200 bài tập hình học tọa Độ phẳng

Trn S Tng PP to trong mt phng

Trang 1

TP 01: NG THNG Cu 1. Trong mt phng vi h to Oxy, cho 2 ng thng d x y1 : 7 17 0- + = ,

d x y2 : 5 0+ - = . Vit phng trnh ng thng (d) qua im M(0;1) to vi d d1 2, mt tam

gic cn ti giao im ca d d1 2, .

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

x y x y x y ( )

x y ( )1

2 2 2 2 2

7 17 5 3 13 03 4 01 ( 7) 1 1

DD

- + + - + - == - - =+ - +

ng thng cn tm i qua M(0;1) v song song vi 1D hoc 2D . KL: x y3 3 0+ - = v x y3 1 0- + = Cu 2. Trong mt phng vi h trc to Oxy, cho cho hai ng thng d x y1 : 2 5 0- + = .

d x y2 : 3 6 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 a1 (2; 1)= -r ; d2 VTCP a2 (3;6)=

r

Ta c: a a1 2. 2.3 1.6 0= - =uur uur

nn d d1 2^ 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 B y Ax By A B: ( 2) ( 1) 0 2 0- + + = + - + =

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

A B A BA AB B

B AA B

0 2 22 2 2 2

2 3cos45 3 8 3 032 ( 1)

- = = - - = = -+ + -

* Nu A = 3B ta c ng thng d x y: 3 5 0+ - = * Nu B = 3A ta c ng thng d x y: 3 5 0- - = Vy c hai ng thng tho mn yu cu bi ton. d x y: 3 5 0+ - = ; d x y: 3 5 0- - = . Cu hi tng t: a) d x y1 : 7 17 0- + = , d x y2 : 5 0+ - = , P(0;1) . S: x y3 3 0+ - = ; x y3 1 0- + = . Cu 3. Trong mt phng Oxy, cho hai ng thng d x y1 : 3 5 0+ + = , d x y2 : 3 1 0+ + = v im

I(1; 2)- . Vit phng trnh ng thng D i qua I v ct d d1 2, ln lt ti A v B sao cho

AB 2 2= . Gi s A a a d B b b d1 2( ; 3 5) ; ( ; 3 1)- - - - ; IA a a IB b b( 1; 3 3); ( 1; 3 1)= - - - = - - +

uur uur

I, A, B thng hng b k aIB kIAb k a1 ( 1)

3 1 ( 3 3) - = - = - + = - -

uur uur

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

Nu a 1 th bb a a ba

13 1 ( 3 3) 3 21

-- + = - - = -

-

AB b a a b t t22 2 2( ) 3( ) 4 2 2 (3 4) 8 = - + - + = + + = (vi t a b= - ).

t t t t2 25 12 4 0 2;5

+ + = = - = -

+ Vi t a b b a2 2 0, 2= - - = - = = - x y: 1 0 D + + =

PP to trong mt phng Trn S Tng

Trang 2

+ Vi t a b b a2 2 4 2,5 5 5 5

- -= - = = = x y: 7 9 0 D - - =

Cu 4. Trong mt phng vi h trc to Oxy, cho hai ng thng d x y1 : 1 0+ + = ,

d x y2 : 2 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 MA MB2 0+ =uuur uuur r

. Gi s: A(a; a1), B(b; 2b 1). T iu kin MA MB2 0+ =

uuur uuur r 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 d x y d x y1 2: 1 0, : 2 2 0+ + = + = ln lt ti A, B sao cho MB = 3MA.

A d A a a MA a aB d B b b MB b b

1

2

( ) ( ; 1 ) ( 1; 1 )( ) (2 2; ) (2 3; )

- - = - - - - = -

uuuruuur .

T A, B, M thng hng v MB MA3= MB MA3=uuur uuur

(1) hoc MB MA3= -uuur uuur

(2)

(1) A d x yB

2 1; ( ) : 5 1 03 3( 4; 1)

- - - - =

- -

hoc (2) ( )A d x yB

0; 1 ( ) : 1 0(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 d x y d x y1 2: 3 5 0, : 4 0- - = + - = ln lt ti A, B sao cho MA MB2 3 0= .

Gi s A a a d1( ;3 5)- , B b b d2( ;4 )- .

V A, B, M thng hng v MA MB2 3= nn MA MBMA MB

2 3 (1)2 3 (2)

=

= -

uuur uuuruuur uuur

+ a b a A Ba b b

5 5 52( 1) 3( 1)(1) ; , (2;2)22(3 6) 3(3 ) 2 22

- = - = - = - =. Suy ra d x y: 0- = .

+ a b a A Ba b b

2( 1) 3( 1) 1(2) (1; 2), (1;3)2(3 6) 3(3 ) 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 OB( 3 )+ nh nht.

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

1+ = (a,b>0)

M(3; 1) d C si

aba b a b3 1 3 11 2 . 12

-= + .

M OA OB a b ab3 3 2 3 12+ = + = a b aOA OB

ba b

min

3 6( 3 ) 12 3 1 1 22

= = + = == =

Phng trnh ng thng d l: x y x y1 3 6 06 2

+ = + - =


Trang 3

Cu 8. Trong mt phng vi h to Oxy, vit phng trnh ng thng D 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.

x y2 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 OA OB2 2

9 4+ 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 B b( ;0); (0; ) vi a b. 0 Phng trnh ca (d) c dng x ya b

1+ = .

V (d) qua M nn a b1 2 1+ = . p dng bt ng thc Bunhiacpski ta c :

a b a b a b

2 2

2 21 2 1 3 2 1 9 41 . 1. 1

3 9

= + = + + +

a b2 29 4 9

10+

OA OB2 29 4 9

10+ .

Du bng xy ra khi a b

1 3 2: 1:3

= v a b1 2 1+ = a b 2010,

9= = d x y: 2 9 20 0+ - = .

Cu 10. Trong mt phng vi h to Oxy, vit phng trnh ng thng D 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). x y x y3 6 0; 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 B b a b( ;0), (0; ) ( , 0) l giao im ca d vi Ox, Oy, suy ra: x yda b

: 1+ = .

Theo gi thit, ta c: a bab

2 1 1

8

+ =

=

b a abab2

8 + = =

.

Khi ab 8= th b a2 8+ = . Nn: b a d x y12; 4 : 2 4 0= = + - = .

Khi ab 8= - th b a2 8+ = - . Ta c: b b b2 4 4 0 2 2 2+ - = = - . + Vi ( ) ( )b d x y2 2 2 : 1 2 2 1 2 4 0= - + - + + - = + Vi ( ) ( )b d x y2 2 2 : 1 2 2 1 2 4 0= - - + + - + = . Cu hi tng t: a) M S(8;6), 12= . S: d x y: 3 2 12 0- - = ; d x y: 3 8 24 0- + = Cu 12. Trong mt phng vi h ta Oxy, cho im A(2; 1) v ng thng d c phng trnh

x y2 3 0+ = . Lp phng trnh ng thng (D) qua A v to vi d mt gc c cos 110

= .

PT ng thng (D) c dng: a x b y( 2) ( 1) 0+ + = ax by a b2 0+ + = a b2 2( 0)+

Ta c: a b

a b2 22 1cos

105( )a -= =

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

(D1): x + y 1 = 0 v (D2): x + 7y + 5 = 0


Trang 4

Cu 13. Trong mt phng vi h ta Oxy, cho im A(2;1) v ng thng d x y: 2 3 4 0+ + = .

Lp phng trnh ng thng D i qua A v to vi ng thng d mt gc 045 .

PT ng thng (D) c dng: a x b y( 2) ( 1) 0+ - = ax by a b(2 ) 0+ + = a b2 2( 0)+ .

Ta c: a b

a b

02 2

2 3cos4513.

+=

+ a ab b2 25 24 5 0- - = a b

a b5

5 = = -

+ Vi a b5= . Chn a b5, 1= = Phng trnh x y: 5 11 0D + - = . + Vi a b5 = - . Chn a b1, 5= = - Phng trnh x y: 5 3 0D - + = . Cu 14. Trong mt phng vi h to Oxy , cho ng thng d x y: 2 2 0- - = v im I(1;1) .

Lp phng trnh ng thng D cch im I mt khong bng 10 v to vi ng thng

d mt gc bng 045 .

Gi s phng trnh ng thng D c dng: ax by c 0+ + = a b2 2( 0)+ .

V d 0( , ) 45D = nn a b

a b2 22 1

2. 5

-=

+ a b

b a3

3 = = -

Vi a b3= D: x y c3 0+ + = . Mt khc d I( ; ) 10D =c4

1010

+ = c

c6

14 = = -

Vi b a3= - D: x y c3 0- + = . Mt khc d I( ; ) 10D =c2

1010

- + = c

c8

12 = - =

Vy cc ng thng cn tm: x y3 6 0;+ + = x y3 14 0+ - = ; x y3 8 0;- - = x y3 12 0- + = . Cu 15. Trong mt phng vi h ta Oxy , cho im M (0; 2) v hai ng thng d1 , d2 c

phng trnh ln lt l x y3 2 0+ + = v x y3 4 0- + = . Gi A l giao im ca d1v d2 .

Vit phng trnh ng thng i qua M, ct 2 ng thng d1v d2 ln lt ti B , C

( B vC khc A ) sao cho AB AC2 2

1 1+ t gi tr nh nht.

A d d A1 2 ( 1;1)= - . Ta c d d1 2^ . Gi D l ng thng cn tm. H l hnh chiu

vung gc ca A trn D . ta c: AB AC AH AM2 2 2 2

1 1 1 1+ = (khng i)

AB AC2 2

1 1+ t gi tr nh nht bng

AM21 khi H M, hay D l ng thng i qua M

v vung gc vi AM. Phng trnh D: x y 2 0+ - = . Cu hi tng t: a) Vi M(1; 2)- , d x y1 : 3 5 0+ + = , d x y2 : 3 5 0- + = . S: x y: 1 0D + + = . Cu 16. Trong mt phng vi h trc ta Oxy, cho ng thng d x y( ) : 3 4 0= v ng

trn C x y y2 2( ) : 4 0+ = . Tm M thuc (d) v N thuc (C) sao cho chng i xng qua im A(3; 1).

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

N (C) (2 3b)2 + (2 b)2 4(2 b) = 0 b b 60;5

= =


Trang 5

Vy c hai cp im: M(4;0) v N(2;2) hoc M N38 6 8 4; , ;5 5 5 5

-

Cu 17. Trong mt phng ta Oxy, cho im A(1; 1) v ng thng D: x y2 3 4 0+ + = . Tm

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

D c PTTS: x ty t

1 32 2

= - = - +

v VTCP u ( 3;2)= -r . Gi s B t t(1 3 ; 2 2 ) D- - + .

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

=uuur r

AB uAB u

. 1. 2

=

uuur rr

tt t

t

2

1513169 156 45 0

313

=

- - = = -

.

Vy cc im cn tm l: B B1 232 4 22 32; , ;13 13 13 13

- -

.

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

bng 152

.

Ta c ON (3;4)=uuur

, ON = 5, PT ng thng ON: x y4 3 0- = . Gi s M m m d(3 6; )+ .

Khi ta c ONMONMS

S d M ON ON d M ONON

21 ( , ). ( , ) 32

DD = = =

m m m m m4.(3 6) 3 133 9 24 15 1;5 3

+ - -= + = = - =

+ Vi m M1 (3; 1)= - - + Vi m M13 137;3 3

- -= -

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

trn ng thng d hai im B, C sao cho tam gic ABC vung B v AB = 2BC . Gi s B b b C c c d(2 2; ), (2 2; )- - .

V DABC vung B nn AB ^ d dAB u. 0=uuur r B 2 6;

5 5

AB 2 55

= BC 55

=

BC c c21 125 300 1805

= - + = 55

c C

c C

1 (0;1)7 4 7;5 5 5

= =

Cu 20. Trong mt phng to Oxy, cho hai ng thng d x y1 : 3 0+ - = , d x y2 : 9 0+ - = v

im A(1;4) . Tm im B d C d1 2, sao cho tam gic ABC vung cn ti A.

Gi B b b d C c c d1 2( ;3 ) , ( ;9 )- - AB b b( 1; 1 )= - - -uuur

, AC c c( 1;5 )= - -uuur

.

DABC vung cn ti A AB ACAB AC

. 0 = =

uuur uuur b c b c

b b c c2 2 2 2( 1)( 1) ( 1)(5 ) 0( 1) ( 1) ( 1) (5 )

- - - + - = - + + = - + -

(*)

V c 1= khng l nghim ca (*) nn


Trang 6

(*)

b cbccb b c c

c

22 2 2 2

2

( 1)(5 )1 (1)1

(5 )( 1) ( 1) ( 1) (5 ) (2)( 1)

+ -- = -

- + + + = - + - -

T (2) b c2 2( 1) ( 1)+ = - b cb c

2 = - = -

.

+ Vi b c 2= - , thay vo (1) ta c c b4, 2= = B C(2;1), (4;5) . + Vi b c= - , thay vo (1) ta c c b2, 2= = - B C( 2;5), (2;7)- . Vy: B C(2;1), (4;5) hoc B C( 2;5), (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: d m x m y m1 : ( 1) ( 2) 2 0+ + = ; d m x m y m2 : (2 ) ( 1) 3 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 x m y mm x m y m

( 1) ( 2) 2(2 ) ( 1) 3 5

- + - = - - + - = - +

.

Ta c m mD m mm m

23 11 2 2 0,

2 1 2 2 - -= = - + > " - -

d d1 2, lun ct nhau. Ta c: A d B d d d1 2 1 2(0;1) , (2; 1) , - ^ D APB vung ti P P

nm trn ng trn ng knh AB. Ta c: PA PB PA PB AB2 2 2 2( ) 2( ) 2 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 (D): x y2 2 0= v hai im A( 1;2)- ,

B(3;4) . Tm im M(D) sao cho MA MB2 22 + c gi tr nh nht.

Gi s M M t t AM t t BM t t(2 2; ) (2 3; 2), (2 1; 4)D+ = + - = - -uuur uuur

Ta c: AM BM t t f t2 2 22 15 4 43 ( )+ = + + = f t f 2min ( )15

= -

M 26 2;

15 15

-

Cu 23. Trong mt phng to Oxy, cho ng thng d x y: 2 3 0- + = v 2 im A B(1;0), (2;1) .

Tm im M trn d sao cho MA MB+ nh nht. Ta c: A A B Bx y x y(2 3).(2 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 y: 5 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 : M 8 17;11 11

-

.


Trang 7

TP 02: NG TRN Cu 1. Trong mt phng vi h to Oxy, gi A, B l cc giao im ca ng thng (d):

x y2 5 0= v ng trn (C): x y x2 2 20 50 0+ - + = . Hy vit phng trnh ng trn (C) i qua ba im A, B, C(1; 1).

A(3; 1), B(5; 5) (C): x y x y2 2 4 8 10 0+ - - + =

Cu 2. 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 DABC nm trn ng thng d x y: 3 8 0= . Vit phng trnh ng trn i qua 3 im A, B, C.

Tm c C (1; 1)1 - , C2( 2; 10)- - .

+ Vi C1(1; 1)- (C): 2 2x y x y11 11 16 0

3 3 3+ - + + =

+ Vi C2( 2; 10)- - (C): 2 2x y x y91 91 416 0

3 3 3+ - + + =

Cu 3. Trong mt phng vi h to Oxy, cho ba ng thng: d x y1 : 2 3 0+ - = ,

d x y2 : 3 4 5 0+ + = , d x y3 : 4 3 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 t( ;3 2 )- d1.

Khi : d I dd I d2 3) ( , )( , = t t t t3 4(3 2 ) 5

54 3(3 2 ) 2

5+ - +

=+ - +

tt24

==

Vy c 2 ng trn tho mn: x y2 2 4925

( 2) ( 1) =- + + v x y2 2 9( 4) ( 5)25

- + + = .

Cu hi tng t: a) Vi d x y1 : 6 10 0= , d x y2 : 3 4 5 0+ + = , d x y3 : 4 3 5 0- - = .

S: x y2 2( 10) 49- + = hoc x y2 2 2

10 70 743 43 43

- + + =

.

Cu 4. Trong mt phng vi h to Oxy, cho hai ng thng D : x y3 8 0+ + = ,

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

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

t t

t t2 22 2

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

3 4

- - - += - - + + -

+ t 3= - I R(1; 3), 5- =

PT ng trn cn tm: x y2 2( 1) ( 3) 25- + + = . Cu 5. Trong mt phng vi h to Oxy, cho hai ng thng x y: 4 3 3 0D - + = v

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

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


Trang 8

aa b a bd I d I a aIM u a b a b

54 34 3 3 3 4 31( , ) ( , ') 4 3 3 6 8545 5(3;4) 3( 6) 4( 9) 0 3 4 54D

D D -- + - - = - + = -= ^ = - + - = + =

uuur r

a a a b

a a bb

25 150 4 6 85 10; 654 3 190; 1564

- = - = = - = - ==

Vy: C x y2 2( ) : ( 10) ( 6) 25- + - = tip xc vi 'D ti N(13;2)

hoc C x y2 2( ) : ( 190) ( 156) 60025+ + - = tip xc vi 'D ti N( 43; 40)- - Cu 6. 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: x a y a a ax a y a a b

2 2 2

2 2 2( ) ( ) ( )( ) ( ) ( )

- + + =

- + - =

a) a a1; 5= = b) v nghim.

Kt lun: x y2 2( 1) ( 1) 1- + + = v x y2 2( 5) ( 5) 25- + + = . Cu 7. Trong mt phng vi h ta Oxy, cho ng thng d x y( ) : 2 4 0- - = . Lp phng

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

Gi I m m d( ;2 4) ( )- l tm ng trn cn tm. Ta c: m m m m 42 4 4,3

= - = = .

m 43

= th phng trnh ng trn l: x y2 2

4 4 163 3 9

- + + =

.

m 4= th phng trnh ng trn l: x y2 2( 4) ( 4) 16- + - = . Cu 8. Trong mt phng vi h ta Oxy, cho im A(1;1) v B(3;3), ng thng (D):

x y3 4 8 0+ = . Lp phng trnh ng trn qua A, B v tip xc vi ng thng (D). 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)=

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

Ta c IA = d(I,D) a a a211 8 5 5 10 10 - = - + 2a2 37a + 93 = 0 a

a

3312

=

=

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

Vi a = 312

I 31; 272

-

, R = 65

2 (C): x y

2231 4225( 27)

2 4

- + + =

Cu 9. Trong h to Oxy cho hai ng thng d x y: 2 3 0+ - = v x y: 3 5 0D + - = . Lp

phng trnh ng trn c bn knh bng 2 105

, c tm thuc d v tip xc vi D .

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

d I R( , )D =a 2 2 10

510

- = a

a6

2 = = -


Trang 9

(C): x y2 2 8( 9) ( 6)5

+ + - = hoc (C): x y2 2 8( 7) ( 2)5

- + + = .

Cu 10. Trong mt phng vi h to Oxy, cho ng trn (C): x y x2 2 4 3 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 ty t

2 32 2

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

AI I A t I12 '( 3;3)2

= = uur uur

(C): x y2 2( 3) ( 3) 4- + - =

Cu 11. Trong mt phng vi h to Oxy, cho ng trn (C): x y y2 2 4 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): x y2 2

8 6 95 5

- + + =

Cu 12. Trong mt phng vi h ta Oxy, cho ng trn (C): x y x y2 2 2 4 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: x y3 4 11 0- - = . AB 3= .

Gi H x y( ; ) l trung im ca AB. Ta c: H IM

IH R AH2 2 32

= - =

x y

x y2 23 4 11 0

9( 1) ( 2)4

- - = - + + =

x y

x y

1 29;5 10

11 11;5 10

= - = -

= = -

H 1 29;5 10

- -

hoc H 11 11;

5 10

-

.

Vi H 1 29;5 10

- -

. Ta c R MH AH2 2 2 43 = + = PT (C): x y2 2( 5) ( 1) 43- + - = .

Vi H 11 11;5 10

-

. Ta c R MH AH2 2 2 13 = + = PT (C): x y2 2( 5) ( 1) 13- + - = .

Cu 13. Trong mt phng vi h ta Oxy, cho ng trn (C): x y2 2( 1) ( 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= . IABSD ln nht DIAB vung ti I AB 2 2= .

M IK 2 2= nn c hai ng trn tho YCBT. + T1( ) c bn knh R R1 2= = T x y

2 21( ) : ( 3) ( 4) 4- + - =


Trang 10

+ T2( ) c bn knh R2 2

2 (3 2) ( 2) 2 5= + = T x y2 2

1( ) : ( 3) ( 4) 20- + - = . Cu 14. Trong mt phng vi h to Oxy, vit phng trnh ng trn ni tip tam gic ABC

vi cc nh: A(2;3), B C1 ;0 , (2;0)4

.

im D(d;0) d1 24

<


Trang 11

y + 7 = 0; 4x + 3y + 27 = 0.

Cu 18. Trong mt phng ta Oxy, cho ng trn ( )C x y x2 2: 2 0+ + = . Vit phng trnh tip tuyn ca ( )C , bit gc gia tip tuyn ny v trc tung bng 30o .

C x y I R2 2( ) : ( 1) 1 ( 1;0); 1+ + = - = . H s gc ca tip tuyn (D) cn tm l 3 .

PT (D) c dng x y b1 : 3 0D - + = hoc x y b2 : 3 0D + + =

+ x y b1 : 3 0D - + = tip xc (C) d I R1( , )D = b b3 1 2 3

2-

= = + .

Kt lun: x y1( ) : 3 2 3 0D - + =

+ x y b2( ) : 3 0D + + = tip xc (C) d I R2( , )D =b b3 1 2 3

2-

= = + .

Kt lun: x y2( ) : 3 2 3 0D + + = .

Cu 19. Trong mt phng vi h to Oxy, cho ng trn (C): x y x y2 2 6 2 5 0+ - - + = v ng thng (d): x y3 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 (D): ax by c c0 ( 0)+ + = .

T: d I

d

( , ) 52cos( , )

2

D

D

=

=

a b ca b c

2, 1, 101, 2, 10

= = - = - = = = -

x yx y

: 2 10 0: 2 10 0

DD

- - = + - =

.

Cu 20. Trong h to Oxy , cho ng trn C x y2 2( ) : ( 1) ( 1) 10- + - = v ng thng d x y: 2 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( ; )=r l VTPT ca tip tuyn D a b2 2( 0)+ ,

V d 0( , ) 45D = nn a b

a b2 22 1

2. 5

-=

+ a b

b a3

3 = = -

Vi a b3= D: x y c3 0+ + = . Mt khc d I R( ; )D =c4

1010

+ = c

c6

14 = = -

Vi b a3= - D: x y c3 0- + = . Mt khc d I R( ; )D =c2

1010

- + = c

c8

12 = - =

Vy c bn tip tuyn cn tm: x y3 6 0;+ + = x y3 14 0+ - = ; x y3 8 0;- - = x y3 12 0- + = . Cu 21. Trong mt phng vi h to Oxy, vit phng trnh tip tuyn chung ca hai ng trn

(C1): x y x y2 2 2 2 2 0+ = , (C2): x y x y2 2 8 2 16 0+ + = . (C1) c tm I1(1; 1) , bn knh R1 = 2; (C2) c tm I2(4; 1) , bn knh R2 = 1.

Ta c: I I R R1 2 1 23= = + (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( ) : ( ) : 0D D= + - + = ta c:


Trang 12

a ba ad I R a b hay

d I R a bb b

a b

2 21 1

2 2

2 2

12 22

( ; ) 4 4( ; ) 4 1 4 7 2 4 7 2

14 4

DD

+ - = = = - = + = + - - + = == +

Vy, c 3 tip tuyn chung: x y x y x1 2 32 4 7 2 2 4 7 2( ) : 3, ( ) : , ( )

4 4 4 4D D D

+ -= = - + = +

Cu 22. Trong mt phng vi h ta Oxy, cho hai ng trn (C): x y2 2( 2) ( 3) 2- + - = v

(C): x y2 2( 1) ( 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 R R' 2 = = - (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) = - -uur

PTTT: x y 7 0+ - =

Cu 23. Trong mt phng vi h ta Oxy, cho hai ng trn C x y y2 21( ) : 2 3 0+ - - = v

C x y x y2 22( ) : 8 8 28 0+ - - + = . Vit phng trnh tip tuyn chung ca C1( ) v C2( ) .

C1( ) c tm I1(0;1) , bn knh R1 2= ; C2( ) c tm I2(4;4) , bn knh R2 2= . Ta c: I I R R1 2 1 25 4= > = + C C1 2( ),( ) ngoi nhau. Xt hai trng hp: + Nu d // Oy th phng trnh ca d c dng: x c 0+ = . Khi : d I d d I d c c1 2( , ) ( , ) 4= = + 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 : d I dd I d d I d

1

1 2

( , ) 2( , ) ( , )

= =

b

ab a b

a a

2

2 2

1 21

1 4 4

1 1

- +=

+ - + - + = + +

a b

a b

a b

3 7;4 23 3;4 2

7 37;24 12

= =

= = - = - =

d x y: 3 4 14 0- + = hoc d x y: 3 4 6 0- - = hoc d x y: 7 24 74 0+ - = . Vy: d x: 2 0- = ; d x y: 3 4 14 0- + = ; d x y: 3 4 6 0- - = ; d x y: 7 24 74 0+ - = .

Cu 24. Trong mt phng vi h ta Oxy, cho hai ng trn C x y y2 21( ) : 4 5 0+ - - = v

C x y x y2 22( ) : 6 8 16 0+ - + + = . Vit phng trnh tip tuyn chung ca C1( ) v C2( ) .

C1( ) c tm I1(0;1) , bn knh R1 3= ; C2( ) c tm I2(3; 4)- , bn knh R2 3= .

Gi s tip tuyn chung D ca C C1 2( ), ( ) c phng trnh: ax by c a b2 20 ( 0)+ + = + .

D l tip tuyn chung ca C C1 2( ), ( ) d I Rd I R

1 1

2 2

( , )( , )

DD

= =

b c a ba b c a b

2 2

2 22 3 (1)

3 4 3 (2)

+ = +

- + = +

T (1) v (2) suy ra a b2= hoc a bc 3 22

- += .

+ TH1: Vi a b2= . Chn b 1= a c2, 2 3 5= = - x y: 2 2 3 5 0D + - =


Trang 13

+ TH2: Vi a bc 3 2

2- +

= . Thay vo (1) ta c: a

a b a ba b

2 20

2 2 43

=- = +

= -

.

y: 2 0D + = hoc x y: 4 3 9 0D - - = .

Cu 25. Trong mt phng Oxy, cho ng trn (C): x y x2 2 4 3 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 ty t

2 32 2

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

(T) tip xc ngoi vi (C) ti A nn AI JA t J12 ( 3;3)2

= = uur uur

.

Vy: T x y2 2( ) : ( 3) ( 3) 4- + - = .

Cu 26. Trong mt phng Oxy, cho ng trn (C): x y2 2 1+ = v phng trnh:

x y m x my2 2 2( 1) 4 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 m( 1; 2 )+ - , bn knh R m m2 2' ( 1) 4 5= + + + ,

(C) c tm O(0; 0) bn knh R = 1, OI m m2 2( 1) 4= + + , ta c OI < R

Vy (C) v (Cm) ch tip xc trong. R R = OI ( v R > R) m m31;5

= - = .

Cu 27. Trong mt phng Oxy, cho cc ng trn c phng trnh C x y2 211( ) : ( 1)2

- + = v

C x y2 22( ) : ( 2) ( 2) 4- + - = . Vit phng trnh ng thng d tip xc vi C1( ) v ct C2( )

ti hai im M N, sao cho MN 2 2= .

C1( ) c tm I1(1;0) , bn knh R112

= ; C2( ) c tm I1(2;2) , bn knh R2 2= . Gi H l

trung im ca MN MNd I d I H R2

22 2 2( , ) 22

= = - =

Phng trnh ng thng d c dng: ax by c a b2 20 ( 0)+ + = + .

Ta c: d I d

d I d

1

2

1( , )2

( , ) 2

=

=

a c a ba b c a b

2 2

2 22

2 2 2

+ = +

+ + = +. Gii h tm c a, b, c.

Vy: d x y d x y: 2 0; : 7 6 0+ - = + - = ; d x y: 2 0- - = ; d x y: 7 2 0- - =

Cu 28. Trong mt phng vi h to Oxy, cho ng trn (C): x y x2 2 6 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 .


Trang 14

(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 AMBAMB

0

060 (1)120 (2)

=

=

V MI l phn gic ca AMB nn:

(1) AMI = 300 IAMI0sin30

= MI = 2R m m2 9 4 7+ = =

(2) AMI = 600 IAMI0sin 60

= MI = 2 33

R m2 4 393

+ = V nghim Vy c

hai im M1(0; 7 ) v M2(0; 7- ) Cu 29. Trong mt phng vi h ta Oxy, cho ng trn (C) v ng thng D nh bi:

C x y x y x y2 2( ) : 4 2 0; : 2 12 0D+ - - = + - = . Tm im M trn D 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 R=2 52= . Nh th im M nm trn ng trn (T) c phng trnh: x y2 2( 2) ( 1) 20- + - = . Mt khc, im M nm trn ng thng D, nn ta ca M nghim ng h phng trnh:

x yx y

2 2( 2) ( 1) 20 (1)2 12 0 (2)

- + - = + - =

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

y y y yy

2 2 23

2 10 1 20 5 42 81 0 275

=- + + - = - + =

=

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

Cu 30. Trong mt phng vi h ta Oxy, cho ng trn (C): x y2 2( 1) ( 2) 9- + + = v 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 mm

m1 53 2 1 6

72

- = -= - = =

Cu hi tng t: a) C x y d x y m2 2( ) : 1, : 0+ = - + = S: m 2= .

Cu 31. Trong mt phng vi h ta Oxy, cho ng trn (C): x y2 2( 1) ( 2) 9- + + = v ng thng d x y m: 3 4 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= . DPAB u PI AI R2 2 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


Trang 15

tuyn ca (T) m md I dm

11 19( , ) 6 6415

+ == = = -.

Cu 32. Trong mt phng vi h to Oxy, cho hai ng trn C x y x y2 2( ) : 18 6 65 0+ - - + =

v C x y2 2( ) : 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

AH 125

= . Suy ra: OH OA AH2 2 95

= - = v OAOMOH

25= = .

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

2 2

2 2( ) 18 6 65 0

5 25

+ - - + = = + = x x

y y4 53 0

= = = =

Vy M(4;3) hoc M(5;0) .

Cu 33. Trong mt phng vi h to Oxy, cho ng trn (C): x y2 2( 1) ( 2) 4- + + = . M l im di ng trn ng thng d y x: 1= + . Chng minh rng t M k c hai tip tuyn MT1 ,

MT2 ti (C) (T1, T2 l tip im) v tm to im M, bit ng thng T T1 2 i qua im A(1; 1)- .

(C) c tm I(1; 2)- , bn knh R 2= . Gi s M x x d0 0( ; 1)+ .

IM x x x R2 2 20 0 0( 1) ( 3) 2( 1) 8 2= - + + = + + > = M nm ngoi (C) qua M k c 2 tip tuyn ti (C).

Gi J l trung im IM x x

J 0 01 1

;2 2

+ -

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

knh IMR1 2= c phng trnh

x x x xT x y

2 2 2 20 0 0 01 1 ( 1) ( 3)( ) :2 2 4

+ - - + +- + - =

T M k c 2 tip tuyn MT1, MT2 n (C) IT M IT M T T T01 2 1 290 , ( )= = T T C T1 2{ , } ( ) ( ) = to T T1 2, tho mn h:

x x x x

x y x x x y xx y

2 22 20 0 0 0

0 0 02 2

1 1 ( 1) ( 3)( ) ( ) (1 ) (3 ) 3 0 (1)2 2 4( 1) ( 2) 4

+ - - + + - + - = - - + - - = - + + =

To cc im T T1 2, tho mn (1), m qua 2 im phn bit xc nh duy nht 1 ng

thng nn phng trnh T T1 2 l x x y x x0 0 0(1 ) (3 ) 3 0- - + - - = . A(1; 1)- nm trn T T1 2 nn x x x0 0 01 (3 ) 3 0- + + - - = x0 1= M(1;2) .

Cu 34. Trong mt phng vi h ta Oxy, cho ng trn (C): x y2 2( 1) ( 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 CP /( ) 27 0= > M nm ngoi (C). (C) c tm I(1;1) v R = 5. Mt khc:

M CP MA MB MB MB BH2

/( ) . 3 3 3= = = =uuur uuur

IH R BH d M d2 2 4 [ ,( )] = - = =


Trang 16

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

aa b

d M da ba b2 2

06 4[ ,( )] 4 4 12

5

=- -= =

= -+ . Vy (d): y 3 = 0 hoc (d): 12x 5y 69 = 0.

Cu 35. 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 x y2 2( 2) ( 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.

( ) a b a bd I d a b a ba b

2 22 2

2 2, 3 3 3- - -= = - = ++

a

a aba b

20

8 6 0 34

= + =

= -

a = 0: chn b = 1 d: y 2 = 0 a = b34

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

Cu hi tng t: a) d i qua O, C x y x y2 2( ) : 2 6 15 0+ - + - = , l 8= . S: d x y: 3 4 0- = ; d y: 0= .

b) d i qua Q(5;2) , C x y x y2 2( ) : 4 8 5 0+ - - - = , l 5 2= . S: d x y: 3 0- - = ; d x y:17 7 71 0- - = .

c) d i qua A(9;6) , C x y x y2 2( ) : 8 2 0+ - - = , l 4 3= .

S: d y x: 2 12= - ; d y x1 21:2 2

= - +

Cu 36. Trong mt phng vi h to Oxy, cho ng trn (C) : x y x y2 2 2 8 8 0+ + - - = . Vit phng trnh ng thng D song song vi ng thng d x y: 3 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 D c dng: x y c c3 0, 2+ + = . V D ct (C) theo mt dy cung c di bng 6 nn:

( ) c cd Ic2

3 4 4 10 1, 44 10 13 1

D- + + = - = =

= - -+.

Vy phng trnh D cn tm l: x y3 4 10 1 0+ + - = hoc x y3 4 10 1 0+ - - = . Cu hi tng t: a) C x y2 2( ) : ( 3) ( 1) 3- + - = , d x y: 3 4 2012 0- + = , l 2 5= . S: x y: 3 4 5 0D - + = ; x y: 3 4 15 0D - - = .

Cu 37. Trong mt phng vi h trc ta Oxy, cho ng trn C x y2 2( ) :( 4) ( 3) 25+ + - = v ng thng x y: 3 4 10 0D - + = . Lp phng trnh ng thng d bit d ( )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 D^ nn PT ca d c dng: x y m4 3 0+ + = .

Ta c: d I 1( ,( ))D = IH = AI AH2 2 2 25 3 4- = - =

mmm2 2

2716 9 4134 3

=- + += = -+


Trang 17

Vy PT cc ng thng cn tm l: x y4 3 27 0+ + = v x y4 3 13 0+ - = .

Cu 38. Trong mt phng vi h to Oxy, cho ng trn (C): x y x y2 2 2 2 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 = IA IH IH IM2 2 2 22 2 5 2 5 2 3- = - - = . Du "=" xy ra H M hay d ^ IM. Vy d l ng thng qua M v c VTPT MI (1; 1)= -

uuur

Phng trnh d: x y 2 0- + = . Cu hi tng t: a) Vi (C): x y x y2 2 8 4 16 0+ - - - = , M(1; 0). S:

d x y: 5 2 5 0+ + = Cu 39. 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 DOAB c din tch ln nht.

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

= .

Gi s phng trnh ng thng d: A x B y A B2 2( 2) ( 6) 0 ( 0)- + - = +

d O d 5 2( , )2

= A B

A B2 22 6 5 2

2- -

=+

B AB A2 247 48 17 0+ - = B A

B A

24 5 5547

24 5 5547

- -=

- + =

+ Vi B A24 5 5547

- -= : chn A = 47 B = 24 5 55- -

d: ( )x y47( 2) 24 5 55 ( 6) 0- - + - =

+ Vi B A24 5 5547

- += : chn A = 47 B = 24 5 55- +

d: ( )x y47( 2) 24 5 55 ( 6) 0- + - + - = Cu hi tng t: a) C x y x y2 2( ) : 4 6 9 0+ + - + = , M(1; 8)- . S: x y x y7 1 0; 17 7 39 0+ + = + + = .

Cu 40. Trong mt phng vi h to Oxy, cho ng trn (C): x y x y2 2 6 2 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: a x b y a b2 2( 3) ( 3) 0, 0- + - = + ax by a b3 3 0+ - - = .

Gi s d qua A ct (C) ti hai im A, B AB = 4 2 . Gi I l tm hnh vung.

Ta c: d I d AD AB1 1( , ) 2 2 ( )2 2

= = =a b a b

a b2 23 3 3

2 2- - -

=+


Trang 18

b a b a b a b2 2 2 24 2 2 = + = = . 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 41. Trong mt phng vi h to Oxy, cho hai ng trn (C1): x y2 2 13+ = v (C2):

x y2 2( 6) 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: a x b y a b2 2( 2) ( 3) 0 ( 0)- + - = + . Gi d d O d d d I d1 2 2( , ), ( , )= = .

T gi thit R d R d2 2 2 21 1 2 2- = - d d2 22 1 12- =

a a b a ba b a b

2 2

2 2 2 2(6 2 3 ) ( 2 3 ) 12- - - -- =

+ +

b ab2 3 0+ = bb a

03

= = -

.

Vi b = 0: Chn a = 1 Phng trnh d: x 2 0- = . Vi b = 3a: Chn a = 1, b = 3 Phng trnh d: x y3 7 0- + = . Cu 42. Trong mt phng vi h ta Oxy, cho ng thng D: mx y4 0+ = , ng trn (C):

x y x my m2 2 22 2 24 0+ - - + - = c tm I. Tm m ng thng D ct ng trn (C) ti hai im phn bit A, B sao cho din tch tam gic IAB bng 12.

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

m m mIH d Im m2 2

4 5( , )16 16

+= D = =

+ +; mAH IA IH

m m

22 2

2 2

(5 ) 202516 16

= - = - =+ +

IABS 12D = m

d I AH m mm

23

( , ). 12 3 25 48 0 163

= D = - + =

=

Cu 43. Trong mt phng ta Oxy, cho ng trn C x y2 2( ) : 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 : OABS OA OB AOB AOB1 1 1. .sin .sin2 2 2

= = . Du "=" xy ra AOB 090= .

Vy AOBS ln nht AOB 090= . Khi d I d 1( ; )

2= m 1 = .

Cu 44. Trong mt phng vi h to Oxy, cho ng thng d( ) : x my2 1 2 0+ + - = v

ng trn c phng trnh C x y x y2 2( ) : 2 4 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( , ) < m m22 2 1 2 3 2 - + - < +


Trang 19

m m m m m m R2 2 21 4 4 18 9 5 4 17 0 - + < + + + >

Ta c: S IA IB AIB IA IBIAB1 1 9. sin .2 2 2

= =

Vy: SIAB ln nht l 92

khi AIB 090= AB = R 2 3 2= d I d 3 2( , )2

=

m m3 2 21 2 22

- = + m m22 16 32 0 + + = m 4 = -

Cu hi tng t: a) Vi d x my m: 2 3 0+ + = , C x y x y2 2( ) : 4 4 6 0+ + + + = . S:

m m 8015

= =

Cu 45. Trong mt phng vi h to Oxy, cho ng trn C x y x y2 2( ) : 4 6 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 b: 8 0+ - + = ( a b2 2 0+ ).

IABS IA IB AIB AIB1 . .sin 2sin2D

= = .

Do : IABSD ln nht AIB 090= d I d IA 2( , ) 2

2= =

b a

a b2 211 3 2- =

+ a ab b2 27 66 118 0- + = a b

a b7

7 17 = =

.

+ Vi b a1 7= = d x y: 7 1 0+ + = + Vi b a7 17= = d x y:17 7 39 0+ + =

Cu 46. Trong mt phng vi h ta Oxy, cho ng trn (C): x y x y2 2 4 4 6 0+ + + + = v ng thng D: x my m2 3 0+ + = vi m l tham s thc. Gi I l tm ca ng trn (C). Tm m D ct (C) ti 2 im phn bit A v B sao cho din tch DIAB ln nht.

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

K ng cao IH ca DIAB, ta c: SDABC = IABS IA IB AIB1 . .sin2

= = AIBsin

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

2= (tha IH < R)

m

m21 4

11

-=

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

15

Cu hi tng t: a) Vi C x y x y2 2( ) : 2 4 4 0+ - + - = , x my: 2 1 2 0D + + - = . S: m 4= - .

b) Vi C x y x y2 2( ) : 2 4 5 0+ - - - = , x my: 2 0D + - = . S: m 2= - Cu 47. Trong mt phng vi h ta Oxy, cho ng thng d: x y 5 2 0= v ng trn (C):

x y x y2 2 2 4 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


Trang 20

tam gic ABC vung B. Ta giao im A, B l nghim ca h phng trnh

y xx y x yy xx y

2 2 0; 22 4 8 01; 35 2 0

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

V ABC 090= 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 48. Trong mt phng vi h ta Oxy , cho ng trn (C ): x y x y2 2 2 4 8 0+ + - - = v ng thng ( D ): x y2 3 1 0- - = . Chng minh rng ( D ) 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 . d I R9( , )13

D = < ng thng ( D ) ct (C) ti

hai im A, B phn bit. Gi M l im nm trn (C), ta c ABMS AB d M1 . ( , )2D

D= . Trong

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

x y3 2 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: x y x yx y

2 2 2 4 8 03 2 1 0

+ + - - = + - = x y

x y1, 1

3, 5 = = - = - =

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

Ta c d P 4( , )13

D = ; d Q 22( , )13

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

Vy ta im M(3; 5).

Cu 49. Trong mt phng vi h to Oxy, cho ng trn (C): x y x y2 2 2 4 5 0+ - - - = v A(0; 1) (C). Tm to cc im B, C thuc ng trn (C) sao cho DABC u.

(C) c tm I(1;2) v R= 10 . Gi H l trung im BC. Suy ra AI IH2.=uur uur

H 3 7;2 2

ABCD u I l trng tm. Phng trnh (BC): x y3 12 0+ - = V B, C (C) nn ta ca B, C l cc nghim ca h phng trnh:

x y x y x y x yx y x y

2 2 2 22 4 5 0 2 4 5 03 12 0 12 3

+ - - - = + - - - = + - = = -

Gii h PT trn ta c: B C7 3 3 3 3 7 3 3 3 3; ; ;2 2 2 2

+ - - +

hoc ngc li.

Cu 50. Trong mt phng vi h to Oxy, cho ng trn (C): x y2 2( 3) ( 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. DABC 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)=uur

(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 a(1; )=r l VTCP ca d. Ta c:


Trang 21

( ) a aIA u

a a2 2 22 2 2cos ,

21 2 1 5 1

+ += = =

+ + +

uur r a a22 2 5 1+ = + a

a

313

=

= -

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

55 3

= + = +

.

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 u 11;3

= -

r Phng trnh ng thng d:

x t

y t

5153

= + = -

.

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 51. Trong mt phng to Oxy, cho ng trn (C): x y2 2 4+ = v cc im A 81;3

-

,

B(3;0) . Tm to im M thuc (C) sao cho tam gic MAB c din tch bng 203

.

AB AB x y64 104 ; : 4 3 12 09 3

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

Ta c: x y x yh AB hx y

4 3 121 20 4 3 8 0. 4 44 3 32 02 3 5

- - - + == = = - - =

+ x y M Mx y2 24 3 8 0 14 48( 2;0); ;

25 754 - + =

- - + = + x y

x y2 24 3 32 0

4 - - = + =

(v nghim)

Cu 52. Trong mt phng to Oxy, cho ng trn C x y x y2 2( ) : 2 6 9 0+ + - + = v ng thng d x y: 3 4 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 R( , ) 2= > d C( ) = . Gi D l ng thng qua I v vung gc vi d x y( ) : 4 3 5 0D + - = .

Gi N d N0 01 7;5 5

D

=

.

Gi M M1 2, l cc giao im ca D v (C) M M1 22 11 8 19; , ;5 5 5 5

- -

MN ngn nht khi M M N N1 0, .

Vy cc im cn tm: M C2 11; ( )5 5

-

, N d1 7;

5 5

.


Trang 22

TP 03: CC NG CNIC

Cu 1. Trong mt phng vi h to Oxy, cho elip (E): x y2 2

125 16

+ = . A, B l cc im trn (E)

sao cho: AF BF1 2 8+ = , vi F F1 2, l cc tiu im. Tnh AF BF2 1+ .

1AF AF a2 2+ = v BF BF a1 2 2+ = 1 2AF AF BF BF a1 2 4 20+ + + = =

M 1AF BF2 8+ = 2AF BF1 12+ = Cu 2. Trong mt phng vi h to Oxy, vit phng trnh elip vi cc tiu im

F F1 2( 1;1), (5;1)- v tm sai e 0,6= .

Gi s M x y( ; ) l im thuc elip. V na trc ln ca elip l cae

3 50,6

= = = nn ta c:

MF MF x y x y2 2 2 21 2 10 ( 1) ( 1) ( 5) ( 1) 10+ = + + - + - + - = x y2 2( 2) ( 1) 1

25 16- -

+ =

Cu 3. Trong mt phng vi h to Oxy, cho im C(2; 0) v elip (E): x y2 2

14 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.

A B2 4 3 2 4 3; , ;7 7 7 7

-


1100 25

+ = . Tm cc im M (E) sao

cho F MF 01 2 120= (F1, F2 l hai tiu im ca (E)).

Ta c: a b10, 5= = c 5 3= . Gi M(x; y) (E) MF x MF x1 23 310 , 10

2 2= - = + .

F F MF MF MF MF F MF2 2 21 2 1 2 1 2 1 22 . .cos= + -

( ) x x x x2 2

2 3 3 3 3 110 3 10 10 2 10 102 2 2 2 2

= - + + - - + -

x = 0 (y= 5). Vy c 2 im tho YCBT: M1(0; 5), M2(0; 5).

Cu 5. Trong mt phng Oxy, cho elip (E) c hai tiu im F F1 2( 3;0); ( 3;0)- v i qua im

A 13;2

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

thc: P F M F M OM F M F M2 2 21 2 1 23 .= + .

(E): x ya b a b

2 2

2 2 2 23 11 1

4+ = + = , a b2 2 3= + x y

2 21

4 1+ =

M M M M MP a ex a ex x y a e x2 2 2 2 2 2 2( ) ( ) 2( ) ( ) 1= + + + - =


Trang 23

Cu 6. Trong mt phng to Oxy, cho elip (E): x y2 24 16 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 x 8:3

D = c gi tr khng i.

Ta c: F2( 12;0) . Gi M x y E0 0( ; ) ( ) x

MF a ex 02 08 3

2-

= - = ,

x

d M x 008 38( , )

3 3D

-= - = (v x04 4- )

MFd M

2 3( , ) 2D

= (khng i).

Cu 7. Trong mt phng vi h to Oxy, cho elip (E): x y2 25 16 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 DMAB.

Phng trnh ng thng (AB): x y2 3 0- + = v AB 2 5=

Gi M x y E x y2 20 0 0 0( ; ) ( ) 5 16 80. + = Ta c: x y x y

d M AB 0 0 0 02 3 2 3

( ; )1 4 5

- + - += =

+

Din tch DMAB: S AB d M AB x y0 01 . . ( ; ) 2 32

= = - -

p dng bt ng thc Bunhiacpxki cho 2 cp s x y0 01 1; , ( 5 ; 4 )

25

-

c:

( )x y x y2

2 20 0 0 0

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

- + + = =

x y x y x y x y0 0 0 0 0 0 0 02 6 6 2 6 3 2 3 9 2 3 9 - - - - - + - +

x yx y

x yx y

x y

0 00 0

0 0

0 0

5 45 81 1max 2 3 9

2 6252 3 9

= = - - + = - - =

- + =

x

y

0

0

83

53

=

= -

Vy, MABS khi M8 5max 9 ;3 3

= -

.

Cu 8. Trong mt phng vi h to Oxy, cho elp x yE2 2

( ) : 19 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: x y2 3 0+ = . Gi C(x; y) (E), vi x y0, 0> > x y2 2

19 4

+ = .

ABCx yS AB d C AB x y1 85 85. ( , ) 2 3 3.

2 13 3 22 13= = + = + x y

2 285 1703 2 313 9 4 13

+ =

Du "=" xy ra x y

xx y

y

2 221 39 4 2

23 2

+ = = = =

. Vy C 3 2 ; 22

.


Trang 24

Cu 9. Trong mt phng ta Oxy , cho elip x yE

2 2( ) : 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 D qua M(1; 1) c PT: y k x( 1) 1= - + . To cc giao im A B, ca D v

E( ) l nghim ca h: x y

y k x

2 21 (1)

25 9( 1) 1 (2)

+ = = - +

k x k k x k k2 2 2(25 9) 50 ( 1) 25( 2 9) 0+ - - + - - = (3)

PT (3) lun c 2 nghim phn bit x x1 2, vi mi k . Theo Viet: k kx xk1 2 2

50 ( 1)25 9

-+ =

+.

Do M l trung im ca AB Mk kx x x kk1 2 2

50 ( 1) 92 22525 9

- + = = = -

+.

Vy PT ng thng D: x y9 25 34 0+ - = . Cu hi tng t:

a) Vi x yE2 2

( ) : 19 4

+ = , M(1;1) S: x y: 4 9 13 0D + - =


18 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 M x y E0 0( ; ) ( ) vi x y x y Z0 0 0 0, 0; , .

Ta c: x y2 20 0 18 2

+ = y20 2 y00 2 y x loaiy x

0 0

0 0

0 2 2 ( )1 2

= = = =

M(2;1) .

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


18 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( ; ) ( ) x y2 2

18 2

+ = . p dng BT Bunhiacpxki, ta c:

x yx y2 2

2( ) (8 2) 108 2

+ + + =

x y10 10- + .

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

x y8 2

10

=

+ =

M 4 10 10;5 5

.

+ x y 10+ - . Du "=" xy ra x y

x y8 2

10

=

+ = -

M 4 10 10;5 5

- -


Trang 25

Cu 12. Trong mt phng vi h to Oxy, cho elip (E): x y

2 21

9 3+ = v im A(3;0) . Tm trn

(E) cc im B, C sao cho B, C i xng qua trc Ox v DABC l tam gic u. Khng mt tnh tng qut, gi s B x y C x y0 0 0 0( ; ), ( ; )- vi y0 0> .

Ta c: x y

x y2 2

2 20 00 01 3 99 3

+ = + = . BC y02= v BC x x0( ) : = d A BC x0( ,( )) 3= -

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

Suy ra: DABC u d A BC BC3( ,( ))2

= x y0 03 3- = y x2 20 03 ( 3)= -

xx xx

2 2 00 0

0

0( 3) 93

=+ - = =

.

+ Vi x0 0= y0 3= B C(0; 3), (0; 3)- . + Vi x0 3= y0 0= (loi).

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


19 4

+ = v cc ng thng

d mx ny1 : 0- = , d nx+my2 : 0= , vi m n2 2 0+ . Gi M, N l cc giao im ca d1 vi (E),

P, Q l cc giao im ca d2 vi (E). Tm iu kin i vi m n, din tch t gic MPNQ t gi tr nh nht.

PTTS ca d d1 2, l: x nt

dy mt

11

1: = =

, x mtdy nt

22

2: = - =

.

+ M, N l cc giao im ca d1 v (E)

n m n mM Nm n m n m n m n2 2 2 2 2 2 2 2

6 6 6 6; , ;9 4 9 4 9 4 9 4

- - + + + +

+ P, Q l cc giao im ca d2 v (E)

m n m nP Qm n m n m n m n2 2 2 2 2 2 2 2

6 6 6 6; , ;4 9 4 9 4 9 4 9

- - + + + +

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

MPNQS S MN PQ OM OP1 . 2 .2

= = = = M M P Pm nx y x y

m n m n

2 22 2 2 2

2 2 2 2

72( )2 .(9 4 )(4 9 )

++ + =

+ +

p dng BT C-si: m n m nm n m n m n2 2 2 2

2 2 2 2 2 2(9 4 ) (4 9 ) 13(9 4 )(4 9 ) ( )2 2

+ + ++ + = +

m nSm n

2 2

2 2

72( ) 14413 13( )2

+ =

+. Du "=" xy ra m n m n m n2 2 2 29 4 4 9+ = + =

Vy: S 144min13

= khi m n= .

Cu 14. Trong mt phng vi h trc to Oxy, cho Hypebol (H) c phng trnh: x y2 2

116 9

- = .


Trang 26

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 F F1 2( 5;0); (5;0)- . HCN c s ca (H) c mt nh l M( 4; 3),

Gi s phng trnh chnh tc ca (E) c dng: x ya b

2 2

2 21+ = ( vi a > b)

(E) cng c hai tiu im F F a b2 2 21 2( 5;0); (5;0) 5 (1)- - =

M E a b a b2 2 2 2(4;3) ( ) 9 16 (2) + =

T (1) v (2) ta c h: a b aa b a b b

2 2 2 2

2 2 2 2 25 40

9 16 15

= + = + = =

. Vy (E): x y2 2

140 15

+ =

Cu 15. Trong mt phng vi h trc to Oxy , cho hypebol (H) c phng trnh x y2 2

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 cbx ay b13

+ = - - =

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

Cu 16. Trong mt phng vi h to Oxy, cho parabol (P): y x2 = v im I(0; 2). Tm to

hai im M, N (P) sao cho IM IN4=uuur uur

.

Gi M x y N x y0 0 1 1( ; ), ( ; ) l hai im thuc (P), khi ta c: x y x y2 2

0 0 1 1;= =

IM x y y y20 0 0 0( ; 2) ( ; 2)= - = -uuur

; IN y y y y IN y y2 21 1 1 1 1 1( ; 2) ( ; 2); 4 (4 ; 4 8)= - = - = -uur uur

Theo gi thit: IM IN4=uuur uur

, suy ra: y yy y

2 20 1

0 1

42 4 8

= - = -

y x y xy x y x

1 1 0 0

1 1 0 0

1 1; 2; 43 9; 6; 36

= = = - = = = = =

Vy, c 2 cp im cn tm: M N(4; 2), (1;1) hay M N(36;6), (9;3) .

Cu 17. Trong mt phng vi h to Oxy, cho parabol (P): y x2 8= . 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 x x1 2, .

Chng minh: AB = x x1 2 4+ + .

Theo cng thc tnh bk qua tiu: FA x1 2= + , FB x2 2= + AB FA FB x x1 2 4= + = + + .

Cu 18. Trong mt phng vi h to Oxy, cho Elip (E): x y2 25 5+ = , Parabol P x y2( ) : 10= . Hy vit phng trnh ng trn c tm thuc ng thng x y( ) : 3 6 0D + - = , 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 D nn: I b b(6 3 ; )- . Ta c: b b b

b bb b b

4 3 16 3 2

4 3 2 - = =

- - = - = - =

(C): x y2 2( 3) ( 1) 1- + - = hoc (C): x y2 2( 2) 4+ - =


Trang 27

TP 04: TAM GIC

Cu 1. Trong mt phng vi h to Oxy, cho DABC bit: B(2; 1), ng cao qua A c phng trnh d1: x y3 4 27 0+ = , phn gic trong gc C c phng trnh d2: x y2 5 0+ = . Tm to im A.

Phng trnh BC: x y2 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 y2 11 2- +

= x y2 5 0 - - =

+ To im I l nghim ca h: x y x Ix y y2 5 0 3 (3;1)

2 5 0 1 - - = = + - = =

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

x x xB

y y y'

'

2 4 (4;3)2 3

= - = = - =

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

+ To im A l nghim ca h: y x Ax y y

3 0 5 ( 5;3)3 4 27 0 3

- = = - - - + = =

Cu 2. Trong mt phng vi h to Oxy, cho tam gic ABC c ng cao AH, trung tuyn CM

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

-

v BD c phng trnh x y 5 0+ - = . Tm

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

Phng trnh AB: x y5 29 0+ - = . B = AB BD B(6; 1)- A 4 ;255

Cu 3. 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 y2 5 0+ - = , ng trung tuyn (AM): x y4 13 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 y9 22 9 1 2

- +=

- - + x y7 5 0+ + = .

Vit phng trnh ng thng Cx // AB (Cx): x y7 25 0+ - =

Cu 4. 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): x y3 4 0= .

PTTS ca d: x ty t4 3

= = - +

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

( )S AB AC A AB AC AB AC22 21 1. .sin . .

2 2= = -

uuur uuur = 3

2 t t24 4 1 3+ + = t

t2

1 = - =

C(2; 10) hoc C(1;1).


Trang 28

Cu 5. Trong mt phng Oxy, cho tam gic ABC bit A(2; 3), B(3; 2), c din tch bng 3

2 v

trng tm G thuc ng thng D : x y3 8 0= . Tm ta nh C.

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

-

. Phng trnh AB: x y 5 0- - = .

ABCS AB d C AB d C AB1 3 3. ( , ) ( , )2 2 2

= = = .

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

= t t(3 8) 5 12 2

- - -= t

t12

= =

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 B(2; 1) , (1; 2)- - , ABCS272

= , G x y: 2 0D + - = . S: C(18; 12)- hoc C( 9;15)-

Cu 6. Trong mt phng vi h to Oxy , cho ng thng d x y: 2 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 a a( 2 3; )- + d. Phng trnh ng thng AB x y: 3 5 0+ - = .

ABCS 2D = AB d C AB1 . ( , ) 22

=a 21 10. 2

2 10

- = a

a6

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 y: 2 1 0- - = , A(1; 0), B(3; -1) , ABCS 6= . S: C(7;3) hoc C( 5; 3)- - . Cu 7. 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: x y3 8 0- - = . Tm to im C.

V CH ^ AB, IK ^ AB. AB = 2 CH = ABCSAB

2 32

D = IK = CH1 13 2

= .

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

d I AB IK( , ) = a3 2 1- = aa

21

= =

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

+ Vi I(2; 2) C(1; 1) + Vi I(1; 5) C(2; 10). Cu 8. Trong mt phng vi h to Oxy, cho tam gic ABC c A B(1;0), (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 x y: 2 2 0+ - = . Gi s I t t d( ; ) C t t(2 1;2 )- .

Theo gi thit: ABCS AB d C AB1 . ( , ) 22D

= = t6 4 4- = t t 40;3

= = .

+ Vi t 0= C( 1;0)- + Vi t 43

= C 5 8;3 3

.

Cu 9. 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 y: 2 8 0+ - = . Lp phng trnh ng trn ngoi tip tam gic ABC.


Trang 29

Gi E l im i xng ca A qua d E BC. Tm c E(1;1) PT ng thng BC: x y4 3 1 0+ + = . C d BC= C( 2;5)- .

Phng trnh ng trn (ABC) c dng: x y ax by c a b c2 2 2 22 2 0; 0+ - - + = + - >

Ta c A, B, C (ABC) a b ca b c a b ca b c

4 10 291 5 996 10 34 ; ;2 8 4

8 6 25

- + = - -- - + = - = = = - + + = -

Vy phng trnh ng trn l: x y x y2 2 5 99 04 4

+ - - - = .

Cu 10. 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 x y2 1 0+ + = . Tm to nh C.

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

lm VTPT: AB x y( ) : 3 0- + = .

To im A l nghim ca h: x yx y

3 02 1 0

- + = + + =

A 4 5;3 3

-

.

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

-

.

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

y t

2 23

73

= - +

= +

Gi s C t t BC2 72 ; ( )3 3

- + +

.

Ta c: IB IC t t2 2 2 2

8 10 8 1023 3 3 3

= - + + = +

t loai v C B

t

0 ( )45

=

=

Vy: C 14 47;15 15

.

Cu 11. 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 y2 3 0+ - = , trng tm ca DABC 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 DABC CG CI23

=uuur uur

G

G

ax

by

2 13

2 13

-=

- =

Do G d nn a b2 1 2 1 2 03 3- -

+ - = To im I l nghim ca h:

a b

a b2 3 0

2 1 2 1 2 03 3

+ - =- - + - =

ab

51

= = -

I(5; 1)- .

Ta c A B AB

IA IB

, ( )5

2

= =

To cc im A, B l cc nghim ca h: x y

x y2 22 3 0

5( 5) ( 1)4

+ - = - + + =


Trang 30

x y

x y

14;236;2

= = -

= = -

A B1 34; , 6;2 2

- -

hoc A B3 16; , 4;

2 2

- -

.

Cu 12. Trong mt phng vi h to Oxy, cho im G(2;1) v hai ng thng

d x y1 : 2 7 0+ - = , d x y2 : 5 8 0+ - = . Tm to im B d C d1 2, sao cho tam gic ABC

nhn im G lm trng tm, bit A l giao im ca d d1 2, .


2 7 05 8 0

+ - = + - =

xy

13

= =

A(1;3) .

Gi s B b b d C c c d1 2(7 2 ; ) ; ( ;8 5 )- - .

V G l trng tm ca DABC nn:A B C

G

A B CG

x x xx

y y yy

3

3

+ +=

+ + =

b cb c2 2

5 8 - = - = -

bc

22

= =

.

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

Cu 13. Trong mt phng vi h to Oxy, cho tam gic ABC c A(2;1) . ng cao BH c phng trnh x y3 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 y( ) : 3 7 0- - = .

To im C l nghim ca h: x yx y

3 7 01 0

- - = + + =

C(4; 5)- .

Trung im M ca AB c: B BM Mx y

x y2 1

;2 2

+ += = . M CM( ) B B

x y2 11 0

2 2+ +

+ + = .

To im B l nghim ca h: B Bx y

x y3 7 0

2 11 0

2 2

- - =+ + + + =

B( 2; 3)- - .

To im H l nghim ca h: x yx y

3 7 03 7 0

- - = + - =

H 14 7;5 5

-

.

BH AC8 10 ; 2 105

= = ABCS AC BH1 1 8 10. .2 10. 162 2 5D

= = = (vdt).

Cu 14. 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- + = , x y3 4 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 b AB( ;2 ) ( )- , C c c CH( ; 2) ( )+ Trung im M ca BC: b c b cM 4;2 2

+ - +

.

V M thuc trung trc ca BC nn: b c b c3( ) 4(4 ) 4 0+ + - + - = b c7 12 0- + + = (1)

BC c b c b( ; )= - +uuur

l 1 VTPT ca trung trc BC nn c b c b4( ) 3( )- = + c b7= (2)

T (1) v (2) c b7 1,4 4

= - = - . Vy B C1 9 7 1; , ;4 4 4 4

- -

.


Trang 31

Cu 15. Trong mt phng Oxy, cho tam gic ABC cn ti A( 1;4)- v cc nh B, C thuc ng thng x y: 4 0D - - = . 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 D H 7 1;2 2

-

AH 9

2=

Theo gi thit: ABCS BC AH BC118 . 18 4 22D

= = = HB HC 2 2= = .

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

x y2 24 0

7 1 82 2

- - =

- + + =

x y

x y

11 3;2 23 5;2 2

= =

= = -

Vy B C11 3 3 5; , ;2 2 2 2

-

hoc B C3 5 11 3; , ;

2 2 2 2

-

.

Cu 16. Trong mt phng vi h trc ta Oxy, cho hai ng thng d1: x y 5 0+ + = , d2:

x y2 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 DABC nn m nm n

2 7 2 3.23 5 3.0

+ + - = - - + =

mn

11

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

PT ng trn ngoi tip DABC: x y x y2 2 83 17 338 027 9 27

+ - + - =

Cu 17. 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 d x y1 : 2 13 0- + = v

d x y2 : 6 13 29 0- + = . Vit phng trnh ng trn ngoi tip tam gic ABC .

ng cao CH : x y2 13 0- + = , trung tuyn CM : x y6 13 29 0- + = C( 7; 1) - - PT ng thng AB: x y2 16 0+ - = . M CM AB= M(6;5) B(8;4) .

Gi s phng trnh ng trn (C) ngoi tip ABC x y mx ny p2 2: 0.D + + + + =

V A, B, C (C) nn m n pm n pm n p

52 4 6 080 8 4 050 7 0

+ + + = + + + = - - + =

mnp

46

72

= - = = -

.

Suy ra PT ng trn: x y x y2 2 4 6 72 0+ - + - = . Cu 18. 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 d x y1 : 5 0 + + = v d x y2 : 2 7 0+ = . Vit phng trnh ng trn c tm C v tip xc vi ng thng BG.

Gi s B b b d C c c d1 2( 5 ; ) ; (7 2 ; )- - - .

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

x xy y

2 63 0

+ + = + + =

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

Phng trnh BG: x y4 3 8 0= . Bn knh R d C BG 9( , )5

= =

Phng trnh ng trn: x y2 2 81( 5) ( 1)25

+ =


Trang 32

Cu 19. Trong mt phng vi h to Oxy, cho tam gic ABC c A( 3;6)- , trc tm H(2;1) ,

trng tm G 4 7;3 3

. Xc nh to cc nh B v C.

Gi I l trung im ca BC. Ta c AG AI I2 7 1;3 2 2

=

uuur uur

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 x y(7 ;1 )- - v B Bx y 3 0- - = .

H l trc tm ca tam gic ABC nn CH AB^ ; B B B BCH x y AB x y( 5 ; ), ( 3; 6)= - + = + -uuur uuur

B B B BB B B B B

x y x xCH AB

x x y y y3 1 6. 0

( 5)( 3) ( 6) 0 2 3 - = = =

= - + + - = = - =

uuur uuur

Vy ( ) ( )B C1; 2 , 6;3- hoc ( ) ( )B C6;3 , 1; 2- Cu 20. 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 x y: 2 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:x yx y

2 5 01 0

+ + =

+ + = x

y4

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 y2 5 0- - = .

Gi I d BN( )= . Gii h: x y

x y2 5 0

2 5 0 + + =

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

+ Phng trnh BC: x y7 25 0+ + = . Gii h: BC x yCH x y

: 7 25 0: 1 0

+ + =

- + =

C 13 9;4 4

- -

.

+ BC2 2

13 9 4504 34 4 4

= - + + + =

, d A BC

2 2

7.1 1( 2) 25( ; ) 3 2

7 1

+ - += =

+.

Suy ra: ABCS d A BC BC1 1 450 45( ; ). .3 2. .2 2 4 4

= = =

Cu 21. Trong mt phng vi h to Oxy, cho ABCD , vi nh A(1; 3) phng trnh ng phn

gic trong BD: x y 2 0+ - = v phng trnh ng trung tuyn CE: x y8 7 0+ - = . Tm to cc nh B, C.

Gi E l trung im ca AB. Gi s B b b BD( ;2 )- b bE CE1 1;2 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 y2 7 0+ - = ; x yC CE BC Cx y

8 7 0: (7;0)2 7 0

+ - == + - =.

Cu 22. 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 d x y1 : 1 0+ - = v

d x y2 : 3 9 0- - = . Tm ta cc nh B, C ca tam gic ABC.


Trang 33

Gi C c c d2( ;3 9)- v M l trung im ca BC M m m d1( ;1 )- .

B m c m c(2 ;11 2 3 )- - - . Gi I l trung im ca AB, ta c m c m cI 2 3 7 2 3;2 2

- + - -

.

V I d2( ) nn m c m c2 3 7 2 33. 9 0

2 2- + - -

- - = m 2= M(2; 1)-

Phng trnh BC: x y 3 0- - = . C BC d C B2 (3;0) (1; 2)= - . Cu 23. 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 m BC( ; 4 )- - C m m( 4 ; )- -

CE m m , AB m m(5 ; 3 ) ( 6; 10 )= + - - = - - -uuur uuur

.

V CE AB^ nn AB CE m m m m. 0 ( 6)( 5) ( 3)( 10) 0= - + + + + =uuur uuur

m m0; 6= = - . Vy: B C(0; 4), ( 4;0)- - hoc B C( 6;2), (2; 6)- - . Cu 24. Trong mt phng vi h ta Oxy, cho tam gic ABC c nh A(2;4) . ng thng D

qua trung im ca cnh AB v AC c phng trnh x y4 6 9 0- + = ; trung im ca cnh BC nm trn ng thng d c phng trnh: x y2 2 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 D, ta tnh c A 40 31' ;13 13

BC x y: 2 3 1 0- + =

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

.

Gi s tC t BC3 1; ( )2

-

. Ta c ABCS d A BC BC BC BC

1 7 1 7( ; ). . 132 2 2 13D

= = =

CM 132

=t t Ct

t C loai

223 6 13 3 (4;3)( 2)

1 (1;1) ( )2 2 - = + - = =

B(1;1) .

Vy: B(1;1) , C(4;3) . Cu 25. Trong mt phng vi h ta Oxy, cho DABC 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 d1 : 2x 5y + 3 = 0 v

d2 : x + y 5 = 0. Tm ta cc nh A v C ca tam gic ABC.

Gi M l trung im AB th M d2 nn M a a( ;5 )- . nh A d1 nn bA b5 3;2

-

.

M l trung im AB: A B MA B M

x x xy y y

22

+ = + =

a b aa b b

4 5 3 22 5 1

- = = + = = A(1; 1).

Phng trnh BC: x y5 2 25 0+ - = ; C d BC2= C(5; 0).

Cu 26. Trong mt phng to vi h to Oxy, cho ABCD vi AB 5,= nh C( 1; 1)- - , phng trnh cnh AB x y: 2 3 0+ - = v trng tm G ca ABCD thuc ng thng


Trang 34

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 DABC

G

G

xxCG CI

yy

2 12 3

2 133

-=

= - =

uuur uur

G d x y: 2 0 + - = nn c: G Gx y 2 0+ - = x y2 1 2 1 2 03 3- -

+ - =

Ta im I tha mn h: x y

Ix y2 3 0

(5; 1)2 1 2 1 2 03 3

+ - = -- - + - =

Gi A A A AABA x y IA x y

22 2 2 5( ; ) ( 5) ( 1)

2 4

= - + + = =

.

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

( ) ( )A A A A

A A A A

x y x x

x y y y2 2

2 3 0 4 65 1 35 14 2 2

+ - = = = - + + = = - = -

Vy: A B1 34, , 6;2 2

- -

hoc B A1 34, , 6;

2 2

- -

.

Cu 27. 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 x y: 3 2 0- + = . To im C khng tho mn phng trnh cnh huyn nn DABC vung cn ti C. Gi I

l trung im ca AB . Phng trnh ng thng CI: x y3 0+ = .

I CI AB= I 3 1;5 5

-

AI BI CI 72

5= = =

Ta c: A B d

AI BI

,725

= =

x y

x y2 2

3 2 0

3 1 725 5 5

- + = + + - =

x y

x y

3 19;5 5

9 17;5 5

= =

= - = -

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

- -

.

Cu 28. Trong mt phng vi h to Oxy, cho im C(2; 5) v ng thng D c phng trnh:

x y3 4 4 0- + = . Tm trn D hai im A v B i xng nhau qua I 52;2

sao cho din tch

tam gic ABC bng 15.

Gi a aA a B a3 4 16 3; 4 ;4 4

D + -

-

ABCS AB d C AB1 . ( , ) 32

D= = AB = 5.

a aAB aa

22 6 3 45 (4 2 ) 25

02 - == - + = =

. Vy hai im cn tm l A(0; 1) v B(4; 4).

Cu 29. Trong mt phng vi h trc ta Oxy cho tam gic ABC vi B(1; 2)- ng cao


Trang 35

AH x y: 3 0- + = . Tm ta cc nh A, C ca tam gic ABC bit C thuc ng thng d x y:2 1 0+ - = v din tch tam gic ABC bng 1.

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

Gi A x y AH x y0 0 0 0( ; ) 3 0 - + = (1); x y

BC AH d A BC 0 01

2, ( , )2

+ += = =

ABCx y x y

S AH BCx y

0 0 0 0

0 0

1 1 2 (2)1 1. 1 . . 2 11 2 (3)2 2 2D

+ + + + == = = + + = -

T (1) v (2) x Ay

0

0

1 ( 1;2)2

= - - =

. T (1) v (3) x Ay

0

0

3 ( 3;0)0

= - - =

Cu 30. 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 C c b c( ;0), (0; ), ( , 0) .

DABC vung ti A AB AC. 0=uuur uuur

c b2 5 0= - + b 502

.

ABCS AB AC1 .2D

= = b c b b b2 2 2 2 21 ( 2) 1. 2 ( 1) ( 2) 1 4 52

- + + - = - + = - +

Do b 502

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

Cu 31. 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 x y: 3 2 4 0+ - = . Vit phng trnh ng trn ngoi tip tam gic ABC.

Gi M l trung im ca BC AM AG32

=uuur uuur

M 13 3;2 2

-

.

AB d^ AB nhn du (2; 3)= -r lm VTPT Phng trnh AB x y: 2 3 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 DABC c dng: x y ax by c2 2 2 2 0+ + + + = ( a b c2 2 0+ - > ).

Khi ta c h: a b ca b ca b c

2 6 1010 2 2616 8 80

+ - =+ + = -

- + = -

a

b

c

7421

237

83

=

= -

=

. Vy: C x y x y2 2 148 46 8( ) : 021 7 3

+ - + + =

Cu 32. 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: x y4 14 0+ + = ; x y2 5 2 0+ - = . Tm ta cc nh A, B, C.

A(4, 2), B(3, 2), C(1, 0) Cu 33. Trong mt phng vi h to Oxy, cho tam gic ABC c trc tm H( 1;6)- , cc im

M N(2;2) (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+ + = .


Trang 36

Gi s C a a CH( ;5 )- CN a a(1 ; 4)= - -uuur

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

V N l trung im ca BC nn B a a(2 ; 3)- -

V H l trc tm DABC nn: AH CN. 0=uuur uuur

a a a a( 5)(1 ) (7 )( 4) 0- - + - - = a

a

3112

=

=

.

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

+ Vi a 112

= C A B11 1 3 9 7 5; , ; , ;2 2 2 2 2 2

- - -

Cu 34. 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 y2 5 0- + = . im M(3;0) thuc on AC tho mn AB AM2= . 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 x y( ) : 2 3 0+ - = .

To im A l nghim ca h: x yx y2 3 0

2 0 + - = + - =

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

Do AB AM2= nn E l trung im ca AB B(3; 3)- .

To im C l nghim ca h: x yx y

2 3 02 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 x y( ) : 2 3 0+ + = , M(0; 1)- . S: A(1;1) ; B( 3; 1)- - ;C 1 ; 22

- -

Cu 35. Trong mt phng vi h to Oxy, cho tam gic ABC cn ti A, ng thng BC c

phng trnh x y2 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 yx y

2 2 04 0

+ - = - + =

B( 2;2)- .

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

x yx y

4 02 1 0

- + = + + =

N( 3;1)- .

Gi I l trung im ca MN I 12;2

-

. Gi E l trung im ca BC IE l ng trung

trc ca BC IE x y: 4 2 9 0- + = .

To im E l nghim ca h: x yx y

2 2 04 2 9 0

+ - = - + =

E 7 17;5 10

-

C 4 7;

5 5

-

.

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

+ - = .

To nh A l nghim ca h: x y

x y

4 2 9 03 05

- + = + - =

A 13 19;10 10

-

.


Trang 37

Vy: A 13 19;10 10

-

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

5 5

-

.

Cu 36. Trong mt phng vi h to Oxy, cho tam gic ABC c nh A thuc ng thng d:

x y 4 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 : xx y A

y x y

22 24 2 0 3 ;

2 3 33

= - - - = - - = = -

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

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

= +

x y xBH BC B Bx yy

3 0 4: ( 4;1)12

4

+ + = = - = - == +

Cu 37. Trong mt phng vi h ta Oxy, cho tam gic ABC c ng cao

BH x y: 3 4 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 x y: 4 3 1 0- - =

A AC AD A(4;5)= . AB i qua M, A PT cnh AB x y: 3 4 8 0- + = B 13;4

- -

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

.

Kim tra iu kin B, C khc pha vi AD, ta c c hai im trn u tha mn. Cu 38. 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: x y2 6 3 0+ + = . Tm to cc nh A, B, C.


2 02 6 3 0

+ - = + + =

A 15 7;4 4

-

.

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

6 - -

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

b c

cb

12

3 226 1

2

+= -

- - - +

=

b

c

14

94

=

= -

B 1 7;4 4

, C 9 1;4 4

-

.

Cu 39. Trong mt phng vi h ta Oxy, cho ABCD 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 x: 3 7( 1)= - . Bit chu


Trang 38

vi ca ABCD bng 18, tm ta cc nh A, B, C.

B AB Ox B(1;0)= , ( )A AB A a a a;3 7( 1) 1 - > (do A Ax y0, 0> > ). Gi AH l ng cao ABC H a C a BC a AB AC a( ;0) (2 1;0) 2( 1), 8( 1)D - = - = = - .

( )Chu vi ABC a C A18 2 (3;0), 2;3 7D = = . Cu 40. 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 x y4 3 4 0+ = ; x y 1 0= . Phn gic trong ca gc A nm trn ng thng x y2 6 0+ = . Tm ta cc nh ca tam gic ABC.

Ta ca A nghim ng h phng trnh: x y x Ax y y4 3 4 0 2 ( 2;4)

2 6 0 4 + - = = - - + - = =

Ta ca B nghim ng h phng trnh ( )x y x Bx y y4 3 4 0 1 1;0

1 0 0 + - = = - - = =

Phng trnh AC qua im A(2;4) c dng: a x b y ax by a b( 2) ( 4) 0 2 4 0+ + - = + + - = Gi x y x y ax by a b1 2 3: 4 3 4 0; : 2 6 0; : 2 4 0D D D+ - = + - = + + - =

T gi thit suy ra ( ) ( )2 3 1 2; ;D D D D= .

Do a b

a b2 3 1 2 2 2

1. 2. 4.1 2.3cos( ; ) cos( ; )25. 55.

D D D D + += =+

aa b a b a a ba b

2 2 02 2 (3 4 ) 03 4 0

= + = + - = - =

a = 0 b 0 . Do y3 : 4 0D - = 3a 4b = 0: Chn a = 4 th b = 3. Suy ra x y3 : 4 3 4 0D + - = (trng vi 1D ). Do vy, phng trnh ca ng thng AC l y 4 = 0.

Ta ca C nghim ng h phng trnh: y x Cx y y

4 0 5 (5;4)1 0 4

- = = - - = =

Cu 41. 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 c( ; 2 3)+ v I m m( ;6 )- l trung im ca BC. Suy ra: B m c m c(2 ; 9 2 2 )- - - .

V C l trung im ca AB nn: m c m cC CC2 5 11 2 2' ; '2 2

- + - -

nn m c m c m2 5 11 2 2 52 3 02 2 6

- + - -- + = = -

I 5 41;

6 6

-

.

Phng trnh BC: x y3 3 23 0- + = C 14 37;3 3

B 19 4;3 3

-

.

Cu 42. 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 M 5 5;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 DABC ta c : GH GI2= -uuur uur

G 4 ;23


Trang 39

Mt khc v GA GM2= -uuur uuur

nn A( 1;1)- . Phng trnh BC: x y3 10 0+ - = . ng trn (C)

ngoi tip D c tm I(1; 2) v bn knh R 4 1 5= + = . Do (C) : x y2 2( 1) ( 2) 5- + - = .

Khi to B ;C l nghim h : x xx yy yx y

2 2 2 3( 1) ( 2) 54 13 10 0

= =- + - = = =+ - =

V B Cx x> nn B(3;1) ; C(2;4). Vy : A(1; 1); B(3; 1) ; C(2; 4). Cu 43. 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 y2 0- = , im I(4; 2) l trung im ca AB, im M 94;2

thuc

cnh BC. Tm ta cc nh A, B, C bit tung im B ln hn hoc bng 3. Gi s B BB y y AB(2 ; ) B BA y y(8 2 ;4 )- - . Phng trnh CI: x y2 10 0+ - = .

Gi C CC x x( ;10 2 )- CCI x5 4= -uur

; BAB y20 2= -uuur

.

ABC B C C BS CI AB y x x y1 . 10 4 2 8 22

= = + - - = C B B CC B B C

x y y xx y y x

4 2 6 (1)4 2 10 (2)

- - = - - - = -

V ( )C B

C B

x k yM BC CM kMB

x k y

4 2 411 922 2

- = - = - + = -

uuur uuur C B B Cx y y x2 6 5 16 0 - - + = (3)

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

x y y x yx y y x y

4 2 6 1 22 6 5 16 0 1 2

- - = - = - - - - + = = - + (loi, v By 3 )

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

x y y x yxx y y x

4 2 10 322 6 5 16 0

- - = - = =- - + =

(tho)

Vy A(2; 1), B(6; 3), C(2; 6). Cu 44. 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: x y3 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 d a( ;2) ,( 2) , C c c BC c( ;2 3) ,( 0)+ .

AB a AC c a c BC c c( ;0), ( ; 3), ( ; 3)= - = - =uuur uuur uuur

AB a AC c a c BC c2 2, ( ) 3 , 2= = - + =

DABC vung A v r 3= AB ACS pr

. 0 = =

uuur uuur

AB ACAB BC ACAB AC

. 01 . . 32 2

= + +

=

uuur uuur

( )a c a

a c a c a c c a c2 2 2 2( ) 0

( ) 3 2 ( ) 3 3

- - =

- + = + + - + c a

a0

3 3 = = +

c a A Cc a A C

3 3 (3 3;2), (3 3;5 3 3)3 3 ( 3 3;2), ( 3 3; 1 3 3)

= = + + + +

= = - - - - - - - -

Cu 45. Trong mt phng vi h to Oxy, tm to cc nh ca tam gic vung cn ABC, c

phng trnh hai cnh AB x y: 2 1 0- + = , AC x y: 2 3 0+ - = v cnh BC cha im

I 8;13

.


Trang 40

Ta c: AB ^ AC DABC 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 y x y

x y x y

x y

2 1 2 35 5

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

- + + -=

- + - + > + - = + - + - >

BCAt BC n (3; 1)^ = -r BC x y: 3 7 0 - - = ; x yB AB BC B

x y2 1 0: (3;2)

3 7 0 - + == - - =

;

x yC AC BC Cx y

2 3 0: (2; 1)3 7 0

+ - == - - - =.

Cu 46. 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 DABC 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 b AC c( 4; 2), ( 3; 4)= - - = - -

uuur uuur.

Ta c: AB ACBC2

. 050

=

=

uuur uuur b c

b c5, 0

1, 6 = = = - =

A B CA B C

(3;2), ( 1;5), (0; 2)(3;2), ( 1; 1), (6; 2)

- - - - -

.

Cu 47. 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 G 14 5;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- - = H 5 1;2 2

-

C(9;6) .

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

= - - = - -

uuur uuur

ABCaS AB CH a aa

265 1 65 0. 8 40 052 2 2

== = - = =V

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

PT ng trn (C) ngoi tip DABC c dng: x y ax by c a b c2 2 2 22 2 0 ( 0)+ + + + = + - >

(C) qua A, B, C nn

ab c

a b c ba b c

c

137264 4 5910 6 342618 12 117 6613

-=

+ = - - - + = - = + + = - =

C x y x y2 2 137 59 66( ) : 013 13 13

+ - - + =


Trang 41

Cu 48. Trong mt phng to Oxy, cho D ABC c phng trnh cnh AB: x y 3 0+ = , phng

trnh cnh AC: x y3 7 0 + = v trng tm G 12;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 m C n n( ;3 ), ( ;7 3 ) .

G 12;3

l trng tm DABC nn: m n mm n n

2 6 11 3 7 3 1 3

+ + = = + - + - = = B(1; 2), C(3; 2)

H l trc tm DABC AH BCBH AC

^

^

uuur uuuruuur uuur H(10;5) .

PT ng trn (S) qua B, C, H c dng: x y ax by c a b c2 2 2 22 2 0 ( 0)+ + + + = + - >

Do B, C, H (S) a b c aa b c b

a b c c

2 4 5 66 4 13 220 10 125 15

+ + = - = - - + = - = - + + = - =

. Vy (S): x y x y2 2 12 4 15 0+ + = .

Cu 49. Trong mt phng vi h to Oxy, cho im A(0; 2) v ng thng d: x y2 2 0+ = .

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

B 2 6;5 5

; C C1 24 7(0;1); ;5 5

Cu 50. Trong mt phng vi h to Oxy, cho tam gic ABC vung cn ngoi tip ng trn

C x y2 2( ) : 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 DABC 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 51. 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)=

uuur VTPT l

BHn BH x y(1; 1) : 2 0= - - - =r

+ AC vung gc vi BH nn AC BHn u AC x y(1;1) : 4 0= = + - =r r

+ AC vung gc vi CD nn DC ACn u DC x y(1; 1) : 6 0= = - - - =r r .

+ C l giao ca AC v DC nn ta C l nghim ca h: x y Cx y

4 0 (5; 1)6 0

+ - = - - - =

+ M l trung im ca BC

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