B(x ; ; )A A B BA x y z y z va .MA k MB=JJJG JJJG

( k 1 ) th

.

1.

1

.1

BM

A BM

A BM

x k xx

k

y k yy

k

z k zzk

=

=

=

120

ac biet : M la trung iem cua AB

2

2

2

BM

BM

A BM

x xx

y yy

z zz

+ =

+=

+

=

BAI TAP NG DUNG:Bai 1: Trong Kg(Oxyz) cho ba iem A(3;1;0), B(-1;2;-1), C(2;-1;3)

Tm iem D sao cho t giac ABCD la hnh bnh hanh

Bai 2: Trong Kg(Oxyz) cho ba iem A(2;-1;6), B(-3;-1;-4), C(5;-1;0)a.Chng minh rang tam giac ABC vuong .b. Tm toa o trong tam G cua tam giac ABCc. Tnh o dai ng trung tuyen ke t A

VI. Tch co hng cua hai vec t:1. nh ngha: Tch co hng cua hai vec t 1 2 3 1 2 3( ; ; ) va ( ; ; )a a a a b b b b= =

G Gla mot vec t c

ky hieu : co toa o la :;a bG G

2 3 3 1 1 2

2 3 3 1 1 2

; ; ;a a a a a a

a bb b b b b b

=

G GCach nh: 1 2 3

1 2 3

( ; ; )

( ; ; )

a a a a

b b b b

=

=

G

G

1 2 3

2. Tnh chat:

; va ;a b a a b b G G G G G G

A

1. ;

2ABCS AB =

JJJG HJJGAC

B C

;ABCDS AB = .JJJG JJJG

A

B

C

D

'A'B

'C'D

AD

' ' ' ' '. ; .ABCD A B C DV AB AD =

JJJGJJJG JJJGAA

AB

CD

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121

1

. ; .6ABCD

V AB AC = JJJG JJJG JJJG

AD

bG G

A

B

C

D

cung phng ; 0a b a = G G G

, , ong phang , . 0a b c a b c = G G G G G G

BAI TAP NG DUNG:Bai 1: Cho bon iem A(-1;-2;4), B(-4;-2;0), C(3;-2;1), D(1;1;1)

a. Chng minh rang bon iem A,B,C,D khong ong phangb. Tnh dien tch tam giac ABCc. Tnh the tch t dien ABCD

Bai 2: Tnh the tch t dien ABCD biet A(-1;-2;0), B(2;-6;3), C(3;-3;-1), D(-1;-5;3)

NG THANG VA MAT PHANG TRONG KHONG GIAN

I. Cac nh ngha:1. Vec t ch phng cua ng thang:1. VTCP cua ng thang :

la VTCP cua ng thang ( )n

0

a co gia song song hoac trung vi ( )

a

G G

G aG

aK

aK )(

Chu y: Mot ng thang co vo so VTCP, cac vec t nay cung phng vi nhau. Mot ng thang ( ) hoan toan c xac nh khi biet mot iem thuoc no va mot VTCP cua

no.

2. Cap VTCP cua mat phang:aK

Cho mat phang xac nh bi hai ng thang cat nhau a va b . Goi la VTCP cua ngG aG

thang a va la VTVP cua ng thang b. Khi o :JG J

b

Cap c goi la cap VTCP cua mat phang( , )a bJG

Chu y : Mot mat phang hoan toan c xac nh khi biet mot iem thuoc no va mot cap VTCP cua

no.

bK

a

b

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122

3. Vec t phap tuyen ( VTPT) cua mat phang : nK

n la VTPT cua mat phangG

n

0

n co gia vuong goc vi mp

n

G G

G

Chu y: Mot mat phang co vo so VTPT, cac vec t nay cung phng vi nhau. Mot mat phang hoan toan c xac nh khi biet mot iem thuoc no va mot cap VTPT cua no

4. Cach tm toa o mot VTPT cua mat phang khi biet cap VTCP cua no:

nh ly: Gia s mat phang co cap VTCP la : 1 2 3

1 2 3

( ; ; )

( ; ; )

a a a a

b b b b

=

=

G

G th mp co mot VTPT la :

2 3 3 1 1 2

2 3 3 1 1 2

; ; ;a a a a a a

n a bb b b b b b

= =

G G G

BAI TAP NG DUNG:Tm mot VTPT cua mat phang biet i qua ba iem A(-2;0;1), B(0;10;3), C(2;0;-1)II. Phng trnh cua mat phang :nh ly 1: Trong Kg(Oxyz) . Phng trnh mat phang i qua iem 0 0 0 0( ; ; )M x y z va co mot

VTPT la:( ; ; )n A B C =G

0 0 0( ) ( ) ( ) 0A x x B y y C z z + + =

nh ly 2: Trong Kg(Oxyz) . Phng trnh dang :

0Ax By Cz D+ + + = vi 2 2 2 0A B C+ +

],[ banKKK

=

aK

bK

);;( CBAn =K

);;( 0000 zyxM);;( CBAn =

K

0M

z

y

la phng trnh tong quat cua mot mat phang . x

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Chu y : Neu ( ) : 0Ax By Cz D + + + = th ( ) co mot VTPT la ( ; ; )n A B C =

G

123

0 0 0 0 0 0 0( ; ; ) ( ) : 0 Ax 0M x y z Ax By Cz D By Cz D + + + = + + + =

Cac trng hp ac biet:1. Phng trnh cac mat phang toa o:

(Oxy):z = 0 (Oyz):x = 0

(Oxz):y = 02. Phng trnh mat phang theo oan chan:

Phng trnh mat phang cat cac truc Ox, Oy, Oz tai

( ; 0; 0)

(0; ; 0) (a,b,c 0)

(0;0; )

A a

B b

C c

)(Oxz

)(Oxy

(Oyz

z

y

O

x

la: 1x y z

a b c+ + =

AB

C

a

b

c

O

BAI TAP AP DUNG:Bai 1: Trong Kg(Oxyz) cho ba iem A(3;1;0), B(-1;2;-1), C(2;-1;3)

Viet phng trnh mat phang (ABC)

Bai 2: Cho iem A(1;3;2), B(1;2;1), C(1;1;3)Viet phng trnh tham so cua ng thang (d) i qua trong tam tam giac ABC va vuong goc mat phang cha tam giac.

III. V tr tng oi cua hai mat phang - Chum mat phang :1. Mot so quy c va ky hieu:

Hai bo n so : c goi la ty le vi nhau neu co so 1 2

1 2

( , ,..., )

( , ,..., )n

n

a a a

b b b

0t sao cho

1 1

2 2

.

.

n n

a tb

a tb

a tb

=

=

=

Ky hieu: hoac1 2 1 2: : ... : : : ... :na a a b b b= n 1 21 2

... n

n

aa a

b b b= = =

2. V tr tng oi cua hai mat phang:nh ly: Trong Kg(Oxyz) cho hai mat phang , xac nh bi phng trnh :

1 1 1 1 1 1 1 1

2 2 2 2 2 2 2

( ) : 0 co VTPT ( ; ; )

( ) : 2; )0 co VTPT ( ;

x B y C z D n A B C

A x B y C

+ + + = =

+ + C z D n A B+ = =

JJG

JJG

1nK

2nK

1nK 2n

K

1

nK

2nK

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1 1 1 1 1 11 1 1 2 2 2

2 2 2 2 2 2

1 1 1 1

2 2 2 2

1 1 1 1

2 2 2 2

A( ) cat ( ) A : : : : (hay: )

A

A( ) // ( )

A

A( ) ( )

A

B B C C AB C A B C hoac hoac

B B C C A

B C D

B C D

B C D

B C D

= =

= = =

ac biet:1 2 1 2 1 2A 0A B B C C + + =

124

3. Chum mat phang :

a. nh ngha: Tap hp cac mat phang cung i qua mot ng thang c goi la mot chum matphang .

goi la truc cua chum Mot mat phang hoan toan c xac nh neu biet

i. Truc cua chumhoac ii. Hai mat phang cua chum

b. nh ly: Trong Kg(Oxyz) cho hai mat phang , cat nhau xac nh bi phng trnh :1 1 1 1

2 2 2 2

( ) : 0

( ) : 0

A x B y C z D

A x B y C z D

+ + + =

+ + + =

Khi o : Moi mat phang qua giao tuyen cua va eu co phng trnh dang:

2 21 1 1 1 2 2 2 2( ) : ( ) ( ) 0 ( 0)A x B y C z D A x B y C z D + + + + + + + = +

Chu y:

0 va 0 th

0 va 0 th

=

=

ac biet :

1 1 1 1 2 2 2 2

1 1 1 1 2

Neu 0 va 0 th va trong trng hp nay

phng trnh co the viet di dang sau:

1. m(A ) (A ) 0

hoac 2. (A ) (A

x B y C z D x B y C z D

x B y C z D n x B

+ + + + + + + =

+ + + + + 2 2 2 ) 0y C z D+ + =

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NG THANG TRONG KHONG GIAN

I. Phng trnh cua ng thang:1.Phng trnh tham so cua ng thang:nh ly: Trong Kg(Oxyz) . Phng trnh tham so cua ng thang ( ) i qua iem 0 0 0 0( ; ; )M x y z

va nhan lam VTCP la :1 2 3( ; ; )a a a a=G

0 1

0 2

0 3

( ) : (t )

x x ta

y y ta

z z ta

= + = +

= +

\

125

2. Phng trnh chnh tac cua ng thang:nh ly: Trong Kg(Oxyz) . Phng trnh chnh tac cua ng thang ( ) i qua iem 0 0 0 0( ; ; )M x y z

va nhan lam VTCP la :1 2 3( ; ; )a a a a=G

0 0 0

1 2 3

( ) :x x y y z z

a a a

= =

3. Phng trnh tong quat cua ng thang :Trong khong gian ta co the xem ng thang la giao tuyen cua hai mat phang nao o.

Xem ( ) = vi 1 1 1 1

2 2 2 2

( ) : 0

( ) : 0

A x B y C z D

A x B y C z D

+ + + =

+ + + =ta co nh ly sau.

nh ly: Trong Kg(Oxyz) he phng trnh:

1 1 1 11 1 1 2 2 2

2 2 2 2

0vi A : : : :

0

A x B y C z DB C A B C

A x B y C z D

+ + + =

+ + + =

la phng trnh tong quat cua mot ng thang.

Chu y: Neu 1 1 1 1 1 1 1

2 2 2 2 2 2 2

( ) : 0 ( ( ; ; ))( ):

( ) : 0 ( ( ; ; ))

A x B y C z D n A B C

x B y C z D n A B C

+ + + = =

+ + + = =

G

G th ( ) co mot VTCP la :

1 1 1 1 1 1

2 2 2 2 2 2

, ; ;B C C A A Ba n nB C C A A B

= =

G G G

O

z

y

x

)

aK

(

0M ),,( zyxM

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II. V tr tng oi cua ng thang va mat phang :1.V tr tng oi cua ng thang va mat phang :

nK

M )(aK

nK

M

)( aK

nK

M )(aK

nh ly: Trong Kg(Oxyz) cho :

ng thang 0 0

1 2

0

3

:( )x x y y z z

a a a

= = co VTCP 1 2 3( ; ; )a a a a=

Gva qua 0 0 0 0( ; ; )M x y z

va mat phang ( ) : 0Ax By Cz D + + + = co VTPT ( ; ; )n A B C =G

Khi o :

1 2 3

1 2 3

0 0 0

1 2 3

0 0 0

( ) cat ( ) Aa 0

Aa 0( ) // ( )

0

Aa 0( ) ( )

0

Ba Ca

Ba Ca

Ax By Cz D

Ba Ca

Ax By Cz D

+ +

+ + = + + + + + =

+ + + =

126

ac biet: 1 2 3( ) ( ) a : : : :a a A B C =

aK

nK

Chu y: Muon tm giao iem M cua ( ) va () ta giai he phng trnh :

( )

( )

pt

pt

tm x,y,z

Suy ra: M(x,y,z)

2. V tr tng oi cua hai ng thang :

0M

'

0MaK 1

2bK

0M uK

'uK

1

2'

0M

0M'

0M uK

'uK

1 2

uK

'uK

0M

'

0M

1

2

nh ly: Trong Kg(Oxyz) cho hai ng thang :0 0 0

1 0 0 0 0

' ' ' ' ' ' ' '0 0 02 0 0 0 0' ' '

( ) : co VTCP ( ; ; ) va qua M ( ; ; )

( ) : co VTCP ( ; ; ) va qua M ( ; ; )

x x y y z zu a b c x y z

a b c

x x y y z zu a b c x y z

a b c

= = =

= = =

G

JG

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=

=

=

JG JJJJJJJGG

JG JJJJJJJGG

' '1 2 0 0

' '0 0

1 2' ' '

1 2

( ) va ( ) ong phang , . 0

, . 0( ) cat ( )

: : : :

( ) // ( ) : :

u u M M

u u M M

a b c a b c

a b c

= =

JG JJJJJJJGG

' ' ' ' ' '0 0 0 0 0 0

' ' ' ' ' '1 2 0 0 0 0 0

' '1 2 0 0

: : ( ) : ( ) : ( )

( ) ( ) : : : : ( ) : ( ) : ( )

( ) va ( ) cheo nhau , . 0

a b c x x y y z z

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

u u M M

0

Chu y: Muon tm giao iem M cua ( )1 2va ( ) ta giai he phng trnh : tm x,y,z1

2

( )

( )

pt

pt

Suy ra: M(x,y,z)

III. Goc trong khong gian:1. Goc gia hai mat phang:

nh ly: Trong Kg(Oxyz) cho hai mat phang , xac nh bi phng trnh :1 1 1 1

2 2 2 2

( ) : 0

( ) : 0

A x B y C z D

A x B y C z D

+ + + =

+ + + =

Goi la goc gia hai mat phang ( ) & ( ) ta co cong thc:

127

1 2 1 2 1 2

2 2 2 2 2 2

1 1 1 2 2 2

cos.

A A B B C C

A B C A B C

+ +=

+ + + +

2. Goc gia ng thang va mat phang:

;;( 222 CBAn =K

);;( 1111 CBAn =K

00 900

)(

nh ly: Trong Kg(Oxyz) cho ng thang = =0 0( ) : 0x x y y z za b c

;(aa =

va mat phang ( ) : 0Ax By Cz D + + + =

Goi la goc gia hai mat phang ( ) & ( ) ta co cong thc:

;( BAnK

=

K

0 900

2 2 2 2 2 2sin

.

Aa Bb Cc

A B C a b c

+ +=

+ + + +

3.Goc gia hai ng thang :nh ly: Trong Kg(Oxyz) cho hai ng thang :

= =

= =

0 01

0 02 ' ' '

( ) :

( ) :

0

0

x x y y z z

a b c

x x y y z z

a b c

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Goi la goc gia hai mat phang 1( ) & ( )2 ta co cong thc:

128

' ' '

2 2 2 '2 '2 '2cos

.

aa bb cc

a b c a b c

+ +=

+ + + +

);;(1 cbaa =K

1

2

)';';'(2 cbaa =K

00900

IV. Khoang cach:

1. Khoang cach t mot iem en mot mat phang:nh ly: Trong Kg(Oxyz) cho mat phang ( ) : 0Ax By Cz D+ + = va iem 0 0 0 0( ; ; )M x y z +Khoang cach t iem M0 en mat phang ( ) c tnh bi cong thc:

0 0 00

2 2 2( ; )

Ax By Cz Dd M

A B C

+ + + =

+ +

);;( 0000 zyxM

H

2. Khoang cach t mot iem en mot ng thang:

nh ly: Trong Kg(Oxyz) cho ng thang ( ) i qua iem 0 0 0 0( ; ; )M x y z va co VTCP. Khi o khoang cach t iem M( ; ; )u a b c=

G

1 en ( ) c tnh bi cong thc:

0 1

1

;( , )

M M ud M

u

=

JJJJJJG G

G

3. Khoang cach gia hai ng thang cheo nhau:

nh ly: Trong Kg(Oxyz) cho hai ng thang cheo nhau :G

1 0

' ' ' ' ' ' ' '2 0

( ) co VTCP ( ; ; ) va qua M ( ; ; )

( ) co VTCP ( ; ; ) va qua M ( ; ; )

u a b c x y z

u a b c x y z

=

=

0 0 0

0 0 0

JG

2va ( )Khi o khoang cach gia ( )1 c tnh bi cong thc

' '0 0

1 2'

, .( , )

,

u u M M

d

u u

=

JG JJJJJJJGG

JGG

H

uK

);;( 0000 zyxM

1

)

M

(

0M

'

0M

uK

'uK

1

2

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BAI TAP REN LUYEN-------------***-------------

Bai 1: Trong Kg(Oxyz) cho hnh lap phng ABCD.A'B'C'D' vi A(0;0;0), B(1;0;0), D(0;1;0), A'(0;01Goi M, N lan lt la trung iem cua AB va CD.1. Tnh khoang cach gia hai ng thang A'C va MN

2. Viet phng trnh mat phang cha A'C va tao vi mat phang Oxy mot goc biet cos =

Bai 2: Trong Kg(Oxyz) cho iem A(0;1;2) va hai ng thang :

+=

=

+=

+

=

=

tz

ty

tx

dzyx

d

2

21

1

:&1

1

1

1

2: 21

1. Viet phng trnh mat phang (P) qua A, ong thi song song vi d1 va d2.2. Tm toa o cac iem M thuoc d1, N thuoc d2 sao cho ba iem A,M,N thang hang

Bai 3: Trong Kg(Oxyz) cho iem A(1;2;3) va hai ng thang :

1

1

2

1

1

1:&

1

3

1

2

2

2: 21

+=

=

=+

= zyx

dzyx

d

1. Tm toa o iem A' oi xng vi iem A qua ng thang d12. Viet phng trnh ng thang i qua A, vuong goc vi d1 va cat d2Bai 4: Trong Kg(Oxyz) cho 4 iem A(0;1;0), B(2;3;1), C(-2;2;2), D(1;-1;2) .

1. Chng minh cac tam giac ABC, ABD, ACD la cac tam giac vuong .2. Tnh the tch t dien ABCD.3. Goi H la trc tam tam giac BCD, viet phng trnh ng thang AH.

Bai 5: Trong Kg(Oxyz) cho 3 iem A(1;1;2), B(-2;1;-1), C(2;-2;1).1. Viet phng trnh mat phang (ABC).2. Xac nh toa o hnh chieu vuong goc cua iem O tren mat phang (ABC).3. Tnh the tch t dien OABC.

Bai 6: Trong Kg(Oxyz) cho hai ng thang:

1 2

12 4 0

: va :2 2 4 0

1 2

2

x tx y z

y tx y z

z t

= + + =

+ + =

= +

= +

1. Viet phng trnh mat phang (P) cha ng thang 1 va song song vi ng thang 2

2. Cho iem M(2;1;4). Tm toa o iem H thuoc ng thang 2 sao cho oan thang MH co

dai nho nhat.

Bai 7: Trong Kg(Oxyz) cho mat phang (P) : 2x-y+2=0 va ng thang(2 1) (1 ) 1 0

: (2 1) 4 2 0mm x m y m

d mx m z m

+ + + = + + + + =

Xac nh m e ng thang dm song song vi mat phang (P)

Bai 8: Trong Kg(Oxyz) cho mat phang (P) :x-y+z+3=0 va hai iem A(-1;-3;-2), B(-5;7;12)1. Tm toa o iem A la iem oi xng vi iem A qua mat phang (P)2. Gia s M la iem chay tren mat phang (P). Tm gia tr nho nhat cua bieu thc : MA+MB

Bai 9: Trong Kg(Oxyz) cho ng thang2 1 0

: va mat phang (P): 4x-2y+z-1=02 0

x y z

x y z

+ + + =

+ + + =

129

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Viet phng trnh hnh chieu vuong goc cua ng thang tren mat phang (P).

Bai 10: Trong Kg(Oxyz) cho hai ng thang: 1 20 3

: va d :1 0 3 6

3 0x az a ax yd

y z x z

= + =

+ = =

1. Tm a e hai ng thang d1 va d2 cat nhau2. Vi a=2, viet phng trnh mat phang (P) cha ng thang d2 va songsong vi ng tha

d1. Tnh khoang cach gia d1 va d2 khi a=2

Bai 11: Trong Kg(Oxyz) cho hnh hop ch nhat ABCD.ABCD co A trung vi goc toa o,B(a;0;0).D(0;a;0), A(0;0;b) (a>0,b>0) . Goi M la trung iem cua canh CC .1. Tnh the tch khoi t dien BDAM theo a va b

2. Xac nh ty soa

be hai mat phang (ABD) va (MBD) vuong goc vi nhau.

Bai 12: Trong Kg(Oxyz) cho t dien ABCD vi A(2;3;2), B(6;-1;-2), C(-1;-4;3), D(1;6;-5). Tnhgoc gia hai ng thang AB va CD . Tm toa o iem M thuoc ng thang CD sao chotam giac ABM co chu vi nho nhat.

Bai 13: 2. Trong khong gian vi he toa do e cac vuong goc Oxyz cho hai ng thang1

:1 1 2 1

x yd

+= =

z 0

0va

3 1:2 2 1

x zd

x y

+ =

+ =

1. Chng minh rang d1, d2 cheo nhau va vuong goc vi nhau.2. Viet phng trnh tong quat cua ng thang d cat ca hai ng thang d1, d2 va song song

vi ng thang4 7

:1 4

3

2

x y z = =

Bai 14: Trong khong gian vi he truc toa o ecac vuong goc Oxyz cho t dien OABC viA(0;0; 3a ), B(a;0;0), C(0; 3a ;0) (a>0). Goi M la trung iem cua BC. Tnh khoang

cach gia hai ng thang AB va OM.

Bai 15: Trong khong gian vi he truc toa o ecac vuong goc Oxyz cho hai iem A(2;1;1),

B(0;-1;3) va ng thang3 2 11

:

3 8 0

0x yd

y z

=

+ =

1. Viet phng trnh mat phang (P) i qua trung iem I cua AB va vuong goc vi AB. Goi Kgiao iem cua ng thang d va mat phang (P), chng minh rang d vuong goc vi IK.2. Viet phng trnh tong quat cua hnh chieu vuong goc cua d tren mat phang co phng trn

1 0x y z+ + =

Bai 16: Trong Kg(Oxyz) cho hai ng thang :

1 2

2 01 2( ) : va (d ) :

1 03 1 1

x y zx y zd

x

+ + = += =

+ =

Lap phng trnh ng thang qua M(0;1;1) sao cho vuong goc vi (d1) va cat (d2).Bai 17: Trong Kg(Oxyz) cho hai ng thang :

1 2

3 2 8 01 3 2( ) : va (d ) :

5 2 123 2 1 0

x yx y zd

x z

=+ + = =

+ + =

1. Chng minh d1 va d2 cheo nhau.2. Tnh khoang cach gia hai ng thang tren.3. Lap phng trnh ng thang qua M(-4;-5;3) sao cho cat ca d1 va d2.

Bai 18: Trong Kg(Oxyz) cho ng thang (d) va mat phang (P) co phng trnh :1 1 2

( ) : va (P):x-y-z-1=02 2 3

x y zd

+ = =

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Lap phng trnh ng thang qua A(1;1;-2) sao cho va //(P)d .


1 2

2 41 1( ): va (d ) :

2 2 12 1 1

0

0

x y zx y zd

x y z

+ = += =

+ + =

va mat phang ( ) .: 1 0P x y z+ + =

Lap phng trnh ng thang sao cho ( )P va cat ca hai ng thang d1 va d2.

Bai 20: Trong Kg(Oxyz) cho ng thang :

1 2( ) :2 1 3

x yd = =

z va iem I(2;-1;3)

Goi K la iem oi xng cua I qua (d) . Tm toa o iem K.

Bai 21: Trong Kg(Oxyz) cho ng thang :1

( ) :3 4 1

3x y zd

+= = va iem A(1;2;1)

Tnh khoang cach t iem A en ng thang (d).


1 2

2 1 0 3 3( ) : va (d ) :

x-y+z-1=0 2 1 0

0x y x yd

x y

+ + = + + =

+ =

z

1. Chng minh rang d1 va d2 cat nhau. Tm toa o giao iem I cua d1 va d2 .2. Lap phng trnh mat phang (P) i qua d1 va d2 .3. Tnh the tch phan khong gian gii han bi (P) va cac mat phang toa o.

Bai 23: Trong Kg(Oxyz) cho hai iem A(1;2;1) , B(2;1;3) va mat phang (P): x-3y+2z-6 = 0.1. Lap phng trnh mat phang (Q) i qua A, B va vuong goc vi (P).2. Viet phng trnh chnh tac cua giao tuyen cua (P) va (Q).3. Goi K la iem oi xng cua A qua (P). Tm toa o iem K.

Bai 24: Trong Kg(Oxyz) cho hai iem A(1;2;-1) va B(7;-2;3) va ng thang (d):2 3 4

4 0

0x y

y z

+ =

+ =

1. Chng minh (d) va AB ong phang .2. Tm toa o giao iem I0 cua ng thang (d) vi mat phang trung trc cua oan AB.3. Tm ( )I d sao cho tam giac ABI co chu vi nho nhat.

Bai 25: Trong Kg(Oxyz) cho hai iem A(0;0;-3) , B(2;0;-1) va mat phang (P): 3x - 8y + 7z -1 = 01. Tm toa o giao iem I cua ng thang AB va mat phang (P).2. Tm iem C thuoc mat phang (P) sao cho tam giac ABC eu.

Bai 26: Trong Kg(Oxyz) cho hai iem A(1;2;3) , B(4;4;5) va mat phang (P): z = 01. Tm M (P) sao cho MA+MB la nho nhat.2. Tm N (P) sao cho NA NB la ln nhat.

Bai 27: Trong Kg(Oxyz) cho hai iem A(3;1;0) , B(-9;4;9) va mat phang (P): 2x - y + z + 1 = 0Tm M (P) sao cho MA MB la ln nhat.

Bai 28: Trong Kg(Oxyz) cho ng thang (d) va mat phang (P) co phng trnh :4 1

( ) : va (P):x-y+3z+8=04 3 2

x y zd

+= =

Viet phng trnh hnh chieu cua (d) len (P)


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1 23 2 02 1

( ) : va (d ) :3 01 1 2

x zx y zd

y

+ = = =

=

1. Chng minh d1 va d2 cheo nhau.2. Lap phng trnh ng vuong goc chung cua hai ng thang d1 va d2 .


1 21 5 1

( ): va (d ) :1 0 1 0 2 3

5x y z x y zd

+ = = = =

1. Chng minh d1 va d2 cheo nhau.2. Tm toa o cac iem A, B cua ng vuong goc chung AB cua d 1 va d2 .

Bai 31: Cho tam giac ABC co toa o cac nh : A(0;1;0); B(2;2;2); C(-2;3;4)

va ng thang1 2

( ) :2 1 2

3x y zd

+ += =

.

1. Tm toa o iem M nam tren (d) sao cho AM AB .2. Tm toa o iem N nam tren (d) sao cho VNABC = 3.

Bai 32: Trong Kg(Oxyz) cho O(0;0;0), A(6;3;0), B(-2;9;1) va S(0;5;8)1. Chng minh rang SB .OA2. Chng minh rang hnh chieu cua canh SB len mat phang (OAB) vuong goc vi OA. Goi

K la giao iem cua hnh chieu o vi OA. Tm toa o iem K.3. Goi P, Q lan lt la trung iem cua cac canh OS va AB.Tm toa o M thuoc SB sao choPQ va KM cat nhau.

Bai 33: Cho hai ng thang :

1 2

2 01 2 3( ) : va (d ) :

2 3 51 2 3 0

x y zx y zd

x y z

+ = = =

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