Giao an Hinh Hoc Co Ban Lop 10 Hoc Ki 2(Suu Tam)

8/6/2019 Giao an Hinh Hoc Co Ban Lop 10 Hoc Ki 2(Suu Tam)

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Bi 2: PHNG TRNH NG THANGSO TIET: 4

I. MUC CH YEU CAU1. Kien thc c ban:

- Hieu vect phap tuyen, vect ch phng cuang thang

- Hieu cach viet phng trnh tong quat, phng trnhtham so cua ng thang- Hieu c ieu kien hai ng thang cat nhau,song song, trung nhau, vuong goc nhau- Biet cong thc tnh khoang cach t mot iemen mot ng thang, goc gia hai ng thang

2. Ky nang:- Viet c phng trnh tong quat, phng trnh thamso cua ng thang d i qua iem M(x0;y0) va co phngcho trc hoac i qua hai iem cho trc

- Tnh c toa o cua vect phap tuyen khi biet toao cua vect ch phng cua mot ng thang vangc lai- Biet chuyen oi gia hai loai phng trnh- S dung c cong thc tnh khoang cach t motiem en mot ng thang- Tnh c so o cua goc gia hai ng thang

3. Trong tam: Viet phng trnh ng thang, cong thctnh goc, khoang cach

II. PHNG PHAP:III. TIEN TRNH:

1. Kiem tra bai cu:i. Hay neu mot dang phng trnh ng thang ma embiet ?

ii. Hay cho biet he so goc cua ng thang co phngtrnh y = ax + b

iii. ng thang nao sau ay song song vi ng thang: y= 2x + 3

a) y = -2x + 1 b) y =1

2x + 1 c) 2x y 12 = 0 d) y

= 3

2. Giang bai mi:HOAT ONG 1I. Vect ch phng cua ng thang

Trong mp Oxy cho ng thang la o th cua ham so: y =1

2x

a) Tm tung o cua 2 iem M0, M nam tren co hoanh olan lt la 2 va 6

b) Cho vect u = (2;1). Hay chng to 0M M cung phng vi

u

GV: Treo hnh 3.2 len bang, neu cau hoi e hng dan HS thc

hien hoat ong nayHoat ong cua GV Hoat ong cua HS+ e tm tung o cua mot Ta thay hoanh o vao


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iem khi biet hoanh o vaphng trnh ng thang iqua no ta can lam nhng g ?+ Hay tm tung o cua M0 vaM

+ Hai vect cung phng khinao ?

+ Tm t thoa:0

M M = tu

Ta noi vect u la vtcp cuang thang

nh ngha

phng trnh cua ngthang

0M

y =1

2.2 = 1, My =

1

2.6 = 3

Hai vect cung phng khi

vect nay bang t lan vectkia

0M M = (4;2) = 2.(2;1) = 2. u

t = 2

nh ngha: Vect u c goi la vect ch phng cua ngthang neu 0u va gia cua u song song hoac trung vi

* Nhan xet: + Neu u la mot vtcp cua th ku (k # 0) cung la vtcpcua . Do omot ng thang co vo so vtcp

+ Mot ng thang hoan toan c xac nh neu bietmot iem va mot

vtcp cua noHOAT ONG 2

II. Phng trnh tham so cua ng thangTrong mp Oxy cho ng thang i qua iem ( )0 0 0;M x y va co vect

ch phng ( )1 2;u u u= . Tm ieu kien can va u e iem M(x;y)

thuoc GV: Treo hnh 3.3 len bang, neu cau hoi e hng dan HS thchien hoat ong nay

Hoat ong cua GV Hoat ong cua HS+ Neu nhan xet ve phngcua vect

0M M va vect u

+ K e0

M M cung phng

vi u

+0

M M = ? ; tu = ?

+ K e hai vect bang nhau ?

0M M = tu

0 1

0 2

x x tu

y y tu

= =

0 10 2

x x tu

y y tu

= + = +

(*)

He (*) goi la pt tham so cuat

nh ngha

Vect0

M M cung phng vi

vect u

0M M = tu

0M M = (x x0;y y0), tu =

(tu1;tu2)Hoanh o = hoanh o, tungo = tung o


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1) nh ngha:Trong mp Oxy, ng thang i qua iem ( )0 0 0;M x y

va nhan vect ( )1 2;u u u= lam vtcp co phng trnh tham so la:

0 1

0 2

x x tu

y y tu

= + = +

, t

* Chu y: Neu iem M thuoc th M co toa o la: M(x0 + tu1;y0 +tu2). ng vi moi gia tr t ta c mot iem M thuoc

V du: Cho ng thang d co pt tham so:5 6

2 8

x t

y t

= = +

a) Tm mot vtcp cua db) Tm 3 iem thuoc dc) Cho A(1;1), tm iem M thuoc d sao cho MA = 2

2) Lien he gia vect ch phng va he so goc cua ngthang

Hoat ong cua GV Hoat ong cua HSCho t co pt tham so:

0 1

0 2

x x tu

y y tu

= + = +

(*)

Vi u1 # 0. Hay bien oi pt(*) ve dangy = ax + b roi t o xac

nh he so goc

0 1

0 2

x x tu

y y tu

= + = +

0

1

0 2

x xt

u

y y tu

= = +

Suy ra y = ( )2 0 01

u x x y

u +

He so goc la k =2

1

u

u

Neu ng thang co vtcp ( )1 2;u u u= vi u1 # 0 th co he so

goc k = 21

u

u

V du 1: Tm he so goc cua cac t d1, d2, d3 co vtcp lan lt la

( )1 1; 3u = , ( )2 0; 2u = , ( )3 5; 0u = V du 2: Viet pt tham so cua ng thang d i qua 2 iem A(2;3),B(3;1). Tnh he so goc cua d

HOAT ONG 3III. Vect phap tuyen cua ng thang

Cho ng thang co phng trnh:5 2

4 3

x t

y t

= +

= +va vect ( )3; 2n = . Hay

chng to n vuong goc vi vtcp cua Hoat ong cua GV Hoat ong cua HS

+ Hay xac nh vtcp cua + Hai vect vuong goc nhaukhi nao ?+ Chng to n vuong goc

vi uVec n nh tren goi la vect

( )2;3u =Khi tch vo hng cua chung= 0

n .u = 3.2 + (-2).3 = 0Suy ra n vuong goc vi u


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phap tuyen cua t nh ngha

nh ngha: Vect n c goi la vect phap tuyen cua ngthang neu 0n va n vuong goc vi vect ch phng cua * Nhan xet:+ Neu n la mot vtpt cua th k n (k # 0) cung la vtpt

cua . Do omot ng thang co vo so vtpt+ Mot ng thang hoan toan c xac nh neu

biet mot iem va motvtpt cua no

HOAT ONG 4IV. Phng trnh tong quat cua ng thangTrong mp Oxy cho ng thang i qua iem ( )0 0 0;M x y va co vect

phap tuyen ( );n a b= . Tm ieu kien can va u e iem M(x;y)thuoc

GV: Treo hnh 3.5 len bang, neu cau hoi e hng dan HS thchien hoat ong nayHoat ong cua GV Hoat ong cua HS

+ Neu nhan xet ve phngcua vect

n va vect 0M M

+0

M M = ?+ ieu kien e hai vectvuong goc ?+

n.

0

M M = 0 ?+ Neu ta at c = ax0 by0 thta co pt:ax + by + c = 0 va goi la pttong quat cua ng thangnh ngha

n vuong goc vi 0M M

0M M = (x x0;y y0)

Tch vo hng cua chung =0

n . 0M M = 0 a(x x0) + b(y

y0) = 0 ax +by ax0 by0 = 0

1) nh ngha: Phng trnh ax + by + c = 0 vi a, b khong cungbang 0 c goi la phng trnh tong quat cua ng thang* Nhan xet: + Neu ng thang co phng trnh ax + by + c = 0th co:

Vect phap tuyen la: ( );n a b=

Vect ch phng la: u = (-b;a)

+ ng thang( )

( )

0 0 0,

:: ;

quaM x y

VTPT n a b

=

r co phng trnh tong

quat la:a(x x0) + b(y y0) = 0

V du: Lap pt tong quat cua ng thang d i qua 2 iem A(2;2),B(4;3)V du: Tm toa o vect ch phng cua ng thang co phngtrnh:3x +4y +5 = 0

GV: Treo hnh 3.6, 3.7, 3.8, 3.9 e gii thieu cac trng hp ac bietsau:2) Cac trng hp ac biet


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Cho ng thang d co phng trnh tong quat ax + by + c = 0 (1)a) Neu a = 0 th (1) tr thanh y = -c/b. Khi o d vuong gocvi Oy tai (0;-c/b)b) Neu b = 0 th (1) tr thanh x = -c/a. Khi o d vuong gocvi Ox tai (-c/a;0)c) Neu c = 0 th (1) tr thanh ax + by = 0. Khi o d i qua

goc toa o Od) Neu a, b, c eu khac 0 th (1) luon a c ve dang:

0 0

1x y

a b+ = va phng trnh nay goi laphng trnh ng thang

theo oan chan . ng thang nay cat Ox, Oy lan lt taiA(a0;0), B(0;b0)

V du: Ve cac ng thang co phng trnh sau:

d1:x = 2; d2:x 2y = 0; d3:y + 1 = 0; d4: 18 4

x y+ =

HOAT ONG 5

V. V tr tng oi cua hai ng thangCho hai ng thang: d1: a1x + b1y + c1 = 0 va d2: a2x + b2y + c2 = 0Toa o giao iem cua d1 va d2 la nghiem cua he phng trnh:

1 1 1

2 2 2

a x + b y + c = 0

a x + b y + c = 0

(*)

Ta co cac trng hp sau:1) He (*) co mot nghiem (x0;y0), khi o d1 cat d2 tai iemM(x0;y0)2) He (*) vo nghiem, khi o d1 song song vi d23) He (*) co vo so nghiem, khi o d1 trung vi d2

V du: Xet v tr tng oi cua ng thang : x 2y + 1 = 0 vimoi ng thang sau: d1: -3x + 6y 3 = 0, d2: y = -2x, d3: 2x + 5 = 4y

HOAT ONG 6VI. Goc gia hai ng thangCho hnh ch nhat ABCD tam I co cac canh AB = 1, AD = 3 . Tnh so

o cac goc AID va DICGV: Ve hnh 3.13 len bang, neu cau hoi e hng dan HS thchien hoat ong nay

Hoat ong cua GV Hoat ong cua HSHay tnh o dai canh BD

Hay tnh cos ADB

ADB = ?

T o hay tnh AID va DIC

Treo hnh 3.14 len bang vahng dan HS i en congthc tnh goc

BD = 2 2 4 2 AB AD+ = =

cos ADB =3

2

AD

DB=

ADB = 300

AID = 1800 (300 + 300) = 1200

DIC = 600

Cho 2 ng thang:1 1 1 1

2 2 2 2

: 0

: 0

a x b y c

a x b y c

+ + = + + = co vtpt lan lt la 1n va 2n


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Goi la goc gia 1 va 2 th ta co:1 2 1 2 1 2

2 2 2 2

1 2 1 1 2 2

.

cos

. .

n n a a b b

n n a b a b

+

= =

+ +

ur uur

Chu y:

1 2 1 2 1 2 1 2 0n n a a b b + =

Neu 1 va 2 lan lt co phng trnh y = k1x + m1 va y = k2x +m2 th: 1 2 1 2 . 1kk=

HOAT ONG 7VII. Khoang cach t mot iem en mot ng thangTrong mat phang Oxy cho ng thang co phng trnh ax + by + c= 0 va iem M0(x0;y0). Khoang cach t iem M0 en ng thang

k hieu d(M0, ) c tnh bi cong thc: ( )0 0

02 2

d M ,ax by c

a b

+ + =

+

V du: Tnh khoang cach t iem M(-2;1) va O(0;0) en ng

thang d co phng trnh 3x 2y 1 = 0CUNG CO : Cho ng thang d co phng trnh 3x + 4y + 1 = 0

a) Tm mot vtpt va mot vtcp cua db) Tm mot iem thuoc d va viet pt tham so cua dc) Xet v tr tng oi cua d va d': -2x + 3y + 3 = 0d) Tnh goc gia d va d'e) Tnh khoang cach t iem A(1;-2) en ng thang d

DAN DO: Lam bai tap SGK trang 80, 81

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Bi 3: PHNG TRNH NG TRONSO TIET: 2

I. MUC CH YEU CAU1. Kien thc c ban:

- Hieu cach viet phng trnh ng tron2. Ky nang:

- Viet c phng trnh ng tron khi biet tam vaban knh- Xac nh c tam va ban knh ng tron khibiet phng trnh- Viet c phng trnh tiep tuyen tai mot iemthuoc ng tron

3. Trong tam: Viet phng trnh ngtronII. PHNG PHAP:III. TIEN TRNH:

1. Kiem tra bai cu:i. Cong thc tnh khoang cach gia 2 iem A(xA;yA), B(xB;yB)

ii. Viet phng trnh ng thang i qua iem ( )0 0 0;M x y vaco vect phap tuyen ( );n a b=

iii. Mot ng tron c xac nh khi biet nhng yeu tonao ?

2. Giang bai mi:HOAT ONG 1

I. Phng trnh ng tron co tam va ban knh cho trcTrong mat phang Oxy cho ng tron (C) co tam I(a;b), ban knh R.Tm ieu kien can va u e iem M(x;y) thuoc (C)GV treo hnh 3.16 len bang, neu cau hoi e hng dan HS thchien hoat ong nay

Hoat ong cua GV Hoat ong cua HSiem M(x;y) thuoc (C) khinao ?IM = ?IM = R ?Pt (*) goi la phng trnhng tron tam I(a;b) banknh R nh ngha

Khi IM = R

( ) ( )2 2

IM x a y b= +

IM = R ( ) ( )2 2

x a y b R + =

( ) ( )2 2

x a y b + = R2 (*)

nh ngha: Trong mp Oxy ng tron tam I(a;b) ban knh R co

phng trnh la: (x a)2 + (y b)2 = R2

* Chu y : ng tron tam O ban knh R co phng trnh: x2 + y2 = R2


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V du: Cho 2 iem A(3;-4), B(-3;4). Viet phng trnh ng tron (C)nhan AB lam ng knh

Hoat ong cua GV Hoat ong cua HSVe ng tron co motng knh la AB va ch ratam I cua ng tron

Hay xac nh ban knhViet phng trnh ng tron(C)

Tam I la trung iem AB=> I(0;0)Ban knh R = AB/2 = 5/2

=> (C): x2 + y2 = 25/4

* Nhan xet: Phng trnh ng tron (x a)2 + (y b)2 = R2 co theviet di dangx2 + y2 2ax 2by + c = 0 trong o c = a2 + b2 R2

Ngc lai phng trnh x2 + y2 2ax 2by + c = 0 la phng trnh cuamot ng tron (C) khi va ch khi a2 + b2 c > 0. Khi o (C) co tamI(a;b) va ban knh 2 2 R a b c= + V du: Trong cac phng trnh sau, phng trnh nao la phng trnh

cua ng tron a) 2x2 + y2 8x + 2y 1= 0

b) x2 + y2 + 2x 4y 4 =0

c) x2 + y2 2x 6y + 20= 0d) x2 + y2 + 6x + 2y + 10= 0

HOAT ONG 2II. Phng trnh tiep tuyen cua ng tronGV treo hnh 3.17 len bang, neu cau hoi e hng dan HS thchien hoat ong nay

Hoat ong cua GV Hoat ong cua HSGoi HS nhac lai cong thc vietpt ng thang i qua motiem va co vtpt cho trcTren hnh 3.17 ng thang c xac nh nh the nao ?Viet pt ng thang ?

a(x x0) + b(y y0) = 0

qua ( )0 0 0;M x y va co vtpt

( )0 0 0; IM x a y b=

(x0 a)(x x0) + (y0 b)(y y0)= 0

Cho ng tron (C) co tam I(a;b) va iem ( )0 0 0;M x y nam tren (C).Tiep tuyen cua (C) tai M0 co phng trnh: (x0 a)(x x0) + (y0 b)(y y0) = 0

V du:Viet pt tiep tuyen tai iem M(3;4) thuoc ng tron(C):(x 1)2+ (y 2)2 = 8CUNG CO: + HS nhac lai 2 dang cua phng trnh ng tron.

+ Cach viet phng trnh tiep tuyen tai mot iem thuocng tronDAN DO: Lam BT trong SGK trang 83,84


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Bi 4 : PHNG TRNH NG ELIP

SO TIET : 2I. MUC CH YEU CAU1. Kien thc c ban:

- Biet nh ngha elip, phng trnh chnh tac va hnhdang cua elip

2. Ky nang:- T phng trnh chnh tac cua elip xac nh c odai cac truc, tieu c, toa o cac tieu iem, giao iemcua elip vi cac truc toa o

3. Trong tam: Viet phng trnh ngelipII. PHNG PHAP:III. TIEN TRNH:

1. Kiem tra bai cu:i. Viet phng trnh ng tron (C) co tam I(1;-2), ban knhR = 5ii. Viet phng trnh tiep tuyen cua (C) tai iem M(2;-4)

2. Giang bai mi:HOAT ONG 1

I. nh ngha ng elipHoat ong cua GV Hoat ong cua HS

Ve hnh 3.19 len bang va

giai thch cach ve cho HSngheng vave tren bangc goi la ng elipnh ngha

Nghe e hieu cach ve

Hnh dung trc quan hnh dangcua ng elip. Tm VD ngelip a gap trong cuoc songhang ngay

Cho hai iem co nh F1, F2 va mot o dai khong oi 2a ln hnF1F2. Elip la tap hp cac iem M trong mat phang sao cho :

F1M + F2M = 2aCac iem F1, F2 goi la cac tieu iem cua elip. o dai F1F2 = 2cgoi la tieu c cua elip

HOAT ONG 2II. Phng trnh chnh tac cua elipCho elip (E) co cac tieu iem F1, F2. iem M thuoc elip khi va chkhiF1M+F2M = 2a


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Chon he truc toa o Oxy sao cho F1(-c;0), F2(c;0). Khi o ngi tachng minh c :

( ) ( )2 2

2 2; 1

x yM x y E

a b + = (1), trong o b2 = a2 c2

Phng trnh (1) goi laphng trnh chnh tac cua elip* Chu y: T ang thc b2 = a2 c2 suy ra b < a

HOAT ONG 3III. Hnh dang cua elipGV treo hnh 3.21 len bang, neu cau hoi e hng dan HS thchien hoat ong nay

Hoat ong cua GV Hoat ong cua HSiem M thuoc (E) th cac iemM1, M2, M3 cung thuoc (E). Do o(E) co cac truc oi xng Ox,Oy va co tam oi xng la OTm giao iem cua (E) vi cac

truc toa o ?

Cac iem A1, A2, B1, B2 goi lacac nh cua elip. oan thangA1A2 goi la truc ln, oan B1B2goi la truc nho cua elip

Thay y = 0 vao phng trnh(1) ta co x a= , suy ra (E) cat

Ox tai 2 iem ( ) ( )1 2;0 , ;0 A a A a. Thay x = 0 vao (1) ta cy b= . Vay (E) cat Oy tai 2iem ( ) ( )1 20; , 0; B b B b

Tom tat:

Phng trnh chnh tac:2 2

2 21,

x ya b

a b+ = >

Quan he gia a, b, c: b2

= a2

c2

Tieu iem: F1(-c;0), F2(c;0) Cac nh: ( ) ( )1 2;0 , ;0 A a A a , ( ) ( )1 20; , 0; B b B b

o dai truc ln: A1A2 = 2a o dai truc nho: B1B2 = 2b Tieu c: F1F2 = 2c

V du 1: Cho elip (E):2 2

19 1

x y+ = . Hay xac nh toa o cac nh, tieu

iem, tieu c, o dai cac truc va ve (E)V du 2: Lap phng trnh chnh tac cua elip (E) trong cac trng hp

sau:a) o dai hai truc la 10 va 8

b) o dai truc ln bang 12 tieu c bang 8

c) (E) i qua 2 iem: M(2;0) va1

3;2

N

IV. Lien he gia ng tron va ng elipa) T he thc b2 = a2 c2 ta thay neu tieu c cua elipcang nho th b cang gan bang a, tc la truc nho cang ganbang truc ln. Luc o elip co dang gan nh ng tron

b) Trong mat phang Oxy cho ng tron (C) co phng trnh:

x2 + y2 = a2. Vi moi iem M(x;y) thuoc (C) ta xet iem M'(x';y')


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sao cho:

'

'

x x

by y

a

=

=(0


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CH BM ST TRONG PHN NY :PHNG PHP TA TRONG MT PHNG

S tit : 4............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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Giao an Hinh Hoc Co Ban Lop 10 Hoc Ki 2(Suu Tam)