elip - leminhansp - vmfer

7/29/2019 Elip - Leminhansp - VMFer

1/10Http://di

endan

toanh

oc.ne

t/Din n ton hc VMF Chun b cho k thi TSH - 2013

CHUYN : ELIPA - TM TT L THUYT

1. nh ngha:Cho hai im c nh F1,F2 vi F1F2 = 2c v mt di khng i 2a(a > c). Elip l tp hp nhng

im M sao cho:F1M+ F2M= 2a

Ta gi: F1,F2: Tiu im, F1F2 = 2c: Tiu c, F1M,F2M: Bn knh qua tiu.

F1 F2

A1 A2

B1

B2

O

M

2. Phng trnh chnh tc ca ElipTrong mt phng ta Oxy vi F1(c; 0),F2(c; 0):

M(x;y)

(E)

x2

a

2+

y2

b

2= 1 (1).

Trong : b2 = a2c2(1) c gi l phng trnh chnh tc ca (E)

3. Hnh dng v tnh cht ca ElipElip c phng trnh (1) nhn cc trc ta l trc i xng v gc ta lm tm i xng.

+ Tiu im: Tiu im tri F1 (c; 0), tiu im phi F2 (c; 0)+ Cc nh: A1 (a; 0) ,A2 (a; 0) ,B1 (0;b) ,B2 (0; b)+ Trc ln: A1A2 = 2a, nm trn trc Ox; Trc nh: B1B2 = 2b, nm trn trc Oy

+ Hnh ch nht c s: L hnh ch nht to bi cc ng thng x =a, y =bT ta thy hnh ch nht c s c chiu di l 2a v chiu rng l 2b+ Tm sai: e =

c

a< 1

+ Bn knh qua tiu ca im M(xM,yM) (E) l:

MF1 = a + exM = a +cxM

a, MF2 = a exM = a axM

c

+ ng chun ca Elip:

ng thng 1 : x +a

e= 0 c gi l ng chun ca elip, ng vi tiu im F1(c; 0)

ng thng 2 : x

a

e

= 0 c gi l ng chun ca elip, ng vi tiu im F2(c; 0)

Tnh cht ca ng chun:

MF1

d(M;1)=

MF2

d(M;2)= e < 1, M (E)

Elip (Cu VIb.1) 1 Gv: L Minh An


2/10Http://di

endan

toanh

oc.ne


B - GII TON

I - VIT PHNG TRNH CHNH TC CA ELIP

Cc bc thc hin:

Bc 1: Gi s phng trnh chnh tc ca elip l:x2

a2+

y2

b2= 1 (a > b > 0) (E)

Bc 2: S dng cc d kin bi ton thit lp cc phng trnh tm a,b

Ch cc kin thc lin quan n a,b, chng hn: ta tiu im, ta nh, tm sai, b2 = a2c2...

V d 1.1: Trong mt phng ta Oxy, cho im M(3; 1), ng elip (E) i qua im M vkhong cch gia hai ng chun ca (E) l 6. Lp phng trnh chnh tc ca (E).

Nhn xt: y l bi tp c bn vi cc yu t c cho kh r rng, lm c bi tonch yu cu thuc cc khc khi nim v k nng bin i gii h phng trnh c bn.

Li gii:

Gi s phng trnh chnh tc ca (E) l:x2

a2+

y2

b2= 1 (a > b > 0)

Hai ng chun ca (E) c phng trnh l: 1 : x + ae

= 0; 2 : x ae

= 0

Do khong cch gia hai ng chun l: 2a

e=

2a2

c

2a2

c

= 6

a4 = 9c2 = 9a2b

2 b2 =

9a2a4

9

(1)

Mt khc: M(3; 1) (E) 3a2

+1

b2= 1 (2)

Th (1) vo (2) v rt gn ta c: a412a2 + 36 = 0 a26= 0 a2 = 6 b2 = 2.p s: (E) :

x2

6+

y2

2= 1.

V d 1.2: Thi th ln 1 - THPT Bm Sn - Thanh HaTrong mt phng ta Oxy, lp phng trnh chnh tc ca elip (E) bit n c mt nh v hai tiu

im to thnh mt tam gic u v chu vi ca hnh ch nht c s ca (E) l 12(2 +3).

F1 F2

A1 A2

B1

B2

O



3/10Http://di

endan

toanh

oc.ne


Nhn nh:D kin: chu vi hnh ch nht c s v tam gic u s thit lp c hai phng trnh tm a v b+ Chu vi ca hnh ch nht c s l: 2(2a + 2b)

+ Cc nh l A1(1; 0),A2(1; 0),B1(0;b),B2(0; b) v cc tiu im l F1(c; 0),F2(c; 0)

B1F1F2 v B2F1F2 u

F2B2 = F1F2

+ a2 c2 = b2Li gii:

Gi phng trnh chnh tc ca elip (E) l:x2

a2+

y2

b2= 1 (a > b > 0)

Khi chu vi ca hnh ch nht c s l 2(2a + 2b)

2(2a + 2b) = 12(2 +3) a + b = 3(2 +3) (1)Do cc nh A1(a; 0), A2(a; 0) v F1(c; 0), F2(c; 0) cng nm trn Ox nn theo gi thit F1, F2 cngvi nh B2(0; b) trn Oy to thnh mt tam gic u B2F2 = F1F2 = B2F1 ()Ta thy: F1,F2 i xng nhau qua Oy nn B2F1F2 lun l tam gic cn ti B2

Do : ()B2F2 = F1F2 c2 + b2 = 2c b2 = 3c2Li c: a2 c2 = b2 3a2 = 4b2 (2)T (1) v (2) suy ra: a = 6 v b = 3

3

p s: (E) :x2

36+

y2

27= 1

V d 1.3:(B-2012) Trong mt phng ta Oxy, cho hnh thoi ABCD c AC = 2BD v ngtrn tip xc vi cc cnh ca hnh thoi c phng trnh x2 +y2 = 4. Vit phng trnh chnh tc

ca elip (E) i qua cc nh A,B,C,D ca hnh thoi. Bit im A nm trn trc OxNhn nh:- Cc c im ca hnh thoi:

ng trn ni tip c phng trnh: x2 +y2 = 4. (Tm O (0; 0), bn knh R = 2)

Tm ng trn ni tip l tm ca hnh thoi Gc ta O (0; 0) l tm ca hnh thoi.A Ox C Ox, BDACB,D Oy- A,B,C,D (E) nn A,B,C,D l cc nh ca (E)!- Nh vy ta xc nh c mi lin h gia nh ca elip vi hnh thoi, vi hai iu kin AC= 2BD

v ng trn ni tip hnh thoi c bn knh R = 2 ta s thit lp c hai phng trnh xc nh

ta cc nh ca elip.

C A

D

B

O

H

Li gii:



4/10Http://di

endan

toanh

oc.ne


Gi s phng trnh ca elip (E) l:x2

a2+

y2

b2= 1(a > b > 0)

Ta c: ng trn (C): x2 +y2 = 4 l ng trn ni tip hnh thoi ABCD, c tm O(0; 0), bn knh

R = 2

V tm ca (C) l tm ca hnh thoi nn AC v BD vung gc vi nhau ti trung im O ca mi

ngM A Ox C Ox v B,D OyLi c: A,B,C,D (E)A,B,C,D l bn nh ca (E)Nu i ch A v C cho nhau hoc B v D cho nhau th Elip khng thay i nn ta c th gi s A,B

ln lt nm na trc dng ca Ox v Oy, khi ta ca chng l A(a; 0),B(0; b)

OA = a,OB = b. V AC= 2BD nn OA = 2OB a = 2bK OH vung gc vi AB ti H OH = R = 2V tam gic ABO vung ti O 1

OH2=

1

OA2+

1

OB2 1

4=

1

a2+

4

a2 a2 = 20 b2 = 5

Vy phng trnh (E) l: x2

20+ y

2

5= 1

II - TM IM THUC ELIP

Cc bc thc hin:

Bc 1: Xc nh cc "t kha" lin quan n im cn tm, c gng chuyn chng thnh cng thctng ng.

Bc 2: T gi thit, thit lp phng trnh tm ta ca im. Ch rng im cn tm lun thuc

(E) nn ta ca n lun tha mn phng trnh (E), y l mt phng trnh.

V d 2.1: Trong mt phng ta Oxy cho elip (E) :x2

9+

y2

1= 1. Tm trn (E) nhng im t/m:

1. C bn knh qua tiu im ny bng 3 ln bn knh qua tiu im kia?

2. Nhn hai tiu im di gc vung.

Li gii:

(E) :x2

9 +y2

1 = 1 a = 3,b = 1 c =a2b2 = 221. T kha cn quan tm "bn knh qua tiu"Gi M(xo,yo) l im phi tm. Khi bn knh qua tiu ca M l:

MF1 = a + exo = a +cxo

a, MF2 = a exo = a cxo

aT gi thit suy ra:

MF1 = 3MF2

MF2 = 3MF1

MF13MF2 = 0MF23MF1 = 0

(MF13MF2) (MF23MF1) = 0 (1)

Khai trin rt gn ta c:

() 16MF1.MF23 (MF1 +MF2)2

= 0 16(a + exo) (a exo)3(2a)2

= 0

x2o =a2

4e2=

a4

4c2=

81

32xo = 9

2

8

Li c: M (E)y2o = 1x2o

9=

23

32yo =

46

8.



5/10Http://di

endan

toanh

oc.ne


p s: M1

9

2

8;

46

8

;M2

9

2

8;

46

8

;M3

9

2

8;

46

8

;M4

9

2

8;

46

8

Nhn xt:

Trong gii ton, ta thng ch quen vi chiu bin i AB = 0 A = 0B = 0 nhng trong nhiutrng hp bin i theo chiu ngc li s gip vic gii bi ton ngn gn hn rt nhiu, m bi

ton trn l mt v d.

bi ton ny, vic bin i rt gn cng l mt cng vic kh vt v nu khng c nhng nhnxt tinh t, cn ch rng MF1 +MF2 = 2a

Khi kt lun cn ch ly nghim, nhiu bn thng nhm ln ch ly hai nghim M1,M4.

2. T kha "gc vung"

F1 F2

M

O

Vi gc F1MF2 = 90o th ta c cc "cng thc" tng ng:

1. MF21 +MF2

2 = F1F2

2 ; 2. MO =F1F2

2

= OF2; 3.MF1.

MF2 = 0

Vi tng "cng thc" ta s c cc hng lm khc nhau tng ng, di y ti trnh by hai cchc th ni l kh ngn gn.

Gi M(xo;yo) l im cn tm. M (E) nn x2o

9+y2o = 1 (1)

Cch 1:Ch rng MF1,MF2 l bn knh qua tiu, nn ta c:

F1MF2 = 90o MF21 +MF22 = F1F22 (a + exo)2 + (a exo)2 = 32 x2 =

(16a2)a2c2

=63

8

T (1) suy ra: y2o =1

8.

Cch 2:im M nhn F1,F2 di mt gc vung nn MF1F2 vung ti M.M d thy O l trung im ca F1F2 nn OM=

F1F2

2x2o +y2o = 8 (2)

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

(I)

x2o9

+y2o = 1

x2o +y2o = 8

x2o =63

8

y2o =1

8

xo = 3

14

4

yo =

2

4

Nhn xt: cch 2 c th gii thch theo cch khc nh sau:

Do M nhn F1,F2 di mt gc vung nn M nm trn ng trn (C) nhn F1F2 lm ng knh.

Tc l (C) c tm O bn knhF1F2

2= 2

2

M l giao im ca (E) v (C) : x2 +y2 = 8. Do ta M l nghim h (I).



6/10Http://di

endan

toanh

oc.ne


p s: M1

3

14

4;

2

4

;M2

3

14

4;

2

4

;M3

3

14

4;

2

4

;M4

3

14

4;

2

4

V d 2.2: Thi th Chuyn Vnh Phc - Ln 2

Trong mt phng vi h ta Oxy cho elip (E) : x2

9+ y

2

4= 1 v cc im A(3; 0), I(1; 0). Tm

ta cc im B,C (E) sao cho I l tm ng trn ngoi tip tam gic ABC.

Li gii:

IA

B

C

O

Gi (C) l phng trnh ng trn ngoi tip tam gic ABC, (C) c tm I(1; 0) bn knh IA = 2.Phng trnh (C) : x2 +y2 + 2x3 = 0.

Do B,C (E) nn ta ca B,C l nghim h:

x2 +y2 + 2x3 = 0x2

9+

y2

4= 1

x =3x =

3

5 Vi x = 3 y = 0 B tc l trng vi A hoc C trng vi A (khng tha mn) Vi x = 3

5y =4

6

5.

p s: B1

3

5;

4

6

5

,C1

3

5; 4

6

5

; B2

3

5;4

6

5

,C2

3

5;

4

6

5



7/10Http://di

endan

toanh

oc.ne


III - BI TP TNG HP LIN QUAN N ELIP

V d 3.1: Trong mt phng ta Oxy, cho elip (E) :x2

8+

y2

4= 1 c cc tiu im F1,F2. ng

thng d i qua F2 v song song vi ng phn gic ca gc phn t th nht ct (E) ti A,B. Tnh

din tch tam gic ABF1.

Li gii:

Do gi thit cho vit c ngay phng trnh ng thng d cha A,B ( i qua mt im

bit v song song vi ng thng cho trc) nn ta nh hng tnh din tch theo cng thc:

SABF1 =1

2AB.d(F1; d).

F1 F2

A

B

O

(E) :

x2

8 +

y2

4 = 1 a2

= 8,b2

= 4 c = a2b2 = 2 F1(2; 0),F2(2; 0)ng phn gic ca gc phn t th nht c phng trnh l: y = x

Ta c d // v F2 d nn phng trnh d l: y = x2.

KhitacaB v Cl nghim h:

y = x2x2

8+

y2

4= 1

A(0;2), B

8

3;

2

3

hocB(0;2), A

8

3;

2

3

.

AB = 8

2

3, d(F1,d) = 2

2

p s: SABF1 =16

3

.

V d 3.2: Thi th Hocmai - Thy L B Trn Phng - 02

Trong mt phng ta Oxy cho ng thng d : 2x +y + 3 = 0 v elip (E) :x2

4+

y2

1= 1. Vit

phng trnh ng thng vung gc vi ng thng d ct (E) ti hai im A,B sao cho din

tch tam gic AOB bng 1.

Li gii:



8/10Http://di

endan

toanh

oc.ne


B

A

B

AO

V d nn phng trnh d c dng: x2y + m = 0. Khi , A,B l nghim h:

x2y + m = 0x2

4+y2 = 1

x = 2ym8y24my + m24 = 0 ()

d ct (E) ti hai im phn bit A,B th h phi c hai nghim phn bit

() c hai nghim phn bit = 324m2 > 0 m 22; 22

(1)

Gi A(2y1m;y1),B(2y2m;y2), trong y1,y2 l nghim ()AB2 = 5(y2y1)2 = 5(y1 +y2)24y1y2 = 54(8m2)

ng cao OH= d(,) =|OH|

5 S2OAB =

1

2OH.AB

2=

1

16m2(8m2) = 1m = 4m =2

p s: 1 : x2y + 2 = 0, 2 : x2y2 = 0.

V d 3.3: Thi th ln 2 - THPT Hn Thuyn - Bc Ninh

Trong mt phng ta Oxy cho elip (E) :x2

16+

y2

9= 1 v im I(1; 2). Vit phng trnh ng

thng (d) i qua I, sao cho d ct (E) ti A v B tha mn I l trung im ca on AB.

Li gii:

I

A

B

O

Gi u = (a,b) l vct ch phng ca d.Ta c d i qua I(1; 2) nn phng trnh d c dng:

x = 1 + at

y = 2 + btt R

Gi s (d) ct (E) ti A(1 + at1; 2 + bt1), B(1 + at2; 2 + bt2)V I l trung im ca AB nn 2xI = xA +yA t1 + t2 = 0 (1)Mt khc: A,B (E) nn t1, t2 l nghim ca phng trnh:



9/10Http://di

endan

toanh

oc.ne


(1 + at)2

16+

(2 + bt)2

9= 1

a2

16+

b2

9

t2 + 2

a

16+

2b

9

t 71

144= 0

Theo nh l Viet ta c: t1 + t2 =2

a

16+

2b

9

:

a2

16+

b2

9

(2)

T (1) v (2) suy raa

16+

2b

9= 0. Chn a = 32,b =

9 ta c u = (32;

9) l mt VTCP ca d

n = (9;32) l mt VTPT ca d.Vy phng trnh d l: 9x + 32y73 = 0

BI TP NGH

Bi 1.1: Trong mt phng ta Oxy, cho im A(0; 5). Lp phng trnh chnh tc ca elip (E)bit (E) i qua A v c hnh ch nht c s ni tip ng trn (C) : x2 +y2 = 41.

p s: (E) :x2

25+

y2

16= 1.

Bi 1.2: Trong mt phng vi h trc ta Oxy, cho ng thng d : x5 = 0. Lp phng trnhchnh tc ca elip (E), bit mt cnh hnh ch nht c s ca (E) nm trn d v hnh ch nht c

di ng cho bng 6.

p s: (E) :x2

5+

y2

4= 1.

Bi 1.3: Trong mt phng ta Oxy, cho im M(3; 1) ng elip (E) ti qua im M v ckhong cch gia hai ng chun l 6. Lp phng trnh chnh tc ca (E).

p s: (E) :x2

6+

y2

2= 1.

Tng t vi M5; 2, khong cch gia hai ng chun l 10. p s: (E) :

x2

15 +

y2

6 = 1.Bi 1.4 Thi th Hocmai - Thy L B Trn Phng - 04Trong mt phng ta Oxy, mt elip (E) i qua im M(2;3) v c phng trnh mt ngchun l x + 8 = 0. Vit phng trnh chnh tc ca (E).

p s: (E1) :x2

16+

y2

12= 1, (E2) :

x2

52+

y2

39/4= 1.

Bi 1.5: A-2012Trong mt phng ta Oxy, cho ng trn (C) : x2 +y2 = 8. Vit phng trnh chnh tc ca elip

(E), bit (E) c di trc ln bng 8 v (E) ct (C) ti 4 im to thnh mt hnh vung.

Bi 2.1: Trong mt phng vi h trc ta Oxy, cho elip (E) c di trc ln l 6 v qua imM

3

2

2;

2

.im N nm trn (E) cch O mt on c di bng

5. Tm ta N?

p s:

3

5

5;4

5

5

;

3

5

5;4

5

5

Bi 2.2: Thi th Din n Truonghocso - Ln 1

Trong mt phng ta Oxy cho elip (E) :x2

100+

y2

25= 1. Tm ta im K nm trn elip sao cho

K nhn cc tiu im di mt gc 120o.

Bi 2.3: Thi th Hocmai - Thy L B Trn Phng - 01Trong mt phng ta Oxy, cho elip (E) :

x2

16+

y2

9= 1 v ng thng d : 3x + 4y12 = 0. Chng

minh rng ng thng d ct (E) ti hai im phn bit A,B. Tm im C thuc (E) sao cho ABC

c din tch bng 6.



10/10Http://di

endan

toanh

oc.ne


p s: C1

2

2; 32

, C2

22; 3

2

Bi 2.4: Trong mt phng ta Oxy, cho im M di ng trn elip (E) :x2

9+

y2

4= 1. Gi H,K l

hnh chiu ca M ln cc trc ta . Xc nh ta ca Mdin tch OHMKt gi tr ln nht.

p s:32

2 ;2

;3

2

2 ;2

Bi 2.5: A-2011Trong mt phng ta Oxy, cho elip (E) :

x2

4+

y2

1= 1. Tm ta cc im A,B (E), c honh

dng sao cho tam gic OAB cn ti O v c din tch ln nht.

Bi 3.1: Thi th Din n K2pi - Ln 5

Trong mt phng ta Oxy cho elip c phng trnh (E) :x2

8+

y2

4= 1 v im I(1;1). Mt ng

thng qua I ct (E) ti hai im phn bit A,B. Tm ta cc im A,B sao cho ln ca tch

IA.IB t gi tr nh nht.

Bi 3.2: Trong mt phng Oxy, cho elip (E) :x2

8+

y2

2= 1. Vit phng trnh ng thng d ct (E)

ti hai im phn bit c ta l cc s nguyn.

Bi 3.3: Trong mt phng Oxy cho elip (E) :x2

9+

y2

4= 1 v im M(1; 1). Lp phng trnh ng

thng i qua M sao cho ct (E) ti hai im phn bit A,B sao cho MA = MB.

p s: : 4x + 9y13 = 0Bi 3.4:Trong mt phng Oxy cho elip (E) :

x2

9+

y2

4= 1 v ng thng d : y = x + m. d ct (E) ti hai im

P,Q. Gi P,Q ln lt l im i xng ca P,Q qua O. Tm m PQPQ l hnh thoi.Bi 3.5: B-2010Trong mt phng ta Oxy cho A(2,sqrt3) v elip (E) :

x2

3+

y2

2= 1. Gi F1,F2 l cc tiu im

ca (E) (F1 c honh m), M l giao im c tung dng ca ng thng AF1 vi (E), N l

im i xng ca F2 qua M. Vit phng trnh ng trn ngoi tip tam gic ANF2.

oOo

TI LIU THAM KHO[1] Trn Thnh Minh - Gii ton hnh hc 12 - NXB Gio Dc - 2001

[2

]Cc thi th i hc nm hc 2012 - 2013

[3] Din n Boxmath - Tuyn tp HHGT trong mt phng

elip - leminhansp - vmfer

Documents