Tam Giac Trong Khao Sat Ham So

Chuyn : TAM GIC TRONG KHO ST HM S

TAM GIC TRONG KHO ST HM STrong mt s bi ton kho st hm s ta hay gp cc trng hp tm tham s sao cho ba im no to thnh mt tam gic c tnh cht c bit v d nh : Tam gic vung , cn , u hay vung cn ....

gip cc em c cch lm tng qut , ta tm ngin cu mt s dng gp trong cc thi i hc hoc thi th tp hp li thng qua mt s v d tiu biu v mt s bi tp ngh cc em v nh t gii .

Trong bi vit ny nu c g cha ng mong cc em ng gp ti sa cha li kha sau hon thin hn . Sau y l ni dung bao gm 5 dng thng gp. .Dng 1. BA IM CC TR TO THNH TAM GICBi ton : Cho hm s y=f(x;m) , tm m hm s cho c 3 im cc tr to thnh mt tam gic : vung , cn , u ...Cch gii

Bc 1. Tm iu kin (*) cho m hm s c 3 im cc tr .

Bc 2. Tm ta 3 im cc tr

Bc 3. Da vo gi thit cho tam gic l tam gic g ? t ta p dng tnh cht ca tam gic thit lp cc phng trnh c lin quan n tham s m

Bc 4. Gii cc phng trnh lp c suy ra tham s m

Bc 5. Kim tra cc gi tr m tm c vi iu kin (*) chn m ph hp .

MT S V D MINH HAV d 1. ( DB-2004 ). Cho hm s (1)

1. Kho st s bin thin v v th (C) vi m=1

2. Tm m d hm s (1) c ba im cc tr l ba nh ca mt tam gic vung cn .

Gii1. Hc sinh t v th (C)

2. Ta c :

- Vi iu kin (*) th hm s (1) c ba im cc tr . Gi ba im cc tr l :

. Do nu ba im cc tr to thnh mt tam gic vung cn , th nh s l A .

- Do tnh cht ca hm s trng phng , tam gic ABC l tam gic cn ri , cho nn tha mn iu kin tam gic l vung , th AB vung gc vi AC.

Tam gic ABC vung khi :

Vy vi m = -1 v m = 1 th tha mn yu cu bi ton .

* Ta cn c cch khc

- Tam gic ABC l tam gic vung khi trung im I ca BC : AI = IB , vi

. Hay

V d 2: Cho hm s (1)

1.Kho st s bin thin v v th (C) ca hm s (1) khi m = 1

2.Tm cc gi tr ca tham s m thi hm s (1) c ba im cc tr v ng trn i qua ba im ny c bn knh bng 1.

Gii1. Hc sinh t v th (C).

2. Ta c

- Hm s c 3 cc tr y i du 3 ln phng trnh y = 0 c 3 nghim phn bit m > 0Khi m > 0 , th hm s (1) c 3 im cc tr l

EMBED Equation.3 - Gi I l tm v R l bn knh ca ng trn i qua 3 im A, B, C.

V 2 im A, B i xng qua trc tung nn I nm trn trc tung.

t I(0; y0). Ta c: IC = R

hoc

* Vi

IA = R

So snh iu kin m > 0, ta c m = 1 v m =

* Vi I(0; 2)

IA = R (*)

Phng trnh (*) v nghim khi m > 0

Vy bi ton tha mn khi m = 1 v m =

BI TP T LUYNBi 1. . Cho hm s (1) , vi m l tham s thc.

1. Kho st s bin thin v v th hm s (1) khi .

2. Xc nh hm s (1) c ba im cc tr, ng thi cc im cc tr ca th to thnh mt tam gic c din tch bng .Bi 2. . Cho hm s (1), trong m l tham s thc.1. Kho st s bin thin v v th ca hm s (1) khi m = 1.

2. Tm gi tr ca tham s m hm s (1) c ba im cc tr l ba nh ca mt tam gic c din tch bng 32.

Bi 3. . Cho hm s (1) , vi l tham s thc.

1. Kho st s bin thin v v th hm s (1) khi .

2. Xc nh m hm s (1) c ba im cc tr, ng thi cc im cc tr ca th to thnh mt tam gic c gc bng 1200.

Bi 4. Cho hm s y = x4 2m2x2 + 1, (1)

a) Kho st v v th ca hm s.

b) Tm m th hm s (1) c ba im cc tr A, B, C v din tch ca tam gic ABC bng 32.

Bi 5. . Cho hm s y = x4 2m2x2 + 1 (1) 1. Kho st hm s (1) khi m = 1 2. Tm m th h/s (1) c 3 im cc tr l 3 nh ca mt tam gic vung cn

Bi 6. . Cho hm s c th (Cm) vi m l tham s .

1. Kho st s bin thin v v th (C).ca hm s khi m = 0 .

2. Chng minh rng vi mi gi tr ca m tam gic c ba nh l ba im cc tr ca th () l mt tam gic vung cn.

Bi 7. . Cho hm s .

1. Kho st v v th ca hm s khi m = 1.

2.Tm m th hm s c cc im cc i v cc tiu to thnh mt tam gic vung cn.

Bi 8. . Cho hm s y = x4 2mx2 + m 1 . (1)

1. Kho st v v v th hm s khi m = 1.

2. Xc nh m hm s (1) c ba cc tr, ng thi cc im cc tr ca th to thnh mt tam gic c bn knh ng trn ngoi tip bng 1.

Bi 9. . Cho hm s (1) vi m l tham s thc.1. Kho st s bin thin v v th ca hm s (1) khi m = ( 1.2 nh m th ca hm s (1) c ba im cc tr l ba nh ca mt tam gic vung.

Bi 10. Cho hm s (1)1. Kho st s bin thin v v th hm s (1) khi

2. Xc nh m hm s (1) c cc i v cc tiu, ng thi cc im cc i v cc tiu ca th hm s (1) lp thnh mt tam gic u.Dng 2. HAI IM CC TR V MT IM KHC TO THNH TAM GICBi ton : Cho hm s y=f(x;m) v mt im c nh . Tm tham s m hm s c 2 im cc tr cng vi M to thnh mt tam gic : Vung , cn , u ....Cch gii :Bc 1. Tm iu kin hm s cho c 2 im cc tr ( gi l iu kin (*) )

Bc 2. Tm ta ca hai im cc tr

Bc 3. Da vo gi thit cho tam gic c c im g ? da vo thit lp cc phng trnh c lin quan n tham s m

Bc 4. Gii cc phng trnh lp c tm m , sau kim tra vi iu kin (*) chn m ph hp .

MT S V D MINH HAV d 1. Cho hm s

1. Kho st v v th (1) vi m=1

2. Tm m hm s (1) c cc i , cc tiu , ng thi cc im cc i v cc tiu cng vi gc ta O to thnh mt tam gic vung ti O.

Gii1. Hc sinh t v th .

2. Ta c :

- hm s c cc i , cc tiu th : =0 c hai nghim phn bit

- Vi iu kin (*) hm s c cc i , cc tiu .Gi l hai im cc i ,cc tiu ca hm s . Nu A, B cng vi O to thnh tam gic vung ti O th OA vung gc vi OB :

- Ta c :

- Bng php chia phng trnh hm s cho o hm ca n , ta c :

- Phng trnh ng thng i qua hai im cc tr :

- Do :

- p dng Vi-t cho (1) , thay vo :

- Vy :

Hay :

Kt lun : Vi m tha mn (*) th hai im cc i , cc tiu ca hm s cng vi O to thnh tam gic vung ti O .

V d 2. Cho hm s

1. Kho st s bin thin v v th (C) vi m = 1 .

2. Tm m hm s c cc i , cc tiu , ng thi cc im cc i v cc tiu cng vi gc ta O to thnh mt tam gic c din tch bng 4 .Gii1. Hc sinh t v th (C).

2. hm s c cc i , cc tiu th phng trnh y = 0 c hai nghim phn bit

x2 2x + (1 m) = 0 c 2 nghim phn bit 1 (1 m) > 0m > 0 (*)

- Vi iu kin (*), hm s c C, CT . Gi l hai im cc tr . Vi l hai nghim ca phng trnh = 0 (1) .

- Bng php chia phng trnh hm s cho o hm ca n , ta c :

. Suy ra phng trnh ng thng i qua hai im cc tr l d : y = -2mx + 2m + 2 ..

- Ta c :

- Gi H l hnh chiu vung gc ca O trn (AB), h l khong cch t O n AB th :

-

- Theo gi thit :

Kt lun : vi m = 1 tha mn yu cu bi ton .

BI TP T LUYNBi 1. Cho hm s y = x3 3x2 + m , (1).

1. Kho st v v th hm s khi m = 2.

2. Tm m hm s (1) c cc i v cc tiu , ng thi cc im cc i, cc tiu v gc ta O to thnh mt tam gic c din tch bng 4.

Bi 2. Cho hm s

EMBED Equation.DSMT4 (1)

1.Kho st s bin thin v v th ca hm s (1) ng vi m=1

2.Tm m hm s (1) c cc tr v cc im cc tr vi gc ta to thnh mt tam gic vung ti O

Bi 3. Cho hm s (1)

1. Kho st s bin thin v v th ca hm s (1) .

2. Gi ln lt l cc im cc i, cc tiu ca th hm s (1). Tm im M thuc trc honh sao cho tam gic MAB c din tch bng 2.

Bi 4. Cho hm s (1), vi m l tham s thc.

1. Kho st s bin thin v v th ca hm s (1) khi .

2. Tm m hm s (1) c cc i v cc tiu, ng thi cc im cc tr ca th cng vi gc to O to thnh mt tam gic vung ti O.

Bi 5. Cho hm s (1) vi m l tham s thc.1. Kho st s bin thin v v th ca hm s (1) khi m = 0.2.nh m hm s (1) c cc tr, ng thi ng thng i qua hai im cc tr ca th hm s to vi hai trc ta mt tam gic cn.Dng 3. GIAO IM CA CC TH VI MT IM KHC TO THNH TAM GIC .Bi ton . Cho 2 hm s y=f(x) v y=g(x;m) ? Tm tham s m f(x) ct g(x) ti 2 im A,B cng vi im C ( c nh ) to thnh mt tam gic c bit Cch gii

Bc 1. Tm iu kin (*) phng trnh f(x)=g(x;m) (1) c 2 nghim phn bit

Bc 2. Tm ta 2 im A,B c honh l 2 nghim ca (1)

Bc 3. Da vo gi thit cho tam gic c bit l tam gic g? wr thit lp cc phng trnh c lin quan n m

Bc 4. Gii cc phng trnh lp suy ra m , sau kim tra vi iu kin (*) chn m ph hp

MT S V D MINH HAV d 1.Cho hm s

1. Kho st s bin thin v v th (C)

2. Gi d l ng thng i qua im A(- 1; 0) vi h s gc l k ( k thuc R). Tm k ng thng d ct (C) ti ba im phn bit v hai giao im B, C ( B, C khc A ) cng vi gc ta O to thnh mt tam gic c din tch bng 1.Gii1. Hc sinh t v th (C)

2. ng thng d i qua A(-1; 0) vi h s gc l k , c phng trnh l : y = k(x+1) = kx+ k .- Nu d ct (C) ti ba im phn bit th phng trnh: x3 3x2 + 4 = kx + k

x3 3x2 kx + 4 k = 0 (x + 1)( x2 4x + 4 k ) = 0

EMBED Equation.3 c ba nghim phn bit g(x) = x2 4x + 4 k = 0 c hai nghim phn bit khc - 1

Vi iu kin : (*) th d ct (C) ti ba im phn bit A, B, C .Vi A(-1;0) , do B,C c honh l hai nghim ca phng trnh g(x) = 0.

- Gi vi l hai nghim ca phng trnh : . Cn .

- Ta c :

- Khong cch t O n ng thng d :

- Vy theo gi thit :

V d 2. Cho hm s

1. Kho st s bin thin v v th (H) vi m = 1

2. Tm m ng thng d : 2x + 2y - 1= 0 ct ti hai im phn bit A, B sao cho tam gic OAB c din tch bng .

Gii1. Hc sinh t v th (H).

2. ng thng d vit li : . d ct ti hai im phn bit A, B th phng trnh: c hai nghim phn bit khc - 2 (*)- Gi

- Khong cch t O n d l h , th :

- Theo gi thit :

Hay : , tha mn iu kin (*) .

- p s : m = .V d 3. Cho hm s


2.Tm tham s m ng thng d : y = - 2x + m ct th ti hai im phn bit A, B sao cho din tch tam gic OAB bng .

Gii1. Hc sinh t v th (C) .

2. Nu d ct (C) ti hai im phn bit A, B th phng trnh:

c hai nghim phn bit khc -1

. Chng t vi mi m d lun ct (C) ti hai im phn bit A,B- Gi : . Vi : l hai nghim ca phng trnh (1)

- Ta c : .

- Gi H l hnh chiu vung gc ca O trn d , th khong cch t O n d l h :

- Theo gi thit :

Vy :

Vi m tha mn iu kin (*) th d ct (C) ti A, B tha mn yu cu bi ton .V d 4 . Cho hm s (1)

1. Kho st s bin thin v v th (C) vi m = 2 .

2. Tm m ng thng d : y = x + 4 ct th hm s (1) ti ba im phn bit A, B, C sao cho tam gic MBC c din tch bng 4 . ( im B, C c honh khc khng ; M(1;3) ).Gii1. Hc sinh t v th (C)

2. th (1) ct d ti ba im A,B,C c honh l nghim ca phng trnh :

Vi m tha mn (*) th d ct (1) ti ba im A(0; 4) , cn hai im B,C c honh l hai nghim ca phng trnh :

- Ta c :

-Gi H l hnh chiu vung gc ca M trn d . h l khong cch t M n d th :

- Theo gi thit : S = 4

Kt lun : vi m tha mn : ( chn ).Bi 5 . Cho hm s (1), m l tham s thc

1.Kho st s bin thin v v th hm s khi .

2.Tm m th hm s ct ng thng ti 3 im phn bit ; B; C sao cho tam gic c din tch , vi

Gii2. Phng trnh honh giao im ca th vi l:

ng thng ct d th hm s (1) ti ba im phn bit A(0;2), B, C

Phng trnh (2) c hai nghim phn bit khc 0.

Gi v , trong l nghim ca (2); v

Ta c

M =

Suy ra =16(tho mn) hoc (tho mn)BI TP T LUYNBi 1. Cho hm s y = x3 + 2mx2 + (m + 3)x + 4, c th (Cm).

1. Kho st v v th ca hm s khi m = 1.

2. Cho d l ng thng c phng trnh y = x + 4 v im K(1 ; 3). Tm m d ct (Cm) ti ba im phn bit A(0 ; 4), B, Csao cho tam gic KBC c din tch bng .

Bi 2. Cho hm s (1), m l tham s thc

1. Kho st s bin thin v v th hm s khi .

2.Tm m th hm s ct ng thng ti 3 im phn bit ; B; C sao cho tam gic c din tch , vi

Bi 3. Cho hm s y = (1)

1. Kho st s bin thin v v th ca hm s (1)

2. nh k ng thng d: y = kx + 3 ct th hm s (1) ti hai im M, N sao cho tam gic OMN vung gc ti O. ( O l gc ta )

Bi 4. Cho hm s c th l , vi l tham s thc.1.Kho st s bin thin v v th ca hm s cho khi .2.Tm m ng thng ct ti hai im cng vi gc ta to thnh mt tam gic c din tch l

Cu 5. Cho hm s c th l (C).1.Kho st s bin thin v v th (C) ca hm s.2. Gi l ng thng i qua im vi h s gc

EMBED Equation.DSMT4 . Tm ng thng ct th (C) ti ba im phn bit A, B, C v 2 giao im B, C cng vi gc to to thnh mt tam gic c din tch bng .

Cu 6 . Cho hm s c th l (C).1. Kho st s bin thin v v th (C) ca hm s.2. Gi E l tm i xng ca th (C). Vit phng trnh ng thng qua E v ct (C) ti ba im E, A, B phn bit sao cho din tch tam gic OAB bng .Cu 7 . Cho hm s (C).1. Kho st s bin thin v v th (C) ca hm s.2. Tm m ng thng d: ct (C) ti hai im phn bit A, B sao cho (OAB vung ti O.Dng 4. TIP TUYN CNG VI CC TRC TA TO THNH MT TAM GIC C BIT

Bi ton : Cho hm s y=f(x) c th l (C), tm cc im M thuc (C) sao cho tip tuyn vi (C) ti M cng vi cc trc ta to thnh mt tam gic c bit Cch gii Bc 1. Gi M thuc (C) sau vit phng trnh tip tuyn d :

y=f'(x)(x-

Bc 2. Tm ta cc giao im ca d vi 2 trc ta

Bc 3. Cn c vo u bi tam gic cho l tam gic g ? da vo tnh cht ca tam gic thit lp cc phng trnh

Bc 4. Gii cc phng trnh lp tm ra . S cp chnh l s im M tm c .

MT S V D MINH HA

V d 1. (KA-2009). Cho hm s

1. Kho st s bin thin v v th hm s (C)

2. Vit phng trnh tip tuyn ca (C) bit tip tuyn ct trc honh , trc tung ln lt tai hai im phn bit A, B v tam gic OAB cn ti gc ta O .


2.Gi d l tip tuyn ca (C) ti

- d ct trc Oy ti B :

- d ct trc Ox ti A :

- Tam gic OAB cn

Vy c hai im M tha mn yu cu bi ton :

V d 2. (KD-2007). Cho hm s


2. Tm ta im M thuc (C) , bit tip tuyn ti M ct hai trc Ox, Oy ti hai im A, B sao cho tam gic OAB c din tch bng .

Giia. Hc sinh t v th (C)

b. Gi

- Tip tuyn ti M l d :

- d ct Ox tiA.

- d ct Oy ti im B :

- Gi H l hnh chiu vung gc ca O trn d, h l khong cch t O n d th :

Vy :

Cho nn

Do c hai im M tha mn yu cu bi ton :

V d 3: Cho hm s c th (C)

1. Kho st s bin thin v v th (C) ca hm s.

2.Vit phng trnh tip tuyn ca (C), bit tip tuyn ct trc Ox, Oy ln lt ti A, B v tam gic OAB cn ti O.Gii1. Hc sinh t v th (C)

2. Gi .- Tip tuyn ti M l d:

- d ct trc Ox ti A:

- d ct trc Oy ti B:

- Tam gic OAB cn ti O nn OA = OB

(1)

(2)

T (1) v (2) ta c :

* Vi (loi)* Vi .V d 4: Cho hm s . (1) 1. Kho st s bin thin v v th (C) ca hm s (1) khi m = 0

2. Tm m tip tuyn ca th (1) ti im c honh bng 1 ct cc trc Ox, Oy ln lt ti cc im A v B sao cho din tch tam gic OAB bng .


2. Vi M(1 ; m 2)

- Tip tuyn ti M l d:

d: y = -3x + m + 2.

- d ct trc Ox ti A:

- d ct trc Oy ti B:

-

Vy: m = 1 v m = - 5BI TP T LUYNBi 1. Cho hm s y =

1.Kho st s bin thin v v th (C).

2.Vit phng trnh tip tuyn ca (C), bit tip tuyn ny ct hai trc ta to thnh mt tam gic c din tch bng .

Bi 2. Cho hm s:

1.Kho st s bin thin v v th (C) ca hm s.

2.Tm nhng im M trn (C) sao cho tip tuyn vi (C) ti M to vi hai trc ta mt tam gic c trng tm nm trn ng thng 4x + y = 0.

Bi 3. Cho hm s c th l (Cm)1. Kho st s bin thin v v th (C) ca hm s khi m = 2.

2. Gi A l giao im ca (Cm) vi trc tung. Tm m sao cho tip tuyn ca (Cm) ti A to vi hai trc ta mt tam gic c din tch bng

Dng 5. TIP TUYN CNG CC TIM CN TAO THNH TAM GIC .Bi ton : Cho hm s y=f(x) c th l (C). ( L hm phn thc mi c tim cn - hoc hm cn thc ), Tm cc im M sao cho tip tuyn ti M ct hai tim cn ti A,B sao cho tam gic IAB c tnh cht no ( vi I l giao 2 tim cn )Cch gii

Bc 1. Vit phng trnh tip tuyn d ti M :

Bc 2. Tm giao ca d vi 2 tim cn l to A,B , tm ta I ( giao ca 2 tim cn ).

Bc 3. Cn c gi thit , lp cc phng trnh ( n l )

Bc 4. Gii cc phng trnh lp tm , tc l tm c ta M

MT S V D MINH HA

V d 1. Cho hm s


2. Tm nhng im M thuc (C) sao cho tip tuyn ti M ct hai tim cn ti A, B sao cho vng trn ngoi tip tam gic IAB c bn knh nh nht . Vi I l giao hai tim cn .

Gii1. Hc sinh t v th (C) .

2.Tip tuyn ca (C) ti l

- d ct tim cn ng : x = -2 ti A

- d ct tim cn ngang : y = 2 ti B

- Nh vy:

- Ta c :

Do :

Du bng xy ra khi :

-Kt lun : C hai im M tha mn yu cu bi ton .

;

V d 2.(DB-2007). Cho hm s


2. Lp phng trnh tip tuyn d ca (C) sao cho d v hai tim cn ct nhau to thnh mt tam gic cn .


2.Gi d l tip tuyn ca (C) ti im , th d :

- Nu d ct tim cn ng : x = -1 ti im B :

- Khi d ct tim cn ngang : y=1 ti im A , th :

- Goi giao hai tim cn l I(-1;1) . Tam gic IAB l tam gic cn khi : IA = IB

Vi x = 0 v y = 0 , ta c tip tuyn : y = x

Vi x = -2 v y = 2/3 , ta c tip tuyn : y = x+8/3 .

V d 3. Cho hm s


2. Cho M l im bt k trn (C). Tip tuyn ca (C) ti M ct cc ng tim cn ca (C) ti A v B. Gi I l giao im ca cc ng tim cn. Tm to im M sao cho ng trn ngoi tip tam gic IAB c din tch nh nht.

Gii1. Hc sinh t v th (C).

2. Ta c: ,

Phng trnh tip tuyn vi ( C) ti M c dng:

To giao im A, B ca v hai tim cn l:

Ta thy , suy ra M l trung im ca AB.

Mt khc I = (2; 2) v tam gic IAB vung ti I nn ng trn ngoi tip tam gic IAB c din tch

S =

Du = xy ra khi

Do c hai im M cn tm l M(1; 1) v M(3; 3)BI TP T LUYNBi 1. Cho hm s


2. Cho M l im bt k trn (C). Tip tuyn ca (C) ti M ct cc ng tim cn ca (C) ti A v B. Gi I l giao im ca cc ng tim cn. Tm to im M sao cho ng trn ngoi tip tam gic IAB c din tch nh nht.

Bi 2. Cho hm s c th (C)

1.Kho st s bin thin v v th (C) ca hm s.

2.Gi M l im bt k trn (C). Tip tuyn ca (C) ti M ct cc ng tim cn ca (C) ti A v B. Gi I l giao im ca cc ng tim cn. Tm ta M sao cho ng trn ngoi tip tam gic IAB c din tch nh nht.

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Documents

Tam Giac Trong Khao Sat Ham So