TUYN TP CC BI TP HNH HC PHNG HAY NHT( Ti liu n thi i hc )Bi 1. Trong mt phng Oxy cho cc im ( ) ( ) ( ) ( ) A 1;0 ,B 2;4 ,C 1;4 ,D 3;5 v ng thng d : 3x y 5 0 . Tm im M trn d sao cho hai tam gic MAB, MCD c din tch bng nhau.Gii- M thuc d thi M(a;3a-5 )- Mt khc :( ) ( )13; 4 5, : 4 3 4 03 4x yAB AB AB x y + uuur( ) ( )1 44;1 17; : 4 17 04 1x yCD CD CD x y+ uuur- Tnh :( )( ) ( )1 24 3 3 5 4 4 3 5 17 13 19 3 11, ,5 5 17 17a a a a a ah MAB h+ - Nu din tich 2 tam gic bng nhau th :1 21113 19 3 11 5. 13 19 17. 3 111 1. .1213 19 11 3 2 2 5 178a a a a aAB h CD ha aa

- Vy trn d c 2 im :( )1 211 27; , 8;1912 12M M _ ,Bi 2. Cho hnh tam gic ABC c din tch bng 2. Bit A(1;0), B(0;2) v trung im I ca AC nm trn ng thng y = x. Tm to nh CGii- Nu C nm trn d : y=x th A(a;a) do suy ra C(2a-1;2a).- Ta c :( )0 2, 22d B d . - Theo gi thit : ( ) ( ) ( )2 2 1 4. , 2 2 2 2 02 2S AC d B d AC a a + 2 21 328 8 8 4 2 2 1 01 32aa a a aa

+

+

- Vy ta c 2 im C : 1 21 3 1 3 1 3 1 3; , ;2 2 2 2C C _ _ + + , ,Bi 3. Trong mt phng ta Oxycho tam gicABC, vi ) 5 ; 2 ( , ) 1 ; 1 ( B A, nh Cnm trn ng thng0 4 x , v trng tm G ca tam gic nm trn ng thng 0 6 3 2 + y x. Tnh din tch tam gic ABC.Gii- Ta C c dng : C(4;a) ,( )( )53; 41 1: 4 3 7 03 4ABABx yAB x y

+ uuurChuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218- Theo tnh cht trng tm ; 1 2 413 31 5 63 3 3A B CG GA B CG Gx x xx xy y y a ay y+ + + ' '+ + + + + - Do G nm trn : 2x-3y+6=0 , cho nn : 62.1 3 6 0 23aa+ _ + ,.- Vy M(4;2) v( ) ( )4.4 3.2 71 1 15, 3 . , 5.32 2 2 16 9ABCd CAB S AB d CAB+ + (vdt)Bi 4. Trong mt phng ta Oxy cho tam gic ABC, vi ) 2 ; 1 ( , ) 1 ; 2 ( B A, trng tm G ca tam gic nm trn ng thng 0 2 + y x. Tm ta nh C bit din tch tam gic ABC bng 13,5 .Gii.- Ta c : M l trung im ca AB th M3 1;2 2 _ ,. Gi C(a;b) , theo tnh cht trng tam tam gic : 3333GGaxby+ '- Do G nm trn d : ( )3 32 0 6 13 3a ba b+ + + - Ta c : ( ) ( ) ( )3 52 11;3 : 3 5 0 ,1 3 10a bx yAB AB x y hCAB uuur- T gi thit :( )2 5 2 51 1. , 10. 13, 52 2 2 10ABCa b a bS AB hCAB 2 5 27 2 322 5 272 5 27 2 22a b a ba ba b a b - Kt hp vi (1) ta c 2 h :( )1 2206 632 32 3 38 3838 20; , 6;1233 3 6 6122 22 3 186ba b a ba b aaC Ca b a bba b aa

+ +

' ' ' _

+ + ,

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Bi 5. Trong mt phng oxy choABC c A(2;1) . ng cao qua nh B c phng trnhx- 3y - 7 = 0 .ng trung tuyn qua nh C c phng trnh :x + y +1 = 0 . Xc nh ta B v C . Tnh din tchABC .Bin son t-6-2012( Ti liu ni b-lu )Trang 2A(2;1)B(1;-2)CM()G d:x+y-2=0Chuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218Gii- ng thng (AC) qua A(2;1) v vung gc vi ng cao k qua B , nn c vc t ch phng( ) ( ) ( )21; 3 :1 3x tn AC t Ry t + ' r- Ta C l giao ca (AC) vi ng trung tuyn k qua C : 21 31 0x ty tx y + '+ + Gii ta c : t=2 v C(4;-5). V B nm trn ng cao k qua B suy ra B(3a+7;a) . M l trung im ca AB 3 9 1;2 2a aM+ + _ ,. - Mt khc M nm trn ng trung tuyn k qua C :( )3 9 11 0 3 1; 22 2a aa B+ + + + - Ta c :( ) ( ) ( )122 11; 3 10, : 3 5 0, ;1 3 10x yAB AB AB x y hCAB uuur- Vy : ( )1 1 12. , 10. 62 2 10ABCS AB hCAB (vdt).Bi 6. Trong mt phng vi h ta 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 catam gic ABCGii- Gi B(a;b) suy ra M5 2;2 2a b + + _ ,. M nm trn trung tuyn nn : 2a-b+14=0 (1).- B,B i xng nhau qua ng trung trc cho nn : ( ) ( ) :x a tBC t Ry b t + ' + .T suy ra ta N :623 626 062a btx a ta by b t xx yb ay + + ' ' + + 3 6 6;2 2a b b aN + _ ,. Cho nn ta c ta C(2a-b-6;6-a )- Do C nm trn ng trung tuyn :5a-2b-9=0 (2) Bin son t-6-2012( Ti liu ni b-lu )Trang 3A(2;1)BCx+y+1=0x-3y-7=0MA(5;2)B Cx+y-6=02x-y+3=0MNChuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218- T (1) v (2) :( ) ( )2 14 0 3737;88 , 20; 315 2 9 0 88a b aB Ca b b + ' ' Bi 7. Trong mt phng vi h ta Oxy cho hai ng thng:3 8 0 x y + + , ' :3 4 10 0 x y + v im A(-2 ; 1). Vit phng trnh ng trn c tm thuc ng thng, i qua im A v tip xc vi ngthng. Gii- Gi tm ng trn l I , do I thuc( )2 3: 2 3 ; 22x tI t ty t + + ' - A thuc ng trn ( ) ( )2 23 3 IA t t R + + (1)- ng trn tip xc vi ( ) ( ) 3 2 3 4 2 10 13 12'5 5t t tR R + + + . (2)- T (1) v (2) :( ) ( ) ( ) ( ) ( )2 2 2 2 213 123 3 25 3 3 13 125tt t t t t+ 1+ + + + + ]Bi 8. Trong mt phng vi h ta Oxy cho ng trn hai ng trn 2 2( ) : 2 21 0, C x y x y + + 2 2( ') : 4 5 0 C x y x + + cng i qua M(1; 0). Vit phng trnh ng thng qua M ct hai ng trn ( ), ( ') C Cln lt ti A, B sao cho MA= 2MBGii* Cch 1. - Gi d l ng thng qua M c vc t ch phng( )1; :x atu a b dy bt + 'r- ng trn( ) ( ) ( ) ( )1 1 1 2 2 2: 1;1 , 1. : 2; 0 , 3 C I R C I R , suy ra :( ) ( ) ( ) ( ) ( )2 2 221 2: 1 1 1, : 2 9 C x y C x y + + + - Nu d ct ( )1C ti A :( )22 2 22 2 2 22 202 22 0 1 ;2t Mab ba b t bt Aba b a b ta b

_

+ +

+ + ,+ - Nu d ct( )2C ti B :( )22 2 22 2 2 22 206 66 0 1 ;6t Ma aba b t at Baa b a b ta b

_

+ +

+ + ,+ - Theo gi thit : MA=2MB( )2 24 * MA MB - Ta c : 2 22 22 22 2 2 2 2 2 2 22 2 6 64ab b a aba b a b a b a b 1 _ _ _ _+ +1 + + + + , ,1 , , ]2 22 22 2 2 26 : 6 6 04 364. 366 : 6 6 0b a d x yb ab ab a d x y a b a b + + +* Cch 2.- S dng php v t tm I t s v t k=12. ( Hc sinh t lm )Bi 9. Trong mt phng vi h to Oxy, hy vit phng trnh cc cnh ca tam gic ABC bit trc tm (1; 0) H, chn ng cao h t nh B l (0; 2) K, trung im cnh AB l (3; 1) M.GiiBin son t-6-2012( Ti liu ni b-lu )Trang 4Chuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218- Theo tnh cht ng cao : HK vung gc vi AC cho nn (AC) qua K(0;2) c vc t php tuyn ( ) ( ) ( ) 1; 2 : 2 2 0 2 4 0 KH AC x y x y + uuur.- B nm trn (BH) qua H(1;0) v c vc t ch phng ( ) ( ) 1; 2 1 ; 2 KH B t t + uuur.- M(3;1) l trung im ca AB cho nn A(5-t;2+2t).- Mt khc A thuc (AC) cho nn : 5-t-2(2+2t)+4=0 , suy ra t=1 . Do A(4;4),B(2;-2)- V C thuc (AC) suy ra C(2t;2+t) ,( ) ( ) 2 2; 4 , 3; 4 BC t t HA + uuur uuur. Theo tnh cht ng cao k t A : ( ) ( ) . 0 3 2 2 4 4 0 1 HABC t t t + + uuur uuur. Vy : C(-2;1).- (AB) qua A(4;4) c vc t ch phng ( ) ( ) ( )4 42; 6 // 1;3 :1 3x yBA u AB uuur r3 8 0 x y - (BC) qua B(2;-2) c vc t php tuyn( ) ( ) ( ) ( ) 3; 4 : 3 2 4 2 0 HA BC x y + + uuur3 4 2 0 x y + + .Bi 10. Trong h ta Oxy, cho hai ng trn c phng trnh ( )2 21: 4 5 0 C x y y + v( )2 22: 6 8 16 0. C x y x y + + + Lp phng trnh tip tuyn chung ca( )1C v( )2. CGii- Ta c :( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 221 1 1 2 2 2: 2 9 0; 2 , 3, : 3 4 9 3; 4 , 3 C x y I R C x y I R + + + - Nhn xt :( )1 2 19 4 13 3 3 6 I I C + < + khng ct( )2C- Gi d : ax+by+c =0 ( 2 20 a b + ) l tip tuyn chung , th th : ( ) ( )1 1 2 2, , , d I d R d I d R ( )( )2 22 2 2 22 223 13 4 2 2 3 42 3 43 4 2 3 43 2b ca b c b c b c a b ca bb c a b ca b c b c a b ca b a ba b + + + + ++ + + '

+ ++ + +23 2 2 0a ba b c + . Mt khc t (1) :( ) ( )22 22 9 b c a b + + - Trng hp : a=2b thay vo (1) :( ) ( )( )22 2 2 2 2 2 22 3 542 9 4 41 4 0. ' 4 41 452 3 54bb cbb c b b b bc c c c ccb

+ + +

+

- Do ta c hai ng thng cn tm :( ) ( )( ) ( ) 12 3 5 2 3 5: 1 0 2 2 3 5 2 3 5 4 02 4d x y x y + + + + Bin son t-6-2012( Ti liu ni b-lu )Trang 5H(1;0)K(0;2)M(3;1)ABCChuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218( ) ( )( ) ( ) 12 3 5 2 3 5: 1 0 2 2 3 5 2 3 5 4 02 4d x y x y+ ++ + + + + + - Trng hp : 2 32b ac, thay vo (1) :2 22 22 3223 2b abb a a ba b+ ++( )22 2 20, 2 022 3 4 044 , 633 6ab a c b cb a a b b abaa a b a cb c

+

- Vy c 2 ng thng : 3: 2 1 0 d x , 4: 6 8 1 0 d x y + Bi 11. Trong h ta Oxy, hy vit phng trnh hyperbol (H) dng chnh tc bit rng (H) tip xc vi ng thng : 2 0 d x y ti im A c honh bng 4.Gii- Do A thuc d : A(4;2) - Gi s (H) :( ) ( ) ( )2 22 2 2 216 41 * 1 1x yA Ha b a b - Mt khc do d tip xc vi (H) th h sau c 12 nghim bng nhau :( ) ( )2 2 2 2 2 2 2 22 2 2 2 2 2 2 2 2 2 24 4 0222 2b a x ax a a b bx ay a b bx a x a by xy x y x + ' ' ' ( ) ( ) ( )4 2 2 2 2 2 2 2 2 4 4 2 2 2 2 2 2 2' 4 4 4 4 0 4aa b a a a b a b a b a b a b b a a b + + + + +- Kt hp vi (1) :( )2 2 2 2 4 2 22 22 2 2 2 216 4 8 16 0 4: 18 44 4 8b a a b b b bx yHa b a b a + ' ' ' + + Bi 12. Trong mt phng to Oxy, cho hnh ch nht ABCD c phng trnh ng thng AB: x 2y + 1 = 0, phng trnh ng thng BD: x 7y + 14 = 0, ng thng AC i qua M(2; 1). Tm to cc nh ca hnh ch nhtGii- D nhn thy B l giao ca BD vi AB cho nn ta d B l nghim ca h : 2 1 021 13;7 14 0 5 5x yBx y + _' + ,- ng thng (BC) qua B(7;3) v vung gc vi (AB) cho nn c vc t ch phng:( ) ( )2151; 2 :1325x tu BCy t + ' r- Ta c :( ) ( ) , 2 2 2 , ACBD BIC ABD ABBD R R R R Bin son t-6-2012( Ti liu ni b-lu )Trang 6ABCD M(2;1)x-7y+14=0x-2y+1=0IChuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218- (AB) c( )11; 2 n ur, (BD) c ( )1 221 2n . 1 14 15 31; 7 os =5 50 5 10 10nn cn n+ uur uuruurur uur- Gi (AC) c( ) ( )22 2a-7b9 4, os AC,BD os2 = 2cos 1 2 110 550n a b c ca b _ ,+r- Do :( ) ( )22 2 2 2 2 25 7 4 50 7 32 31 14 17 0 a b a b a b a b a ab b + + + - Suy ra : ( ) ( ) ( )( )17 17: 2 1 0 17 31 3 031 31: 2 1 0 3 0a b AC x y x ya b AC x y x y

+

+ +

- (AC) ct (BC) ti C 21513 7 14 52 ;5 15 3 33 0x ty t t Cx y + _ ' , - (AC) ct (AB) ti A :( )2 1 0 77; 43 0 4x y xAx y y + ' ' - (AD) vung gc vi (AB) ng thi qua A(7;4) suy ra (AD) : 74 2x ty t + ' - (AD) ct (BD) ti D : 77 98 464 2 ;15 15 157 14 0x ty t t Dx y + _ ' , + - Trng hp (AC) : 17x-31y-3=0 cc em lm tng t .Bi 13. 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 thngd1: x + y + 5 = 0v d2: x + 2y 7 = 0. Vit phng trnh ng trn c tm C v tip xc vi ng thng BGGii- B thuc d suy ra B :5x ty t ' , C thuc d' cho nn C: 7 2 x my m '.- Theo tnh cht trng tm : ( ) 2 922, 03 3G Gt mm tx y + - Ta c h : 2 12 3 1m t mt m t ' ' Bin son t-6-2012( Ti liu ni b-lu )Trang 7A(2;3)BCx+y+5=0x+2y-7=0G(2;0)MChuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218- Vy : B(-1;-4) v C(5;1) . ng thng (BG) qua G(2;0) c vc t ch phng( ) 3; 4 u r, cho nn (BG):( )20 15 82 134 3 8 0 ;3 4 5 5x yx y d CBG R - Vy ng trn c tm C(5;1) v c bn knh R= ( ) ( ) ( )2 2 13 169: 5 15 25C x y + Bi 14. Tam gic cn ABC c y BC nm trn ng thng : 2x 5y + 1 = 0, cnh bn AB nm trn ng thng : 12x y 23 = 0 . Vit phng trnh ng thng AC bit rng n i qua im (3;1) Gii- ng (AB) ct (BC) ti B 2 5 1 012 23 0x yx y + ' Suy ra : B(2;-1). . (AB) c h s gc k=12, ng thng (BC) c h s gc k'=25 , do ta c :2125tan 221 12.5B +. Gi (AC) c h s gc l m th ta c : 22 55tan25 215mmCmm ++. V tam gic ABC cn ti A cho nn tanB=tanC, hay ta c :82 5 4 102 52 2 5 2 2 592 5 4 10 5 212m m m mm mm m mm

+

+

+ - Trng hp :( ) ( )9 9: 3 1 9 8 35 08 8m AC y x x y + + - Trng hp : m=12 suy ra (AC): y=12(x-3)+1 hay (AC): 12x-y-25=0 ( loi v n //AB ).- Vy (AC) : 9x+8y-35=0 .Bi 15. Vit phng trnh tip tuyn chung ca hai ng trn : (C1) : (x - 5)2 + (y + 12)2 = 225 v (C2) :(x 1)2 + ( y 2)2 = 25Gii : .- Ta c (C) vi tm I(5;-12) ,R=15. (C') c J(1;2) v R'=5. Gi d l tip tuyn chung c phng trnh : ax+by+c=0 (2 20 a b + ).- Khi ta c :( ) ( ) ( ) ( )2 2 2 25 12 2, 15 1 , , 5 2a b c a b ch I d h Jda b a b + + + + +- T (1) v (2) suy ra : 5 12 3 6 35 12 3 25 12 3 6 3a b c a b ca b c a b ca b c a b c + + + + + + + 9322a b ca b c

+ . Thay vo (1) : 2 22 5 a b c a b + + +ta c hai trng hp :- Trng hp : c=a-9b thay vo (1) :( ) ( )22 2 2 22 7 25 21 28 24 0 a b a b a ab b + + Bin son t-6-2012( Ti liu ni b-lu )Trang 8AB C2x-5y+1=0M(3;1)H12x-y-23=0Chuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218Suy ra :14 10 7 14 10 7 175 10 7: 021 21 2114 10 7 14 10 7 175 10 7: 021 21 21a d x ya d x y _ + +

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_+ +

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, - Trng hp :( ) ( ) ( )22 2 2 232 1 : 7 2 100 96 28 51 02c a b b a a b a ab b + + + + . V nghim . ( Ph hp v :16 196 212 ' 5 15 20 400 IJ R R + < + + . Hai ng trn ct nhau ) .Bi 16. Trong mt phng vi h to Oxy, cho ng trn (C) : 2 2x y 2x 8y 8 0 + + . Vit phng trnh ng thng song song vi ng thng d: 3x+y-2=0 v ct ng trn theo mt dy cung c di bng 6.Gii- ng thng d' song song vi d : 3x+y+m=0 - IH l khong cch t I n d' : 3 4 15 5m mIH + + + - Xt tam gic vung IHB : 22 225 9 164ABIH IB _ ,( )219 ' : 3 19 0 116 1 2021 ' : 3 21 0 25m d x y mmm d x y + + + + + Bi 17. Vit phng trnh cc cnh ca tam gic ABC bit B(2; -1), ng cao v ng phn gic trong qua nh A, C ln lt l : (d1) : 3x 4y + 27 = 0 v (d2) : x + 2y 5=0Gii- ng thng (BC) qua B(2;-1) v vung gc vi (AH) suy ra (BC):2 31 4x ty t + ' , hay :( )2 14 3 7 0 4; 33 4x yx y n + + r- (BC) ct (CK) ti C : ( )2 31 4 1 1;32 5 0x ty t t Cx y + '+ - (AC) qua C(-1;3) c vc t php tuyn( ) ; n a b rSuy ra (AC): a(x+1)+b(y-3)=0 (*). Gi 4 6 10 2os =5 16 9 5 5 5KCB KCA c + +R R- Tng t :( ) ( )22 22 2 2 2a+2b a+2b2os = 2 455 5c a b a ba b a b + ++ +Bin son t-6-2012( Ti liu ni b-lu )Trang 9I(-1;4)ABHB(2;-1)ACx+2y-5=03x-4y+27=0HKChuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218( )( ) ( )20 3 0 3 03 4 04 41 3 0 4 3 5 03 3a b y ya abba x y x y

+ + +

- (AC) ct (AH) ti A :( )1 233 053 4 27 031 582315;3 , ;25 25 4 3 5 0253 4 27 0 58225yyxx yA Axx yx yy '

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- Lp (AB) qua B(2;-1) v 2 im A tm c trn . ( hc sinh t lp ).Bi 18. Trong mt phng vi h ta cc vung gc Oxy , xt tam gic ABC vung ti A, phng trnh ng thng BC l :3 x y -3= 0, cc nh A v B thuc trc honh v bn knh ng trn ni tiptam gic ABC bng 2 . Tm ta trng tm G ca tam gic ABC .Gii- ng thng (BC) ct Ox ti B : Cho y=0 suy ra x=1 , B(1;0) . Gi A(a;0) thuc Ox l nh ca gc vung ( a khc 1 ).. ng thng x=a ct (BC) ti C :( )( ); 3 1 a a .- di cc cnh : 2 2 21 , 3 1 2 1 AB a AC a BC AB AC BC a + - Chu vi tam gic : 2p=( )( )3 3 11 3 1 2 1 3 3 12aa a a a p+ + + + - Ta c : S=pr suy ra p=Sr.(*)Nhng S=( )2 1 1 3. 1 3 1 12 2 2AB AC a a a . Cho nn (*) tr thnh :( )( )( )23 2 31 33 3 1 1 1 1 2 3 12 41 2 3aa a aa

++ + - Trng tm G : ( )( )( )12 3 2 3 12 17 4 33 7 4 3 2 3 63 3;3 3 3 13 2 2 32 3 633 3GGGGaxxGayy+ ++ + _+ + ' ' + , + ( )( )( )22 1 2 3 12 11 4 33 1 4 3 2 3 63 3;3 3 3 13 2 2 32 3 633 3GGGGaxxGayy ++ + _+ + ' ' , + Bi 19. Trong mt phng vi h ta Oxy. Cho ng trn (C) :0 1 2 42 2 + y x y xv ng thng d : 0 1 + + y x. Tm nhng im M thuc ng thng d sao cho t im M k c n (C) hai tip tuyn hp vi nhau gc 090GiiBin son t-6-2012( Ti liu ni b-lu )Trang 10Chuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218- M thuc d suy ra M(t;-1-t). . Nu 2 tip tuyn vung gc vi nhau th MAIB l hnh vung ( A,B l 2 tip im ). Do AB=MI= IA 2 =R 2 = 6 2 2 3 .- Ta c : ( ) ( )2 222 2 2 8 2 3 MI t t t + + + - Do : ( )( )12 222 2; 2 12 8 12 22 2; 2 1t Mt tt M

+

.* Ch : Ta cn cch khc - Gi d' l ng thng qua M c h s gc ksuy ra d' c phng trnh : y=k(x-t)-t-1, hay : kx-y-kt-t-1=0 (1) .- Nu d' l tip tuyn ca (C) k t M th d(I;d')=R 22 261k kt tk +( ) ( ) ( ) ( ) ( ) ( )22 2 2 22 2 6 1 4 2 2 2 2 4 2 0 t k t k t t k t t k t t + + + + + 1 ]- T gi thit ta c iu kin :( ) ( ) ( )22 2 2224 2 0' 4 2 4 2 4 04 214 2t tt t t t tt tt t + >'+ - ( )1 2 2 21 221 22 61' 19 0 2 ; 212tk kt t t k k Mk kt t+ t > t ' ' Bi 20. Trong mt phng vi h ta Oxy. Cho elip (E) : 0 4 42 2 + y x.Tm nhng im N trn elip (E)sao cho : 02 160 F N F ( F1 , F2 l hai tiu im ca elip (E) )Gii- (E) : 22 2 2 21 4, 1 3 34xy a b c c + - Gi( ) ( )2 20 00 0 1 0 2 01 24 43 3; 2 ; 22 22 3x yN x y E MF x MF xFF + + '. Xt tam gic 1 2FMF theo h thc hm s cos :( )22 2 01 2 1 2 1 22 os60 FF MF MF MFMF c + ( )2 220 0 0 03 3 3 32 3 2 2 2 22 2 2 2x x x x _ _ _ _ + + +

, , , ,Bin son t-6-2012( Ti liu ni b-lu )Trang 11Mx+y+1=0ABI(2;1)Chuyn : HNH HC PHNGNguyn nh S -T: 0985.270.2180 02 2 2 2 20 0 0 0 0004 2 13 3 9 32 13 312 8 4 81 2 4 4 9 94 233x yx x x x yyx

_

+

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-Nh vy ta tm c 4 im :1 2 3 44 2 1 4 2 1 4 2 1 4 2 1; , ; , ; , ;3 3 3 3 3 3 3 3N N N N _ _ _ _ , , , ,Bi 21. Trong m t ph ng to a Oxy cho i m A(1;1) v ng thng: 2x + 3y + 4 =0Ti m ta im B thuc ng thng sao cho ng thng AB v hp vi nhau gc 450.Gii- Gi d l ng thng qua A(1;1) c vc t php tuyn( ) ; n a b r th d c phng trnh dng : a(x-1)+b(y-1)=0 (*). Ta c( ) 2;3 n uur.- Theo gi thit :( ) ( ) ( )20 2 22 22 3 1os d, os45 2 2 3 13213a bc c a b a ba b+ + ++( ) ( )( ) ( )2 21 1: 1 1 0 5 4 05 5 5 24 5 05 : 5 1 1 0 5 6 0a b d x y x ya ab ba b d x y x y

+ +

+ +

- Vy B l giao ca d vi cho nn : 1 1 2 25 4 0 5 6 032 4 22 32; , : ;2 3 4 0 2 3 4 0 13 13 13 13x y x yB B B Bx y x y + + _ _ ' ' + + + + , , Bi 22. Trong mt phng vi h trc to Oxy cho cho hai ng thng 0 5 2 :1 + y x d.d2: 3x +6y 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.Gii- Trc ht lp phng trnh 2 ng phn gic to bi 2 ng thng ct nhau : 3 6 7 2 59 3 8 03 5 53 6 7 2 5 3 9 22 03 5 5x y x yx yx y x y x y+ +

+ +

+ + +

- Lp ng thng 1 qua P(2;-1) v vung gc vi tip tuyn : 9x+3y+8=0 . 12 1: 3 5 09 3x yx y + - Lp 2 qua P(2;-1) v vung gc vi : 3x-9y+22=0 22 1: 3 5 03 9x yx y + + Bi 23. Trong mt phng vi h trc to Oxy cho Hypebol (H) c phng trnh: 19 162 2 y x. Vit Bin son t-6-2012( Ti liu ni b-lu )Trang 12P(2;-1)d:2x-y+5=0d':3x+6y-7=0Chuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218phng 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).Gii- (H) c( ) ( )2 2 21 216, 9 25 5 5; 0 , 5; 0 a b c c F F . V hnh ch nht c s ca (H) c cc nh :( ) ( ) ( ) ( ) 4; 3 , 4; 3 , 4; 3 , 4;3 .- Gi s (E) c : 2 22 21x ya b+ . Nu (E) c tiu im trng vi tiu im ca (H) th ta c phng trnh :( )2 2 225 1 c a b - (E) i qua cc im c honh 216 x v tung ( )22 216 99 1 2 ya b + - T (1) v (2) suy ra :( )2 22 240, 15 : 140 15x ya b E + Bi 24. Trong mt phng vi h to Oxy, cho ng trn (C) c phng trnh: 2 24 3 4 0 x y x + + Tia Oy ct (C) ti A. Lp phng trnh ng trn (C), bn knh R = 2 v tip xc ngoi vi (C) ti AGii- (C) c I( 2 3; 0 ), R= 4 . Gi J l tm ng trn cn tm : J(a;b) ( ) ( ) ( )2 2' : 4 C x a y b + -Do (C) v (') tip xc ngoi vi nhau cho nn khong cch IJ =R+R'( )22 2 22 3 4 2 6 4 3 28 a b a a b + + + + + - V A(0;2) l tip im cho nn :( ) ( ) ( )2 20 2 4 2 a b + - Do ta c h : ( )( )222 22 2222 3 364 3 244 02 4a ba a ba b ba b+ + + + ' ' + + - Gii h tm c : b=3 v a= ( )( )( )223 ' : 3 3 4 C x y + . * Ch : Ta c cch gii khc .- Gi H l hnh chiu vung gc ca J trn Ox suy ra OH bng a v JH bng b - Xt cc tam gic ng dng : IOA v IHJ suy ra : 4 2 3 2IJ 6 2 3IA IO OAIH HJ b a +- T t s trn ta tm c : b=3 v a=3 .Bi 25. Trong mt phng vi h to Oxy, cho hnh ch nht ABCD c cnh AB: x -2y -1 =0, ng cho BD: x- 7y +14 = 0 v ng cho AC i qua im M(2;1). Tm to cc nh ca hnh ch nhtGii- Hnh v : ( Nh bi 12 ).Bin son t-6-2012( Ti liu ni b-lu )Trang 13I(-2;0)A(0;2)yxChuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218- Tm ta B l nghim ca h :( )2 1 07;37 14 0x yBx y ' + . - ng thng (BC) qua B(7;3) v( ) ( ) ( )71; 2 :3 2BCx tAB u BCy t + ' uuur 12 17 02BCx y k + . Mt khc : 1 11 1 17 2, tan1 17 2 317 2BD ABk k +- Gi (AC) c h s gc l k 21 27 1 2tan 37 3tan 217 1 tan 41 17 9kkkk + + - Do : 1728 4 3 214 7 1 3 73128 4 3 211k k kk kk kk

+

+- Trng hp : k=1 suy ra (AC) : y=(x-2)+1 , hay : x-y-1=0 .- C l giao ca (BC) vi (AC) :( )73 2 1, 6; 51 0x ty t t Cx y + ' - A l giao ca (AC) vi (AB) :( )73 2 0, 1; 02 1 0x ty t t Ax y + ' - (AD) //(BC) suy ra (AD) c dng : 2x+y+m=0 (*) , do qua A(1;0) : m= -2 . Cho nn (AD) c phng trnh : 2x+y-2=0 . - D l giao ca (AD) vi (BD) :( )2 2 00; 27 14 0x yDx y+ ' + - Trng hp : k=- 1731 cch gii tng t ( Hc sinh t lm ).Bi 26. Trong mp (Oxy) cho ng thng () c phng trnh: x 2y 2 = 0 v hai im A (-1;2); B (3;4). Tm im M() sao cho 2MA2 + MB2 c gi tr nh nhtGii- M thuc suy ra M(2t+2;t ) - Ta c :( ) ( )2 22 2 2 22 3 2 5 8 13 2 10 16 26 MA t t t t MA t t + + + + + +Tng t :( ) ( )2 22 22 1 4 5 12 17 MB t t t t + +- Do d : f(t)=( )2215 4 43 ' 30 4 015t t f t t t + + + . Lp bng bin thin suy ra min f(t) = 64115 t c ti 2 26 2;15 15 15t M _ ,Bi 27. Cho ng trn (C): x2 + y2 2x 6y + 6 = 0 v im M (2;4)Bin son t-6-2012( Ti liu ni b-lu )Trang 14Chuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218Vit phng trnh ng thng i qua M ct ng trn ti 2 im A v B, sao cho M l trung im ca ABGii- ng trn (C) : ( ) ( ) ( )2 2/( )1 3 4 1; 3 , 2, 1 1 4 2 0M Cx y I R P M + + < nm trong hnh trn (C) .- Gi d l ng thng qua M(2;4) c vc t ch phng( )2; :4x atu a b dy bt + ' +r- Nu d ct (C) ti A,B th :( ) ( ) ( ) ( ) ( )2 22 2 21 1 4 2 2 0 1 at bt a b t a bt + + + + + + ( c 2 nghim t ) . V vy iu kin :( ) ( ) ( )22 2 2 2' 2 3 2 3 0 * a b a b a ab b + + + + + >- Gi( ) ( )1 1 2 22 ; 4 , 2 ; 4 A at bt B at bt + + + + M l trung im AB th ta c h :( )( )( )( )1 2 1 21 21 2 1 24 4 008 8 0at t at tt tbt t bt t+ + + + ' '+ + + . Thay vo (1) khi p dng vi t ta c :( )1 2 2 222 40 0 : : 6 01 1a bx yt t a b a b d d x ya b+ + + + + Bi 28. Vit phng trnh cc tip tuyn ca e lp (E): 2 2116 9x y+ , bit tip tuyn i qua imA(4;3) Gii- Gi s ng thng d c vc t php tuyn( ) ; n a b r qua A(4;3) th d c phng trnh l :a(x-4)+b(y-3)=0 (*) , hay : ax+by-4a-3b (1) .- d l tip tuyn ca (E) th iu kin cn v l :( )22 2.16 .9 4 3 a b a b + +2 2 2 20 : 3 016 9 16 24 9 24 00 : 4 0a d ya b a ab b abb d x + + + Bi 29. Trong mt phng vi h ta Oxy, cho ng trn (C): x2 + y2 - 2x - 2my + m2 - 24 = 0 c tm I v ng thng : mx + 4y= 0. Tm m bit ng thng ct ng trn (C) ti hai im phn bit A,B tha mn din tch tam gic IAB bng 12.Gii- (C) :( ) ( )2 21 25 (1; ), 5 x y m I m R + .- Nu d : mx +4y=0 ct (C) ti 2 im A,B th ( )2 22 2416 42 24 0 116 4my xm mx x m ' _ _ + + + , , - iu kin : 2' 25 0 m m R + > . Khi gi 1 1 2 2; , ;4 4m mA x x B x x _ _ , , ( ) ( )2 2 22 22 1 2 1 2 1216 25816 416m m mAB x x x x x xm+ + + +Bin son t-6-2012( Ti liu ni b-lu )Trang 15Chuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218- Khong cch t I n d = 2 24 516 16m m mm m++ +- T gi thit : 2 222 251 1 25 25. .8 . 4 5 122 2 1616 16mm mS AB d mmm m+ + ++ +( ) ( )222 2 22255 3 25 25 9 1616mm m m mm+ + ++- Ta c mt phng trnh trng phng , hc sinh gii tip . Bi 30. Trong mt phng vi h ta Oxy, cho tam gic ABC c phng trnh cnh AB: x - y - 2 = 0, phng trnh cnh AC: x + 2y - 5 = 0. Bit trng tm ca tam gic G(3; 2). Vit phng trnh cnh BCGii- (AB) ct (AC) ti A :( )2 03;12 5 0x yAx y '+ - B nm trn (AB) suy ra B(t; t-2 ), C nm trn (AC) suy ra C(5-2m;m) - Theo tnh cht trng tm : ( )( )2 832 1; 22 131 7 5 5;323GGt mxm Ct mt m t m t By + ' ' '+ + Bi 31. Vit phng trnh ng trn i qua hai im A(2; 5), B(4;1) v tip xc vi ng thng c phng trnh 3x y + 9 = 0.Gii- Gi M l trung im AB suy ra M(3;3 ) . d' l ng trung trc ca AB th d' c phng trnh : 1.(x-3)-2(y-3)=0 , hay : x-2y+3=0 . - Tm I ca (C) nm trn ng thng d' cho nn I(2t-3;t) (*)- Nu (C) tip xc vi d th( )( ) 3 2 3 9 510,2 10 10t t th I d R t R + . (1)- Mt khc : R=IA=( ) ( )2 25 2 5 t t + . (2) .- Thay (2) vo (1) : ( ) ( ) ( )2 22 2105 2 5 4 5 30 50 102t t t t t t + + 26 3412 2 06 34tt tt

+ + . Thay cc gi tr t vo (*) v (1) ta tm c ta tm I v bn knh R ca (C) .* Ch : Ta c th s dng phng trnh (C) : 2 22 2 0 x y ax by c + + ( c 3 n a,b,c)- Cho qua A,B ta to ra 2 phng trnh . Cn phng trnh th 3 s dng iu kin tip xc ca (C) v d : khong cch t tm ti d bng bn knh R .Bi 32. Cho ng trn (C): x2 + y2 2x + 4y + 2 = 0. Vit phng trnh ng trn (C') tm M(5, 1) bit (C') ct (C) ti cc im A, B sao cho3 AB .Gii- ng trn (C) :Bin son t-6-2012( Ti liu ni b-lu )Trang 16I MABHChuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218( ) ( ) ( )2 21 2 3 1; 2 , 3 x y I R + + .- Gi H l giao ca AB vi (IM). Do ng trn (C') tm M c bn knh R' = MA . Nu AB=3 IA R , th tam gic IAB l tam gic u , cho nn IH=3. 3 32 2( ng cao tam gic u ) . Mt khc : IM=5 suy ra HM= 3 752 2 . - Trong tam gic vung HAM ta c 22 2 249 313 '4 4 4ABMA IH R + + - Vy (C') :( ) ( )2 25 1 13 x y + .Bi 33. Trong mt phng vi h ta Oxy cho ng trn (C) c ph-ng trnh (x-1)2 + (y+2)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.Gii- (C) c I(1;-2) v bn knh R=3 . Nu tam gic ABC vung gc ti A ( c ngha l t A k c 2 tip tuyn ti (C) v 2 tip tuyn vung gc vi nhau ) khi ABIC l hnh vung . Theo tnh cht hnh vung ta c IA= IB 2 (1) .- Nu A nm trn d th A( t;-m-t ) suy ra : ( ) ( )2 21 2 IA t t m + + . Thay vo (1) :( ) ( )2 21 2 3 2 t t m + + ( )2 22 2 1 4 13 0 t m t m m + (2). trn d c ng 1 im A th (2) c ng 1 nghim t , t ta c iu kin : ( ) ( )2210 25 0 5 0 5 m m m m + + + .Khi (2) c nghim kp l :( )1 2 01 5 13 3;82 2mt t t A Bi 34. Trong mt phng to Oxy cho hai ng thng (d1) : 4x - 3y - 12 = 0 v (d2): 4x + 3y - 12 = 0. Tm to tm v bn knh ng trn ni tip tam gic c 3 cnh nm trn (d1), (d2), trc Oy.Gii- Gi A l giao ca( )1 24 3 12 0, : 3; 0 Ox4 3 12 0x yd d A Ax y '+ - V (BC) thuc Oy cho nn gi B l giao ca 1d vi Oy : cho x=0 suy ra y=-4 , B(0;-4) v C l giao ca 2d vi Oy : C(0;4 ) . Chng t B,C i xng nhau qua Ox , mt khc A nm trn Ox v vy tam gic ABC l tam gic cn nh A . Do tm Ing trn ni tip tam gic thuc Ox suy ra I(a;0). - Theo tnh cht phn gic trong : 5 5 4 94 4 4IA AC IA IO OAIO AO IO IO+ + 4 4.3 49 9 3OAIO . C ngha l I(4; 03) Bin son t-6-2012( Ti liu ni b-lu )Trang 17I(1;-2)BCAx+y+m=0Chuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218- Tnh r bng cch : ( ) ( ) 5 8 51 1 15 1 1 18 6. .5.32 2 2 2 2 15 5AB BC CAS BC OA rr r+ + + + .Bi 35. Trong mt phng to Oxy cho im C(2;-5 ) v ng thng : : 3 4 4 0 x y + . Tm trn hai im A v B i xng nhau qua I(2;5/2) sao cho din tch tam gic ABC bng15Gii- Nhn xt I thuc, suy ra A thuc : A(4t;1+3t) . Nu B i xng vi A qua I th B c ta B(4-4t;4+3t) ( ) ( )2 216 1 2 9 1 2 5 1 2 AB t t t + - Khong cch t C(2;-5) n bng chiu cao ca tam gic ABC : 6 20 465+ + - T gi thit : ( ) ( )( ) ( )0 0;1 , 4; 41 1. 5. 1 2 .6 15 1 2 12 2 1 4; 4 , 0;1t A BS AB h t tt A B

Bi 36. Trong mt phng vi h to Oxy cho elp 2 2( ) : 19 4x yE + 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.Gii- A,B c honh l honh ca 2 nh ca 2 bn trc ln ca (E) , chng nm trn ng thng y-2=0 . C c honh v tung dng th C nm trn cung phn t th nht - Tam gic ABC c AB=6 c nh . V th tam gic c din tch ln nht khi khong cch t C n AB ln nht . - D nhn thy C trng vi nh ca bn trc ln (3;0) Bi 37. Trong mt phng Oxy cho tam gic ABC bit A(2; - 3), B(3; - 2), c din tch bng 32 v trng tm thuc ng thng: 3x y 8 = 0. Tm ta nh C.Gii- Do G thuc suy ra G(t;3t-8). (AB) qua A(2;-3) c vc t ch phng( ) 1;1 u AB r uuur, cho nn (AB) : 2 35 01 1x yx y + . Gi M l trung im ca AB : M5 5;2 2 _ ,.- Ta c : 5 5 5 11; 3 8 ; 32 2 2 2GM t t t t _ _ + , ,uuuur. Gi s C( )0 0; x y, theo tnh cht trng tm ta c :( ) ( )0000525 2 22 2 5;9 19 19 19 113 8 2 32x t tx tGC GM C t ty ty t t _ + , ' ' _ + ,uuur uuuur- Ngoi ra ta cn c : AB= 2 ,( )( ) ( ) 3 2 5 9 19 8 4 3,10 10t t thC - Theo gi thit :( )4 31 1 3. , 2 2 4 3 3 102 2 2 10tS AB hC t Bin son t-6-2012( Ti liu ni b-lu )Trang 18Chuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218( )224 3 5 7 6 5; 7 9 53 32 4 3 90 9 24 29 04 3 5 6 5 7;9 5 73 3t Ct t tt C _ +

,

_+

, Bi 38. Trong mt phng Oxy cho elip (E): 2 214 3x y+ v ng thng :3x + 4y =12. T im M bt k trn k ti (E) cc tip tuyn MA, MB. Chng minh rng ng thng AB lun i qua mt im c nhGiiBi 39. Trong mt phng ta Oxy cho hnh ch nht ABCD c tm 1( ;0)2Ing thng AB c phng trnh: x 2y + 2 = 0, AB = 2AD v honh im A m. Tm ta cc nh ca hnh ch nht Gii- Do A thuc (AB) suy ra A(2t-2;t) ( do A c honh m cho nn t + + >

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, 2 22 2 221 22 22 2 24925 : 025: 025 2425244925 : 025 : 02525 24x yxx y x xOM MFMFx yx y x xx _

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Bi 79. Trong mat phang Oxy cho Hyperbol (H) : 12x2 16y2 = 192 va iemP(2 ; 1). Viet phng trnhng thang i qua P va cat (H) tai 2 iem M, N sao cho P la trung iem cua MN.Gii(H): ( ) ( )2 21 21 4, 2 3, 2 7 2 7; 0 , 2 7; 016 12x ya b c F F . Gi M(x;y) thuc (H) v N i xng vi M qua P(2;1) th N(4-x; 2-y) . tha mn yu cu bi ton th N phi thuc (H)., do ta c h : ( )( ) ( )( )2 22 21 116 124 21 216 12x yx y ' . Ly (2)-(1) ta c phng trnh rt gn : 3x-2y-4=0 . cng chnh l phng trnh ng thng qua P .Bi 80. Trong mat phang Oxy cho (E) : 4x2 + y2 = 4.a. Tnh o dai truc ln, truc nho, toa o hai tieu iem, tam sai cua (E).b. Tm cac gia tr cua m e ng thang y = x + m cat (E) tai 2 iem phan biet M, N khi m thay oi. Tm tap hp cac trung iem cua MNGiia/ (E): ( ) ( )2 21 21 1, 2, 3 0; 3 , 0; 31 4x ya b c F F + . Tiu im thuc Oy .b/ ng thng y=x+m ct (E) ti 2 im M,N c t l nghim ca h :( )( )( )2 2 22 2 25 2 4 0 14 4 4 42x mx mx y x x my x m y x my x m + + + + + ' ' ' + + + - Nh vy honh ca M,N l 2 nghim ca (1) vi iu kin : 2' 4 20 0 m + > , hay :( ) 5 * m ( ) ( )2 22 22 2 2 294 ( '; ) ( ; ) 35 4. 35a bdI d dI da b a b + +2 22 22 23635 36a ba ba b +D thy0 b nn chn 616 aba.Kim tra iu kinIA IH >ri thay vo (*) ta c hai ng thng tho m n.Bi 95. .Tam gic cn ABC c y BC nm trn ng thng : 2x 5y + 1 = 0, cnh bn AB nm trn ng thng : 12x y 23 = 0 . Vit phng trnh ng thng AC bit rng n i qua im (3;1) HDBi 96. . Trong mt phng vi h ta Oxy. Cho elip (E) : 0 4 42 2 + y x.Tm nhng Bin son t-6-2012( Ti liu ni b-lu )ng thng AC i qua im (3 ; 1) nn c phng trnh : a(x 3) + b( y 1) = 0 (a2 + b2 0)Gc ca n to vi BC bng gc ca AB to vi BC nn :2 2 2 2 2 2 2 22a 5b 2.12 5.12 5 . a b 2 5 . 12 1 ++ + + + 2 22a 5b295a b +( ) ( )22 25 2a 5b 29 a b + 9a2 + 100ab 96b2 = 0a 12b8a b9

Nghim a = -12b cho ta ng thng song song vi AB ( v im ( 3 ; 1) khng thuc AB) nn khng phi l cnh tam gic . Vy cn li : 9a = 8b hay a = 8 v b = 9 Phng trnh cn tm l :8x + 9y 33 = 0Trang 45Chuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218HD+ (C) c tm I(2 , 1) v bn knh R =6 +B A B M A , ( 900 l cc tip im ) suy ra : 12 2 . 2 . R MA MIVy M thuc ng trn tm I bn knh R/ =12v M thuc d nn M( x , y) c ta tha h:

( ) ( )'+ ' ' + + + 2 122 120 112 1 22 2yxyxy xy xVy c 2 im tha yu cu bi ton c ta nu trnBi 97. . Trong mp (Oxy) cho ng thng () c phng trnh: x 2y 2 = 0 v hai im A (-1;2); B (3;4). Tm im M() sao cho 2MA2 + MB2 c gi tr nh nhtHDM (2 2; ), (2 3; 2), (2 1; 4) M t t AM t t BM t t + + uuuur uuuur2 2 22 15 4 43 ( ) AM BM t t f t + + + Min f(t) = 215f _ ,=> M26 2;15 15 _ ,Bi 98. . Trong mt phng vi h ta Oxy, cho ng trn (C): x2 + y2 - 2x - 2my + m2 - 24 = 0 c tm I v ng thng : mx + 4y= 0. Tm m bit ng thng ct ng trn (C) ti hai im phn bit A,B tha mn din tch tam gic IAB bng 12.HDng trn (C) c tm I(1; m), bn knh R = 5.Gi H l trung im ca dy cung AB. Ta c IH l ng cao ca tam gic IAB.IH = 2 2| 4 | | 5 |( , )16 16m m mdIm m+ + +22 222(5 ) 20251616mAH IA IHmm ++Din tch tam gic IAB l 12 2 12 SIAB IAHS 23( , ). 12 25 | | 3( 16)163mdI AH m mm t

+

tBi 99. . Cho ng trn (C): x2 + y2 2x + 4y + 2 = 0. Vit phng trnh ng trn (C') tm M(5, 1) bit (C') ct (C) ti cc im A, B sao cho 3 AB .HD Phng trnh ng trn (C): x2 + y2 2x + 4y + 2 = 0 c tm I(1, 2)3 R ng trn (C') tm M ct ng trn (C) ti A, B nn AB IM ti trung im H ca on AB. Ta c 232ABBH AH . C 2 v tr cho AB i xng qua tm I.Gi A'B' l v tr th 2 ca AB, Gi H' l trung im ca A'B'Bin son t-6-2012( Ti liu ni b-lu )Trang 46IA BH5Chuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218Ta c: 22 23 3IH' IH IA AH 32 2 _ ,, ( ) ( )2 2MI 5 1 1 2 5 + + Vy c 2 ng trn (C') tha ycbt l: (x 5)2 + (y 1)2 = 13hay (x 5)2 + (y 1)2 = 43Bi 100. . Trong mt phng Oxy cho elip (E): 2 214 3x y+ v ng thng:3x + 4y =12. T im M bt k trn k ti (E) cc tip tuyn MA, MB. Chng minh rng ng thng AB lun i qua mt im c nh.HDGi M(x0 ;y0 ), A(x1;y1), B(x2;y2) . Tip tuyn ti A c dng : 1 114 3xx yy+ Tip tuyn i qua M nn : 0 1 0 114 3x x yy+ (1)Ta thy ta ca A v B u tha mn (1) nn ng thng AB c pt : 0 014 3xx yy+ do M thuc nn 3x0 + 4y0 =12 4y0 =12-3x0 0 04 444 3xx yy+ , 0 04 (12 3 )44 3xx y x + Gi F(x;y) l im c nh m AB i qua vi mi M th : (x- y)x0 + 4y - 4 = 0{ {0 14 4 0 1x y yy x . Vy AB lun i qua im c nh F(1;1)Bi 101. . Trong mt phng Oxy cho ba ng thng 1: 4 9 0, d x y +2: 2 6 0, d x y + 3: 2 0 d x y +. Tm ta cc nh ca hnh thoi ABCD, bit hnh thoi ABCD c din tch bng15, cc nh A,C thuc3d, B thuc1d v D thuc2d. HDng cho (BD) vung gc vi (AC) cho nn (BD c dng : x+y+m=0 (BD) ct 1d ti B c ta l nghim ca h : 09 4 9;4 9 0 3 3x y mm mBx y+ + + + _ ' + ,(BD) ct 2d ti D c ta l nghim ca h : 06 2 6;2 6 0 3 3x y mm mDx y+ + + + _ ' + ,Trung im I ca BD l tm hnh thoi c ta l : 1 2 1;2 2mI+ _ ,Theo gi thit I thuc (AC) :( )1 2 12 0 3 : 3 02 2mm BD x y++ + + v ta cc im B(2;1),D(-1;4) v I1 5;2 2 _ ,. Gi A(t;t+2) thuc (AC) . Suy ra : ( )2 3, ( )2t th A AC+ + ( )( ) ( )( ) ( )3 3;5 2; 02 1 2 112 . , 3 2 152 2 2; 0 3;5 2 2t A Ct tS BD h AACt A C

Bin son t-6-2012( Ti liu ni b-lu )Trang 47Chuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218Bi 102. Trong (Oxy) cho ng trn (C): 2 232x y + v ( )2: P y x . Tm trn (P) cc im M m t k c 2 tip tuyn n (C) v 2 tip tuyn to vi nhau mt gc 060GiiGi M( ) ( )20 0 0 0; x y P y x . d l ng thng tip tuyn ca (P) ti M th d c phng trnh :( )0 0 0 012 02yy x x x yy x + + . d l tip tuyn k t M n (C) th iu kin cn v l : Bi 103. Trong (Oxy) cho . thng d: 3x-y+5=0 v ng trn (C): 2 22 6 9 0 x y x y + + + . Tm im M thuc (C) v im N thuc d sao cho MN c di nh nht ?Gii(C) : ( ) ( ) ( )2 21 3 1 1;3 , 1 x y I R + + - Gi d' //d th d': 3x-y+m=0 . d' tip xc vi (C) ti M ( M l im cch d nh nht ) , khi : ( )6 10 ' : 3 6 10 0 3 3; ' 1 6 10106 10 ' : 3 6 10 0m d x y mh I d R mm d x y

+ + + + + Gi s N' thuc d ta lun c : 2 2' MN MN . Du bng ch xy ra khi N' trng vi N . Vy ta ch cn lp ng thng qua I(-1;3) v vung gc vi d suy ra ng thng 1 3:3x ty t + ' . Khi ct d' ti 2 im : ( ) ( )13 1 3 3 6 10 010t t t + + + .V ( ) ( )13 1 3 3 6 10 010t t t + + .Do vy ta tm c 2 im M : 13 11;310 10M _ ,, v 23 11 ;310 10M _ + ,. Tng t ct d ti N c ta l nghim : 1 31 7 293 ;10 10 103 5 0x ty t t Nx y + _ ' , + . Ta chn M bng cch tnh 1 2, M NMN, sau so snh : Nu 1 2M N MN > th M l 2M. Cn 1 2M N MN < th M l 1M.Bi 104. Trong (Oxy) cho( ) ( ) ( )2 2: 1 3 1 C x y + + v im 1 7;5 5M _ ,. Tm trn (C) im N sao cho MN c di ln nht ?Gii(C) vit di dng tham s :( ) ( )1 sin1 sin ;3 ost3 ostx tN C N t cy c + + +' +Bin son t-6-2012( Ti liu ni b-lu )Trang 48d'd'd:3x-y+5=0MMNN'I(-1;3)Chuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218Khi : 2 22 26 8 12 16sin ost sin os sin ost+45 5 5 5MN t c t c t t c _ _ + + + + , ,( )12 16 12 16sin ost+5 5 4 sin ost *5 5 20 20t c t c _ + ,. V : 2 212 16120 20 _ _+ , ,,12 3 16 4os ;sin =20 5 20 5c th (*) tr thnh :( ) 5 4sin 5 4 1 t + Du ng thc xy ra khi :( ) sin 1 22t t k + + +Do vy : 3 3 2sin sin os = 1 sin 12 5 5 5t c x t _ + + ,Tng t : 4 4 19 2 19ost=cos sin 3 ost=3+ ;2 5 5 5 5 5c y c N _ _ + , ,Bi 105. Tnh din tch tam gic u ni tip (E): 2 2116 4x y+ , nhn A(0;2) lm nh v trc Oy lm trc i xng ?GiiDo ABC l tam gic u , A(0;2) thuc Oy l trc i xng cho nn B,C phi nm trn ng thng y=m (//Ox) ct (E) . V th ta ca B,C l nghim ca h : 2 2 2 24 16 64 16 4 0y m y mx y x m ' '+ 22 016 4y mmx m ' t . Ta c : 2 2 2 24 16 4 20 , 2 16 AC x m m BC m + + Do vy ABC u khi :AC=BC( )2 2 2 2 24420 2 16 20 4 163m m m m m Vy : 2 333m , suy ra( )2 21 1 1 1. . 3 3 3.4 162 2 2 2ABCS BC AH BC BC BC m Hay :44 8 32 3. 163 3ABCS _ , Bi 106. Tnh din tch tam gic u ni tip (P): 22 y x , nhn nh ca (P) lm nh v trc Ox lm trc i xng ?Gii (P) c tiu im F1; 02 _ ,. Nu Ox lm trc i xng th B,C nm trn ng thng : x=m ( song song vi Oy) . Do vy ta ca B,C l nghim ca h : 2 22 2 y x y mx m x m ' ' ( ) ( )2; 2 , ; 2 2 20y mBm m Cm m BC mx m t ' >Bin son t-6-2012( Ti liu ni b-lu )Trang 49A(0;2)B Cy=mHOxyChuyn : HNH HC PHNGNguyn nh S -T: 0985.270.218V OBC l tam gic u :( ) ( )2 2 2 22 4 2 6 0 6 0 OB BC m m m m m m m + >Vy ( ) ( )21 1 3 3. . 3 4 2 4 2.6 24 32 2 2 2OBCS BC OH BC m (vdt)Bi 107. Trong (Oxy) cho im M(1;2) . Lp phng trnh ng thng d qua M ct Ox,Oy ln lt ti A,B sao cho din tch tam gic OAB nh nht .Giing thng dng : x=1 v y=2 khng ct 2 trc ta . Cho nn gi d l ng thng qua M(1;2) c h s gc k( khc 0) th d : y=k(x-1)+2 , hay y=kx+2-k .ng thng d ct Ox ti A2; 0kk _ , v ct Oy ti B(0;2-k) Do : 21 2 1 4 4 1 42 42 2 2OABk k kS k kk k k + +(1)Xt f(k)= ( )24 44 ' 1 0 2 k f k kk k + tTa c bng bin thin : k --2 0 2 +f'(k) + 0 - - 0 +f(k)--16-6+Cn c vo bng bin thin ta c macx ( ) 16 f k t c khi k=-2 . Khi ng thng d : y=-2(x-1)+2 , hay y=-2x+4 v A(2;0) v B(0;4) .Bi 108. Trong (Oxy) cho tam gic ABC, bit ba chn ng cao tng ng vi 3 nh A,B,C l A'(1;1),B'(-2;3),C'(2;4). Vit phng trnh ng thng cha cnh (BC).GiiDo l cc ng cao cho nn t gic AC'IB' l t gic ni tip trong ng trn c ng knh l AI , C'B' l mt dy cung v vy AA' vung gc vi C'B' . Vy (BC) qua A'(1;1) v c vc t php tuyn ( ) ( ) ( ) ( ) ' ' 4; 1 // 4;1 : 4 1 1 0 CB n BC x y + uuuuur r4 5 0 x y + .Tng t nh lp lun trn ta tm ra phng trnh cc cnh ca tam gic ABC : (AB) : 3x-2y+2=0 ....Bi 109. Trong (Oxy) cho hai im ( ) ( )2 3; 2 , 2 3; 2 A B a/ Chng t tam gic OAB l tam gic u b/ Chng minh rng tp hp cc im M sao cho : 2 2 232 MO MA MB + + l mt ng trn (C). c/ Chng t (C) l ng trn ngoi tip tam gic OAB. Giia/ Ta c : ( )222 3 2 4, 4, 4 OA OB AB + . Chng t OAB l tam gic u .b/ Gi M(x;y) th ng thc gi thit cho tng ng vi biu thc :Ta c : 2 2 2 2 2 2 2 2 2, 4 3 4 16, 4 3 4 16 MO x y MA x y x y MB x y x y + + + + + +Bin son t-6-2012( Ti liu ni b-lu )Trang 50AA'(1;1) BB'(-2;3)CC'(2;4)IChuyn : HNH HC PHNGNguyn nh S -T: 0985.270.2182 2 2 2 2 2 28 332 3 3 8 3 32 32 03MO MA MB x y x x y x + + + + + 2 224 3 4 33 3x y _ _ + , ,. Chng t l ng trn (C) c tm 4 3 4 3; 0 ,3 3I R _ ,c/ Thay ta O,A,B vo (1) ta thy tha mn , chng t (C) l ng trn ngoi tip tam gic OAB.Bi 110. Vit phng trnh cc cnh hnh vung ABCD bit AB,CD,ln lt i qua cc im P(2;1) v Q(3;5), cn BC v AD qua cc im R(0;1) v S(-3;-1)GiiGi (AB) c dng y=kx+b v (AD) : y=-1/kx+b' .Cho AB v AD qua cc im tng ng ta c : 2k+b=1 (1) v( )3' 1 2 bk + Ta c :( ) ( )2 23 5 0 ', ; ,1 1k b k kbhQAB hRADk k + + + +. Theo tnh cht hnh vung :( ) ( )2 23 5 0 ', , 3 5 '1 1k b k kbhQAB h RAD k b k kbk k + + + + +T ta c h : 2 11 1 4' 3 , , ' 10 , 7, 15, '3 3 73 5 'k bk kb k b b k b bk b k kb + _ _+ '

, , + Do : : 3 1 0, : 3 10 0, : 3 12 0, : 3 1 0 AB x y AD x y CDx y BC x y + + + + + Hoc : : 7 15 0, : 7 4 0, : 7 26 0, : 7 7 0 AB x y ADx y CD x y BC x y + + + Bin son t-6-2012( Ti liu ni b-lu )Trang 51