Chun kin thc HH 10

CHUYN NG THNG - NG TRN - LIP

PHNG PHP TA TRONG MT PHNGI. CC H THC LNG TRONG TAM GIAC V GII TAM GIC

Cho tam gic ABC c AB = c, BC = a, CA = b, ng cao AH = ha v cc ng trung tuyn AM = ma, BN = mb, CP = mc.

a. nh l cosin.

a2 = b2 + c2 2bccosA

b2 = a2 + c2 2accosB

c2 = a2 + b2 2abcosC H qu

b. nh l sin.

: bn knh ng trn ngoi tip tam gic ABC)

c. di ng trung tuyn ca tam gic.

4. Cc cng thc tnh din tch tam gic.

Din tch S ca tam gic c tnh theo cc cng thc:

* *

* ( R : bn knh ng trn ngoi tip tam gic ABC)

* vi v r l bn knh ng trn ni tip tam gic ABC.

* vi (cng thc H- rng)

II. TA 1. H trc to oxy gm ba trc ox, oy i mt vung gc vi nhau vi ba vect n v .

2. ; M(x;y)(

3. Ta ca vect: cho

a.

b.

c.

d.

e.

f.

g. .

4. Ta ca im: cho A(xa;ya), b(xb;yb)a.

b.

c. G l trng tm tam gic ABC ta c:

;

d. M chia AB theo t s k:

c bit: M l trung im ca AB:

III. PHNG TRNH NG THNG

A. CC KIN THC CN NH.

1. Phng rnh tham s.

* Phng trnh tham s ca ng thng i qua im M0(x0 ; y0), c vec t ch phng l

* Phng trnh ng thng i qua M0(x0 ; y0) v c h s gc k l: y y0 = k(x x0).

2. Phng trnh tng qut.* Phng trnh ca ng thng i qua im M0(x0 ; y0) v c vec t php tuyn l:

a(x x0) + b(y y0) = 0 ( a2 + b2

* Phng trnh ax + by + c = 0 vi a2 + b2 l phng trnh tng qut ca ng thng nhn lm vc t php tuyn v ( b; -a ) lm vect ch phng * ng thng ct Ox v Oy ln lt ti A(a ; 0) v B(0 ; b) c phng trnh theo on chn l :

* Cho (d) : ax+by+c=0 Nu // d th phng trnh l ax+by+m=0 (m khc c) Nu vung gc d th phnh trnh l : bx-ay+m=0

3. V tr tng i ca hai ng thng.

Cho hai ng thng

xt v tr tng i ca hai ng thng ta xt s nghim ca h phng trnh

(I)

( Ch : Nu a2b2c2 th :

4. Gc gia hai ng thng. Gc gia hai ng thng c VTPT c tnh theo cng thc:

5. Khonh cch t mt im n mt ng thng.

Khong cch t mt im M0(x0 ; y0) n ng thng : ax + by + c = 0 cho bi cng thc:

d(M0,) =

B. BI TP.1) Cho tam gic ABC vi A(-1;2);B(2;-4);C(1;0).Tm phng trnh cc ng thng cha ng cao tam gic ABC

2) Vit phng trnh cc trung trc cc cnh tam gic ABC bit trung im 3 cnh l M(-1;1) ; N(1;9) v P(9;1)

3) Cho A(-1;3) v d: x-2y +2=0.Dng hnh vung ABCD c B v C thuc d, C c ta l s dnga) Tm ta d A,B,C,D

b) Tm chu vi v din tch hnh vung ABCD

4) Cho d1: 2x-y-2=0 v d2:x+y+3=0 ; M(3;0)

a) Tm giao im d1 v d2b) Tm phng trnh ng thng d qua M ct d1 v d2 ti A v B sao cho M l trung im on AB

5) a) Vit phng trnh tng qut ng thng d: t

b)Vit phng trnh tham s ng thng d: 3x-y +2 = 0

6) Xt v tr tng i cp ng thng sau : t v d2:

7) Cho d1 v d2:

a) Tm giao im ca d1 v d2 gi l M

b) Tm phn trnh tng qut ng thng d i qua M v vung gc d18) Lp phng trnh sau y M( 1;1) ; d : 3x +2y-1 = 0

a) ng thng di qua A( -1;2) song song ng thng d

b) ng thng i qua M vung gc d

c) ng thng i qua M v c h s gc k = 3

d) ng thng i qua M v A

9) Cho d v M (3;1) a) Tm A thuc d sao cho AM = 3

b) Tm B thuc d sao cho MB t gi tr nh nht

10) Cho d c 1 cnh c trung im M( -1;1) ; 2 cnh kia nm trn cc ng thng: 2x + 6y+3 = 0 v Tm phng trnh cnh th 3 ca tam gic

11) Cho tam gic ABC c pt BC : Pt ng trung tuyn BM v CN c pt : 3x + y 7 = 0 v x + y 5 =0 vit pt cc cnh AB v AC

12) Cho A ( -1; 2 ) ; B(3;1) v d : . Tm C thuc d sao choABC cn

13) Cho A( -1;2) v d : Tm d (A;d) . Tm din tch hnh trn tm A tip xc d

14/ Vit pt ng thng : Qua A( -2; 0) v to vi : d : x + 3y + 3 = 0 mt gc 450 15/ Vit pt ng thng : Qua B(-1;2) to vi ng thng d: mt gc 600

16/ a) Cho A(1;1) ; B(3;6) . Tm pt ng thng i qua A v cch B mt khong bng 2

b) Cho d: 8x 6y 5 = 0 tm pt d sao cho d song song d v d cch d mt khong bng 5

17) A(1;1); B(2;0); C(3;4) .Tm pt ng thng qua A cch u B v C

18) Cho hnh vung c nh A (-4;5) pt mt ng cho l 7x y + 3 = 0 lp pt cc cnh hnh vung v ng cho cn liIV. PHNG TRNH NG TRN

A. CC KIN THC CN NH.

1. phng trnh ng trn.

* Phng trnh ng trn tm I(a; b), bn knh R l: (x a)2 + (y b)2 = R2.

* Nu a2 + b2 c > 0 th phng trnh x2 + y2 2ax 2by + c = 0 l phng trnh ca ng trn tm

I(a ; b), bn knh R =

* Nu a2 + b2 c = 0 th ch c mt im I(a ; b) tha mn phng trnh: x2 + y2 2ax 2by + c = 0

* Nu a2 + b2 c < 0 th khng c im M(x ; y) no tha mn phng trnh: x2 + y2 2ax 2by + c = 0

2. Phng trnh tip tuyn ca ng trn.

Tip tuyn ti im M0(x0 ; y0) ca ng trn tm I(a ; b) c phng trnh

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

B. BI TP.19) Tm pt ng trn (C) trong cc trng hp sau

a) C ng knh AB vi A ( 7;3); B(1;7)

b) Ngoi tip tam gic ABC vi A(1;3);B(5;6) v C(7;0)

c) i qua A(2;-1) tip xc cc trc ta

d) C tm thuc d: 3x 5y 8 = 0 v tip xc cc trc ta

e) i qua A(-1;0) ; B(1;2) tip xc d: x y 1 = 0

f) Tip xc 0x ti A(6;0) v i qua B(9;9) g) C tm I(1;3) tip xc d: x + y + 2 = 0

20/ Tm tm I v bn knh R ca cc ng trn sau :

a) x2 + y2 4x 2y + 1 = 0

b) 3x2 + 3y2 6x + 4y 1 = 0

21/ Cho (C) : x2 + y2 2x + 6y + 5 = 0 v d: 2x + y 1 = 0 .Tm pttt d ca (C) bit d song song d. Tm ta tip im

22/ Cho ( C) : x2 + y2 + 4x + 4y 17 = 0

a) Tm tm I v bn knh R ca (C)

b) Tm pttt d vi (C) ti M (2;1)

c) Tm pttt d vi (C) bit d song song d : 4x 3y +1 = 0

d) Tm pttt d vi (C) bit d vung gc d : 4x 3y + 1 = 0

e) Tm pttt d vi (C) bit d i qua A(2;6) V. ELIPA. CC KIN THC CN NH.1. Phng trnh chnh tc: , (a>b>0).

2. Cc yu t: , c>0.

Tiu c: F1F2=2c;

di trc ln A1A2=2a

di trc b B1B2=2b.

Hai tiu im .

Bn nh: nh trn trc ln ,

nh trn trc b .

Bn knh qua tiu im:

Tm sai:

ng chun:

Khong cch gia hai ng chun: .

3. iu kin ng thng Ax+By+C=0 tip xc vi elip l: A2a2+B2b2=C2.

B. BI TP

23/ Xc nh di hai trc, tiu c, tm sai, ta cc tiu im v cc nh ca elip sau:

a) b) 4x2 + 16y2 1 = 0 c) x2 + 4y2 = 1 d) x2 + 3y2 = 2

24/ Lp phng trnh chnh tc ca elip (E) bit.

a) A(0; - 2) l mt nh v F(1; 0) l mt tiu im ca (E).

b) F1(-7; 0) l mt tiu im v (E) i qua M(-2; 12)

c) Tiu c bng 6, tm sai bng 3/5.

d) Phng trnh cc cnh ca hnh ch nht c s l x =

25/ Tm nhng im trn elip (E): tha mn:

a) C bn knh qua tiu im bn tri bng hai ln bn knh qua tiu im bn phi.b) Nhn hai tiu im di mt gc vung.

26/ Cho elip (E): .

a) Tm ta cc tiu im, cc nh; tnh tm sai v v (E).

b) Xc nh m ng thng d: y = x + m v (E) c im chung.

x

y

F

2

F

1

B

2

B

1

A

2

A

1

O

M

GV: Trng Cng Hng

2

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