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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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7/25/2019 [VNMATH.COM]-Duong thang-duong tron_Nguyen Tuan Lam.pdf
3/5
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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7/25/2019 [VNMATH.COM]-Duong thang-duong tron_Nguyen Tuan Lam.pdf
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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:>