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  • 7/25/2019 [VNMATH.COM]-Duong thang-duong tron_Nguyen Tuan Lam.pdf

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    GIUPBANQNTAp

    ~Sl1Jilf jNCI O

    G IllA D UO NG

    THANG vA

    DUONG TRON

    B

    ai toan vi su tuong giao giiia duong

    thdng va duong trim thuang xudt hien

    trong cdc di thi tuyen sinh vao Dai h9C, Cao

    acing va co thi tcng dung di giai mot s6 bai

    todn

    dai

    s6.

    I. LiTHUYET

    Cho dirong

    trim

    C) co tam 1, ban kinh

    R va

    dirong thang

    L l.

    D~t

    h

    =

    d

    I,

    L l).

    TrUOng hQ1> h = R. Ll la tiep tuyen cua

    dirong

    trim C

    M~ T~~

    Hinh 1

    N~u tir di~m M ngoai C) ke hai tiep tuyen

    MA va ME

    d~n duong

    trim

    C) h. 1) thi

    Ll

    MAl =

    Ll

    MEl, 1M2 =R2 +AM- =R2 +BM2.

    AB ..11M

    tai trung di~m

    H

    cua

    AB.

    SIAMB

    =

    2S

    IAM

    =

    Al.AM

    =

    AH.IM.

    :TrU Ong hQ1>h < R. L l dt C) tai hai diem

    phan biet

    P

    va

    Q

    h. 2 .

    Hinh 2

    GQi

    H

    la trung diem cua doan

    PQ

    thi

    Tam giac

    IPQ

    can tai 1,

    SIPQ

    = ~

    R2

    sinPIQ.

    NGUYEN TUAN

    LAM

    GV THPT Thanh Nhan, TP H6 Chi Minh

    IH..l PQ, R2

    =

    IH2

    +

    HP2

    =

    IH2

    +

    PQ2 .

    4

    DQ dai doan PQ IO n nhelt khi

    Ll

    di qua tam

    1 cua dirong tron C).

    .:. Truong hQ Ph > R.

    Ll

    va C) khong co di~m

    chung.

    Tir mot diem belt ki tren Ll luon ke duoc hai

    tiep tuyn d~n

    C . .

    Lely diem K n~m tren duong tron C) thi

    h - R ~ d(K,Ll) ~ h + R .

    II.

    cAc

    THi D V MINH HQA

    .:. Truong hQ1>L l ti~p xuc

    V Ol

    C)

    * Thi du 1. Cho duong tron (C): .xl +Y = 2.

    n phuong trinh tiip tuyen Ll cua

    C

    sao

    cho

    Ll

    ctit cac tia Ox, Oy fdn luot tai A, B va

    dien tich tam giac OAB nho nhat.

    Lai giiii.

    Duong tron C) co tam trung g6c toa

    dQ va ban kinh

    R

    = ..fi.

    Ti~p tuyen Ll qua

    A(a;O) ,B(O;b)(a>O, b>O)

    co

    P T x +y =lbx+ay-ab=O .

    a b

    Ta co

    d(

    O,Ll) =

    R d, J labl =..fi

    a

    2

    +b

    2

    ab

    =

    ~2(a2 +b

    2

    ) ~

    2M

    =>

    ab ~

    4, nen

    SOAB= OA .GB = ab ~

    2. Dang thirc xay ra

    khi

    a=b=2.

    V~yPT Ll:

    x+y-2=O.D

    *Thi dl}2. (Ciiu VI.a. Khbi B 2009)

    Cho duong tron

    C):

    (x - 2)2 + y2

    =.

    va cac

    5

    diarng thang Ll1:X-y=O,Ll2:X-7y=O.

    Xac dinh toa d(J tam K va tinh ban kinh cua

    duong

    iron

    (C

    l

    ,

    biit rang

    duang iron

    (C

    l

    tiip xuc vai cac duong thdng

    Lll ,Ll2

    va tam

    K

    thuoc

    duang iron C .

    l r C >~~ic:>

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    Lai

    giai.

    GQi K(a;b)E(C)~(a-2)2 +b

    2

    =~;

    (C])

    ti~p XUC

    fl],fl2 ~ la ~bl

    =

    la ~bl .

    \12

    5 2

    T

    d {5(a-2)2 +5b

    2

    =4

    ir ta co

    5la-bl

    =

    la-7bl

    Gifti

    M

    nay ta duoc

    a;

    b)

    = ( ~; ~ ).

    Ban

    kinh dirong trim

    (C])

    la

    R

    =

    la ~bl

    =

    2J2 .0

    \12 5

    Thi d1}.3.

    ellu V.a.l Kh8i B 2006)

    Cho

    duong

    trim

    C):

    x

    2

    +y2 -2x-6y+6=0

    va diim

    M -3; 1).

    G9i TJ va T2 fa cac tiep

    diim

    cua

    cac tii

    p

    tuyen ke

    tit

    M

    din

    C .

    Viit

    phuong

    trinh

    duong

    thdng TJT

    2

    .

    Liri giai. Duong tron C) c6 tam 1 1;3 ban

    kinh

    R

    = 2

    ,IM

    =

    215

    >

    R

    nen

    M nfun ngoai

    C).

    Ta c6

    MI; =

    MI;.

    =.J MJ2 - R2 =

    4 nen

    T;,T2

    thuoc dirong tron (c,) c6 tam M va ban kinh

    R =4.

    PT dirong tron

    (C )

    la

    r+y+6x-2y-6=0.

    Tir d6 C) ciit C,) tai hai di~m T;

    ,T2 .

    Xet he PT {X2 +y2 - 2x - 6y +6 = 0

    . x

    2

    + y2 +6x -

    2

    y -

    6 = 0

    ~ 2x+ y-3

    = 0 1). Do

    T;,T2

    la giao diem

    cua C) va (c,) nen toa dQ cua cac diem

    T;

    ,T2

    thoa man d~ng thirc 1).

    Do d6 PT duong thang

    T;T

    2

    la

    2x+y-3=0.0

    Thi

    d }

    4.

    Cho duirng trim (Q:(x_l)2 +(y_2)2=4

    va diim

    A 2 ; 1 ). DU Cmg thdng

    d thay dJi

    di

    qua A ciit

    C)

    tai hai diim T] va h

    Hai

    tiep

    tuyen cua C) tai hai diim T] va T2 cdt nhau

    tai diim M Tim quy tich cua diim M

    Liri giai: Duong tron C) c6 tam 1 1; 2) ban

    kinh

    R

    = 2,

    lA

    =

    J2

    R nen Mn~m

    ngoai, duOn~ tr~n. Vi vay illM luon ke diroc

    hai tiep tuyen den duong tron

    C .

    Do

    Met:

    nen

    M(t,6-tj, Me =MI2 =.JMJ2 _R2 .

    Tuong tu Thi du 3, ta c6

    T;,

    T2 I~ giao diem

    cua dirong tron (C) va dirong tron

    C )

    (C c6

    tam M

    va

    ban kinh R

    =

    MJj ).

    Tir d6 ta c6 PT cua duong thang T;T

    2

    la

    tx+(6-t)y-4=0.

    Tir day suy ra dirong thang

    T;T

    2

    luon di qua

    d

    ;:. ;. dinh H 2 2

    rem co.

    3;3 .

    b) Tac6 SOTlM72 =l...T;T

    2

    .GM=0T;.MJj

    2

    = R ..JM02 - R2.

    Suy ra

    r~r ic:>

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    4J5

    OM

    = 2.JM02 -

    4 =>

    OM = 2J5

    5

    ::> t

    2

    + (6 -

    t)2 =

    20

    ::> t =

    4 hoac

    t =

    2 .

    V~y co hai

    diem

    M,

    4;2)

    va M2

    2;4).

    c) Ta co

    S01\MI2

    = OTj.MY; = R ..JM02 - R2

    = 2.JM02 - 4. Tir do

    S01\MI2

    nho nhat khi

    dQ dai MO nho nhat. Khi do diem M la hinh

    chien cua 0 len ~,

    suy ra

    M(3;3).0

    *Thi dl}

    6.

    (Ciiu V.a.2 Kh6i

    D

    2007)

    Cho duong trim C): (x_l)2 +(y+2)2 =9

    va

    dutmg

    thdng t:.:

    3x -

    4

    y

    +

    m

    = O.

    Tim m

    di

    tren t:.

    co duy nhdt

    mot

    diim P

    ma

    tir do

    co thi

    ke

    duac

    hai tiep tuyen

    PA, PB d~n

    duong trim

    C)

    (A, B

    la

    cac

    tiep

    diim)

    sao

    cho tam

    giac

    PAB d~u.

    Loi gidi. Duong trim

    C)

    co tam

    I

    1;-2 ) va

    ban kinh R = 3. Tam giac PAB deu nen IP

    21A

    =

    2R

    = 6 nen

    P

    thuoc duong trim (c,)

    tam I ban kinh R = 6.

    Tren ~ co duy nhat dim P thoa man yeu du

    bai toan

    ::>

    dJ,~

    =

    6

    =>

    m

    =

    19;m

    =

    -41.0

    :TfUOng hQP ~ dt C) tlili hai di~m

    phan bi~t

    *Thi dl}7. Cho

    iIuOng tMng ~:1n(+y-2m-l=O

    ill duong trim

    (C) :

    (x_l)2 +(y - 2)2

    = 4.

    a) Tim m di ~ cat C tai hai diim phdn

    biet

    A

    va

    B

    sao

    cho d(J

    dai doan thang

    AB

    ngan nhdt.

    b)

    Tim quy tich trung

    diem H cua doan

    thang AB khi duong thang ~ thay d6i.

    Lai giai. a) Duong trim

    C)

    co tam

    I

    1; 2)

    va

    ban kinh R = 2. Ta thfty ~ luon di qua dim

    M(2;1) va 1M = J2

    IH

    max =

    1M. Khi ~ .L 1M , suy ra

    m = -1. V~y PT ~: x - y -1 = 0 .

    b) Dim

    H

    la

    trung dim

    cua doan

    AB

    nen

    IH .L HM. Do do H nam tren duong trim

    (e ) co dirong kinh la 1M.

    PT dirong trim

    e):

    x-~J

    ~-~J

    ~.O

    *Thi du 8.

    Xet hai

    s6

    thuc x y thoa man

    x

    2

    +

    4x

    +

    y2 -

    5 = O.

    Tim gia tri Ian nhdt,

    va

    gia tri nho nhdt cua biiu thiec T = 3x +

    4

    y.

    Lai giai. Tren mat phang toa dQ

    Oxy,

    Ifty

    dim M(X,Y)E~:3x+4y-T=0.

    Hai s6 x,y thoa man x2

    +

    4x

    +

    y2 - 5

    =

    0

    nen M(X,Y)E(C):X2+4x+y2-5=0 la dirong

    tron

    tam

    1(-2;0) va

    ban kinh

    R

    =

    3.

    MIa diem chung cua e) va t:. ::>d l,~ ~R

    ::> 16 + TI ~ 15 ::> -21 ~ T < 9.

    D~ng

    thirc xay

    ra khi ~ Ia tiep tuyen cua C).

    Nhtr v~y max(T)

    = 9

    khi

    {

    X2 +4x+ y2

    -5

    =

    0

    ::> {x = ~

    3x

    +

    4y - 9 = 0

    12

    y=-

    5

    rnin(T)

    = -21

    khi

    {

    X2 +4x+ y2

    -5 ~0

    ::>

    {x

    =

    _1:

    3x+4y+21=0 y=-g.D

    5

    *Thi dl}9. (Ciiu VI.h.1 Kh6iA 2009)

    Cho Quang tron

    (C):

    X2+y2+4x+4y+6=;0

    va

    duong thang ~: x

    +

    my - 2m

    + 3 = O.

    G9i I

    la

    tam duong tron C). Tim m d~ ~ dt C)

    tai hai di~m ph an biet A, B sao cho dien tich

    tam giac lAB Ian nhdt.

    Lili giiii. Duong tron

    (C)

    co tam

    1(-2;-2) va

    1 --- R2

    R

    = J2. Ta co SlAB

    =-IA.IB.sinAIB~- =1.

    loAN HOC ~SIln/slif5

    C T u O i t J ; e

    V

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  • 7/25/2019 [VNMATH.COM]-Duong thang-duong tron_Nguyen Tuan Lam.pdf

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    Yay

    SlAB

    IOn nh~t khi IA1JB=>d(I,tl) = ~ =1

    ~

    1-2-2m-2m+31

    8

    ::>

    =1=>m=O hoac m=-.D

    Jl +nr- . 15

    *Thi du 10. Cau VI.h.1 KhaiD20JI

    Cho diim A(O;I) va duang iron

    C):

    X2+ y2 - 2x + 4y - 5

    =

    O. n phuong trinh

    duong thdng tl eftt C) tai hai diim M va N

    saG cho tam giac AMN vuong can tai A.

    Loi giai. Duong trim C) co tam 1(1;-2)

    ve R = J O; IA =

    0;-2).

    Ta co IM= IN va

    AM = AN

    =>

    AI ..1MN nen PT tl: y = m .

    G9i hai giao diem M (Xl;m), N (X2;m).

    Khi do Xl va X2 la nghiem cua PT

    X2 - 2x +

    m?

    + 4m - 5 = 0 3)

    De PT 3) co hai nghiem Xl;X2 phan biet thi

    m

    2

    +4m-6 AM.AN = 0

    ::> (Xl

    -1)

    X2

    -1)

    +m

    2

    =

    0

    ::>

    XlX2- (XI + X2) + 1+

    m?

    = 0

    Ap dung dinh li Viete d6i voi PT 3), SUY ra

    2m

    2

    + 4m - 6 = 0 ::> m = - 3 hoac m = 1 ,

    thoa man 4).

    V~yphuongtrinh

    tl: y = 1 hoac tl: y = -3.0

    *Thi du 11.

    Cau

    VI.a.1 Khai

    A

    2010

    Cho hai duong thdng d,: 13.x

    +

    y

    = 0

    va

    d

    2

    :13.x - y = O. G9i 1) fa duang tron ti~p

    xuc vai dl tai A, eftt d2 tai hai diem B,

    C

    saG

    cho tam giac ABC vuong tai B. n phuong

    trinh cua 1 , bi~t rdng tam giac ABC co dien

    , h b ~

    13

    d ;: A

    i

    h h ~~

    d.

    tic ang 2 va iem co oan ao uong.

    un gidi.

    Ta co

    A E d => A

    a ; aJ3

    (a>

    0).

    Tir

    AC.l a.

    =>

    AC: x-13y-4a =

    0;

    C = d

    2

    \ AC => c(2a; 2J3a ;

    BA.l d

    2

    =>BA:x+13y+2a=0;

    B~d, nBA=> +~;_~}

    Ta co

    SABC

    =

    BA.BC

    =

    13

    ::>

    13a.3a = 13

    2 2

    =>a= ~ =>A(~;-I} C(~;-2}

    Duong tron (1) co tam 1(- 2~ ;- ) (Ila

    trung diem cua A C) va ban kinh R = IA = 1.

    Tir do PT cua duong tron 1) la

    x+ 2~ + Y+H ~LD

    .:. Truimg hQ1>tl khong dt C)

    *Thi du 12. Cho duong trim (C):r+y2-2x

    -2y

    -7

    =

    0

    va duong thdng tl:3x+4y+13=0.

    Tim gia tri tan nhdt va gia tri nho nhdt cua

    khoang each tir

    mot aiim

    M tren C

    a~n

    duong thdng tl.

    Loi

    giiii.

    Duong tron C) co tam

    1(1;

    1)va

    ban kinh R = 3; d(I,tl)=4>R nen dirong

    thang tl khong d.t duong tron C).

    Ta vi~t PT tiep tuyen cua C) va song song

    voi tl. Co hai tiep tuyen la tll :3x+4y+8=0

    voi tiep diem M,

    .-4 ;

    -7) va ~

    :3x+4y- 22=0

    55

    r ,~

    d ~ M(1417)

    VO I tiep tern 2 5;5 .

    Khi d6 d(tl,tll)=I,d(tl,tl2)=7. Voi diem

    M thuoc C) suy ra

    l=d(tl,tll):S;d(M,tl):S;d(tl,tl2)=7.

    Nhu vay

    , 14 17)

    ax d(M, tl) =

    7

    khi M

    =-

    M2 5;5

    , ,(~~)

    md(M,tl)

    =1 khi

    M

    =-

    MJ 5;5

    .0

    ToAN HOC ~

    ~StVllsi if5

    'CTuOitte ~

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    Nh~n xet. Hai diem MJ va M2 la giao diem

    cua C) va dirong thang d qua tam I, vuong

    goc voi f::...;

    maxd(M,f::...) = d(I,f::...)+ R,

    mind(M,f::...) = d(I,/),.)-R.

    Thi d1}13 Cho duong tr im (Q:r+j-a-t4y+l= R nen f::... khong c~t C).

    T dat gia tri nho nh~t khi doan AB ngan nhat.

    Theo Thi

    du

    l

    Zi

    ta co minAB,=d(I,f::...)-R=l.

    Khi do B la hinh chieu cua I len f::... A lei giao

    diem cua doan thang

    IB

    va duong tron C).

    -13 19

    Tir do a = -2; b = 3, c = -' d = -. Yay

    5 5 .

    minT

    =

    1

    BAI T~P

    1. Cho duong tron C):

    X2+ y2 - 6x

    -10 = O.

    Duong thang

    d

    qua diem

    A

    dt C) tai hai

    di~mM,

    N.

    a) ViSt phuong trinh duong th~ngd trong cac

    tnrong hQ Pdoan MN nho nhat, Ian nhat.

    b) Tim quy tich trung diem H cua doan MN.

    A ,{1m+3

    y

    +m+3=0

    2 Cho h~ phirong tnnh

    X2+

    y2 _ 2x -15 = 0

    m

    la tham so).

    a) Chung minh h~ da cho co hai nghiem

    phan biet,

    b) GQi

    (Xl; Yl )

    va

    (X2; Y2)

    la hai nghiem cua

    he. Tim gia tri Ian nh~t, gia tri nho nh~t cua

    bi~u tlnrc F = (Xl - X2)2+(Yl - Y2)2.

    3 Cho duong thang f::...: X

    +

    Y

    +

    2 = 0 va

    duong tron C):

    X2 -ta y2 - 4x -

    2Y

    =

    O. GQi Ila

    tam cua C),

    M

    la d~~m thuoc dirong thang f:: ....

    Qua Mke cac tiep tuyen MA va ME dSn C (A,

    B la cac tiSp diem). Tim toa dQ diem M, biSt

    rang ill giac MAlB co dien tich bang 10.

    4 Cho duong tron (C):X2

    +y2

    -8x+6y+21=0

    va dirong thang

    d:

    X

    +Y

    -1 = 0 . Xac dinh toa

    dQ cac dinh hinh vuong ABCD ngoai tiep

    duong tron C), biSt rang diem

    A

    nam tren

    d.

    5 Cho dirong tron(C):r+y-4x-6y-12=O.

    Tim toa dQ diem M thuoc duong thang :

    d: 2x - Y + 3 = 0 sao cho MI = 2R, trong do I

    la tam va RIa ban kinh cua duong tron C).

    lr~~ic:>