Bài tập về phép biến hình – toán 11 nâng cao

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Bài tập về phép biến hình – Toán 11 nâng cao CHƯƠNG I : PHÉP DỜI HÌNH VÀ PHÉP ĐỒNG DẠNG TRONG MẶT PHẲNG Vấn đề 1 : PHÉP DỜI HÌNH A. KIẾN THỨC CƠ BẢN 1 Phep �bien�h� nh . ��N : Phep�bien�h� nh la� mot�quy ta� c ��� ev � i moi��iem � M cua�mat�phang � xac��� nh ���� c mot��iem � duy nhat� M� cua�mat�phang � , �iem � M� goi�la� anh � cua�M qua phe� p bien �h� nh �� o. �K�hieu �: f la� mot�phep �bien�h� nh nao ��� o va� M� la� anh � cua�M qua phe� p f th�ta viet�: M� = f(M) hay f f(M) = M�hay f : M I�� � M� hay M I�� � M� . �ie� m M goi�la�� tao anh �. f la� phep �bien �h� nh �� ong nhat�� f(M) = M ,  M �H . �iem � M goi�la� �iem � bat��� ong , kep �, bat�bien �. f1 ,f2 la� cac�phep�bien�h� nh th�f2 of1 la� phep �bien �h� nh . �Neu �H la� mot�� h nh nao ���� o th tap �h�� p cac��iem � M� = f(M), v�� i M �H, tao�thanh � mo� t h� nh H� ���� c goi�la� anh � cua�H qua phep�bien�h� nh f va� ta viet�: H� = f(H) . 2 Phep �d��� i h nh . �N : Phep �d��� i h nh la� phep �bien �h� nh khong � lam � thay �� oi khoang � cach � gi�� a hai �iem � bat�� k , t�� c la� v�� i � � �� hai �iem � bat�� k M,N va� anh � M , N cu� a chung � , ta luon �co�M N = MN . ( Bao �toan �khoang � cach �). 3 T� nh chat�: ( cua �phep �d��� i h nh ) . g�L : Phep�d��� i h nh bien �ba �iem � tha� ng hang � thanh � ba �iem � thang � hang � , ba �iem � khong � thang � hang � thanh � ba �iem � khong � thang � hang �. gHQ : Phep�d��� i h nh bien �: 1. ���� ng thang � thanh � ���� ng thang �. 2. Tia thanh � tia . 3. �oan �thang � thanh � �oan �thang � ba� ng no� . 4. Tam giac�thanh � tam giac�bang � no� . ( Tr�� c tam � I�� � tr� � c tam � , trong � tam � I�� � trong � tam �) 5. ���� ng tron �thanh � ���� ng tron �ba� ng no� . ( Tam � bien �thanh � tam � : I I�� � I� , R� =R) 6. Goc�thanh � goc�bang � no� . B . BÀI TẬP � x� = 2x  1 1 Trong mpOxy cho phep �bien �h� nh f : M (x;y) I�� � M� = f(M) = � . � y � =y+3 T� m anh � cua �cac �� iem � sau : a) A(1 ;2) b) B(  1;2) c) C(2;  4) Giai�: a) A� = f(A) = (1;5) � b) B = f(B) = (  7;6) c) C� = f(C) = (3;  1) � x� = 2x  y  1 2 Trong mpOxy cho phep �bien �h� nh f : M (x;y) I�� � M� = f(M) = � . y� = x  2y + 3 � T� m anh � cua �cac �� iem � sau : a) A(2 ;1) b) B(  1;3) c) C(  2;4) Giai�: a) A� = f(A) = (4;3) b) B� = f(B) = (  4;  4) c) C� = f(C) = (  7;  7) 3 Trong mpOxy cho phep �bien �h� nh f : M(x;y) I�� � M� = f(M) = (3x; y) . �ay �co� pha� i la� phep �d� � i h� nh hay khong �? -1- Bài tập về phép biến hình – Toán 11 nâng cao Giai�: Lay �hai � iem � bat�� k M(x1; y1),N(x 2 ; y 2 ) Khi �� o f : M(x1; y1 ) I�� � M� = f(M) = (3x1; y1) . f : N(x 2 ; y 2 ) I�� � N� = f(N) = (3x 2 ; y 2 ) Ta co� : MN = (x 2  x1 )2  (y 2  y1)2 , M�� N = 9(x 2  x1)2  (y2  y1 )2 Neu � x1 �x2 th� M�� N �MN . Vay �: f khong � phai�la� phep �d� �� i h nh . (V�co� 1 so� � iem � f khong � bao �toan �khoang � cach) � . 4 Trong mpOxy cho 2 phep �bien �h� nh : a) f : M(x;y) I�� � M� = f(M) = ( y{ ; x{ 2 ) x� b) g : M(x;y) I�� � M� = g(M) = ( 2x . { ; y+1) { y� x� y� Phep �bien �h� nh nao �tren ��� ay la� phep �d� �� i h nh ? HD : a) f la� phep �d� �� i h nh b) g khong � phai�la� phep �d� �� i h nh ( v�x1 �x 2 th� M�� N �MN ) 5 Trong mpOxy cho 2 phep �bien �h� nh : a) f : M(x;y) I�� � M� = f(M) = (y + 1 ;  x) b) g : M(x;y) I�� � M� = g(M) = ( x ; 3y ) . Phep �bien �h� nh nao �tren ��� ay la� phep �d� �� i h nh ? Giai�: a) f la� phep �d� �� i h nh b) g khong � phai�la� phep �d� �� i h nh ( v�y1 �y 2 th� M�� N �MN ) 6 Trong mpOxy cho phep �bien �h� nh f : M(x;y) I�� � M� = f(M) = (2x ; y  1) . T� m anh � cu� a� � � � ng thang � () : x  3y  2 = 0 qua phep �bien �h� nh f . Giai�: Cach � 1: Dung � bieu �th� � c toa� �� o �  x� � x� =  2x � x Ta co� f : M(x;y) I�� � M� = f(M) = � �� 2 y�  y 1 � � y  y� 1 �  x� V�M(x;y) �() � ( )  3(y�  1)  2  0 � x�  6y�  2  0 � M��� (x ;y ) �(� ) : x  6y  2  0 2 Cach � 2 : Lay �2 � iem � bat�� k M,N �() : M �N . gM �() : M(2;0) I�� � M�  f(M)  ( 4;1) gN �( ) : N(  1;  1) I�� � N�  f(N)  (2; 0) � gQua M� (4;1) x+ 4 y  1 uuuuur (� ) �(M�� N ):� � PTCtac�(� ):  � PTTQ (� ) : x  6y  2  0 6 1 gVTCP : M�� N  (6; 1) � 7 Trong mpOxy cho phep �bien �h� nh f : M(x;y) I�� � M� = f(M) = (x  3 ; y  1) . a) CMR f la� phep �d� �� i h nh . b) T� m anh � cua �� � � � ng tron �(C) : (x + 1)2 + (y  2)2 = 4 . -2- I�� � (C� ) : (x  2)2 + (y  3)2 = 4 Bài tập về phép biến hình – Toán 11 nâng cao 8 Trong mpOxy cho phep �bien �h� nh f : M(x;y) I�� � M� = f(M) = (x  3 ; y  1) . a) CMR f la� phep �d� �� i h nh . b) T� m anh � cua ��� � � ng thang � () : x + 2y  5 = 0 . c) T� m anh � cua �� � � � ng tron �(C) : (x + 1)2 + (y  2)2 = 2 . x2 y2 + =1. 3 2 Giai�: a) Lay �hai � iem � bat�� k M(x1; y1),N(x2 ; y2 ) d ) T� m anh � cua �elip (E) : Khi �� o f : M(x1;y1) I�� � M� = f(M) = (x1  3; y1  1) . f : N(x2 ; y2 ) I�� � N� = f(N) = (x2  3; y 2  1) Ta co� : M�� N = (x 2  x1 )2  (y 2  y1)2 = MN Vay �: f la� phe� p d� �� i h nh . b) Cach � 1: Dung � bieu �th� � c toa� �� o � x� = x 3 � x  x� 3 Ta co� f : M(x;y) I�� � M� = f(M) = � �� y�  y 1 y  y� 1 � � V�M(x;y) �() � (x�  3)  2(y�  1)  5  0 � x�  2y�  4  0 � M��� (x ;y ) �(� ) : x  2y  4  0 Cach � 2 : Lay �2 � iem � bat�� k M,N �() : M �N . gM �() : M(5 ;0) I�� � M�  f(M)  (2;1) gN �() : N(3 ; 1) I�� � N�  f(N)  (0;2) � gQua M� (2;1) x  2 y 1 uuuuur (� ) �(M�� N ): � � PTCtac�(� ):  � PTTQ (� ) : x  2y  4  0 2 1 gVTCP : M�� N  (2;1) � Cach � 3 : V�f la� phep �d� �� i h nh nen �f bien ��� �� ng thang � () thanh � ��� � ng thang � (� ) // ( ) . gLay �M �() : M(5 ;0) I�� � M�  f(M)  (2;1) � � gV�( ) // () � ( ) : x + 2y  m = 0 (m �5) . Do : (� )  M� (2;1) � m =  4 � (� ) : x  2y  4  0 c) Cach � 1: Dung � bieu �th� � c toa� �� o �x� = x  3 �x  x� 3 Ta co� f : M(x;y) I�� � M� = f(M) = � �� � � �y  y  1 �y  y  1 V�M(x;y) �(C) : (x + 1)2 + (y  2)2 = 2 � (x�  4)2  (y�  3)2  2 � � M��� (x ;y ) �(C� ) : (x  4)2  (y  3)2  2 � � + Tam � I(  1;2) f + Tam � I� = f [ I(  1;2)]  (4;3) Cach � 2 : (C) � �� � (C� ) �  BK : R� =R= 2 � BK : R = 2 � � (C� ) : (x  4)2  (y  3)2  2 d) Dung � bieu �th� � c toa� �� o � x� = x3 � x  x� 3 Ta co� f : M(x;y) I�� � M� = f(M) = � �� y�  y 1 y  y� 1 � � x2 y2 (x� + 3)2 (y�  1)2 (x + 3)2 (y  1)2 ��� � V� M(x;y) �(E) : + =1 � + = 1 � M (x ;y ) �(E ) : + =1 3 2 3 2 3 2 -3- Bài tập về phép biến hình – Toán 11 nâng cao 9 Trong mpOxy cho phep �bien �h� nh f : M(x;y) I�� � M� = f(M) = (x  1; y  2) . a) CMR f la� phep �d� �� i h nh . b) T� m anh � cua �� � � � ng thang � () : x  2y  3 = 0. c) T� m anh � cua �� � � � ng tron �(C) : (x + 3)2 + (y  1)2 = 2 . d) T� m anh � cua �parabol (P) : y 2 = 4x . �S : b) x  2y  2 = 0 c) (x + 2)2 + (y  1)2 = 2 d) (y + 2)2 = 4(x  1) 10 Trong mpOxy cho phep �bien �h� nh f : M(x;y) I�� � M� = f(M) = (x ; y) . Khang �� � nh nao �sau �� ay sai ? A. f la� 1 phep �d� �� i h nh B. Neu �A(0 ; a) th� f(A) = A C. M va� f(M) ��� oi x � ng nhau qua truc�hoanh � D. f [M(2;3)] �� �ng � � thang � 2x + y + 1 = 0 �S : Chon �C . V� M va� f(M) ��� oi x � ng nhau qua truc�tung � C sai . 12 Trong mpOxy cho 2 phep �bien �h� nh : f1 : M(x;y) I�� � M� = f1(M) = (x + 2 ; y  4) ; f2 : M(x;y) I�� � M� = f2 (M) = (  x ;  y) . T� m toa� �� o anh � cua �A(4;  1) qua f1 roi�f2 , ngh� a la� t� m f2 [f1(A)] . f f 1 � A� 2 � A� � �S : A(4;  1) I�� (6;  5) I�� (6 ; 5 ) . x 11 Trong mpOxy cho phep �bien �h� nh f : M(x;y) I�� � M� = f(M) = ( ; 3y) . Khang � �� nh na� o sau �� ay sai ? 2 A. f (O) = O (O la� �iem � bat�ien) b � B. Anh � cua �A �Ox th�anh � A� = f(A) �Ox . C. Anh � cua �B �Oy th�� anh B� = f(B) �Oy . D. M� = f [ M(2 ;  3)] = (1;  9) �S : Chon �D . V�M� = f [ M(2 ;  3)] = (1; 9) Vấn đề 2 : PHÉP TỊNH TIẾN A. KIẾN THỨC CƠ BẢN uuuuu r r r 1 �N : Phep �� t nh tien �theo vect�u la� mot�phep �d� �� i h nh bien �� iem � M tha� nh � iem � M� sao cho MM�  u. uuuuu r r K�hieu �: T hay Tur .Khi �� o : Tur (M)  M� � MM � u gPhep �� t nh tien �hoan �toan �� � � � c xac�� � nh khi biet�vect�� t nh tien �cua �no � . r r gNeáu To (M)  M , M thì To laø pheùp ñoàng nhaát . r 2 Bieåu thöùc toïa ñoä : Cho u = (a;b) vaø pheùp tònh tieán Tur �x� =x+a M(x;y) I�� � M� =Tur (M)  (x�� ; y ) th�� � �y = y + b 3 T� nh chat�: g�L : Phep �� t nh tien �bao �toan �khoang � c ach � gi� � a hai � iem � bat�� k . gHQ : 1. Bao �toan �� t nh thang � hang � va�� th � t� � cua �cac�� iem � t�� ng �� ng . 2. Bien �mot�tia thanh � tia . 3. Bao �toan �� t nh thang � hang � va�� th � t� � cua �cac�� iem � t�� ng � � ng . 5. Bien �mot�� oan �thang � thanh �� oan �thang � bang � no� . 6. Bien �mot��� � � ng thang � thanh � mot�� �� � ng thang � song song hoa� c trung � v� � i� � �� ng thang � �� a cho . 7. Bien �tam giac �thanh � tam giac�bang � no� . (Tr� � c tam � I�� � tr� � c tam � , trong � tam � I�� � trong � tam �) 8. �� � � ng tron �thanh �� � � � ng tron �ba� ng no� . (Tam � bien �thanh � tam � : I I�� � I� , R� =R)  PHƯƠNG PHÁP TÌM ẢNH CỦA MỘT ĐIỂM -4- Bài tập về phép biến hình – Toán 11 nâng cao �x� =x+a M(x;y) I�� � M� =Tur (M)  (x�� ; y ) th�� =y+b �y�  PHƯƠNG PHÁP TÌM ẢNH CỦA MỘT HÌNH (H) . Cach � 1 : Dung � t� nh chat�(cung � ph� �g n cua �� thang � , ban �k� nh � �� � ng tron �: khong � �� oi ) 1. Lay �M ξ��� (H) I M� (H� ) 2. g(H) ��� � � ng thang � �� � (H� ) �� � � � ng tha� ng cung � ph� � ng � �  Tam �I  Tam � I� g(H) �(C) � I�� � (H� ) �(C� )� (can �t� m I� ). + bk : R + bk : R� =R � � Cach � 2 : Dung � bieu �th� � c toa ��� o. T� m x theo x� , t� m y theo y� roi�hay t vao �bieu �th� � c toa ��� o. Cach � 3 : Lay �hai � iem � phan �biet�: M, N ξ��� (H) I M� , N� (H� ) B, BÀI TẬP r 1 Trong mpOxy . T� m anh � cua �M� cua �� ie� m M(3;  2) qua phep �� t nh tien �theo vect�u = (2;1) . Giai� uuuuu r r � � x� 3 2 x� 5 Theo � � nh ngh� a ta co� : M� = Tur (M) � MM�  u � (x �  3; y�  2)  (2;1) � � �� y� 2 1 y�  1 � � � M� (5; 1) r 2 T� m anh � cac�� iem � ch� ra qua phep �t� nh tien �theo vect�u : r a) A(  1;1) , u = (3;1) � A� (2;3) r b) B(2;1) , u = (  3;2) � B� (  1;3) r c) C(3;  2) , u = (  1;3) � C� (2;1) r 3 Trong mpOxy . T� m anh � A�� ,B lan �� l �� t cua ��iem � A(2;3), B(1;1) qua phep �� t nh tien �theo vect�u = (3;1) . uuur uuuur T� nh �� o dai�AB , A�� B . Giai� uuur uuuur Ta co� : A� = Tur (A)  (5; 4) , B� = Tur (B)  (4;2) , AB = |AB |  5 , A�� B = |A �� B | 5 . r r r 4 Cho 2 vect�u1; u2 . G� a s�� M1  Tur (M),M 2  Tur (M1). T� m v �� e M2  Tvr (M) . 1 2 Giai� uuuuur r uuuuuuur r r r Theo �� e : M1  Tu (M) � MM1  u1 , M2  Tu (M1) � M1M2  u2 . 1 uuuuuu r r r uuuuuur 2uuuuur uuuuuuur r r r r r r Neu �: M2  Tv (M) � MM2  v � v  MM2  MM1  M1M2  u1+ u2 .Vay �: v  u1 + u2 5 �� � � ng thang �  cat�Ox tai�A(  1;0) , c at�Oy tai�B(0;2) . Hay �viet�ph� � ng tr� nh � � � � ng thang � � la� anh � r cua � qua phep �� t nh tien �theo vect�u = (2;  1) . -5- Bài tập về phép biến hình – Toán 11 nâng cao Giai� V� : A�  Tur (A)  (1; 1) , B�  Tur (B)  (2;1) . � gqua A� (1;uuuu 1) ur Mat�khac�: �  Tur () � � � i qua A �� ,B . Do �� o : � � �� g VTCP : A B = (1;2) � � x  1 t � ptts � :� y  1  2t � 6 �� � � ng thang �  cat�Ox tai�A(1;0) , c at�Oy tai�B(0;3) . Hay �viet�ph� � ng tr� nh � � � � ng thang � � la� anh � r cua � qua phep �� t nh tien �theo vect�u = (  1;  2) . Giai� r (A)  (0; 2) , B� r (B)  (1;1) . V� : A�  Tu  Tu � gqua A� (0;  2) � x  t r (  ) � � uuuuu r Mat�khac �: �  Tu � i qua A �� ,B . Do �� o : � � ptts � :� � y  2  3t gVTCP : A� B� = (  1;3) � � r 7 T� � ng t� � : a)  : x  2y  4 = 0 , u = (0 ; 3) � � : x  2y  2  0 r b)  : 3x  y  3 = 0 , u = (  1 ;  2) � � : 3x  y  2  0 r 8 T� m anh � cua �� � � � ng tron �(C) : (x + 1)2  (y  2)2  4 qua phep �� t nh tien �theo vect�u = (1;  3) . Giai� � � x� =x+1 x = x� 1 r la� Bieu �th� � c toa� �� o cua �phep �� t nh tien �Tu : � �� y� = y 3 y = y� +3 � � 2  (y� V�: M(x;y) �(C) : (x + 1)2  (y  2)2  4 � x�  1)2  4 � M��� (x ;y ) �(C� ) : x 2  (y  1)2  4 Vay �: Anh � cua �(C) la� ( C� ) : x2  (y  1)2  4 9 Trong mpOxy cho phep �bien �h� nh f : M(x;y) I�� � M� = f(M) = (x  1; y  2) . a) CMR f la� phep �d� �� i h nh . b) T� m anh � cua �� � � � ng thang � () : x  2y  3 = 0. c) T� m anh � cua �� � � � ng tron �(C) : (x + 3)2 + (y  1)2 = 2 . d) T� m anh � cua �parabol (P) : y 2 = 4x . �S : b) x  2y  2 = 0 c) (x + 2)2 + (y  1)2 = 2 d) (y + 2)2 = 4(x  1) 10 Trong mpOxy cho phep �bien �h� nh f : M(x;y) I�� � M� = f(M) = (x ; y) . Khang �� � nh nao �sau �� ay sai ? A. f la� 1 phep �d� �� i h nh B. Neu �A(0 ; a) th�f(A) = A C. M va� f(M) ��� oi x � ng nhau qua truc�hoanh � D. f [ M(2;3)] �� � � � ng thang � 2x + y + 1 = 0 �S : Chon � C . V�M va� f(M) ��� oi x � ng nhau qua truc�tung � C sai . r 9 T� m anh � cua �� � � � ng tron �(C) : (x  3)2  ( y  2)2  1 qua phep �� t nh tien �theo vect�u = (  2;4) . � � x� = x2 x = x� +2 Giai�: Bieu �th� � c toa� �� o cua �phe� p t� nh tien �Tur la� : � �� � � y = y  4 y = y 4 � � V�: M(x;y) �(C) : (x  3)2  (y  2)2  1 � (x�  1)2  (y�  2)2  1 � M��� (x ;y ) �(C� ) : (x�  1)2  (y�  2)2  1 Vay �: Anh � cua �(C) la� (C� ) : (x  1)2  (y  2)2  1 -6- Bài tập về phép biến hình – Toán 11 nâng cao r BT T�� ng t� � : a) (C) : (x  2)2  (y  3)2  1, u = (3;1) r b) (C) : x2  y2  2x  4y  4  0, u = (  2;3) � (C� ) : (x  1)2  (y  2)2  1 (C� ) : x2  y2  2x  2y  7  0 10 Trong he� truc�toa� �� o Oxy , xac�� � nh toa� �� o cac�� � nh C va� D cua �h� nh b� nh hanh � ABCD biet�� � nh A(  2;0), � � nh B(  1;0) va� giao � iem � cac�� � � � ng cheo �la� I(1;2) . Giai� uur uur uur gGoi�C(x;y) .Ta co� : IC  (x  1; y  2),AI  (3;2) ,BI  (2; 1) gV�I la� trung � iem � cua �AC nen �: uur uur �x  1  3 � x4 C = Tuur (I) � IC  AI � � �� � C(4; 4) AI y4 �y  2  2 � gV�I la� trung � iem � cua �AC nen �: uur uur x 1  2 x 3 � � D = Tuur (I) � ID  BI � � D � �D � D(3; 4) BI yD  2  2 yD  4 � � Bai�tap �� t � ng t� � : A(  1;0),B(0;4),I(1;1) � C(3;2),D(2;  2) . 11 Cho 2 � � �� ng thang � song song nhau d va� d� . Hay �ch�ra mot�phep �� t nh tien � bien �d thanh � d� . Hoi�co� bao nhie� u phep �� t nh tien �nh�the� ? Giai�: Chon �2 � iem � co� � � nh A �d , A� �d � uuuuu r uuur Lay �� iem � tuy� y� M �d . G� a s� � : M� = Tuuur (M) � MM�  AB AB uuuu r uuuur � MA  M� B � M� B / /MA � M� �d� � d� = Tuuur (d) AB Nhan �xet�: Co� vo� so� phep �� t nh tien �bien �d thanh � d� . 12 Cho 2 � � � � ng tron �(I,R) va� (I� ,R� ) .Hay �ch�ra mot�phep �� t nh tien � (I� ,R� ). uuu uu r�bien uu r �(I,R) thanh u u r Giai�: Lay �� iem � M tuy� y� tren �(I,R) . G� a s� � : M� = T (M) � MM�  II� II� uuu r uuuur r [(I,R) ] � IM  I�� M � I�� M  IM  R � M� �(I� ,R� ) � (I� ,R� ) = Tuu II� 13 Cho h� nh b� nh hanh � ABCD , hai � � nh A,B co� � � nh , tam � I thay �� oi di �� ong tren �� � � � ng tron �(C) .T� m quy�ch t� trung � iem � M cua �canh � BC. Giai� uuu r uur Goi�J la� trung � iem � canh � AB . Khi � o� de� thay �J co� � � nh va� IM  JB . Vay �M la� anh � cua �I qua phep �� t nh tien �Tuur . Suy ra : Quy� t� ch cua �M la� JB uur anh � cua �� � � � ng tron �(C) trong phep �t� nh tien �theo vect�JB -7- Bài tập về phép biến hình – Toán 11 nâng cao r 14 Trong he� truc�toa� �� o Oxy , cho parabol (P) : y = ax2 . Goi�T la� phep �� t nh tien �theo vect�u = (m,n) va� (P� ) la� anh � cua �(P) qua phe� p t� nh tien ��� o . Hay �viet�ph� � ng tr� nh cua �(P� ). Giai�: uuuuu r r uuuuu r Tur gM(x;y) I��� � M��� (x ;y ) , ta co� : MM � = u , v�� i MM� = (x�  x ; y�  y) uuuuu r r � � x� x = m x = x� m V�MM� =u�� �� y� y = n y = y� n � � Ma� : M(x; y) �(P) : y  ax 2 � y�  n = a(x �  m)2 � y � = a(x�  m)2  n � M��� (x ;y ) �(P � ) : y = a(x  m)2  n Vay �: Anh � cua �(P) qua phep �� t nh tie � n Tur la� (P� ) : y = a(x  m)2  n � y = ax 2  2amx  am 2  n . r r 15 Cho � t  : 6x + 2y  1= 0 . T� m vect�u �0 �� e  = Tur ( ) . r r r r Giai�: VTCP cua � la� a = (2;  6) . �e� :  = Tur ( ) � u cung � ph�� ng a . Khi �� o : a = (2;  6)  2(1; 3) r � chon �u = (1;  3) . r r 16 Trong he� truc�toa� �� o Oxy , cho 2 � iem � A(  5;2) , C(  1;0) . Biet�: B = Tur (A) , C = Tvr (B) . T� m u va� v �� e co� the�� th � c hien �phep �bie � n �� oi A thanh �C? Giai� Tur Tvr A(  5;2) I��� � B I�� � � C(1; 0) . uuur r uuur r uuur uuur uuur r r Ta coù : AB  u, BC  v � AC  AB  BC  u  v  (4; 2) Tur + vr -8- Bài tập về phép biến hình – Toán 11 nâng cao r r 17 Trong he� truc�toa� �� o Oxy , cho 3 � iem � K(1;2) , M(3;  1),N(2; 3) va� 2 vect�u = (2;3) ,v = (  1;2) . T� m anh � cua �K,M,N qua phep �� t nh tien �Tur roi�Tvr . uuur r uuur r uuur uuur uuur r r Tur Tvr HD : G� a s� � : A(x;y) I��� � B I��� � C(x�� ; y ) . Ta co� : AB  u, BC  v � AC  AB  BC  u  v  (1;5) uuuur � � x� 1  1 x� 2 Do �� o : K� =Tur  vr (K) � KK�  (1;5) � � �� � K� (2;7) . y� 2  5 � y� 7 � T� � ng t� � : M� (4;4) , N� (3;2) . 18 Trong he� tru� c toa� �� o Oxy , cho ABC : A(3;0) , B(  2;4) , C(  4;5) . G la� trong � tam � ABC va� phep � r r r t� nh tien �theo vect�u �0 bien �A than �h G . T� m G� = Tu (G) . Giai� Tur Tur A(3;0) I��� � G(1;3) I��� � G��� (x ; y ) uuur u u u u r r r � � x�  1  4 x�  5 V�AG  (4;3)  u . Theo �� e : GG� u�� �� � G� (5;6). � � y 3  3 y 6 � � 19 Trong mat�phang � Oxy , cho 2 � � � � ng tron �(C) : (x  1)2  (y  3)2  2,(C� ) : x2  y 2  10x  4y  25  0. r Co� hay khong � phe�� p t nh tien �vect�u bien �(C) thanh � (C� ). HD : (C) co�� tam I(1;  3), ban �k� nh R = 2 ; (C� ) co�� tam I� (5;  2), ban �k� nh R� =2. r Ta thay �: R = R� = 2 nen �co� phep �t� nh tien �theo vect�u = (4;1) bien �(C) thanh � (C� ). 20 Trong he� truc�toa� �� o Oxy , cho h� nh b� nh hanh � OABC v� � i A(  2;1) va� B � :2x  y  5 = 0 . T� m tap � h� � p� � nh C ? Giai� uuur uuur r gV�OABC la� h� nh b� nh hanh � nen �: BC  AO  (2; 1) � C  Tur (B) v� � i u = (2; 1) uuur r Tur � � x� x 2 x  x� 2 gB(x;y) I��� � C(x�� ; y ) . Do : BC  u � � �� y�  y  1 � y  y� 1 � gB(x;y) � � 2x  y  5 = 0 � 2x�  y�  10 = 0 � C(x�� ; y ) � � : 2x  y  10 = 0 21 Cho ABC . Goi�A1,B1,C1 lan �� l � � t la� trung � iem � cac�canh � BC,CA,AB. Goi�O1,O2 ,O3 va� I1,I2 ,I3 t� � ng � � ng la� cac�tam �� �� � ng tro� n ngoai�tiep �va� cac�tam �� � � � ng tro� n noi�tiep �cua �ba tam giac�AB1C1, BC1A1, va� CA1B1 . Ch� � ng minh rang � : O1O2O3  I1I 2I3 . HD : wXet�phep �� t nh tien �: T1 uuur bien �A I�� � C,C1 I�� � B, B1 I�� � A1 . AB 2 T1 uuur T1 uuur T1 uuur AB AB AB 2 2 2 � AB1C1 I���� � C1BA1;O1 I���� � O2 ;I1 I���� � I2 . uuuuuur uuuur � O1O2  I1I2 � O1O2  I1I2 . wLy� luan �� t �ng t� � : Xet�cac �phep �t� nh tien �T1 uuur ,T1 uuur suy ra : BC CA 2 2 uuuuuur uuuur uuuuuu r uuuu r O2O3  I2 I3 va� O3O1  I3I1 � O2O3  I2 I3 ,O3O1  I3I1 � O1O2O3  I1I2 I3 (c.c.c). �  60o,B �  150ova� �  90o. 22 Trong t� � giac�ABCD co� AB = 6 3cm ,CD  12cm , A D T� nh �� o dai�cac�canh � BC va� DA . HD : uuuu r uuur Tuuur �  30o(v�B �  150o) BC � M � AM  BC.Ta co� wXet�: A I��� : ABCM la� h� nh b � nh hanh � va� BCM -9- Bài tập về phép biến hình – Toán 11 nâng cao �  360o  (90o 60o 150o)  60o � MCD �  30o. Lai�co� : BCD �� nh ly� ham � cos trong MCD : 3 MD2  MC2  DC2  2MC.DC.cos30o  (6 3)2 (12)2  2.6 3.12.  36 2 � MD = 6cm . 1 Ta co� : MD = CD va� MC = MD 3 � MDC la�m ta giac��� eu 2 �  90o va� � � MCD la� n� � a tam giac��� eu � DMC MDA  30o. � � �  30o � AMD la� Vay �: MDA  MAD  MAB tam giac�ca� n tai�M . 6 3 Döïng MK  AD � K laø trung ñieåm cuûa AD � KD=MDcos30o  cm � AD  6 3cm 2 Toùm laïi : BC = AM = MD = 6cm , AD = AB = 6 3cm Vấn đề 3 : PHÉP ĐỐI XỨNG TRỤC A , KIẾN THỨC CƠ BẢN 1 �N1: �iem � M� goi�la� ��� oi x � ng v� �iem i � � M qua � � � � ng thang � a neu �a la� �� � � ng trung tr� � c cua �� oan � MM� . Phep ���� oi x � ng qua � � � � ng tha� ng con �goi�la� phep ���� oi x � ng truc�. �� � � ng thang � a goi�la� tru� c ��� oi x � ng. �N2 : Phep ���� oi x � ng qua � �� � ng tha� ng a la� phep �bien �h� nh bien �moi�� iem � M thanh �� iem � M� ��� oi x � ng v� � i M qua � � � � ng tha� ng a . uuuuuur uuuuuu r K�hieu �: �a (M)  M� � M o M�  M oM , v�� i Mo la� h� nh chieu �cua �M tren �� � �� ng thang �a. Khi đó : gNeu �M �a th��a (M)  M : xem M la� ��� oi x � ng v� � i ch� nh no� qua a . ( M con �goi�la� � iem � bat��� ong ) gM �a th��a (M)  M� � a la� ���� ng trung tr�� c cu� a MM� gÑa (M)  M� thì Ña (M� )M gÑa (H)  H� thì Ña (H� )  H , H� laø aûnh cuûa hình H . g�N : d la� truc���� oi x � ng cua �h� nh H � �d (H)  H . gPhep ���� oi x � ng truc �hoan �toan �xac�� � nh khi biet�truc���� oi x � ng cua �no� . Chu� y� : Mot�� h nh co� the� khong � co� truc���� oi x � ng ,co� the� co� mot�hay nhieu �truc���� oi x � ng . 2 Bieu �th�� c toa ��� o : M(x;y) I�� � M�  �d (M)  (x�� ;y ) �x� �x� =x =x �d �Ox : � �d �Oy : � � � y =  y y � � =y 3 �L : Phep ���� oi x � ng truc�la� mot�hep p �d� �� i h nh . gHQ : 1.Phep ���� oi x � ng truc�bien �ba � iem �thang � hang � thanh � ba � iem � thang � han �g va� bao �toan �th� � t� � cua �cac� � iem � t� �ng � � ng . 2. �� � � ng thang � thanh �� � � � ng thang �. 3. Tia thanh � tia . 4. �oan �thang � thanh �� oan �thang � ba� ng no� . 5. Tam giac�thanh � tam giac�bang � no� . (Tr� � c tam �I�� � tr� � c tam � , tron �g tam �I�� � trong � tam �) 6. �� � � ng tron �thanh �� � � � ng tron �ba� ng no� . (Tam � bien �thanh � tam � : I I�� � I� , R� =R) 7. Goc�thanh � goc�bang � no� . - 10 - Bài tập về phép biến hình – Toán 11 nâng cao �PP : Tìm aûnh M� = Ña (M) 1. (d)  M , d  a 2. H = d �a 3. H laø trung ñieåm cuûa MM� � M� ? �PP : T� m anh � cua �� � � � ng thang � : � = �a ( ) wTH1: () // (a) 1. Lay �A,B �() : A �B 2. T� m anh � A� = �a (A) 3. �  A�� , // (a) � � w TH2 :  // a 1. T� m K =  �a 2. Lay �P � : P �K .T� m Q = �a (P) 3. � �(KQ) m M �() : (MA + MB)min . �PP : T� T� m M �() : (MA+ MB)min wLoai�1 : A, B nam � cung � ph���� a oi v � i ( ) : 1) goi�A� la� ��� oi x � ng cua �A qua () 2) M �(), th�MA + MB  MA� + MB �A� B Do �� o: (MA+MB)min= A� B � M = (A� B) �() wLoai�2 : A, B nam � khac�ph���� a oi v � i ( ) : M �( ), th�MA + MB �AB Ta co:�(MA+MB)min = AB � M = (AB) �() B . BÀI TẬP 1 Trong mpOxy . T� m anh � cua �M(2;1) �� o i x� � ng qua Ox , roi���� oi x � ng qua Oy . � � Oy Ox � M� � HD : M(2;1) I��� (2;  1) I��� � M� (2; 1) 2 Trong mpOxy . T� m anh � cua �M(a;b) �� o i x� � ng qua Oy , roi���� oi x � ng qua Ox . � � Oy Ox � M� � HD : M(a;b) I��� � M� (  a;b) I��� (a;  b) � � a� � 3 Cho 2 � � � � ng thang � (a) : x  2 = 0 , (b) : y + 1 = 0 va� � iem � M(  1;2) . T� m : M I�� � M� I��b� � M� . � � a� � HD : M(  1;2) I�� � M� (5;2) I��b� � M� (5; 4) [ ve� h� nh ] . 4 Cho 2 � � � � ng thang � (a) : x  m = 0 (m > 0) , (b) : y + n = 0 (n > 0). � � a� b � M� � �� T� m M� : M(x;y) �� � M��� (x ; y ) ��� (x �� ;y� ). � �a �b � � x�  2m  x x�  2m  x � HD : M(x;y) I���� � M� I������ � M� �� �� t� (m;y) t � ( 2m  x;  n) y y y�  2n  y � � 5 Cho � iem � M(  1;2) va� � � � � ng thang � (a) : x + 2y + 2 = 0 . HD : (d) : 2x Ǯy+�4 = 0 , H = d a H( 2;0) , H la� trung � iem � cua �MM� M� ( 3; 2) 6 Cho � iem � M(  4;1) va� � � � � ng thang � (a) : x + y = 0 . � M� = �a (M)  (1; 4) 7 Cho 2 � � � � ng thang � () : 4x  y + 9 = 0 , (a) : x  y + 3 = 0 . T� m anh � � = �a () . HD : 4 1 gV� � �� cat  �aǮ K a K( 2;1) 1 1 gM(  1;5) � � d  M,  a � d : x  y  4  0 � H(1/ 2; 7 / 2) : t� iem � c ua �MM� � M�  �a (M)  (2;2) g� �KM� : x  4y + 6 = 0 - 11 - Bài tập về phép biến hình – Toán 11 nâng cao 8 T� m b = �a (Ox) v� � i� � � � ng thang � (a) : x + 3y + 3 = 0 . HD : ga �Ox = K(  3;0) . 3 9 gM �O(0;0) �Ox : M� = �a (M) = (  ;  ) . 5 5 � gb �KM : 3x + 4y  9 = 0 . 9 T� m b = �a (Ox) v� � i� � � � ng thang � (a) : x + 3y  3 = 0 . HD : ga �Ox = K(3;0) . gP �O(0;0) �Ox . � + Qua O(0;0) g � �  : 3x  y  0 + a � 3 9 3 9 gE = a � � E( ; ) la� trung � iem �OQ � Q( ; ) . 10 10 5 5 gb �KQ : 3x + 4y  9 = 0 . 10 T � m b = �Ox (a) v� � i� � � � ng thang � (a) : x + 3y  3 = 0 . Giai�: Cach � 1: Dung � bieu �th� � c toa� �� o (rat�hay) Cach � 2 : gK= a Ǯ Ox K(3;0) gP(0;1) �a � Q = �Ox (P) = (0;  1) gb �KQ : x  3y  3 = 0 . 11 Cho 2 � � � � ng thang � () : x  2y + 2 = 0 , (a) : x  2y  3 = 0 . T� m anh � � = �a ( ) . PP :  / /a Cach � 1 : T� m A,B � � A�� ,B �� � � �A� B� Cach � 2 : T� m A � � A� �� � � / / ,  �  A� Giai� : gA(0;1) � � A�  �a (A)  (2; 3) g�  A�� , / /  � � : x  2y  8  0 12 Cho � � � � ng tron �(C) : (x+3)2  (y  2)2  1 , � � � � ng thang � (a) : 3x  y + 1= 0 . T� m (C� ) = �a [(C)] HD : (C� ) : (x  3)2  y2  1 . 13 Trong mpOxy cho ABC : A(  1;6),B(0;1) va� C(1;6) . Khang �� � nh nao �sau �� ay sai ? A. ABC can �� � B B. ABC co� 1 truc��� oi x� � ng C. ABC  �Ox (ABC) D. Trong � tam � : G = �Oy (G) HD : Chon �D 14 Trong mpOxy cho � iem � M(  3;2), ���� ng thang � () : x + 3y  8 = 0, ���� ng tron �(C) : (x+3)2  (y  2)2  4. T� m anh � cua �M, () va� (C) qua phe� p ��� oi x � ng truc�(a) : x  2y + 2 = 0 . Giai�: Goi�M� , ( � ) va� (C� ) la� anh � cua �M, () va� (C) qua phep ���� oi x � ng truc�a . � g Qua M(  3;2) a) T� m anh � M� : Goi����� ng thang � (d) : � ga � + (d)  (a) � (d) : 2x  y + m = 0 . V�(d)  M(  3 ;2) � m = 4 � (d) : 2x  y  4 = 0 - 12 - Bài tập về phép biến hình – Toán 11 nâng cao � 1 x H  (x M  x M� ) � 2 � + H = (d) �(a) � H(  2;0) � H la� trung � iem � cu � a M,M � H � 1 � y H  (y M  y M� ) � 2 � 1 2  (3  x M� ) � � x  1 2 �� � � M� � M� ( 1; 2) 1 y   2 � � M � 0  (2  y M� ) � 2 b) T� m anh � ( � ): 1 3 gV� � � ( ) cat� (a) � K= ( ) �(a) 1 2 � x + 3y  8 = 0 � Toa� �� o cua �K la� nghiem � cua �he � : � � K(2; 2) x  2y + 2 = 0 � gLay �P �K � Q = �a [P(  1;3)] = (1; 1) . ( Lam � t� � ng t� � nh�cau � a) ) � g Qua P(  1;3) Goi�� � � � ng thang � (b) : � g a � + (b)  (a) � (b) : 2x  y + m = 0 . V� (b)  P(  1;3) � m =  1 � (b) : 2x  y  1 = 0 + E = (b) �(a) � E(0;1) � E la� trung � ie� m cua �P,Q � � � 1 1 x  (x  xQ ) 0  (1  xQ ) � xQ  1 � � �E 2 P � � � E� �� 2 �� � Q(1; 1) 1 1 y Q  1 � � � y  (y  yQ ) 1  (3  y Q ) �E 2 P � 2 � gQua K(2;2) x2 y2 uuur + ( � ) �(KQ) : � � ( � ):  � 3x  y  4  0 1 3 gVTCP : KQ  (1; 3)  (1;3) � c) + T� m anh � cua �tam � I(  3;2) nh�cau � a) . �a �a � I I��� � I� .T� + V�phep ���� oi x � ng truc�la� phe� p d� �� i h nh nen � (C): gTam �(C� ) : gTam m I I��� � I� � gR  2 gR  R  2 � 2 2 �a � � + Tam � I(  3;2) + Tam � I = � [ I(  3;  2)]  (  ; ) a � Vay �: (C) I��� � (C ) � 5 5  BK : R = 2 �  BK : R� =R=2 � 2 2 � (C� ) : (x  )2  (y  )2  4 5 5    15 Trong mpOxy cho � iem � M(3;  5), ���� ng thang � () : 3x + 2y  6 = 0, ���� ng tron �(C) : (x+1)2  (y  2)2  9. T� m anh � cua �M, () va� (C) qua phe� p ��� oi x � ng truc�(a) : 2x  y + 1 = 0 . HD : �a 33 1 9 13 a) M(3;  5) I��� � M� ( ;  ),(d) : x  2y  7  0,t� iem � H(  ;  ) 5 5 5 5 4 15 b) + K= Ǯ (a) K( ; ) 7 7 + P �() : P(2;0) �K , Q = �a[P(2;0)] = (  2;2) � (� ) �(KQ) : x  18y  38  0 �a 9 8 9 8 c) + I(1;  2) I��� � I� (  ; ) , R� =R=3 � (C� ) : (x + )2  (y  )2  9 5 5 5 5 16 Cho � iem � M(2;  3), � � � � ng thang � () : 2x + y  4 = 0, � � � � ng tron �(C) : x 2  y 2  2x  4y  2  0. T� m anh � cua �M, () va� (C) qua phe� p ��� oi x � ng qua Ox . �Ox �x  x� �x� x HD : Ta co� : M(x;y) ��� � M� (1) � � (2) �  y �y  y� �y� � Ox � M� gThay vao �(2) : M(2;  3) ��� (2;3) - 13 - Bài tập về phép biến hình – Toán 11 nâng cao gM(x;y) �() � 2x�  y�  4 = 0 � M��� (x ;y ) �(� ) : 2x  y  4 = 0 . 2  y� 2  2x� gM(x;y) �(C) : x2  y 2  2x  4y  2  0 � x�  4y � 2  0 � (x�  1)2  (y�  2)2  3 � M��� (x ;y ) �(C� ) : (x  1)2  (y  2)2  3 17 Trong mpOxy cho � � � � ng thang � (a) : 2x  y+3 = 0 . T� m anh � cua �a qua �Ox . �Ox � � x� x x  x� Giai�: Ta co� : M(x;y) I��� � M� �� �� y  y � y   y� � V�M(x;y) �(a) : 2x  y+3 = 0 � 2(x� )  ( y � )+3 = 0 � 2x�  y� +3 = 0 � M� (x�� ; y ) �(a� ) : 2x  y + 3 = 0 � Oy Vay �: (a) I���� (a� ) : 2x  y + 3 = 0 18 Trong mpOxy cho � � � � ng tron �(C) : x 2  y 2  4y  5 = 0 . T� m anh � cua �a qua �Oy . �Oy � x�  x � x   x� Giai�: Ta co� : M(x;y) I���� M� �� � � y  y y � �  y� 2  4(y � 2  y� 2  4y  5 = 0 V�M(x;y) �(C) : x 2  y2  4y  5 = 0 � (  x� )2  y � )  5 = 0 � x� � M��� (x ; y ) �(C� ) : x2  y 2  4y  5 = 0 � Oy Vay �: (C) I���� (C� ) : x 2  y 2  4y  5 = 0 19 Trong mpOxy cho � thang � (a) : 2x  y  3 = 0 , ( ) : x  3y  11 = 0 , (C) : x2  y 2  10x  4y  27 = 0 . a) Viet�bieu �th� � c giai�� t ch cua �phep ���� oi x � ng truc��a . b) T� m anh � cua �� iem � M(4;  1) qua �a . c) T� m anh � : (� ) = �a ( ),(C� )  �a (C) . Giai� a) Tong � quat�(a) : Ax + By + C=0 , A 2  B2 �0 uuuuu r uuuuu r r �a r Goi�M(x;y) I��� � M��� (x ; y ) , ta co� : MM�  (x�  x; y�  y) c ung � ph� � ng VTPT n = (A;B) � MM�  tn x  x�y  y� � x�  x  At �x�  x  At �� �� (t ��) . Goi�I la� trung � iem � cua �MM� ne� n I( ; ) �(a) y�  y  Bt � y�  y  Bt 2 2 � x  x� y  y� x  x  At y  y  Bt � A( )  B( )  C  0 � A( )  B( )C  0 2 2 2 2 2(Ax + By + C) � (A 2  B2 )t  2(Ax + By + C) � t  A 2  B2 � 2A(Ax + By + C) 2B(Ax + By + C) �� x� x ; y�  y � A2  B2 A2  B2 � � 4(2x  y  3) 3 4 12 x�  x x�   x y � � � � 5 5 5 5 Ap �dung � ket�qua� tren �ta co� : � �� 2(2x  y  3) 4 3 6 � �y� y�  y  y y 5 5 5 � � 5 �a 4 7 b) M(4;  1) I��� � M� ( ; ) 5 5 � a � � c)  I��� : 3x  y  17  0 � a � (C� d) (C) I��� ) : (x  1)2  (y  4)2  2 - 14 - Bài tập về phép biến hình – Toán 11 nâng cao 20 Trong mpOxy cho ñöôøng thaúng () : x  5y  7 = 0 vaø (� ) : 5x  y  13 = 0 . Tìm pheùp ñoái xöùng qua truïc bieán () thaønh (� ). Giaûi 1 5 Vì � � () vaø (� ) caét nhau . Do ñoù truïc ñoái xöùng (a) cuûa pheùp ñoái xöùng bieán () thaønh (� ) chính 5 1 laø ñöôøng phaân giaùc cuûa goùc taïo bôûi () vaø (� ). � x y  5  0 (a1) �� x  y  1  0 (a2 ) 1  25 25 + 1 � Vaäy coù 2 pheùp ñoái xöùng qua caùc truïc (1) : x  y  5  0 , ( 2 ) : x  y  1  0 Töø ñoù suy ra (a) : | x  5y  7 |  | 5x  y  13| 21 Qua phep ���� oi x � ng truc��a : 1. Nh� � ng tam giac�nao �bien �thanh � ch� nh no� ? 2. Nh� � ng � � � � ng tron �nao �bien �tha� nh ch� nh no� ? HD : 1. Tam giac�co�� 1 � nh �truc�a , hai � � nh con �lai�� oi�� x � ng qua truc �a . 2. �� � � ng tron �co�� tam �a . 22 T� m anh � cua �� � � � ng tron �(C) : (x  1 )2  (y  2)2  4 qua phep ���� oi x � ng truc�Oy. PP : Dung � bieu �th� � c toa� �� o � �S : (C� ) : (x  1)2  (y  2)2  4 23 Hai ABC va� A� B�� C cung � nam � trong mat�phang � toa� �� o va� ��� oi x � ng nhau qua truc�Oy . Biet�A(  1;5),B(4;6),C� (3;1) . Hay �� t m toa� �� o cac�� � nh A� , B� va� C. �S : A� (1;5), B� (4;6) va� C(  3;1) 24 Xet�cac�h� nh vuong � , ngu� giac��u e�va�� luc giac��� eu . Cho biet�so� truc ���� oi x � ng t� � ng � � ng cua �moi� loai�� a giac��� eu �� o va� ch�ra cach � ve� cac�truc���� oi x � ng �� o. - 15 - Bài tập về phép biến hình – Toán 11 nâng cao �S : gH� nh vuong � co� 4 truc ���� oi x � ng , ��� o la cac�� � � � ng thang �� i qua 2 � � nh �� oi dien �va� cac�� � � � ng thang � � i qua trung � iem � cua �cac�cap �canh � �� oi dien �. gNgu� giac��� eu co� 5 truc���� oi x � ng ,��� o la cac�� � � � ng thang �� i qua � � nh �� oi dien �va�� tam cua �ngu� giac ��� eu gLuc�giac��� eu co� 6 truc���� oi x � ng , ��� o la cac�� � � � ng thang �� i qua 2 � � nh �� oi dien �va� cac�� � � � ng thang �� i qua trung � iem � cua �cac�cap �canh � �� oi dien �. 25 Goi�d la� phan �giac�trong tai�A cua �ABC , B� la� anh � cua �B qua phep ��oi�� x � ng truc��d . Khang �� � nh nao �sau �� ay sai ? A. Neu �AB < AC th�B� �� tren �canh � AC . B. B� la� trung � iem � canh � AC . C. Neu �AB = AC th�B� �C . D. Neu �B� la� trung � iem � canh � AC th�AC = 2AB . �S : Neu �B� = �d (B) th�B� �AC . gA �� ung . V�AB < AC ma� AB� = AB nen �AB� < AC � B�� � tren �canh � AC . 1 gB sai . V�gia� thiet�bai�toan �kho � ng �� u khang �� � nh AB = AC. 2 � � � gC �� ung . V�AB = AB ma� AB = AC nen �AB = AC B C . gD �� ung . V�Neu �B� la� trung � iem � ca� nh AC th�AC=2AB� ma� AB� =AB nen �AC=2AB . 26 Cho 2 � �� � ng thang � a va� b cat�nhau tai�O . Xet�2 phep ���� oi x � ng truc��a va� �b : � � a � B I��� b � C . Khang A I��� �� � nh nao �sau �� ay khong � sai ? A. A,B,C ��� � � ng tron �(O, R = OC) . B. T�� giac�OABC noi�tiep �. C. ABC can ��� B D. ABC vuong � �� B HD : gA. Khong � sai . V� d1 la� trung tr�� c cua �AB � OA = OB , d 2 la� trung tr�� c cua �BC � OB = OC � OA = OB = OC � A,B,C ��� � � ng tron �(O, R = OC) . gCac�cau �B,C,D co� the � sai . 27 Cho ABC co� hai truc���� oi x � ng . Khang �� � nh nao �sau �� ay �� ung ? A. ABC la�  vuong � B. ABC la�  vuong � can � C. ABC la�  �� eu HD : G� a s� � ABC co� 2truc���� oi x � ng la� AC va� BC � AB = AC �� � AB  AB  BC � ABC �� eu . BC = BA � �  110o. T� � va� � �� 28 Cho ABC co� A nh B C e ABC co� truc ���� oi x � ng . � � � = 45o va� �  25o o A. B = 50 va� C  20o B. B C D. ABC la�  can �. � = 40o va� �  30o C. B C HD : Choïn D . Vì : ABC coù truïc ñoái xöùng khi ABC caân hoaëc ñeàu �  110o  90o � ABC caân taïi A , khi ñoù : Vì A � 180o  110o 180o  A � � BC   35o 2 2 29 Trong cac�h� nh sau , h� nh nao �co� nhieu �truc���� oi x � ng nhat�? A. H� nh ch� � nhat� B. H� nh vuong � C. H� nh thoi �S : Chon �B. V�: H� nh vuong � co� 4 truc���� oi x � ng . 30 Trong cac�h� nh sau , h� nh nao �co� � t truc���� oi x � ng nhat�? A. H� nh ch� � nhat� B. H� nh vuong � C. H� nh thoi �S : Chon �D. V� : H� nh thang can �co�truc 1 ���� oi x � ng . - 16 - �=C �  35o D. B D. H� nh thang can �. D. H� nh thang can �. Bài tập về phép biến hình – Toán 11 nâng cao 31 Trong cac�h� nh sau , h� nh nao �co� 3 truc���� oi x � ng ? A. H� nh thoi B. H� nh vuong � C.  �� eu D.  vuon �g can �. �S : Chon �C. V�:  �� eu co� 3 truc�� oi�� x � ng . 32 Trong cac�h� nh sau , h� nh nao �co � n hieu �h� n 4 truc���� oi x � ng ? A. H� nh vuong � B. H� nh thoi C. H� nh tron � �S : Chon �C. V� : H� nh tron �co� vo� so� truc���� oi x � ng . D. H� nh thang can �. 33 Trong cac�h� nh sau , h� nh nao �khon �g co� truc���� oi x � ng ? A. H� nh b� nh ha� nh B.  �� eu C.  can � D. H� nh thoi . �S : Chon �A. V�: H� nh b� nh hanh � kho� ng co� truc���� oi x � ng . 34 Cho hai h� nh vuo� ng ABCD va� AB��� C D co� canh � �� eu bang � a va� co� � � nh A chung . Ch� � ng minh : Co� the�� th � c hien �mot�phep ���� oi x � ng truc�bien �h� nh vuon �g ABCD thanh �� AB��� CD . HD : G� a s� � : BC �B�� C =E. �B �� Ta co� : AB = AB� ,B  90o,AE chung . �ABE ��� = AB  � F � � EB = EB� B I AE � �biet�AB = AB� B� � � EC = EC� Mat�khac�: � ��� C I  AE C� AC = AC� =a 2 � � � BAB � � o Ngoai�ra : AD� = AD va� D� AE  DAE  90  2 �A �AE ���� D I  ��  D� ABCD I AB��� CD 35 Goïi H laø tröïc taâm ABC . CMR : Boán tam giaùc ABC , HBC , HAC , HAC coù ñöôøng troøn ngoaïi tieáp baèng nhau . HD : � =C � (cung � ) Ta co� :A � chan �cung BK 1 2 � =C � (goc�co� � =C � A canh � t� � ng � � ng  ) � C 1 1 1 2 � CHK can �� K ��� oi x � ng v� � i H qua BC . Xet�phep ���� oi x � ng truc�BC . � � � BC H ; B I���� BC B ; C I���� BC C Ta co� : K I���� � BC �� Vay �: �� � � ng tron �ngoai�tiep �KBC I���� �� ng tron �ngoai�tiep �HBC 36 Cho ABC va� � � � � ng thang �a� i qua � � nh A nh� ng khong �� i qua B,C . a) T� m anh � ABC qua phep ���� oi x � ng �a . b) Goi�G la� trong � tam � ABC , Xac�� � nh G� la� anh � cua �G qua phep ���� oi x � ng �a . Giaûi a) Vì a laø truïc cuûa pheùp ñoái xöùng Ña neân : gA �a � A  Ña (A) . gB,C Ͼ����� a neân Ña : B I b) Vì G Ͼ�� a neân Ña : G I B� ,C I C� sao cho a laø trung tröïc cuûa BB� ,CC� G� sao cho a laø trung tröïc cuûa GG� . - 17 - Bài tập về phép biến hình – Toán 11 nâng cao 37 Cho � � � � ng thang � a va� hai � iem � A,B nam � cung � ph���� a oi v � i a . T� m tren �� � � � ng thang �a� iem � M sao cho MA+MB ngan �nha� t. Giai�: Xet�phep ���� oi x � ng �a : A I�� � A� . M �a th�MA = MA� . Ta co� : MA + MB = MA � + MB �A� B �e� MA + MB ngan �nhat�th� chon �M,A,B thang � hang � Vay �: M la� giao � iem � cua �a va� A� B. 38 (SGK-P13)) Cho goc�nhon �xOy va� M la� mot�� iem � ben �trong goc��� o . Hay � t� m� iem � A tren �Ox va� � iem � B tren �Oy sao cho MBA co� chu vi nho� nhat�. Giai� Goi�N = �Ox (M) va� P = �Ox (M) . Khi �� o : AM=AN , BM=BP T� � �� o : CVi = MA+AB+MB = NA+AB+BP �NP (� � � � ng gap �khuc��� � � � ng tha� ng ) MinCVi = NP Khi A,B lan �� l � � t la� giao � iem � cua �NP v� � i Ox,Oy . 39 Cho ABC can �tai�A v� � i� � � � ng cao AH . Biet�A va� H co� � � nh . T� m tap �h� � p � iem � C trong moi�tr� � � ng h� � p sau : a) B di �� ong tren �� � � � ng thang �. b) B di �� ong tren �� � � � ng tro� n tam � I, ban �k� nh R . Giai� a) V� : C = �AH (B) , ma� B � nen �C �� v� � i � = �AH ( ) Vay �: Tap �h� � p cac�� iem � C la� �� � � ng thang � � b) T� � ng t� � : Tap �h� � p cac�� iem � C la� � � � � ng tron �tam � J , ban �k� nh R la� anh � cua � � � � � ng tron �(I) qua �AH . Vấn đề 4 : PHÉP ĐỐI XỨNG TÂM 1 �N : Phep ���� oi x � ng tam � I la� mot�phep �d� �� i h nh bien �moi�� iem � M than �h � iem � M� ��� oi x � ng v� � i M qua I. Phep ���� oi x � ng qua mot�� iem � con �goi�la� phep ��� oi tam �. �iem � I goi�la�m ta� cua �cua �phep ���� oi x � ng hay � � n gian �la����� tam oi x � ng . uuur uuu r Kí hieäu : ÑI (M)  M� � IM�  IM . gNeu �M �I th�M� �I gNeu �M �I th�M�  �I (M) � I la� trung tr�� c cua �MM� . g�N :�iem � I la����� tam oi x � ng cua�h� nh H � �I (H)  H. Chu�� y : Mot�� h nh co� the� khong � co� tam ���� oi x � ng . �I 2 Bieu �th� � c toa ��� o : Cho I(x o ; y o ) va� phep ���� oi x � ng tam � I : M(x;y) I��� � M�  �I (M)  (x�� ; y ) th� x� = 2xo  x � �� y  2yo  y � 3 T� nh chat�: 1. Phep ���� oi x � ng tam � bao �toan �khoang � cach � gi� � a hai � iem � bat�� k . 2. Bien �mot�tia thanh � tia . 3. Bao �toan �� t nh thang � hang � va� th� � t� � cua �cac�� iem � t� � ng � � ng . 4. Bien �mot�� oan �thang � thanh �� oan �thang � bang � no� . 5. Bien �mot�� � � � ng thang � thanh � mot�� � � � ng thang � song song hoac�trung � v� � i� � � � ng thang � �� a cho . 6. Bien �mot�goc�thanh � goc�co� so� � o bang � no� . 7. Bien �tam giac�thanh � tam giac�ban �g no� . ( Tr� � c tam � � tr� � c tam � , trong � tam � � trong � tam �) - 18 - Bài tập về phép biến hình – Toán 11 nâng cao 8. �� � � ng tron �thanh �� � � � ng tron �ba� ng no� . ( Tam � bien �thanh � tam � : I I�� � I� , R� =R) B . BÀI TẬP 1 T� m anh � cua �cac�� iem � sau qua phep ���� oi x � ng tam �I : 1) A(  2;3) , I(1;2) � A� (4;1) � 2) B(3;1) , I(  1;2) � B (5;3) 3) C(2;4) , I(3;1) � C� (4; 2) Giaûi : uur uur x� 1  3 x� 4 a) Gæa söû : A�  ÑI (A) � IA  IA � (x�  1; y�  2)  (3;1) � � � A� (4;1) y�  2  1 y� 1 Caùch �: Duøng bieåu thöùc toaï ñoä   2 T� m anh � cua �cac�� � � � ng thang � sau qua phep ���� oi x � ng tam �I : 1) () : x  2y  5  0,I(2; 1) � (� ) : x  2y  5  0 2) () : x  2y  3  0, I(1; 0) � (� ) : x  2y  1  0 3) () : 3x  2y  1  0, I(2; 3) � (� ) : 3x  2y  1  0 Giai� PP : Co� 3 cach � Cach � 1: Dung � bieu �th� � c toa� �� o � Cach � 2 : Xac �� � nh dang �  //  , roi�dung � cong � th� �� c t nh khoang � cach � d(;� ) � � . Cach � 3 : Lay �bat�ky� A,B � , roi�� t m a� nh A�� ,B �� � � �A� B� �I � � x�  4x x  4  x� 1) Cach � 1: Ta co� : M(x;y) I��� � M� �� �� y  2  y � y  2  y� � V�M(x;y) � � x  2y  5  0 � (4  x� )  2(2  y� )  5  0 � x�  2y� 5 0 � M��� (x ;y ) �� : x  2y  5  0 �I Vay �: () I��� � (� ) : x  2y  5  0 Cach � 2 : Goi�� = �I () � � song song  � � : x + 2y + m = 0 (m �5) . |5| |m| � m  5 (loai) � Theo �� e : d(I;) = d(I;� )�  � 5  | m | � � m  5 � 12  22 12  22 � ( � ) : x  2y  5  0 � � � �� Cach � 3 : Lay �: A(  5;0),B(  1;  2) � � A (9; 2),B (5; 0) �  �A B : x  2y  5  0 3 T� m anh � cua �cac �� � � � ng tron �sau qua phep ���� oi x � ng tam �I : 1) (C) : x2  (y  2)2  1,E(2;1) 2) (C) : x2  y2  4x  2y  0,F(1; 0) 3) (P) : y = 2x2  x  3 , tam � O(0;0) . � (C� ) : (x  4)2  y2  1 � (C� ) : x2  y2  8x  2y  12  0 �/ nghia� hay bieu �th� � c toa� �� o ��������������(P� ) : y =  2x 2  x  3 HD : a) Co� 2 cach � giai�: Cach � 1: Dung � bieu �th� � c toa� �� o. �E Cach � 2 : T� m tam � I I��� � I� ,R�  R  (�� a cho) . b) T� � ng t� � . 4 Cho hai � iem � A va� B .Cho biet�phep �bien ��� oi M than �h M� sao cho AMBM� la� mot�� h nh b� nh han �h . - 19 - Bài tập về phép biến hình – Toán 11 nâng cao HD : uuuu r uuuur � � MA  BM� Neu �AMBM� la� h� nh b� nh hanh � � �uuur uuuur MB  AM� uuuuu r uuuu r uuuur uuuu r uuur � V� : MM�  MA  AM�  MA  MB (1) uur uur Goi�I la� trung � iem � cua � AB . Ta co � : IA   IB uuuuu r uuu r uur uuu r uur uuuuu r uuu r � � T� �(1) �uMM  MI  IA  MI  IB � MM  2MI uu r uuur � MI  IM� � M�  �I (M) . 5 Cho ba � � � � ng tron �bang � nhau (I1; R),(I2 ; R),(I3; R) t� � ng �� oi tiep � xuc�nhau tai�A,B,C . G� a s� � M la� mo� t� iem � tren �(I1; R) , ngoai�ra : �I �C �A �B 1 �Q . M I��� � N ; N I��� � P ; P I��� � Q . CMR : M I��� HD : �Do (I1; R) tiep �xuc�v� � i (I2 ; R) tai�A , nen �: uuuur uuuur �A �A �A M I����� N ; I1 I�������   I2  MI  1 I NI2 MI1 NI2 (1) �Do (I2 ; R) tiep �xuc�v� � i (I3; R) tai�B , nen �: uuuur uuur �B �B �B N I����� P ; I2 I �������   I3  NI  2 I PI3 NI 2 PI3 (2) �Do (I3; R) tiep �xuc�v� � i (I1; R) tai�C , nen �: uuur uuur �C � �C P I��  ������� Q; I3 I  C I1 PI3 I �� QI1 � PI3  QI1 (3) uuuur uuur T� � (1),(2),(3) suy ra : MI1  QI1 � M  �I (Q) . 1 5 Cho ABC la� tam giac�vuong � tai�A . Ke� � � � � ng cao AH . Ve� ph� a ngoai�tam giac�hai h� nh vuong � ABDE va� ACFG . a) Ch� � ng minh tap �h� � p6� iem �  B,C,F,G,E,D co� mot�truc���� oi x � ng . b) Goi�K la� trung � iem � cua �EG . Ch� � ng minh K � � tren �� � � � ng than �g AH . c) Goi�P = DE �FG . Ch� � ng minh P � �en tr �� � � � ng thang � AH . d) Ch� � ng minh : CD  BP, BF  CP . e) Ch� � ng minh : AH,CD,BF �� ong qui . - 20 -
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