692 Bài Hình Học Luyện Thi Đại Học

________________________________________________________________________ TRUNG TAM LTH 17 QUANG TRUNG T: 07103.751.929 Trang - 1 -

TRUNG TM GIO DC V O TO

17 QUANG TRUNG

Cn Th 2013

a ch: 17 Quang Trung Xun Khnh Ninh Kiu Cn Th in thoi: 0939.922.727 0915.684.278 (07103)751.929

200 BAI TOA O TRONG MAT PHANG 200 TOA O TRONG KHONG GIAN 200 BAI HNH HOC KHONG GIAN


I. ng thang II. ng tron III. Cac ng conic IV. Tam giac V. T giac


I. NG THNG Cu 1. Trong mt phng vi h to Oxy, cho 2 ng thng 1d : x 7y 17 0 ,

2d : x y 5 0 . Vit phng trnh ng thng (d) qua im M(0;1) to vi 1 2d ,d mt tam gic cn ti giao im ca 1 2d ,d .

Phng trnh ng phn gic gc to bi d1, d2 l:

12 2 2 2

2

x 3y 13 0 ( )x 7y 17 x y 53x y 4 0 ( )1 ( 7) 1 1

ng thng cn tm i qua M(0;1) v song song vi 1 hoc 2 . KL: x 3y 3 0 v 3x y 1 0 Cu 2. Trong mt phng vi h trc to Oxy, cho cho hai ng thng 1d : 2x y 5 0 .

2d :3x 6y 7 0 . Lp phng trnh ng thng i qua im P(2; 1) sao cho ng thng ct hai ng thng d1 v d2 to ra mt tam gic cn c nh l giao im ca hai ng thng d1, d2.

d1 VTCP 1a (2; 1) ; d2 VTCP 2a (3;6)

Ta c: 1 2a .a 2.3 1.6 0

nn 1 2d d v d1 ct d2 ti mt im I khc P. Gi d l

ng thng i qua P( 2; 1) c phng trnh: d : A(x 2) B(y 1) 0 Ax By 2A B 0

d ct d1, d2 to ra mt tam gic cn c nh I khi d to vi d1 ( hoc d2) mt gc 450

0 2 2

2 2 2 2

A 3B2A Bcos 45 3A 8AB 3B 0

B 3AA B 2 ( 1)

* Nu A = 3B ta c ng thng d :3x y 5 0 * Nu B = 3A ta c ng thng d : x 3y 5 0 Vy c hai ng thng tho mn yu cu bi ton. d :3x y 5 0 ; d : x 3y 5 0 . Cu hi tng t: a) 1d : x 7y 17 0 , 2d : x y 5 0 , P(0;1) . S: x 3y 3 0 ; 3x y 1 0 . Cu 3. Trong mt phng Oxy, cho hai ng thng 1d :3x y 5 0 , 2d :3x y 1 0 v

im I(1; 2) . Vit phng trnh ng thng i qua I v ct 1 2d ,d ln lt ti A v B

sao cho AB 2 2 . Gi s 1 2A(a; 3a 5) d ; B(b; 3b 1) d ; IA (a 1; 3a 3); IB (b 1; 3b 1)

I, A, B thng hng b 1 k(a 1)

IB kIA3b 1 k( 3a 3)

Nu a 1 th b 1 AB = 4 (khng tho).

Nu a 1 th b 13b 1 ( 3a 3) a 3b 2a 1

22 2 2AB (b a) 3(a b) 4 2 2 t (3t 4) 8 (vi t a b ).

2 25t 12t 4 0 t 2; t5

+ Vi t 2 a b 2 b 0,a 2 : x y 1 0

+ Vi 2 2 4 2t a b b ,a5 5 5 5

: 7x y 9 0


Cu 4. Trong mt phng vi h trc to Oxy, cho hai ng thng 1d : x y 1 0 ,

2d : 2x y 1 0 . Lp phng trnh ng thng (d) i qua M(1;1) ct (d1) v (d2)

tng ng ti A v B sao cho 2MA MB 0

. Gi s: A(a; a1), B(b; 2b 1). T iu kin 2MA MB 0

tm c A(1; 2), B(1;1) suy ra (d): x 1 = 0

Cu 5. Trong mt phng vi h ta Oxy, cho im M(1; 0). Lp phng trnh ng

thng (d) i qua M v ct hai ng thng 1 2d : x y 1 0, d : x 2y 2 0 ln lt ti A, B sao cho MB = 3MA.

Ta c 12

A (d ) A(a; 1 a) MA (a 1; 1 a)B (d ) B(2b 2;b) MB (2b 3;b)

.

T A, B, M thng hng v MB 3MA MB 3MA

(1) hoc MB 3MA

(2)

(1) 2 1A ;

(d) : x 5y 1 03 3B( 4; 1)

hoc (2) A 0; 1

(d) : x y 1 0B(4;3)

Cu 6. Trong mt phng vi h ta Oxy, cho im M(1; 1). Lp phng trnh ng

thng (d) i qua M v ct hai ng thng 1 2d :3x y 5 0, d : x y 4 0 ln lt ti A, B sao cho 2MA 3MB 0 .

Gi s 1A(a;3a 5) d , 2B(b;4 b) d .

V A, B, M thng hng v 2MA 3MB nn 2MA 3MB (1)

2MA 3MB (2)

+ 52(a 1) 3(b 1) a 5 5(1) A ; ,B(2; 2)2

2(3a 6) 3(3 b) 2 2b 2

. Suy ra d : x y 0 .

+ 2(a 1) 3(b 1) a 1

(2) A(1; 2),B(1;3)2(3a 6) 3(3 b) b 1

. Suy ra d : x 1 0 .

Vy c d : x y 0 hoc d : x 1 0 . Cu 7. Trong mt phng vi h to Oxy, cho im M(3; 1). Vit phng trnh ng

thng d i qua M ct cc tia Ox, Oy ti A v B sao cho (OA 3OB) nh nht.

PT ng thng d ct tia Ox ti A(a;0), tia Oy ti B(0;b): x y 1a b (a,b>0)

M(3; 1) d C si3 1 3 11 2 . ab 12

a b a b

.

M OA 3OB a 3b 2 3ab 12 mina 3b a 6

(OA 3OB) 12 3 1 1 b 2a b 2

Phng trnh ng thng d l: x y 1 x 3y 6 06 2


Cu 8. Trong mt phng vi h to Oxy, vit phng trnh ng thng i qua im M(4;1) v ct cc tia Ox, Oy ln lt ti A v B sao cho gi tr ca tng OA OB nh nht. S: x 2y 6 0

Cu 9. Trong mt phng vi h to Oxy, vit phng trnh ng thng d i qua im

M(1; 2) v ct cc trc Ox, Oy ln lt ti A, B khc O sao cho 2 29 4

OA OB nh nht.

ng thng (d) i qua M(1;2) v ct cc trc Ox, Oy ln lt ti A, B khc O, nn

A(a;0);B(0;b) vi a.b 0 Phng trnh ca (d) c dng x y 1a b .

V (d) qua M nn 1 2 1a b . p dng bt ng thc Bunhiacpski ta c :

2 2

2 2

1 2 1 3 2 1 9 41 . 1. 1a b 3 a b 9 a b

2 29 4 9a b 10

2 29 4 9

OA OB 10 .

Du bng xy ra khi 1 3 2: 1:3 a b

v 1 2 1a b 20a 10, b

9

d : 2x 9y 20 0 . Cu 10. Trong mt phng vi h to Oxy, vit phng trnh ng thng i qua im

M(3;1) v ct cc trc Ox, Oy ln lt ti B v C sao cho tam gic ABC cn ti A vi A(2;2). S: x 3y 6 0;x y 2 0

Cu 11. Trong mt phng vi h ta (Oxy). Lp phng trnh ng thng d qua M(2;1)

v to vi cc trc ta mt tam gic c din tch bng S 4 .

Gi A(a;0), B(0;b) (a, b 0) l giao im ca d vi Ox, Oy, suy ra: x yd : 1a b .

Theo gi thit, ta c: 2 1 1a bab 8

2b a abab 8

.

Khi ab 8 th 2b a 8 . Nn: 1b 2;a 4 d : x 2y 4 0 .

Khi ab 8 th 2b a 8 . Ta c: 2b 4b 4 0 b 2 2 2 . + Vi b 2 2 2 d : 1 2 x 2 1 2 y 4 0 + Vi b 2 2 2 d : 1 2 x 2 1 2 y 4 0 . Cu hi tng t: a) M(8;6),S 12 . S: d :3x 2y 12 0 ; d :3x 8y 24 0 Cu 12. Trong mt phng vi h ta Oxy, cho im A(2; 1) v ng thng d c phng

trnh 2x y 3 0 . Lp phng trnh ng thng () qua A v to vi d mt gc c

cos 110

.

Ptt () c dng: a(x 2) b(y 1) 0 ax by 2a b 0 2 2(a b 0)


Ta c: 2 2

2a b 1cos105(a b )

7a2 8ab + b2 = 0. Chon a = 1 b = 1; b = 7.

(1): x + y 1 = 0 v (2): x + 7y + 5 = 0 Cu 13. Trong mt phng vi h ta Oxy, cho im A(2;1) v ng thng

d : 2x 3y 4 0 . Lp phng trnh ng thng i qua A v to vi ng thng d mt gc 045 .

Ptt () c dng: a(x 2) b(y 1) 0 ax by (2a b) 0 2 2(a b 0) .

Ta c: 02 2

2a 3bcos 4513. a b

2 25a 24ab 5b 0

a 5b5a b

+ Vi a 5b . Chn a 5,b 1 Phng trnh : 5x y 11 0 . + Vi 5a b . Chn a 1,b 5 Phng trnh : x 5y 3 0 . Cu 14. Trong mt phng vi h to Oxy , cho ng thng d : 2x y 2 0 v im

I(1;1) . Lp phng trnh ng thng cch im I mt khong bng 10 v to vi ng thng d mt gc bng 045 .

Gi s phng trnh ng thng c dng: ax by c 0 2 2(a b 0) .

V 0(d, ) 45 nn 2 2

2a b 12a b . 5

a 3bb 3a

Vi a 3b : 3x y c 0 . Mt khc d(I; ) 10 4 c

1010

c 6c 14

Vi b 3a : x 3y c 0 . Mt khc d(I; ) 10 2 c

1010

c 8c 12

Vy cc ng thng cn tm: 3x y 6 0; 3x y 14 0 ; x 3y 8 0; x 3y 12 0 .

Cu 15. Trong mt phng vi h ta Oxy , cho im M (0; 2) v hai ng thng 1d ,

2d c phng trnh ln lt l 3x y 2 0 v x 3y 4 0 . Gi A l giao im ca

1d v 2d . Vit phng trnh ng thng i qua M, ct 2 ng thng 1d v 2d ln lt

ti B , C ( B v C khc A ) sao cho 2 21 1

AB AC t gi tr nh nht.

Ta c 1 2A d d A( 1;1) . Ta c 1 2d d . Gi l ng thng cn tm. H l hnh

chiu vung gc ca A trn . ta c: 2 2 2 21 1 1 1

AB AC AH AM (khng i)

2 21 1

AB AC t gi tr nh nht bng 2

1AM

khi H M, hay l ng thng i qua

M v vung gc vi AM. Phng trnh : x y 2 0 . Cu hi tng t: a) Vi M(1; 2) , 1d : 3x y 5 0 , 2d : x 3y 5 0 . S: : x y 1 0 . Cu 16. Trong mt phng vi h trc ta Oxy, cho ng thng (d) : x 3y 4 0 v

ng trn 2 2(C) : x y 4y 0 . Tm M thuc (d) v N thuc (C) sao cho chng i xng qua im A(3; 1).


V M (d) M(3b+4; b) N(2 3b; 2 b)

N (C) (2 3b)2 + (2 b)2 4(2 b) = 0 6b 0; b5

Vy c hai cp im: M(4;0) v N(2;2) hoc 38 6 8 4M ; , N ;5 5 5 5

Cu 17. Trong mt phng ta Oxy, cho im A(1; 1) v ng thng : 2x 3y 4 0 .

Tm im B thuc ng thng sao cho ng thng AB v hp vi nhau gc 045 .

c PTTS: x 1 3ty 2 2t

v VTCP u ( 3;2) . Gi s B(1 3t; 2 2t) .

0(AB, ) 45 1cos(AB;u)2

AB.u 1

AB. u 2

2

15t13169t 156t 45 0

3t13

Vy cc im cn tm l: 1 232 4 22 32B ; , B ;13 13 13 13

.

Cu 18. Trong mt phng vi h ta Oxy, cho ng thng d : x 3y 6 0 v im

N(3;4) . Tm ta im M thuc ng thng d sao cho tam gic OMN (O l gc ta

) c din tch bng152

.

Ta c ON (3;4)

, ON = 5, PT ng thng ON: 4x 3y 0 . Gi s M(3m 6;m) d .

Khi ta c ONMONM2S1S d(M,ON).ON d(M,ON) 3

2 ON

4.(3m 6) 3m 133 9m 24 15 m 1; m

5 3

+ Vi m 1 M(3; 1)

+ Vi 13 13m M 7;3 3

Cu 19. Trong mt phng to Oxy, cho im A(0;2) v ng thng d : x 2y 2 0 .

Tm trn ng thng d hai im B, C sao cho tam gic ABC vung B v AB = 2BC .

Gi s B(2b 2;b),C(2c 2;c) d .

V ABC vung B nn AB d dAB.u 0 2 6B ;

5 5

2 5AB5

5BC5

21BC 125c 300c 1805

= 55

c 1 C(0;1)

7 4 7c C ;5 5 5


Cu 20. Trong mt phng to Oxy, cho hai ng thng 1d : x y 3 0 , 2d : x y 9 0 v im A(1;4) . Tm im 1 2B d ,C d sao cho tam gic ABC vung cn ti A.

Gi 1 2B(b;3 b) d , C(c;9 c) d AB (b 1; 1 b)

, AC (c 1;5 c)

.

ABC vung cn ti A AB.AC 0AB AC

2 2 2 2

(b 1)(c 1) (b 1)(5 c) 0(b 1) (b 1) (c 1) (5 c)

(*)

V c 1 khng l nghim ca (*) nn

(*) 22 2 2 2

2

(b 1)(5 c)b 1 (1)c 1

(5 c)(b 1) (b 1) (c 1) (5 c) (2)(c 1)

T (2) 2 2(b 1) (c 1) b c 2b c

.

+ Vi b c 2 , thay vo (1) ta c c 4, b 2 B(2;1), C(4;5) . + Vi b c , thay vo (1) ta c c 2, b 2 B( 2;5), C(2;7) . Vy: B(2;1), C(4;5) hoc B( 2;5), C(2;7) . Cu 21. Trong mt phng to Oxy, cho cc im A(0; 1) B(2; 1) v cc ng thng c

phng trnh: 1d : (m 1)x (m 2)y 2 m 0 ; 2d : (2 m)x (m 1)y 3m 5 0 . Chng minh d1 v d2 lun ct nhau. Gi P = d1 d2. Tm m sao cho PA PB ln nht.

Xt H PT: (m 1)x (m 2)y m 2(2 m)x (m 1)y 3m 5

.

Ta c 2m 1 m 2 3 1D 2 m 0, m

2 m m 1 2 2

1 2d ,d lun ct nhau. Ta c: 1 2 1 2A(0;1) d , B(2; 1) d , d d APB vung ti P P nm trn ng trn ng knh AB. Ta c: 2 2 2 2(PA PB) 2(PA PB ) 2AB 16

PA PB 4 . Du "=" xy ra PA = PB P l trung im ca cung AB P(2; 1) hoc P(0; 1) m 1 hoc m 2 .

Vy PA PB ln nht m 1 hoc m 2 . Cu 22. Trong mt phng to Oxy, cho ng thng (): x 2y 2 0 v hai im

A( 1;2) , B(3;4) . Tm im M() sao cho 2 22MA MB c gi tr nh nht.

Gi s M M(2t 2; t) AM (2t 3; t 2), BM (2t 1; t 4)

Ta c: 2 2 22AM BM 15t 4t 43 f (t) 2min f (t) f15

26 2M ;15 15

Cu 23. Trong mt phng to Oxy, cho ng thng d : 2x y 3 0 v 2 im A(1;0), B(2;1) . Tm im M trn d sao cho MA MB nh nht.

Ta c: A A B B(2x y 3).(2x y 3) 30 0 A, B nm cng pha i vi d. Gi A l im i xng ca A qua d A ( 3;2) Phng trnh A B : x 5y 7 0 . Vi mi im M d, ta c: MA MB MA MB A B . M MA MB nh nht A, M, B thng hng M l giao im ca AB vi d.

Khi : 8 17M ;11 11

.


II. NG TRN Cu 24. Trong mt phng vi h to Oxy, gi A, B l cc giao im ca ng thng (d):

2x y 5 0 v ng trn (C): 2 2x y 20x 50 0 . Hy vit phng trnh ng trn (C) i qua ba im A, B, C(1; 1). S: A(3; 1), B(5; 5) (C): 2 2x y 4x 8y 10 0

Cu 25. Trong mt phng vi h to Oxy, cho tam gic ABC c din tch bng 32

, A(2;

3), B(3; 2), trng tm ca ABC nm trn ng thng d :3x y 8 0 . Vit phng trnh ng trn i qua 3 im A, B, C.

Tm c C (1; 1)1 , 2C ( 2; 10) .

+ Vi 1C (1; 1) (C): 2 2 11 11 16x y x y 0

3 3 3

+ Vi 2C ( 2; 10) (C): 2 2 91 91 416x y x y 0

3 3 3

Cu 26. Trong mt phng vi h to Oxy, cho ba ng thng: 1d : 2x y 3 0 ,

2d : 3x 4y 5 0 , 3d : 4x 3y 2 0 . Vit phng trnh ng trn c tm thuc d1 v tip xc vi d2 v d3.

Gi tm ng trn l I(t;3 2t) d1.

Khi : 2 3) d(I,d )d(I,d 3t 4(3 2t) 5

54t 3(3 2t) 2

5

t 2t 4

Vy c 2 ng trn tho mn: 2 2 4925

(x 2) (y 1) v 2 2 9(x 4) (y 5)25

.

Cu hi tng t a) Vi 1d : x 6y 10 0 , 2d : 3x 4y 5 0 , 3d : 4x 3y 5 0 .

S: 2 2(x 10) y 49 hoc 2 2 210 70 7x y

43 43 43

.

Cu 27. Trong mt phng vi h to Oxy, cho hai ng thng : x 3y 8 0 ,

' :3x 4y 10 0 v im A(2; 1). Vit phng trnh ng trn c tm thuc ng thng , i qua im A v tip xc vi ng thng .

Gi s tm I( 3t 8; t) .. Ta c: d(I, ) IA

2 22 2

3( 3t 8) 4t 10( 3t 8 2) (t 1)

3 4

t 3 I(1; 3), R 5

PT ng trn cn tm: 2 2(x 1) (y 3) 25 . Cu 28. Trong mt phng vi h to Oxy, cho hai ng thng : 4x 3y 3 0 v

' : 3x 4y 31 0 . Lp phng trnh ng trn (C) tip xc vi ng thng ti im c tung bng 9 v tip xc vi '. Tm ta tip im ca (C) v ' .

Gi I(a;b) l tm ca ng trn (C). (C) tip xc vi ti im M(6;9) v (C) tip xc vi nn


54 3a4a 3b 3 3a 4b 31d(I, ) d(I, ') 4a 3 3 6a 85

45 5IM u (3; 4) 3(a 6) 4(b 9) 0 3a 4b 54

25a 150 4 6a 85 a 10; b 6

54 3a a 190; b 156b4

Vy: 2 2(C) : (x 10) (y 6) 25 tip xc vi ' ti N(13;2) hoc 2 2(C) : (x 190) (y 156) 60025 tip xc vi ' ti N( 43; 40) Cu 29. Trong mt phng vi h to Oxy, vit phng trnh ng trn i qua A(2; 1) v

tip xc vi cc trc to .

Phng trnh ng trn c dng: 2 2 2

2 2 2

(x a) (y a) a (a)(x a) (y a) a (b)

a) a 1; a 5 b) v nghim. Kt lun: 2 2(x 1) (y 1) 1 v 2 2(x 5) (y 5) 25 . Cu 30. Trong mt phng vi h ta Oxy, cho ng thng (d) : 2x y 4 0 . Lp

phng trnh ng trn tip xc vi cc trc ta v c tm trn ng thng (d).

Gi I(m;2m 4) (d) l tm ng trn cn tm. Ta c: 4m 2m 4 m 4, m3

.

4m3

th phng trnh ng trn l: 2 24 4 16x y

3 3 9

.

m 4 th phng trnh ng trn l: 2 2(x 4) (y 4) 16 . Cu 31. Trong mt phng vi h ta Oxy, cho im A(1;1) v B(3;3), ng thng ():

3x 4y 8 0 . Lp phng trnh ng trn qua A, B v tip xc vi ng thng (). Tm I ca ng trn nm trn ng trung trc d ca on AB d qua M(1; 2) c VTPT l AB (4;2)

d: 2x + y 4 = 0 Tm I(a;4 2a)

Ta c IA = d(I,D) 211a 8 5 5a 10a 10 2a2 37a + 93 = 0 a 3

31a2

Vi a = 3 I(3;2), R = 5 (C): (x 3)2 + (y + 2)2 = 25

Vi a = 312

31I ; 272

, R = 652

(C): 2

231 4225x (y 27)2 4

Cu 32. Trong h to Oxycho hai ng thng d : x 2y 3 0 v : x 3y 5 0 . Lp

phng trnh ng trn c bn knh bng 2 105

, c tm thuc d v tip xc vi .

Tm I d I( 2a 3;a) . (C) tip xc vi nn:

d(I, ) R a 2 2 10

510

a 6a 2


(C): 2 2 8(x 9) (y 6)5

hoc (C): 2 2 8(x 7) (y 2)5

.

Cu 33. Trong mt phng vi h to Oxy, cho ng trn (C): 2 2x y 4 3x 4 0 . Tia

Oy ct (C) ti A. Lp phng trnh ng trn (C), bn knh R = 2 v tip xc ngoi vi (C) ti A.

(C) c tm I( 2 3;0) , bn knh R= 4; A(0; 2). Gi I l tm ca (C).

PT ng thng IA : x 2 3ty 2t 2

, I ' IA I (2 3t;2t 2) .

1AI 2I A t I '( 3;3)2

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

Cu 34. Trong mt phng vi h to Oxy, cho ng trn (C): 2 2x y 4y 5 0 . Hy

vit phng trnh ng trn (C) i xng vi ng trn (C) qua im M 4 2;5 5

(C) c tm I(0;2), bn knh R = 3. Gi I l im i xng ca I qua M

I 8 6;5 5

(C): 2 28 6x y 9

5 5

Cu 35. Trong mt phng vi h ta Oxy, cho ng trn (C): 2 2x y 2x 4y 2 0 .

Vit phng trnh ng trn (C) tm M(5; 1) bit (C) ct (C) ti hai im A, B sao cho AB 3 .

(C) c tm I(1; 2), bn knh R 3 . PT ng thng IM: 3x 4y 11 0 . AB 3 . Gi H(x; y) l trung im ca AB. Ta c:

2 2

H IM3IH R AH2

2 2

3x 4y 11 09(x 1) (y 2)4

1 29x ; y5 10

11 11x ; y5 10

1 29H ;5 10

hoc 11 11H ;5 10

.

Vi 1 29H ;5 10

. Ta c 2 2 2R MH AH 43 PT (C): 2 2(x 5) (y 1) 43 .

Vi 11 11H ;5 10

. Ta c 2 2 2R MH AH 13 PT (C): 2 2(x 5) (y 1) 13 .

Cu 36. Trong mt phng vi h ta Oxy, cho ng trn (C): 2 2(x 1) (y 2) 4 v

im K(3;4) . Lp phng trnh ng trn (T) c tm K, ct ng trn (C) ti hai im A, B sao cho din tch tam gic IAB ln nht, vi I l tm ca ng trn (C).

(C) c tm I(1;2) , bn knh R 2 . IABS ln nht IAB vung ti I AB 2 2 .

M IK 2 2 nn c hai ng trn tho YCBT. + 1(T ) c bn knh 1R R 2

2 21(T ) : (x 3) (y 4) 4


+ 2(T ) c bn knh 2 2

2R (3 2) ( 2) 2 5 2 2

1(T ) : (x 3) (y 4) 20 . Cu 37. Trong mt phng vi h to Oxy, vit phng trnh ng trn ni tip tam gic

ABC vi cc nh: A(2;3), 1B ;0 , C(2;0)4

.

im D(d;0) 1 d 24

thuc on BC l chn ng phn gic trong ca gc A

khi v ch khi

22

22

91 3d 4DB AB 4 4d 1 6 3d d 1.DC AC 2 d 4 3

Phng trnh AD: x 2 y 3 x y 1 03 3

; AC: x 2 y 3 3x 4y 6 04 3

Gi s tm I ca ng trn ni tip c tung l b. Khi honh l 1 b v bn knh cng bng b. V khong cch t I ti AC cng phi bng b nn ta c:

2 2

3 1 b 4b 6b b 3 5b

3 4

4b 3 5b b31b 3 5b b2

R rng ch c gi tr 1b2

l hp l.

Vy, phng trnh ca ng trn ni tip ABC l: 2 21 1 1x y

2 2 4

Cu 38. Trong mt phng to Oxy, cho hai ng thng (d1): 4x 3y 12 0 v (d2):

4x 3y 12 0 . Tm to tm v bn knh ng trn ni tip tam gic c 3 cnh nm trn (d1), (d2) v trc Oy.

Gi 1 2 1 2A d d , B d Oy,C d Oy A(3;0), B(0; 4),C(0;4) ABC cn nh A v AO l phn gic trong ca gc A. Gi I, R l tm v bn knh ng trn ni

tip ABC 4 4I ;0 , R3 3

.

Cu 39. Trong mt phng vi h to Oxy, cho ng thng d: x y 1 0 v hai ng

trn c phng trnh: (C1): 2 2(x 3) (y 4) 8 , (C2): 2 2(x 5) (y 4) 32 . Vit phng trnh ng trn (C) c tm I thuc d v tip xc ngoi vi (C1) v (C2).

Gi I, I1, I2, R, R1, R2 ln lt l tm v bn knh ca (C), (C1), (C2). Gi s I(a;a 1) d .

(C) tip xc ngoi vi (C1), (C2) nn 1 1 2 2 1 1 2 2II R R , II R R II R II R

2 2 2 2(a 3) (a 3) 2 2 (a 5) (a 5) 4 2 a = 0 I(0; 1), R = 2 Phng trnh (C): 2 2x (y 1) 2 . Cu 40. Trong mt phng vi h to Oxy, cho tam gic ABC vi A(3; 7), B(9; 5), C(5;

9), M(2; 7). Vit phng trnh ng thng i qua M v tip xc vi ng trn ngoi tip ABC.


S: y + 7 = 0; 4x + 3y + 27 = 0. Cu 41. Trong mt phng ta Oxy, cho ng trn 2 2C : x y 2x 0 . Vit phng

trnh tip tuyn ca C , bit gc gia tip tuyn ny v trc tung bng 30 . 2 2(C) : (x 1) y 1 I( 1;0);R 1 . H s gc ca tip tuyn () cn tm l 3 . PT () c dng 1 : 3x y b 0 hoc 2 : 3x y b 0

+ 1 : 3x y b 0 tip xc (C) 1d(I, ) R b 3 1 b 2 3

2

.

Kt lun: 1( ) : 3x y 2 3 0

+ 2( ) : 3x y b 0 tip xc (C) 2d(I, ) R b 3 1 b 2 3

2

.

Kt lun: 2( ) : 3x y 2 3 0 . Cu 42. Trong mt phng vi h to Oxy, cho ng trn (C): 2 2x y 6x 2y 5 0 v

ng thng (d): 3x y 3 0 . Lp phng trnh tip tuyn vi ng trn (C), bit tip tuyn khng i qua gc to v hp vi ng thng (d) mt gc 045 .

(C) c tm I(3; 1), bn knh R = 5 . Gi s (): ax by c 0 (c 0) .

T: d(I, ) 5

2cos(d, )2

a 2,b 1,c 10a 1, b 2,c 10

: 2x y 10 0: x 2y 10 0

.

Cu 43. Trong h to Oxy , cho ng trn 2 2(C) : (x 1) (y 1) 10 v ng thng

d : 2x y 2 0 . Lp phng trnh cc tip tuyn ca ng trn (C) , bit tip tuyn to vi ng thng d mt gc 045 .

(C) c tm I(1;1) bn knh R 10 . Gi n (a;b) l VTPT ca tip tuyn 2 2(a b 0) ,

V 0( ,d) 45 nn 2 2

2a b 12a b . 5

a 3bb 3a

Vi a 3b : 3x y c 0 . Mt khc d(I; ) R 4 c

1010

c 6c 14

Vi b 3a : x 3y c 0 . Mt khc d(I; ) R 2 c

1010

c 8c 12

Vy c bn tip tuyn cn tm: 3x y 6 0; 3x y 14 0 ; x 3y 8 0; x 3y 12 0 .

Cu 44. Trong mt phng vi h to Oxy, vit phng trnh tip tuyn chung ca hai

ng trn (C1): 2 2x y 2x 2y 2 0 , (C2): 2 2x y 8x 2y 16 0 . (C1) c tm 1I (1; 1) , bn knh R1 = 2; (C2) c tm 2I (4; 1) , bn knh R2 = 1. Ta c: 1 2 1 2I I 3 R R (C1) v (C2) tip xc ngoi nhau ti A(3; 1) (C1) v (C2) c 3 tip tuyn, trong c 1 tip tuyn chung trong ti A l x = 3 // Oy. * Xt 2 tip tuyn chung ngoi: ( ) : y ax b ( ) :ax y b 0 ta c:


2 2

1 1

2 2

2 2

a b 1 2 22 a ad(I ; ) R a b 4 4hayd(I ; ) R 4a b 1 4 7 2 4 7 21 b b

4 4a b

Vy, c 3 tip tuyn chung:

1 2 32 4 7 2 2 4 7 2( ) : x 3, ( ) : y x , ( ) y x

4 4 4 4

Cu 45. Trong mt phng vi h ta Oxy, cho hai ng trn (C): 2 2(x 2) (y 3) 2

v (C): 2 2(x 1) (y 2) 8 . Vit phng trnh tip tuyn chung ca (C) v (C).

(C) c tm I(2; 3) v bn knh R 2 ; (C) c tm I(1; 2) v bn knh R ' 2 2 . Ta c: II ' 2 R R (C) v (C) tip xc trong Ta tip im M(3; 4). V (C) v (C) tip xc trong nn chng c duy nht mt tip tuyn chung l ng

thng qua im M(3; 4), c vc t php tuyn l II ( 1; 1)

PTTT: x y 7 0 Cu 46. Trong mt phng vi h ta Oxy, cho hai ng trn 2 21(C ) : x y 2y 3 0 v

2 22(C ) : x y 8x 8y 28 0 . Vit phng trnh tip tuyn chung ca 1(C ) v 2(C ) .

1(C ) c tm 1I (0;1) , bn knh 1R 2 ; 2(C ) c tm 2I (4;4) , bn knh 2R 2 . Ta c: 1 2 1 2I I 5 4 R R 1 2(C ),(C ) ngoi nhau. Xt hai trng hp: + Nu d // Oy th phng trnh ca d c dng: x c 0 . Khi : 1 2d(I , d) d(I , d) c 4 c c 2 d : x 2 0 . + Nu d khng song song vi Oy th phng trnh ca d c dng: d : y ax b .

Khi : 11 2

d(I ,d) 2d(I ,d) d(I ,d)

2

2 2

1 b 2a 11 b 4a 4 ba 1 a 1

3 7a ; b4 23 3a ; b4 2

7 37a ;b24 12

d :3x 4y 14 0 hoc d :3x 4y 6 0 hoc d : 7x 24y 74 0 . Vy: d : x 2 0 ; d :3x 4y 14 0 ; d :3x 4y 6 0 ; d : 7x 24y 74 0 . Cu 47. Trong mt phng vi h ta Oxy, cho hai ng trn 2 21(C ) : x y 4y 5 0 v

2 22(C ) : x y 6x 8y 16 0 . Vit phng trnh tip tuyn chung ca 1(C ) v 2(C ) .

1(C ) c tm 1I (0;1) , bn knh 1R 3 ; 2(C ) c tm 2I (3; 4) , bn knh 2R 3 . Gi s tip tuyn chung ca 1 2(C ), (C ) c phng trnh:

2 2ax by c 0 (a b 0) .

l tip tuyn chung ca 1 2(C ), (C ) 1 1

2 2

d(I , ) Rd(I , ) R

2 2

2 2

2b c 3 a b (1)

3a 4b c 3 a b (2)

T (1) v (2) suy ra a 2b hoc 3a 2bc2

.


+ TH1: Vi a 2b . Chn b 1 a 2,c 2 3 5 : 2x y 2 3 5 0

+ TH2: Vi 3a 2bc2

. Thay vo (1) ta c: 2 2

a 0a 2b 2 a b 4a b

3

.

: y 2 0 hoc : 4x 3y 9 0 . Cu 48. Trong mt phng Oxy, cho ng trn (C): 2 2x y 4 3x 4 0 . Tia Oy ct (C) ti

im A. Lp phng trnh ng trn (T) c bn knh R = 2 sao cho (T) tip xc ngoi vi (C) ti A.

(C) c tm I( 2 3;0) , bn knh R 4 . Tia Oy ct (C) ti A(0;2) . Gi J l tm ca (T).

Phng trnh IA: x 2 3ty 2t 2

. Gi s J(2 3t;2t 2) (IA) .

(T) tip xc ngoi vi (C) ti A nn 1AI 2JA t J( 3;3)2

.

Vy: 2 2(T) : (x 3) (y 3) 4 . Cu 49. Trong mt phng Oxy, cho ng trn (C): 2 2x y 1 v phng trnh:

2 2x y 2(m 1)x 4my 5 0 (1). Chng minh rng phng trnh (1) l phng trnh ca ng trn vi mi m. Gi cc ng trn tng ng l (Cm). Tm m (Cm) tip xc vi (C).

(Cm) c tm I(m 1; 2m) , bn knh 2 2R ' (m 1) 4m 5 ,

(C) c tm O(0; 0) bn knh R = 1, OI 2 2(m 1) 4m , ta c OI < R

Vy (C) v (Cm) ch tip xc trong. R R = OI ( v R > R) 3m 1; m5

.

Cu 50. Trong mt phng Oxy, cho cc ng trn c phng trnh 2 211(C ) : (x 1) y2

v

2 22(C ) : (x 2) (y 2) 4 . Vit phng trnh ng thng d tip xc vi 1(C ) v ct

2(C ) ti hai im M, N sao cho MN 2 2 .

1(C ) c tm 1I (1;0) , bn knh 11R2

; 2(C ) c tm 1I (2;2) , bn knh 2R 2 . Gi H

l trung im ca MN 2

22 2 2

MNd(I ,d) I H R 22

Phng trnh ng thng d c dng: 2 2ax by c 0 (a b 0) .

Ta c: 1

2

1d(I ,d)2

d(I ,d) 2

2 2

2 2

2 a c a b

2a 2b c 2 a b

. Gii h tm c a, b, c.

Vy: d : x y 2 0; d : x 7y 6 0 ; d : x y 2 0 ; d : 7x y 2 0


Cu 51. Trong mt phng vi h to Oxy, cho ng trn (C): 2 2x y 6x 5 0 . Tm im M thuc trc tung sao cho qua M k c hai tip tuyn ca (C) m gc gia hai tip tuyn bng 060 .

(C) c tm I(3;0) v bn knh R = 2. Gi M(0; m) Oy

Qua M k hai tip tuyn MA v MB

0

0

AMB 60 (1)

AMB 120 (2)

V MI l phn gic ca AMB nn: (1) AMI = 300 0

IAMIsin 30

MI = 2R 2m 9 4 m 7

(2) AMI = 600 0IAMI

sin 60 MI = 2 3

3R 2 4 3m 9

3 v nghim

Vy c hai im M1(0; 7 ) v M2(0; 7 ) Cu 52. Trong mt phng vi h ta Oxy, cho ng trn (C) v ng thng nh bi:

2 2(C) : x y 4x 2y 0; : x 2y 12 0 . Tm im M trn sao cho t M v c vi (C) hai tip tuyn lp vi nhau mt gc 600.

ng trn (C) c tm I(2;1) v bn knh R 5 . Gi A, B l hai tip im. Nu hai tip tuyn ny lp vi nhau mt gc 600 th IAM l

na tam gic u suy ra IM 2R=2 5 . Nh th im M nm trn ng trn (T) c phng trnh: 2 2(x 2) (y 1) 20 . Mt khc, im M nm trn ng thng , nn ta ca M nghim ng h phng

trnh: 2 2(x 2) (y 1) 20 (1)

x 2y 12 0 (2)

Kh x gia (1) v (2) ta c:

2 2 2y 3

2y 10 y 1 20 5y 42y 81 0 27y5

Vy c hai im tha mn bi l: M 6;3 hoc 6 27M ;5 5


ng thng d : x y m 0 . Tm m trn ng thng d c duy nht mt im A m t k c hai tip tuyn AB, AC ti ng trn (C) (B, C l hai tip im) sao cho tam gic ABC vung.

(C) c tm I(1; 2), R = 3. ABIC l hnh vung cnh bng 3 IA 3 2

m 5m 1

3 2 m 1 6m 72


ng thng d :3x 4y m 0 . Tm m trn d c duy nht mt im P m t c th k c hai tip tuyn PA, PB ti ng trn (C) (A, B l hai tip im) sao cho PAB l tam gic u.

(C) c tm I(1; 2) , bn knh R 3 . PAB u PI 2AI 2R 6 P nm trn


ng trn (T) c tm I, bn knh r 6 . Do trn d c duy nht mt im P tho YCBT

nn d l tip tuyn ca (T) m 1911 md(I,d) 6 6m 415

.

Cu 55. Trong mt phng vi h to Oxy, cho hai ng trn

2 2(C) : x y 18x 6y 65 0 v 2 2(C ) : x y 9 . T im M thuc ng trn (C) k hai tip tuyn vi ng trn (C), gi A, B l cc tip im. Tm ta im M, bit di on AB bng 4,8 . (C) c tm O 0;0 , bn knh R OA 3 . Gi H AB OM H l trung im ca

AB 12AH5

. Suy ra: 2 2 9OH OA AH5

v 2OAOM 5

OH .

Gi s M(x;y) . Ta c: 2 2

2 2

M (C) x y 18x 6y 65 0OM 5 x y 25

x 4 x 5y 3 y 0

Vy M(4;3) hoc M(5;0) . Cu 56. Trong mt phng vi h to Oxy, cho ng trn (C): 2 2(x 1) (y 2) 4 . M l

im di ng trn ng thng d : y x 1 . Chng minh rng t M k c hai tip tuyn 1MT , 2MT ti (C) (T1, T2 l tip im) v tm to im M, bit ng thng

1 2T T i qua im A(1; 1) . (C) c tm I(1; 2) , bn knh R 2 . Gi s 0 0M(x ; x 1) d .

2 2 20 0 0IM (x 1) (x 3) 2(x 1) 8 2 R M nm ngoi (C) qua M k c 2 tip tuyn ti (C).

Gi J l trung im IM 0 0x 1 x 1J ;2 2

. ng trn (T) ng knh IM c tm J

bn knh 1IMR2

c phng trnh

2 2 2 20 0 0 0x 1 x 1 (x 1) (x 3)(T) : x y2 2 4

T M k c 2 tip tuyn MT1, MT2 n (C) 01 2 1 2IT M IT M 90 T ,T (T) 1 2{T ,T } (C) (T) to 1 2T , T tho mn h:

2 22 20 0 0 0

0 0 02 2

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

(x 1) (y 2) 4

To cc im 1 2T , T tho mn (1), m qua 2 im phn bit xc nh duy nht 1 ng thng nn phng trnh 1 2T T l 0 0 0x(1 x ) y(3 x ) x 3 0 .

A(1; 1) nm trn 1 2T T nn 0 0 01 x (3 x ) x 3 0 0x 1 M(1;2) . Cu 57. Trong mt phng vi h ta Oxy, cho ng trn (C): 2 2(x 1) (y 1) 25 v

im M(7; 3). Lp phng trnh ng thng (d) i qua M ct (C) ti hai im A, B phn bit sao cho MA = 3MB.


M/(C)P 27 0 M nm ngoi (C). (C) c tm I(1;1) v R = 5. Mt khc: 2M/(C)P MA.MB 3MB MB 3 BH 3

2 2IH R BH 4 d[M,(d)]

Ta c: pt(d): a(x 7) + b(y 3) = 0 (a2 + b2 > 0).

2 2

a 06a 4bd[M,(d)] 4 4 12a ba b 5

.

Vy (d): y 3 = 0 hoc (d): 12x 5y 69 = 0. Cu 58. Trong mt phng vi h to Oxy, lp phng trnh ng thng d i qua im

A(1; 2) v ct ng trn (C) c phng trnh 2 2(x 2) (y 1) 25 theo mt dy cung c di bng l 8 .

d: a(x 1)+ b(y 2) = 0 ax + by a 2b = 0 ( a2 + b2 > 0) V d ct (C) theo dy cung c di l 8 nn khong cch t tm I(2; 1) ca (C) n

d bng 3.

2 22 2

2a b a 2bd I,d 3 a 3b 3 a ba b

2

a 08a 6ab 0 3a b

4

a = 0: chn b = 1 d: y 2 = 0

a = 3 b4

: chn a = 3, b = 4 d: 3x 4 y + 5 = 0.

Cu 59. Trong mt phng vi h to Oxy, cho ng trn (C) : 2 2x y 2x 8y 8 0 .

Vit phng trnh ng thng song song vi ng thng d : 3x y 2 0 v ct ng trn (C) theo mt dy cung c di l 6 .

(C) c tm I(1; 4), bn knh R = 5. PT ng thng c dng: 3x y c 0, c 2 . V ct (C) theo mt dy cung c di bng 6 nn:

2

c 4 10 13 4 cd I, 4

3 1 c 4 10 1

.

Vy phng trnh cn tm l: 3x y 4 10 1 0 hoc 3x y 4 10 1 0 . Cu hi tng t: a) 2 2(C) : (x 3) (y 1) 3 , d :3x 4y 2012 0 , l 2 5 . S: : 3x 4y 5 0 ; : 3x 4y 15 0 . Cu 60. Trong mt phng vi h trc ta Oxy, cho ng trn 2 2(C) :(x 4) (y 3) 25

v ng thng : 3x 4y 10 0 . Lp phng trnh ng thng d bit d ( ) v d ct (C) ti A, B sao cho AB = 6.

(C) c tm I( 4; 3) v c bn knh R = 5. Gi H l trung im AB, AH = 3. Do d nn PT ca d c dng: 4x 3y m 0 .

Ta c: 1d(I, ( )) = IH = 2 2 2 2AI AH 5 3 4

2 2

m 2716 9 m 4m 134 3

Vy PT cc ng thng cn tm l: 4x 3y 27 0 v 4x 3y 13 0 .


Cu 61. Trong mt phng vi h to Oxy, cho ng trn (C): 2 2x y 2x 2y 3 0 v im M(0; 2). Vit phng trnh ng thng d qua M v ct (C) ti hai im A, B sao cho AB c di ngn nht.

(C) c tm I(1; 1) v bn knh R = 5 . IM = 2 5 M nm trong ng trn (C). Gi s d l ng thng qua M v H l hnh chiu ca I trn d. Ta c: AB = 2AH = 2 2 2 22 IA IH 2 5 IH 2 5 IM 2 3 . Du "=" xy ra H M hay d IM. Vy d l ng thng qua M v c VTPT

MI (1; 1)

Phng trnh d: x y 2 0 . Cu hi tng t: a) Vi (C): 2 2x y 8x 4y 16 0 , M(1; 0). S:

d :5x 2y 5 0 Cu 62. Trong mt phng vi h to Oxy, cho ng trn (C) c tm O, bn knh R = 5 v

im M(2; 6). Vit phng trnh ng thng d qua M, ct (C) ti 2 im A, B sao cho OAB c din tch ln nht.

Tam gic OAB c din tch ln nht OAB vung cn ti O. Khi 5 2d(O,d)2

.

Gi s phng trnh ng thng d: 2 2A(x 2) B(y 6) 0 (A B 0)

5 2d(O,d)2

2 2

2A 6B 5 22A B

2 247B 48AB 17A 0

24 5 55B A47

24 5 55B A47

+ Vi 24 5 55B A47

: chn A = 47 B = 24 5 55

d: 47(x 2) 24 5 55 (y 6) 0

+ Vi 24 5 55B A47

: chn A = 47 B = 24 5 55

d: 47(x 2) 24 5 55 (y 6) 0 Cu hi tng t: a) 2 2(C) : x y 4x 6y 9 0 , M(1; 8) . S: 7x y 1 0; 17x 7y 39 0 . Cu 63. Trong mt phng vi h to Oxy, cho ng trn (C): 2 2x y 6x 2y 6 0 v

im A(3;3) . Lp phng trnh ng thng d qua A v ct (C) ti hai im sao cho khong cch gia hai im bng di cnh hnh vung ni tip ng trn (C).

(C) c tm I(3; 1), R = 4. Ta c: A(3 ;3) (C). PT ng thng d c dng: 2 2a(x 3) b(y 3) 0, a b 0 ax by 3a 3b 0 . Gi s d qua A ct (C) ti hai im A, B AB = 4 2 . Gi I l tm hnh vung.

Ta c: 1 1d(I,d) 2 2 ( AD AB)2 2

2 2

3a b 3a 3b2 2

a b


2 2 2 24b 2 2 a b a b a b . Chn b = 1 th a = 1 hoc a = 1. Vy phng trnh cc ng thng cn tm l: x y 6 0 hoc x y 0 . Cu 64. Trong mt phng vi h to Oxy, cho hai ng trn (C1): 2 2x y 13 v (C2):

2 2(x 6) y 25 . Gi A l mt giao im ca (C1) v (C2) vi yA > 0. Vit phng trnh ng thng d i qua A v ct (C1), (C2) theo hai dy cung c di bng nhau.

(C1) c tm O(0; 0), bn knh R1 = 13 . (C2) c tm I2(6; 0), bn knh R2 = 5. Giao im A(2; 3). Gi s d: 2 2a(x 2) b(y 3) 0 (a b 0) . Gi 1 2 2d d(O,d), d d(I ,d) .

T gi thit 2 2 2 21 1 2 2R d R d 2 22 1d d 12

2 2

2 2 2 2

(6a 2a 3b) ( 2a 3b) 12a b a b

2b 3ab 0 b 0b 3a

.

Vi b = 0: Chn a = 1 Phng trnh d: x 2 0 . Vi b = 3a: Chn a = 1, b = 3 Phng trnh d: x 3y 7 0 . Cu 65. Trong mt phng vi h ta Oxy, cho ng thng : mx 4y 0 , ng trn

(C): 2 2 2x y 2x 2my m 24 0 c tm I. Tm m ng thng ct ng trn (C) ti hai im phn bit A, B sao cho din tch tam gic IAB bng 12.

(C) c tm I(1;m) , bn knh R = 5. Gi H l trung im ca dy cung AB.

2 2

m 4m 5mIH d(I, )m 16 m 16

; 2

2 22 2

(5m) 20AH IA IH 25m 16 m 16

IABS 12 2

m 3d(I, ).AH 12 3m 25 m 48 0 16m

3

Cu 66. Trong mt phng ta Oxy, cho ng trn 2 2(C) : x y 1 , ng thng

(d) : x y m 0 . Tm m (C) ct (d) ti A v B sao cho din tch tam gic ABO ln nht.

(C) c tm O(0; 0) , bn knh R = 1. (d) ct (C) ti A, B d(O;d) 1

Khi : OAB1 1 1S OA.OB.sin AOB .sin AOB2 2 2

. Du "=" xy ra 0AOB 90 .

Vy AOBS ln nht 0AOB 90 . Khi 1d(I;d)2

m 1 .

Cu 67. Trong mt phng vi h to Oxy, cho ng thng (d) : 2x my 1 2 0 v

ng trn c phng trnh 2 2(C) : x y 2x 4y 4 0 . Gi I l tm ng trn (C) . Tm m sao cho (d) ct (C) ti hai im phn bit A v B. Vi gi tr no ca m th din tch tam gic IAB ln nht v tnh gi tr .

(C) c tm I (1; 2) v bn knh R = 3.

(d) ct (C) ti 2 im phn bit A, B d(I,d) R 22 2m 1 2 3 2 m


2 2 21 4m 4m 18 9m 5m 4m 17 0 m R

Ta c: 1 1 9S IA.IBsin AIB IA.IBIAB 2 2 2

Vy: SIAB ln nht l 92

khi 0AIB 90 AB = R 2 3 2 3 2d(I,d)2

3 2 21 2m 2 m2

22m 16m 32 0 m 4

Cu hi tng t: a) Vi d : x my 2m 3 0 , 2 2(C) : x y 4x 4y 6 0 .

S: 8m 0 m15

Cu 68. Trong mt phng vi h to Oxy, cho ng trn 2 2(C) : x y 4x 6y 9 0 v

im M(1; 8) . Vit phng trnh ng thng d i qua M, ct (C) ti hai im A, B phn bit sao cho tam gic ABI c din tch ln nht, vi I l tm ca ng trn (C).

(C) c tm I( 2;3) , bn knh R 2 . PT ng thng d qua M(1; 8) c dng: d : ax by a 8b 0 ( 2 2a b 0 ).

IAB1S IA.IB.sin AIB 2sin AIB2

.

Do : IABS ln nht 0AIB 90 2d(I,d) IA 2

2

2 2

11b 3a 2a b

2 27a 66ab 118b 0 a 7b7a 17b

.

+ Vi b 1 a 7 d : 7x y 1 0 + Vi b 7 a 17 d :17x 7y 39 0

Cu 69. Trong mt phng vi h ta Oxy, cho ng trn (C): 2 2x y 4x 4y 6 0 v

ng thng : x my 2m 3 0 vi m l tham s thc. Gi I l tm ca ng trn (C). Tm m ct (C) ti 2 im phn bit A v B sao cho din tch IAB ln nht.

(C) c tm l I (2; 2); R = 2 . Gi s ct (C) ti hai im phn bit A, B.

K ng cao IH ca IAB, ta c: SABC = IAB1S IA.IB.sin AIB2

= sin AIB

Do IABS ln nht sinAIB = 1 AIB vung ti I IH = IA 1

2 (tha IH < R)

2

1 4m1

m 1

15m2 8m = 0 m = 0 hay m = 8

15

Cu hi tng t: a) Vi 2 2(C) : x y 2x 4y 4 0 , : 2x my 1 2 0 . S: m 4 . b) Vi 2 2(C) : x y 2x 4y 5 0 , : x my 2 0 . S: m 2 Cu 70. Trong mt phng vi h ta Oxy, cho ng thng d: x 5y 2 0 v ng

trn (C): 2 2x y 2x 4y 8 0 . Xc nh ta cc giao im A, B ca ng trn (C) v ng thng d (cho bit im A c honh dng). Tm ta C thuc ng trn (C) sao cho tam gic ABC vung B.


Ta giao im A, B l nghim ca h phng trnh

2 2 y 0; x 2x y 2x 4y 8 0

y 1; x 3x 5y 2 0

. V Ax 0 nn ta c A(2;0), B(3;1).

V 0ABC 90 nn AC l ng knh ng trn, tc im C i xng vi im A qua tm I ca ng trn. Tm I(1;2), suy ra C(4;4).

Cu 71. Trong mt phng vi h ta Oxy , cho ng trn ( C ): 2 2x y 2x 4y 8 0

v ng thng ( ): 2x 3y 1 0 . Chng minh rng ( ) lun ct ( C ) ti hai im phn bit A, B . Tm to im M trn ng trn (C ) sao cho din tch tam gic ABM ln nht.

(C) c tm I(1; 2), bn knh R = 13 . 9d(I, ) R13

ng thng ( ) ct (C)

ti hai im A, B phn bit. Gi M l im nm trn (C), ta c ABM1S AB.d(M, )2

.

Trong AB khng i nn ABMS ln nht d(M, ) ln nht. Gi d l ng thng i qua tm I v vung gc vi ( ). PT ng thng d l

3x 2y 1 0 . Gi P, Q l giao im ca ng thng d vi ng trn (C). To P, Q l nghim ca

h phng trnh: 2 2x y 2x 4y 8 0

3x 2y 1 0

x 1, y 1x 3, y 5

P(1; 1); Q(3; 5)

Ta c 4d(P, )13

; 22d(Q, )13

. Nh vy d(M, ) ln nht M trng vi Q.

Vy ta im M(3; 5). Cu 72. Trong mt phng vi h to Oxy, cho ng trn (C): 2 2x y 2x 4y 5 0 v

A(0; 1) (C). Tm to cc im B, C thuc ng trn (C) sao cho ABC u.

(C) c tm I(1;2) v R= 10 . Gi H l trung im BC. Suy ra AI 2.IH

3 7H ;2 2

ABC u I l trng tm. Phng trnh (BC): x 3y 12 0 V B, C (C) nn ta ca B, C l cc nghim ca h phng trnh:

2 2 2 2x y 2x 4y 5 0 x y 2x 4y 5 0

x 3y 12 0 x 12 3y

Gii h PT trn ta c: 7 3 3 3 3 7 3 3 3 3B ; ;C ;2 2 2 2

hoc ngc li.

Cu 73. Trong mt phng vi h to Oxy, cho ng trn (C): 2 2(x 3) (y 4) 35 v

im A(5; 5). Tm trn (C) hai im B, C sao cho tam gic ABC vung cn ti A.

(C) c tm I(3; 4). Ta c: AB ACIB IC

AI l ng trung trc ca BC. ABC vung

cn ti A nn AI cng l phn gic ca BAC . Do AB v AC hp vi AI mt gc 045 .

Gi d l ng thng qua A v hp vi AI mt gc 045 . Khi B, C l giao im ca d vi (C) v AB = AC. V IA (2;1)

(1; 1), (1; 1) nn d khng cng phng vi cc


trc to VTCP ca d c hai thnh phn u khc 0. Gi u (1;a) l VTCP ca d. Ta c:

2 2 2

2 a 2 a 2cos IA,u21 a 2 1 5 1 a

22 2 a 5 1 a a 3

1a3

+ Vi a = 3, th u (1;3) Phng trnh ng thng d: x 5 ty 5 3t

.

Ta tm c cc giao im ca d v (C) l: 9 13 7 3 13 9 13 7 3 13; , ;

2 2 2 2

+ Vi a = 13

, th 1u 1;3

Phng trnh ng thng d: x 5 t

1y 5 t3

.

Ta tm c cc giao im ca d v (C) l: 7 3 13 11 13 7 3 13 11 13; , ;

2 2 2 2

+V AB = AC nn ta c hai cp im cn tm l: 7 3 13 11 13 9 13 7 3 13; , ;

2 2 2 2

v 7 3 13 11 13 9 13 7 3 13; , ;2 2 2 2

Cu 74. Trong mt phng to Oxy, cho ng trn (C): 2 2x y 4 v cc im 8A 1;3

, B(3;0) . Tm to im M thuc (C) sao cho tam gic MAB c din tch

bng 203

.

64 10AB 4 ; AB : 4x 3y 12 09 3

. Gi M(x;y) v h d(M, AB) .

Ta c: 4x 3y 8 04x 3y 121 20h.AB h 4 44x 3y 32 02 3 5

+ 2 2

4x 3y 8 0 14 48M( 2;0); M ;25 75x y 4

+ 2 2

4x 3y 32 0x y 4

(v nghim)

Cu 75. Trong mt phng to Oxy, cho ng trn 2 2(C) : x y 2x 6y 9 0 v ng thng d :3x 4y 5 0 . Tm nhng im M (C) v N d sao cho MN c di nh nht.

(C) c tm I( 1;3) , bn knh R 1 d(I,d) 2 R d (C) . Gi l ng thng qua I v vung gc vi d ( ) : 4x 3y 5 0 .

Gi 0 01 7N d N ;5 5

.

Gi 1 2M , M l cc giao im ca v (C) 1 22 11 8 19M ; , M ;5 5 5 5

MN ngn nht khi 1 0M M , N N .

Vy cc im cn tm: 2 11M ; (C)5 5

, 1 7N ; d5 5

.


III. CC NG CNIC

Cu 76. Trong mt phng vi h to Oxy, cho elip (E): 2 2x y 1

25 16 . A, B l cc im trn

(E) sao cho: 1 2AF BF 8 , vi 1 2F,F l cc tiu im. Tnh 2 1AF BF . 1 2AF AF 2a v 1 2BF BF 2a 1 2 1 2AF AF BF BF 4a 20 M 1 2AF BF 8 2 1AF BF 12 Cu 77. Trong mt phng vi h to Oxy, vit phng trnh elip vi cc tiu im

1 2F ( 1;1),F (5;1) v tm sai e 0,6 .

Gi s M(x;y) l im thuc elip. V na trc ln ca elip l c 3a 5e 0,6

nn ta c:

2 2 2 21 2MF MF 10 (x 1) (y 1) (x 5) (y 1) 10

2 2(x 2) (y 1) 1

25 16

Cu 78. Trong mt phng vi h to Oxy, cho im C(2; 0) v elip (E): 2 2x y 1

4 1 . Tm

to cc im A, B thuc (E), bit rng hai im A, B i xng vi nhau qua trc honh v tam gic ABC l tam gic u.

S: 2 4 3 2 4 3A ; , B ;7 7 7 7


100 25 . Tm cc im M

(E) sao cho 01 2FMF 120 (F1, F2 l hai tiu im ca (E)). Ta c: a 10, b 5 c 5 3 . Gi M(x; y) (E)

1 23 3MF 10 x, MF 10 x

2 2 .

2 2 21 2 1 2 1 2 1 2FF MF MF 2MF.MF .cos FMF

2 2

2 3 3 3 3 110 3 10 x 10 x 2 10 x 10 x2 2 2 2 2

x = 0 (y= 5). Vy c 2 im tho YCBT: M1(0; 5), M2(0; 5). Cu 80. Trong mt phng Oxy, cho elip (E) c hai tiu im 1 2F ( 3;0); F ( 3;0) v i qua

im 1A 3;2

. Lp phng trnh chnh tc ca (E) v vi mi im M trn elip, hy

tnh biu thc: 2 2 21 2 1 2P FM F M 3OM FM.F M .

(E): 2 2

2 2 2 2

x y 3 11 1a b a 4b

, 2 2a b 3 2 2x y 1

4 1

2 2 2 2 2 2 2M M M M MP (a ex ) (a ex ) 2(x y ) (a e x ) 1


Cu 81. Trong mt phng to Oxy, cho elip (E): 2 24x 16y 64 . Gi F2 l tiu im bn

phi ca (E). M l im bt k trn (E). Chng t rng t s khong cch t M ti tiu

im F2 v ti ng thng 8: x3

c gi tr khng i.

Ta c: 2F ( 12;0) . Gi 0 0M(x ;y ) (E) 02 08 3xMF a ex

2

,

008 3x8d(M, ) x

3 3

(v 04 x 4 ) 2MF 3

d(M, ) 2

(khng i).

Cu 82. Trong mt phng vi h to Oxy, cho elip (E): 2 25x 16y 80 v hai im A(5;

1), B(1; 1). Mt im M di ng trn (E). Tm gi tr ln nht ca din tch MAB. Phng trnh ng thng (AB): x 2y 3 0 v AB 2 5 Gi 2 20 0 0 0M(x ; y ) (E) 5x 16y 80. Ta c:

0 0 0 0x 2y 3 x 2y 3d(M; AB)1 4 5

Din tch MAB: 0 01S .AB.d(M; AB) x 2y 32

p dng bt ng thc Bunhiacpxki cho 2 cp s 0 01 1; , ( 5x ; 4y )

25

c:

2

2 20 0 0 0

1 1 1 1 9. 5x .4y 5x 16y .80 362 5 4 205

0 0 0 0 0 0 0 0x 2y 6 6 x 2y 6 3 x 2y 3 9 x 2y 3 9

0 00 00 0

0 0

5x 4y5x 8y1 1

max x 2y 3 92 x 2y 65

x 2y 3 9

0

0

8x3

5y3

Vy, MAB8 5maxS 9 khi M ;3 3

.

Cu 83. Trong mt phng vi h to Oxy, cho elp 2 2x y(E) : 1

9 4 v hai im A(3;2),

B(3; 2) . Tm trn (E) im C c honh v tung dng sao cho tam gic ABC c din tch ln nht.

PT ng thng AB: 2x 3y 0 . Gi C(x; y) (E), vi x 0, y 0 2 2x y 1

9 4 .

ABC1 85 85 x yS AB.d(C, AB) 2x 3y 3.2 13 3 22 13

2 285 x y 1703 2 3

13 9 4 13

Du "=" xy ra

2 2x y 21 x 39 4 2x y

y 23 2

. Vy 3 2C ; 22

.


Cu 84. Trong mt phng ta Oxy , cho elip 2 2x y(E) : 1

25 9 v im M(1;1) . Vit

phng trnh ng thng i qua M v ct elip ti hai im A,B sao cho M l trung im ca AB .

Nhn xt rng M Ox nn ng thng x 1 khng ct elip ti hai im tha YCBT. Xt ng thng qua M(1; 1) c PT: y k(x 1) 1 . To cc giao im A,B ca

v (E) l nghim ca h:

2 2x y 1 (1)25 9y k(x 1) 1 (2)

2 2 2(25k 9)x 50k(k 1)x 25(k 2k 9) 0 (3)

PT (3) lun c 2 nghim phn bit 1 2x , x vi mi k . Theo Viet: 1 2 250k(k 1)x x25k 9

.

Do M l trung im ca AB 1 2 M 250k(k 1) 9x x 2x 2 k25k 9 25

.

Vy PT ng thng : 9x 25y 34 0 . Cu hi tng t:

a) Vi 2 2x y(E) : 1

9 4 , M(1;1) S: : 4x 9y 13 0


8 2 . Tm im M (E)

sao cho M c to nguyn. Trc ht ta c nhn xt: Nu im (x; y) (E) th cc im ( x;y), (x; y), ( x; y)

cng thuc (E). Do ta ch cn xt im 0 0M(x ; y ) (E) vi 0 0 0 0x , y 0; x , y Z .

Ta c: 2 20 0x y 1

8 2 20y 2 00 y 2

0 0

0 0

y 0 x 2 2 (loai)y 1 x 2

M(2;1) . Vy cc im tho YCBT l: (2;1), ( 2;1), (2; 1), ( 2; 1) .


8 2 . Tm im M (E)

sao cho tng hai to ca M c gi tr ln nht (nh nht).

Gi s M(x; y) (E) 2 2x y 1

8 2 . p dng BT Bunhiacpxki, ta c:

2 2

2 x y(x y) (8 2) 108 2

10 x y 10 .

+ x y 10 . Du "=" xy ra x y8 2x y 10

4 10 10M ;5 5

.

+ x y 10 . Du "=" xy ra x y8 2x y 10

4 10 10M ;5 5



9 3 v im A(3;0) . Tm

trn (E) cc im B, C sao cho B, C i xng qua trc Ox v ABC l tam gic u. Khng mt tnh tng qut, gi s 0 0 0 0B(x ; y ),C(x ; y ) vi 0y 0 .

Ta c: 2 2

2 20 00 0

x y 1 x 3y 99 3 . 0BC 2y v 0(BC) : x x 0d(A, (BC)) 3 x

Do A Ox , B v C i xng qua Ox nn ABC cn t A

Suy ra: ABC u 3d(A,(BC)) BC2

0 03 x 3y 2 20 03y (x 3)

02 20 00

x 0x (x 3) 9

x 3

.

+ Vi 0x 0 0y 3 B(0; 3), C(0; 3) . + Vi 0x 3 0y 0 (loi).

Vy: B(0; 3), C(0; 3) .


9 4 v cc ng thng

1d : mx ny 0 , 2d : nx+my 0 , vi 2 2m n 0 . Gi M, N l cc giao im ca 1d

vi (E), P, Q l cc giao im ca 2d vi (E). Tm iu kin i vi m, n din tch t gic MPNQ t gi tr nh nht.

PTTS ca 1 2d ,d l: 1

11

x ntd :

y mt

, 222

x mtd :

y nt

.

+ M, N l cc giao im ca 1d v (E)

2 2 2 2 2 2 2 2

6n 6m 6n 6mM ; , N ;9m 4n 9m 4n 9m 4n 9m 4n

+ P, Q l cc giao im ca 2d v (E)

2 2 2 2 2 2 2 2

6m 6n 6m 6nP ; , Q ;4m 9n 4m 9n 4m 9n 4m 9n

+ Ta c: MN PQ ti trung im O ca mi ng nn MPNQ l hnh thoi.

MPNQ1S S MN.PQ 2OM.OP2

= 2 2

2 2 2 2M M P P 2 2 2 2

72(m n )2 x y . x y(9m 4n )(4m 9n )

p dng BT C-si: 2 2 2 2

2 2 2 2 2 2(9m 4n ) (4m 9n ) 13(9m 4n )(4m 9n ) (m n )2 2

2 2

2 2

72(m n ) 144S 13 13(m n )2

. Du "=" xy ra 2 2 2 29m 4n 4m 9n m n

Vy: 144minS13

khi m n .


Cu 89. Trong mt phng vi h trc to Oxy, cho Hypebol (H) c phng trnh: 2 2x y 1

16 9 . Vit phng trnh chnh tc ca elip (E) c tiu im trng vi tiu im

ca (H) v ngoi tip hnh ch nht c s ca (H). (H) c cc tiu im 1 2F ( 5;0);F (5;0) . HCN c s ca (H) c mt nh l M( 4; 3),

Gi s phng trnh chnh tc ca (E) c dng: 2 2

2 2

x y 1a b

( vi a > b)

(E) cng c hai tiu im 2 2 21 2F ( 5;0); F (5;0) a b 5 (1) 2 2 2 2M(4;3) (E) 9a 16b a b (2)

T (1) v (2) ta c h:2 2 2 2

2 2 2 2 2

a 5 b a 409a 16b a b b 15

. Vy (E): 2 2x y 1

40 15

Cu 90. Trong mt phng vi h trc to Oxy , cho hypebol (H) c phng trnh

2 2x y 19 4 . Gi s (d) l mt tip tuyn thay i v F l mt trong hai tiu im ca

(H), k FM (d). Chng minh rng M lun nm trn mt ng trn c nh, vit phng trnh ng trn

(H) c mt tiu im F ( 13;0) .Gi s pttt (d): ax + by + c = 0.Khi :9a2 4b2 = c2 (*) Phng trnh ng thng qua F vung gc vi (d) l (D): b( x 13) a y = 0

To ca M l nghim ca h: ax by c

bx ay 13b

Bnh phng hai v ca tng phng trnh ri cng li v kt hp vi (*), ta c x2 + y2 = 9

Cu 91. Trong mt phng vi h to Oxy, cho parabol (P): 2y x v im I(0; 2). Tm to

hai im M, N (P) sao cho IM 4IN

. Gi 0 0 1 1M(x ; y ), N(x ; y ) l hai im thuc (P), khi ta c:

2 20 0 1 1x y ; x y

20 0 0 0IM (x ; y 2) (y ; y 2)

; 2 21 1 1 1 1 1IN (y ; y 2) (y ; y 2); 4IN (4y ; 4y 8)

Theo gi thit: IM 4IN

, suy ra: 2 20 1

0 1

y 4yy 2 4y 8

1 1 0 0

1 1 0 0

y 1 x 1; y 2; x 4y 3 x 9; y 6; x 36

Vy, c 2 cp im cn tm: M(4; 2), N(1;1) hay M(36;6), N(9;3) . Cu 92. Trong mt phng vi h to Oxy, cho parabol (P): 2y 8x . Gi s ng thng d

i qua tiu im ca (P) v ct (P) ti hai im phn bit A, B c honh tng ng l 1 2x , x . Chng minh: AB = 1 2x x 4 .

Theo cng thc tnh bk qua tiu: 1FA x 2 , 2FB x 2

1 2AB FA FB x x 4 .


Cu 93. Trong mt phng vi h to Oxy, cho Elip (E): 2 2x 5y 5 , Parabol 2(P) : x 10y . Hy vit phng trnh ng trn c tm thuc ng thng

( ) : x 3y 6 0 , ng thi tip xc vi trc honh Ox v ct tuyn chung ca Elip (E) vi Parabol (P).

ng thng i qua cc giao im ca (E) v (P): x = 2

Tm I nn: I(6 3b;b) . Ta c: 4 3b b b 1

6 3b 2 b4 3b b b 2

(C): 2 2(x 3) (y 1) 1 hoc (C): 2 2x (y 2) 4

IV. TAM GIC

Cu 94. Trong mt phng vi h to Oxy, cho ABC bit: B(2; 1), ng cao qua A c phng trnh d1: 3x 4y 27 0 , phn gic trong gc C c phng trnh d2: x 2y 5 0 . Tm to im A.

Phng trnh BC: x 2 y 13 4

To im C( 1;3)

+ Gi B l im i xng ca B qua d2, I l giao im ca BB v d2.

phng trnh BB: x 2 y 11 2

2x y 5 0

+ To im I l nghim ca h: 2x y 5 0 x 3

I(3;1)x 2y 5 0 y 1

+ V I l trung im BB nn: B' I BB' I B

x 2x x 4B (4;3)

y 2y y 3

+ ng AC qua C v B nn c phng trnh: y 3 =0.

+ To im A l nghim ca h: y 3 0 x 5

A( 5;3)3x 4y 27 0 y 3

Cu 95. Trong mt phng vi h to Oxy, cho tam gic ABC c ng cao AH, trung

tuyn CM v phn gic trong BD. Bit 17H( 4;1), M ;125

v BD c phng trnh

x y 5 0 . Tm ta nh A ca tam gic ABC. ng thng qua H v vung gc vi BD c PT: x y 5 0 . BD I I(0;5) Gi s AB H ' . BHH ' cn ti B I l trung im ca HH ' H '(4;9) .

Phng trnh AB: 5x y 29 0 . B = AB BD B(6; 1) 4A ;255

Cu 96. Trong mt phng vi h to Oxy, cho tam gic ABC c nh C(4; 3). Bit phng

trnh ng phn gic trong (AD): x 2y 5 0 , ng trung tuyn (AM): 4x 13y 10 0 . Tm to nh B.

Ta c A = AD AM A(9; 2). Gi C l im i xng ca C qua AD C AB.

Ta tm c: C(2; 1). Suy ra phng trnh (AB): x 9 y 22 9 1 2

x 7y 5 0 .

Vit phng trnh ng thng Cx // AB (Cx): x 7y 25 0


Cu 97. Trong mt phng vi h to Oxy, cho tam gic ABC c din tch bng 32

,

A(2;3), B(3;2). Tm to im C, bit im C nm trn ng thng (d): 3x y 4 0 .

PTTS ca d: x ty 4 3t

. Gi s C(t; 4 + 3t) d.

22 21 1S AB.AC.sin A AB .AC AB.AC

2 2

= 3

2 24t 4t 1 3

t 2t 1

C(2; 10) hoc C(1;1). Cu 98. Trong mt phng Oxy, cho tam gic ABC bit A(2; 3), B(3; 2), c din tch bng

32

v trng tm G thuc ng thng : 3x y 8 0 . Tm ta nh C.

Ta c: AB = 2 , trung im 5 5M ;2 2

. Phng trnh AB: x y 5 0 .

ABC1 3 3S AB.d(C, AB) d(C, AB)2 2 2

.

Gi G(t;3t 8) 1d(G, AB)2

t (3t 8) 5 1

2 2

t 1t 2

Vi t 1 G(1; 5) C(2; 10) Vi t 2 G(2; 2) C(1; 1) Cu hi tng t:

a) Vi A(2; 1) , B(1; 2) , ABC27S2

, G : x y 2 0 .

S: C(18; 12) hoc C( 9;15) Cu 99. Trong mt phng vi h to Oxy , cho ng thng d : x 2y 3 0 v hai im

A( 1;2) , B(2;1) . Tm to im C thuc ng thng d sao cho din tch tam gic ABC bng 2.

AB 10 , C( 2a 3;a) d. Phng trnh ng thng AB: x 3y 5 0 .

ABCS 2 1 AB.d(C, AB) 22

a 21 10. 2

2 10

a 6a 2

Vi a 6 ta c C( 9;6) Vi a 2 ta c C(7; 2) . Cu hi tng t: a) Vi d : x 2y 1 0 , A(1; 0), B(3; 1) , ABCS 6 . S: C(7;3) hoc C( 5; 3) . Cu 100. Trong mt phng vi h to Oxy, cho tam gic ABC c A(2; 3), B(3; 2), din

tch tam gic bng 1,5 v trng tm I nm trn ng thng d: 3x y 8 0 . Tm to im C.

V CH AB, IK AB. AB = 2 CH = ABC2S 3AB 2 IK = 1 1CH

3 2 .

Gi s I(a; 3a 8) d. Phng trnh AB: x y 5 0 .


d(I,AB) IK 3 2a 1 a 2a 1

I(2; 2) hoc I(1; 5).

+ Vi I(2; 2) C(1; 1) + Vi I(1; 5) C(2; 10). Cu 101. Trong mt phng vi h to Oxy, cho tam gic ABC c A(1;0), B(0;2) , din tch

tam gic bng 2 v trung im I ca AC nm trn ng thng d: y x . Tm to im C.

Phng trnh AB : 2x y 2 0 . Gi s I(t; t) d C(2t 1;2t) .

Theo gi thit: ABC1S AB.d(C,AB) 22

6t 4 4 4t 0; t3

.

+ Vi t 0 C( 1;0)

+ Vi 4t3

5 8C ;3 3

.

Cu 102. Trong mt phng vi h to Oxy, cho tam gic ABC c A(3; 5); B(4; 3), ng

phn gic trong v t C l d : x 2y 8 0 . Lp phng trnh ng trn ngoi tip tam gic ABC.

Gi E l im i xng ca A qua d E BC. Tm c E(1;1) PT ng thng BC: 4x 3y 1 0 . C d BC C( 2;5) . Phng trnh ng trn (ABC) c dng: 2 2 2 2x y 2ax 2by c 0; a b c 0

Ta c A, B, C (ABC) 4a 10b c 29

1 5 996a 10b c 34 a ; b ; c2 8 4

8a 6b c 25

Vy phng trnh ng trn l: 2 2 5 99x y x y 04 4

.

Cu 103. Trong mt phng vi h to Oxy, cho tam gic ABC c trung im cnh AB l

M( 1;2) , tm ng trn ngoi tip tam gic l I(2; 1) . ng cao ca tam gic k t A c phng trnh 2x y 1 0 . Tm to nh C.

PT ng thng AB qua M v nhn MI (3; 3)

lm VTPT: (AB) : x y 3 0 .

To im A l nghim ca h: x y 3 02x y 1 0

4 5A ;3 3

.

M( 1;2) l trung im ca AB nn 2 7B ;3 3

.

ng thng BC qua B v nhn n (2;1) lm VTCP nn c PT:

2x 2t3

7y t3

Gi s 2 7C 2t; t (BC)3 3

.


Ta c: 2 2 2 28 10 8 10IB IC 2t t

3 3 3 3

t 0 (loai v C B)

4t5

Vy: 14 47C ;15 15

.

Cu 104. Trong mt phng vi h to Oxy, cho tam gic ABC vi AB 5 , nh

C( 1; 1) , ng thng AB c phng trnh x 2y 3 0 , trng tm ca ABC thuc ng thng d : x y 2 0 . Xc nh to cc nh A, B ca tam gic ABC.

Gi I(a;b) l trung im ca AB, G l trng tm ABC 2CG CI3

G

G

2a 1x3

2b 1y3

Do G d nn 2a 1 2b 1 2 03 3

To im I l nghim ca h:

a 2b 3 02a 1 2b 1 2 0

3 3

a 5b 1

I(5; 1) . Ta c A, B (AB)

5IA IB2

To cc im A, B l cc nghim ca h: 2 2

x 2y 3 05(x 5) (y 1)4

1x 4; y23x 6; y2

1 3A 4; , B 6;2 2

hoc 3 1A 6; , B 4;2 2

.

Cu 105. Trong mt phng vi h to Oxy, cho im G(2;1) v hai ng thng

1d : x 2y 7 0 , 2d : 5x y 8 0 . Tm to im 1 2B d ,C d sao cho tam gic ABC nhn im G lm trng tm, bit A l giao im ca 1 2d , d .

To im A l nghim ca h: x 2y 7 05x y 8 0

x 1y 3

A(1;3) .

Gi s 1 2B(7 2b;b) d ; C(c;8 5c) d .

V G l trng tm ca ABC nn:

A B CG

A B CG

x x xx3

y y yy3

2b c 2b 5c 8

b 2c 2

.

Vy: B(3;2), C(2; 2) .

Cu 106. Trong mt phng vi h to Oxy, cho tam gic ABC c A(2;1) . ng cao BH c phng trnh x 3y 7 0 . ng trung tuyn CM c phng trnh x y 1 0 . Xc nh to cc nh B, C. Tnh din tch tam gic ABC.

AC qua A v vung gc vi ng cao BH (AC) : x 3y 7 0 .


To im C l nghim ca h: x 3y 7 0x y 1 0

C(4; 5) .

Trung im M ca AB c: B BM M2 x 1 yx ; y

2 2

. M (CM)

B B2 x 1 y 1 02 2

.

To im B l nghim ca h: B B

x 3y 7 02 x 1 y 1 0

2 2

B( 2; 3) .

To im H l nghim ca h: x 3y 7 03x y 7 0

14 7H ;5 5

.

8 10BH ; AC 2 105

ABC1 1 8 10S AC.BH .2 10. 162 2 5

(vdt).

Cu 107. Trong mt phng vi h to Oxy, cho tam gic ABC c A(4; 2) , phng trnh

ng cao k t C v ng trung trc ca BC ln lt l: x y 2 0 , 3x 4y 2 0 . Tm to cc nh B v C.

ng thng AB qua A v vung gc vi ng cao CH (AB) : x y 2 0 . Gi B(b;2 b) (AB) , C(c;c 2) (CH) Trung im M ca BC:

b c 4 b cM ;2 2

.

V M thuc trung trc ca BC nn: 3(b c) 4(4 b c) 4 0 b 7c 12 0 (1)

BC (c b;c b)

l 1 VTPT ca trung trc BC nn 4(c b) 3(c b) c 7b (2)

T (1) v (2) 7 1c , b4 4

. Vy 1 9 7 1B ; , C ;4 4 4 4

.

Cu 108. Trong mt phng Oxy, cho tam gic ABC cn ti A( 1;4) v cc nh B, C thuc

ng thng : x y 4 0 . Xc nh to cc im B, C, bit din tch tam gic ABC bng 18.

Gi H l trung im ca BC H l hnh chiu ca A trn 7 1H ;2 2

9AH2

Theo gi thit: ABC1S 18 BC.AH 18 BC 4 22

HB HC 2 2 .

To cc im B, C l cc nghim ca h:

2 2

x y 4 0

7 1x y 82 2

11 3x ; y2 23 5x ; y2 2

Vy 11 3 3 5B ; ,C ;2 2 2 2

hoc 3 5 11 3B ; , C ;2 2 2 2

.


Cu 109. Trong mt phng vi h trc ta Oxy, cho hai ng thng d1: x y 5 0 , d2: x 2y 7 0 v tam gic ABC c A(2; 3), trng tm l im G(2; 0), im B thuc d1 v im C thuc d2 . Vit phng trnh ng trn ngoi tip tam gic ABC.

Do B d1 nn B(m; m 5), C d2 nn C(7 2n; n)

Do G l trng tm ABC nn 2 m 7 2n 3.23 m 5 n 3.0

m 1n 1

B(1; 4), C(5; 1)

PT ng trn ngoi tip ABC: 2 2 83 17 338x y x y 027 9 27

Cu 110. Trong mt phng vi h to Oxy, cho tam gic ABC c A(4;6) , phng trnh

cc ng thng cha ng cao v trung tuyn k t nh C ln lt l 1d : 2x y 13 0 v 2d : 6x 13y 29 0 . Vit phng trnh ng trn ngoi tip

tam gic ABC . ng cao CH : 2x y 13 0 , trung tuyn CM : 6x 13y 29 0 C( 7; 1) PT ng thng AB: x 2y 16 0 . M CM AB M(6;5) B(8;4) . Gi s phng trnh ng trn (C) ngoi tip 2 2ABC : x y mx ny p 0.

V A, B, C (C) nn 52 4m 6n p 080 8m 4n p 050 7m n p 0

m 4n 6p 72

.

Suy ra PT ng trn: 2 2x y 4x 6y 72 0 . Cu 111. Trong mt phng to Oxy, cho tam gic ABC, c im A(2; 3), trng tm G(2; 0).

Hai nh B v C ln lt nm trn hai ng thng 1d : x y 5 0 v

2d : x 2y 7 0 . Vit phng trnh ng trn c tm C v tip xc vi ng thng BG.

Gi s 1 2B( 5 b;b) d ; C(7 2c;c) d .

V G l trng tm ABC nn ta c h: B CB C

x x 2 6y y 3 0

B(1;4) , C(5; 1).

Phng trnh BG: 4x 3y 8 0 . Bn knh 9R d(C, BG)5

Phng trnh ng trn: 2 2 81(x 5) (y 1)25

Cu 112. Trong mt phng vi h to Oxy, cho tam gic ABC c A( 3;6) , trc tm

H(2;1) , trng tm 4 7G ;3 3

. Xc nh to cc nh B v C.

Gi I l trung im ca BC. Ta c 2 7 1AG AI I ;3 2 2

ng thng BC qua I vung gc vi AH c phng trnh: x y 3 0 V I l trung im ca BC nn gi s B BB(x ;y ) th B BC(7 x ;1 y ) v

B Bx y 3 0 . H l trc tm ca tam gic ABC nn CH AB ; B B B BCH ( 5 x ; y ),AB (x 3; y 6)


B B B BB B B B B

x y 3 x 1 x 6CH.AB 0

(x 5)(x 3) (y 6) 0 y 2 y 3

Vy B 1; 2 ,C 6;3 hoc B 6;3 ,C 1; 2 Cu 113. Trong mt phng vi h to Oxy, cho tam gic ABC vi A(1; 2), ng cao

CH : x y 1 0 , phn gic trong BN : 2x y 5 0 . Tm to cc nh B, C v tnh din tch tam gic ABC.

Do AB CH nn phng trnh AB: x y 1 0 . + B = AB BN To im B l nghim ca h:

2x y 5 0x y 1 0

x 4y 3

B( 4;3) .

+ Ly A i xng vi A qua BN th A ' BC . Phng trnh ng thng (d) qua A v vung gc vi BN l (d): x 2y 5 0 .

Gi I (d) BN . Gii h: 2x y 5 0x 2y 5 0

. Suy ra: I(1; 3) A '( 3; 4)

+ Phng trnh BC: 7x y 25 0 . Gii h: BC : 7x y 25 0CH : x y 1 0

13 9C ;4 4

.

+ 2 213 9 450BC 4 3

4 4 4

, 2 2

7.1 1( 2) 25d(A;BC) 3 2

7 1

.

Suy ra: ABC1 1 450 45S d(A;BC).BC .3 2. .2 2 4 4

Cu 114. Trong mt phng vi h to Oxy, cho ABC , vi nh A(1; 3) phng trnh ng

phn gic trong BD: x y 2 0 v phng trnh ng trung tuyn CE: x 8y 7 0 . Tm to cc nh B, C.

Gi E l trung im ca AB. Gi s B(b;2 b) BD b 1 1 bE ; CE2 2

b 3

B( 3;5) . Gi A l im i xng ca A qua BD A BC. Tm c A(5; 1)

Phng trnh BC: x 2y 7 0 ; x 8y 7 0

C CE BC : C(7;0)x 2y 7 0

.

Cu 115. Trong mt phng vi h ta Oxy , cho tam gic ABC c nh A(3; 4). Phng

trnh ng trung trc cnh BC, ng trung tuyn xut pht t C ln lt l 1d : x y 1 0 v 2d :3x y 9 0 . Tm ta cc nh B, C ca tam gic ABC.

Gi 2C(c;3c 9) d v M l trung im ca BC 1M(m;1 m) d . B(2m c;11 2m 3c) . Gi I l trung im ca AB, ta c

2m c 3 7 2m 3cI ;2 2

.

V I 2(d ) nn 2m c 3 7 2m 3c3. 9 0

2 2

m 2 M(2; 1)

Phng trnh BC: x y 3 0 . 2C BC d C(3;0) B(1; 2) .


Cu 116. Trong mt phng vi h ta Oxy, cho tam gic ABC cn ti A c nh A(6; 6), ng thng d i qua trung im ca cc cnh AB v AC c phng trnh x + y 4 = 0. Tm ta cc nh B v C, bit im E(1; 3) nm trn ng cao i qua nh C ca tam gic cho.

Gi H l chn ng cao xut pht t A H i xng vi A qua d H( 2; 2) PT ng thng BC: x y 4 0 . Gi s B(m; 4 m) BC C( 4 m;m) CE (5 m; 3 m), AB (m 6; 10 m)

.

V CE AB nn AB.CE 0 (m 6)(m 5) (m 3)(m 10) 0

m 0; m 6 . Vy: B(0; 4), C( 4;0) hoc B( 6;2), C(2; 6) . Cu 117. Trong mt phng vi h ta Oxy, cho tam gic ABC c nh A(2;4) . ng

thng qua trung im ca cnh AB v AC c phng trnh 4x 6y 9 0 ; trung im ca cnh BC nm trn ng thng d c phng trnh: 2x 2y 1 0 . Tm ta

cc nh B v C, bit rng tam gic ABC c din tch bng 72

v nh C c honh

ln hn 1.

Gi A l im i xng ca A qua , ta tnh c 40 31A ' ;13 13

BC : 2x 3y 1 0

Ta gi M l trung im ca BC, th M l giao ca ng thng d v BC nn 5M ;22

.

Gi s 3t 1C ; t (BC)2

. Ta c

ABC1 7 1 7S d(A;BC).BC .BC BC 132 2 2 13

13CM2

2

2 t 3 C(4;3)3t 6 13(t 2)t 1 C(1;1) (loai)2 2

B(1;1) .

Vy: B(1;1) , C(4;3) . Cu 118. Trong mt phng vi h ta Oxy, cho ABC c ta nh B(3; 5) , phng trnh

ng cao h t nh A v ng trung tuyn h t nh C ln lt l 1d : 2x 5y + 3 = 0 v 2d : x + y 5 = 0. Tm ta cc nh A v C ca tam gic ABC.

Gi M l trung im AB th M 2d nn M(a;5 a) . nh A 1d nn 5b 3A ;b

2

.

M l trung im AB: A B MA B M

x x 2xy y 2y

4a 5b 3 a 22a b 5 b 1

A(1; 1).

Phng trnh BC: 5x 2y 25 0 ; 2C d BC C(5; 0). Cu 119. Trong mt phng to vi h to Oxy, cho ABC vi AB 5, nh

C( 1; 1) , phng trnh cnh AB : x 2y 3 0 v trng tm G ca ABC thuc ng thng d : x y 2 0 . Xc nh ta cc nh A,B ca tam gic.

Gi I(x; y) l trung im AB , G GG(x ; y ) l trng tm ca ABC


G

G

2x 1x2 3CG CI

2y 13 y3

G d : x y 2 0 nn c: G Gx y 2 0 2x 1 2y 1 2 0

3 3

Ta im I tha mn h: x 2y 3 0

I(5; 1)2x 1 2y 1 2 03 3

Gi 2

2 2 2A A A A

AB 5A(x ; y ) IA (x 5) (y 1)2 4

.

Hn na A AB : x 2y 3 0 suy ra ta im A l nghim ca h:

A A A A

2 2A A A A

x 2y 3 0 x 4 x 65 1 3x 5 y 1 y y4 2 2

Vy: 1 3A 4, , B 6;2 2

hoc 1 3B 4, , A 6;2 2

.

Cu 120. Trong mt phng vi h to Oxy , tm to cc nh ca mt tam gic vung

cn, bit nh C(3; 1) v phng trnh ca cnh huyn l d :3x y 2 0 . To im C khng tho mn phng trnh cnh huyn nn ABC vung cn ti C.

Gi I l trung im ca AB . Phng trnh ng thng CI: x 3y 0 .

I CI AB 3 1I ;5 5

72AI BI CI5

Ta c: A, B d

72AI BI5

2 23x y 2 0

3 1 72x y5 5 5

3 19x ; y5 5

9 17x ; y5 5

Vy to 2 nh cn tm l: 3 19 9 17; , ;5 5 5 5

.

Cu 121. Trong mt phng vi h to Oxy, cho im C(2; 5) v ng thng c phng

trnh: 3x 4y 4 0 . Tm trn hai im A v B i xng nhau qua 5I 2;2

sao cho

din tch tam gic ABC bng 15.

Gi 3a 4 16 3aA a; B 4 a;4 4

ABC

1S AB.d(C, ) 3AB2

AB = 5.

2

2 a 46 3aAB 5 (4 2a) 25a 02

. Vy hai im cn tm l A(0; 1) v

B(4; 4).


Cu 122. Trong mt phng vi h trc ta Oxy cho tam gic ABC vi B(1; 2) ng cao AH :x y 3 0 . Tm ta cc nh A, C ca tam gic ABC bit C thuc ng thng d :2x y 1 0 v din tch tam gic ABC bng 1.

Phng trnh BC : x y 1 0 . C = BC d C(2; 3) .

Gi 0 0 0 0A(x ; y ) AH x y 3 0 (1); 0 0x y 1BC 2, AH d(A,BC)

2

0 00 0ABC0 0

x y 1 2 (2)x y 11 1S AH.BC 1 . . 2 1x y 1 2 (3)2 2 2

T (1) v (2) 00

x 1A( 1;2)

y 2

. T (1) v (3) 00

x 3A( 3;0)

y 0

Cu 123. Trong mt phng vi h trc ta Oxy cho tam gic ABC vung ti A(2;1) , im

B nm trn trc honh, im C nm trn trc tung sao cho cc im B, C c to khng m. Tm to cc im B, C sao cho tam gic ABC c din tch ln nht.

Gi s B(b;0), C(0;c), (b,c 0) .

ABC vung ti A AB.AC 0

c 2b 5 0 50 b2

.

ABC1S AB.AC2

= 2 2 2 2 21 (b 2) 1. 2 (c 1) (b 2) 1 b 4b 52

Do 50 b2

nn ABCS t GTLN b 0 B(0;0),C(0;5) .

Cu 124. Trong mt phng vi h to Oxy, cho tam gic ABC c nh A( 1; 3) , trng tm

G(4; 2) , trung trc ca AB l d :3x 2y 4 0 . Vit phng trnh ng trn ngoi tip tam gic ABC.

Gi M l trung im ca BC 3AM AG2

13 3M ;2 2

.

AB d AB nhn du (2; 3) lm VTPT Phng trnh AB : 2x 3y 7 0 .

Gi N l trung im ca AB N = AB d N(2; 1) B(5;1) C(8; 4) . PT ng trn (C) ngoi tip ABC c dng: 2 2x y 2ax 2by c 0

( 2 2a b c 0 ).

Khi ta c h: 2a 6b c 1010a 2b c 2616a 8b c 80

74a21

23b7

8c3

.

Vy: 2 2 148 46 8(C) : x y x y 021 7 3

Cu 125. Trong mt phng vi h to Oxy, cho tam gic ABC c trng tm G(2, 0) v

phng trnh cc cnh AB, AC theo th t l: 4x y 14 0 ; 2x 5y 2 0 . Tm ta cc nh A, B, C. S: A(4, 2), B(3, 2), C(1, 0)


Cu 126. Trong mt phng vi h to Oxy, cho tam gic ABC c trc tm H( 1;6) , cc

im M(2;2) N(1;1) ln lt l trung im ca cc cnh AC, BC. Tm to cc nh A, B, C.

ng thng CH qua H v vung gc vi MN CH : x y 5 0 . Gi s C(a;5 a) CH CN (1 a;a 4)

V M l trung im ca AC nn A(4 a;a 1) AH (a 5;7 a)

V N l trung im ca BC nn B(2 a;a 3) V H l trc tm ABC nn: AH.CN 0

(a 5)(1 a) (7 a)(a 4) 0 a 3

11a2

.

+ Vi a 3 C(3;2), A(1;2),B( 1;0)

+ Vi 11a2

11 1 3 9 7 5C ; , A ; , B ;2 2 2 2 2 2

Cu 127. Trong mt phng vi h to Oxy, cho tam gic ABC c phn gic trong AD v

ng cao CH ln lt c phng trnh x y 2 0 , x 2y 5 0 . im M(3;0) thuc on AC tho mn AB 2AM . Xc nh to cc nh A, B, C ca tam gic ABC.

Gi E l im i xng ca M qua AD E(2; 1) . ng thng AB qua E v vung gc vi CH (AB) : 2x y 3 0 .

To im A l nghim ca h: 2x y 3 0x y 2 0

A(1;1) PT (AM) : x 2y 3 0

Do AB 2AM nn E l trung im ca AB B(3; 3) .

To im C l nghim ca h: x 2y 3 0x 2y 5 0

C( 1;2)

Vy: A(1;1) , B(3; 3) , C( 1;2) . Cu hi tng t: a) (AD) : x y 0 , (CH) : 2x y 3 0 , M(0; 1) .

S: A(1;1) ; B( 3; 1) ; 1C ; 22

Cu 128. Trong mt phng vi h to Oxy, cho tam gic ABC cn ti A, ng thng BC

c phng trnh x 2y 2 0 . ng cao k t B c phng trnh x y 4 0 , im M( 1;0) thuc ng cao k t C. Xc nh to cc nh ca tam gic ABC.

To nh B l nghim ca h: x 2y 2 0x y 4 0

B( 2;2) .

Gi d l ng thng qua M v song song vi BC d : x 2y 1 0 . Gi N l giao im ca d vi ng cao k t B To ca N l nghim ca h:

x y 4 0x 2y 1 0

N( 3;1) .


Gi I l trung im ca MN 1I 2;2

. Gi E l trung im ca BC IE l ng

trung trc ca BC IE : 4x 2y 9 0 .

To im E l nghim ca h: x 2y 2 04x 2y 9 0

7 17E ;5 10

4 7C ;5 5

.

ng thng CA qua C v vung gc vi BN 3CA : x y 05

.

To nh A l nghim ca h: 4x 2y 9 0

3x y 05

13 19A ;10 10

.

Vy: 13 19A ;10 10

, B( 2;2) , 4 7C ;5 5

.

Cu 129. Trong mt phng vi h to Oxy, cho tam gic ABC c nh A thuc ng thng

d: x 4y 2 0 , cnh BC song song vi d, phng trnh ng cao BH: x y 3 0 v trung im ca cnh AC l M(1; 1). Tm to cc nh A, B, C.

Ta c AC vung gc vi BH v i qua M(1; 1) nn c phng trnh: y x .

To nh A l nghim ca h :

2xx 4y 2 0 2 23 A ;y x 2 3 3y

3

V M l trung im ca AC nn 8 8C ;3 3

V BC i qua C v song song vi d nn BC c phng trnh: xy 24

x y 3 0 x 4

BH BC B: B( 4;1)x y 1y 24

Cu 130. Trong mt phng vi h ta Oxy, cho tam gic ABC c ng cao BH :3x 4y 10 0 , ng phn gic trong gc A l AD c phng trnh l x y 1 0 , im M(0; 2) thuc ng thng AB ng thi cch C mt khong bng

2 . Tm ta cc nh ca tam gic ABC. Gi N i xng vi M qua AD . Ta c N AC v N (1;1) PT cnh

AC : 4x 3y 1 0 A AC AD A(4;5) . AB i qua M, A

PT cnh AB :3x 4y 8 0 1B 3;4

Gi C(a;b) AC 4a 3b 1 0 , ta c MC 2 C(1;1) hoc 31 33C ;25 25

.

Kim tra iu kin B, C khc pha vi AD, ta c c hai im trn u tha mn. Cu 131. Trong mt phng vi h to Oxy, cho tam gic ABC c im M(1; 1) l trung

im ca cnh BC, hai cnh AB, AC ln lt nm trn hai ng thng d1: x y 2 0 v d2: 2x 6y 3 0 . Tm to cc nh A, B, C.


To im A l nghim ca h: x y 2 02x 6y 3 0

15 7A ;4 4

.

Gi s: B(b;2 b) d1, 3 2cC c;

6

d2. M(1; 1) l trung im ca BC

b c 12

3 2c2 b6 1

2

1b4

9c4

1 7B ;4 4

, 9 1C ;4 4

.

Cu 132. Trong mt phng vi h ta Oxy, cho ABC cn c y l BC. nh A c ta

l cc s dng, hai im B v C nm trn trc Ox, phng trnh cnh AB : y 3 7(x 1) . Bit chu vi ca ABC bng 18, tm ta cc nh A, B, C.

B AB Ox B(1;0) , A AB A a;3 7(a 1) a 1 (do A Ax 0, y 0 ). Gi AH l ng cao

ABC H(a;0) C(2a 1;0) BC 2(a 1),AB AC 8(a 1) .

Chu vi ABC 18 a 2 C(3;0), A 2;3 7 . Cu 133. Trong mt phng vi h ta Oxy, cho tam gic ABC bit phng trnh cc ng

thng cha cc cnh AB, BC ln lt l 4x 3y 4 0 ; x y 1 0 . Phn gic trong ca gc A nm trn ng thng x 2y 6 0 . Tm ta cc nh ca tam gic ABC.

Ta ca A nghim ng h phng trnh: 4x 3y 4 0 x 2

A( 2;4)x 2y 6 0 y 4

Ta ca B nghim ng h phng trnh 4x 3y 4 0 x 1

B 1;0x y 1 0 y 0

Phng trnh AC qua im A(2;4) c dng: a(x 2) b(y 4) 0 ax by 2a 4b 0

Gi 1 2 3: 4x 3y 4 0; : x 2y 6 0; : ax by 2a 4b 0

T gi thit suy ra 2 3 1 2; ; .

Do 2 3 1 2 2 21.a 2.b 4.1 2.3cos ( ; ) cos ( ; )

25. 55. a b

2 2a 0

a 2b 2 a b a(3a 4b) 03a 4b 0

a = 0 b 0 . Do 3 : y 4 0 3a 4b = 0: Chn a = 4 th b = 3. Suy ra 3 : 4x 3y 4 0 (trng vi 1 ). Do vy, phng trnh ca ng thng AC l y 4 = 0.

Ta ca C nghim ng h phng trnh:y 4 0 x 5

C(5;4)x y 1 0 y 4

Cu 134. Trong mt phng vi h to Oxy, cho tam gic ABC bit A(5; 2). Phng trnh

ng trung trc cnh BC, ng trung tuyn CC ln lt l x + y 6 = 0 v 2x y + 3 = 0. Tm ta cc nh ca tam gic ABC.

Gi C(c; 2c 3) v I(m;6 m) l trung im ca BC. Suy ra: B(2m c; 9 2m 2c) .


V C l trung im ca AB nn: 2m c 5 11 2m 2cC ' ; CC'2 2

nn 2m c 5 11 2m 2c 52 3 0 m2 2 6

5 41I ;6 6

.

Phng trnh BC: 3x 3y 23 0 14 37C ;3 3

19 4B ;3 3

.

Cu 135. Trong mt phng vi h to Oxy, bit to trc tm, tm ng trn ngoi tip

tam gic ABC ln lt l H(2;2), I(1;2) v trung im 5 5M ;2 2

ca cnh BC. Hy tm

to cc nh A, B, C bit B Cx x ( Bx , Cx ln lt honh im B v C).

Gi G l trng tm ABC ta c : GH 2GI

4G ; 23

Mt khc v GA 2GM

nn A( 1;1) . Phng trnh BC: 3x y 10 0 . ng trn (C) ngoi tip c tm I(1; 2) v bn knh R 4 1 5 . Do (C) :

2 2(x 1) (y 2) 5 .

Khi to B ;C l nghim h : 2 2 x 2 x 3(x 1) (y 2) 5

y 4 y 13x y 10 0

V B Cx x nn B(3;1) ; C(2;4). Vy : A(1; 1); B(3; 1) ; C(2; 4). Cu 136. Trong mt phng vi h ta Oxy, cho tam gic ABC cn ti C c din tch bng

10, phng trnh cnh AB l x 2y 0 , im I(4; 2) l trung im ca AB, im 9M 4;2

thuc cnh BC. Tm ta cc nh A, B, C bit tung im B ln hn

hoc bng 3. Gi s B BB(2y ;y ) AB B BA(8 2y ;4 y ) . Phng trnh CI: 2x y 10 0 .

Gi C CC(x ;10 2x ) CCI 5 4 x

; BAB 20 y 2

.

ABC B C C B1S CI.AB 10 4y 2x x y 8 22

C B B C

C B B C

x y 4y 2x 6 (1)x y 4y 2x 10 (2)

V C B

C B

4 x k 2y 4M BC CM kMB 11 92x k y

2 2

C B B C2x y 6y 5x 16 0

(3)

T (1) v (3): C B B C BC B B C B

x y 4y 2x 6 y 1 22x y 6y 5x 16 0 y 1 2

(loi, v By 3 )

T (2) v (3): C B B C BCC B B C

x y 4y 2x 10 y 3x 22x y 6y 5x 16 0

(tha)

Vy A(2; 1), B(6; 3), C(2; 6).


Cu 137. Trong mt phng vi h to Oxy, cho tam gic ABC vung ti A, cc nh A, B thuc ng thng d: y = 2, phng trnh cnh BC: 3x y 2 0 . Tm to cc nh A, B, C bit bn knh ng trn ni tip tam gic ABC bng 3 .

B d BC B(0; 2). Gi s A(a;2) d, (a 2) , C(c;2 c 3) BC, (c 0) . AB ( a;0), AC (c a;c 3), BC (c;c 3)

2 2AB a ,AC (c a) 3c ,BC 2 c

ABC vung A v r 3 AB.AC 0S pr

AB.AC 01 AB BC ACAB.AC . 32 2

2 2 2 2a(c a) 0

a (c a) 3c a 2 c (c a) 3c 3

c a 0

a 3 3

c a 3 3 A(3 3;2),C(3 3;5 3 3)

c a 3 3 A( 3 3;2),C( 3 3; 1 3 3)

Cu 138. Trong mt phng vi h to Oxy, tm to cc nh ca tam gic vung cn ABC,

c phng trnh hai cnh AB : x 2y 1 0 , AC : 2x y 3 0 v cnh BC cha im 8I ;13

.

Ta c: AB AC ABC vung cn ti A A(1;1) .

Gi M(x ; y) thuc tia phn gic At ca gc BAC . Khi M cch u hai ng thng AB, AC. Hn na M v I cng pha i vi ng thng AB v cng pha i vi

ng thng AC, tc l:

x 2y 1 2x y 35 5

8(x 2y 1) 2 1 0 x 3y 4 0316(2x y 3) 1 3 03

BCAt BC n (3; 1) BC :3x y 7 0 ;

x 2y 1 0B AB BC : B(3;2)

3x y 7 0

;

2x y 3 0

C AC BC : C(2; 1)3x y 7 0

.

Cu 139. Trong mt phng vi h to Oxy, cho tam gic ABC vung cn ti A, bit cc

nh A, B, C ln lt nm trn cc ng thng d: x y 5 0 , d1: x 1 0 , d2: y 2 0 . Tm to cc nh A, B, C, bit BC = 5 2 .

Ch : d1 d2 v ABC vung cn ti A nn A cch u d1, d2 A l giao im ca d v ng phn gic ca gc to bi d1, d2 A(3; 2).

Gi s B(1; b) d1, C(c; 2) d2. AB ( 4;b 2), AC (c 3; 4)

.


Ta c: 2

AB.AC 0BC 50

b 5, c 0b 1, c 6

A(3;2), B( 1;5), C(0; 2)A(3;2), B( 1; 1), C(6; 2)

.

Cu 140. Trong mt phng vi h ta Oxy , cho tam gic ABC cn ti nh C bit phng

trnh ng thng AB l: x y 2 0 , trng tm ca tam gic ABC l 14 5G ;3 3

v

din tch ca tam gic ABC bng 652

. Vit phng trnh ng trn ngoi tip tam

gic ABC.

Gi H l trung im ca AB CH AB CH: x y 3 0 5 1H ;2 2

C(9;6) .

Gi A(a;2 a) AB B(5 a;a 3) 13 13AB (5 2a;2a 5); CH ;2 2

2ABCa 065 1 65S AB.CH 8a 40a 0a 52 2 2

Vi a 0 A(0;2);B(5; 3) Vi a 5 A(5; 3), B(0;2)

PT ng trn (C) ngoi tip ABC c dng: 2 2 2 2x y 2ax 2by c 0 (a b c 0)

(C) qua A, B, C nn

137a264b c 45910a 6b c 34 b

2618a 12b c 117 66c

13

2 2 137 59 66(C) : x y x y 013 13 13

Cu 141. Trong mt phng to Oxy, cho ABC c phng trnh cnh AB: x y 3 0 ,

phng trnh cnh AC: 3x y 7 0 v trng tm 1G 2;3

. Vit phng trnh ng

trn i qua trc tm H v hai nh B, C ca tam gic ABC. A AB AC A(2;1) . Gi s B(m;3 m), C(n;7 3n) .

1G 2;3

l trng tm ABC nn: 2 m n 6 m 11 3 m 7 3n 1 n 3

B(1; 2), C(3; 2)

H l trc tm ABC AH BC

BH AC

H(10;5) .

PT ng trn (S) qua B, C, H c dng: 2 2 2 2x y 2ax 2by c 0 (a b c 0)

Do B, C, H (S) 2a 4b c 5 a 66a 4b c 13 b 220a 10b c 125 c 15

.

Vy (S): 2 2x y 12x 4y 15 0 .


Cu 142. Trong mt phng vi h to Oxy, cho im A(0; 2) v ng thng d: x 2y 2 0 . Tm trn d hai im B, C sao cho tam gic ABC vung ti B v AB = 2BC.

S: 2 6B ;5 5

; 1 24 7C (0;1); C ;5 5

Cu 143. Trong mt phng vi h to Oxy, cho tam gic ABC vung cn ngoi tip ng

trn 2 2(C) : x y 2 . Tm to 3 nh ca tam gic, bit im A thuc tia Ox. A l giao ca tia Ox vi (C) A(2;0) . Hai tip tuyn k t A n (C) l: x y 2 0 v x y 2 0 . V ABC vung cn nn cnh BC tip xc vi (C) ti trung im M ca BC M l giao ca tia i tia Ox vi (C) M 2;0 . Phng trnh cnh BC: x 2 . B v C l cc giao im ca BC vi 2 tip tuyn trn To 2 im B, C l: 2;2 2 , 2; 2 2 . Cu 144. Trong mt phng vi h to Oxy, cho tam gic ABC c trung im ca cnh BC

l im M(3; 1) , ng thng cha ng cao k t nh B i qua im E( 1; 3) v ng thng cha cnh AC i qua im F(1;3) . Tm ta cc nh ca tam gic ABC, bit rng im i xng ca nh A qua tm ng trn ngoi tip tam gic ABC l im D(4; 2) .

Gi H l trc tm ca tam gic ABC, ta chng minh c BDCH l hnh bnh hnh nn M l trung im ca HD suy ra H(2;0) . ng thng BH c VTCP l EH (3;3)

VTPT l BHn (1; 1) BH : x y 2 0

+ AC vung gc vi BH nn AC BHn u (1;1) AC : x y 4 0

+ AC vung gc vi CD nn DC ACn u (1; 1) DC : x y 6 0 .

+ C l giao ca AC v DC nn ta C l nghim ca h: x y 4 0

C(5; 1)x y 6 0

+ M l trung im ca BC nn B(1; 1) . AH vung gc vi BC AH : x 2 0 + A l giao im ca HA v AC nn ta A l nghim ca h

x 2 0A(2; 2)

x y 4 0

.

Vy: A(2;2) , B(1; 1) , C(5; 1) . Cu 145. Trong mt phng vi h to Oxy, cho tam gic ABC vung ti A, bit B v C i

xng nhau qua gc ta . ng phn gic trong ca gc ABC l d : x 2y 5 0 . Tm ta cc nh ca tam gic bit ng thng AC i qua im K(6;2)

Gi s B(5 2b;b),C(2b 5; b) d , O(0;0) BC

Gi I i xng vi O qua phn gic trong gcABC nn I(2;4) v I AB Tam gic ABC vung ti A nn BI (2b 3;4 b)

vung gc vi CK (11 2b;2 b)

2b 1

(2b 3)(11 2b) (4 b)(2 b) 0 5b 30b 25 0b 5

+ Vi b 1 B(3;1),C( 3; 1) A(3;1) B (loi)


+ Vi b 5 B( 5;5),C(5; 5) 31 17A ;5 5

Vy 31 17A ; ; B( 5;5);C(5; 5)5 5

Cu 146. Trong mt phng vi h to Oxy, cho tam gic ABC c nh 4 7A ;5 5

v phng

trnh hai ng phn gic trong BB: x 2y 1 0 v CC: x 3y 1 0 . Chng minh tam gic ABC vung.

Gi A1, A2 ln lt l im i xng ca A qua BB, CC A1, A2 BC. Tm c: A1(0; 1), A2(2; 1) Phng trnh BC: y 1 B(1; 1), C(4; 1)

AB AC

A vung. Cu 147. Trong mt phng vi h to Oxy, cho phng trnh hai cnh ca mt tam gic l

5x 2y 6 0 v 4x 7y 21 0 . Vit phng trnh cnh th ba ca tam gic , bit rng trc tm ca n trng vi gc to .

Gi s: (AB) :5x 2y 6 0 , (AC) : 4x 7y 21 0 A(0;3) . ng cao BO i qua B v vung gc vi AC (BO) : 7x 4y 0 B( 4; 7) . Cnh BC i qua B v vung gc vi OA (BC) : y 7 0 .

Cu 148. Trong mt phng vi h to Oxy, cho tam gic ABC c nh A(1; 3) v hai ng

trung tuyn ca n c phng trnh l: x 2y 1 0 v y 1 0 . Hy vit phng trnh cc cnh ca ABC. S: (AC): x + 2y 7 = 0; (AB): x y + 2 = 0; (BC): x 4y 1 = 0.

Cu 149. Trong mt phng vi h to Oxy, cho tam gic ABC c nh B( 12;1) , ng

phn gic trong gc A c phng trnh d : x 2y 5 0 . 1 2G ;3 3

l trng tm tam

gic ABC. Vit phng trnh ng thng BC. Gi M l im i xng ca B qua d M( 6;13) (AC) . Gi s A(5 2a;a) d C(8 2a;1 a) . Do MA, MC

cng phng

a 2 C(4;3) Vy: (BC) : x 8y 20 0 . Cu 150. Trong mt phng vi h to Oxy, cho tam gic ABC c nh B(2; 1) , ng cao

xut pht t A v ng phn gic trong gc C ln lt l 1d : 3x 4y 27 0 ,

2d : x 2y 5 0 . Vit phng trnh cc cnh ca tam gic ABC. ng thng BC qua B v vung gc vi 1d (BC) : 4x 3y 5 0 .

To nh C l nghim ca h: 4x 3y 5 0x 2y 5 0

C( 1;3) .

Gi B l im i xng ca B qua 2d B (4;3) v B (AC) . ng thng AC i qua C v B (AC) : y 3 0 .


To nh A l nghim ca h: y 3 03x 4y 27 0

A( 5;3) .

ng thng AB qua A v B (AB) : 4x 7y 1 0 . Vy: (AB) : 4x 7y 1 0 , (BC) : 4x 3y 5 0 , (AC) : y 3 0 . Cu 151. Trong mt phng vi h to Oxy, cho tam gic ABC cn, cnh y BC c phng

trnh d1: x y 1 0 . Phng trnh ng cao v t B l d2: x 2y 2 0 . im M(2; 1) thuc ng cao v t C. Vit phng trnh cc cnh bn ca tam gic ABC.

B(0; 1). BM (2; 2)

MB BC. K MN // BC ct d2 ti N th BCNM l hnh ch nht.

PT ng thng MN: x y 3 0 . N = MN d2 8 1N ;3 3

.

NC BC PT ng thng NC: 7x y 03

. C = NC d1 2 5C ;3 3

.

AB CM PT ng thng AB: x 2y 2 0 . AC BN PT ng thng AC: 6x 3y 1 0 Cu 152. Trong mt phng vi h ta Oxy, cho tam gic c phng trnh hai cnh l

AB :5x 2y 6 0 v AC : 4x 7y 21 0 . Vit phng trnh cnh BC, bit rng trc tm ca n trng vi gc ta O.

AB: 5x 2y + 6 = 0; AC: 4x + 7y 21 = 0 A(0;3) Phng trnh ng cao BO: 7x 4y = 0 B(4; 7) A nm trn Oy, vy ng cao A

692 Bài Hình Học Luyện Thi Đại Học

Documents

Transcript of 692 Bài Hình Học Luyện Thi Đại Học