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Appendix D Rotation and the General Second-Degree Equation D1
D Rotation and the General Second-Degree Equation
Rotate the coordinate axes to eliminate the xy-term in equations of conics.Use the discriminant to classify conics.
Rotation of AxesEquations of conics with axes parallel to one of the coordinate axes can be written inthe general form
Horizontal or vertical axes
In this appendix, you will study the equations of conics whose axes are rotated so thatthey are not parallel to either the -axis or the -axis. The general equation for such conics contains an -term.
Equation in -plane
To eliminate this -term, you can use a procedure called rotation of axes. The objective is to rotate the - and -axes until they are parallel to the axes of the conic.The rotated axes are denoted as the -axis and the -axis, as shown in Figure D.1.After the rotation, the equation of the conic in the new -plane will have the form
Equation in -plane
Because this equation has no -term, you can obtain a standard form by completingthe square.
The next theorem identifies how much to rotate the axes to eliminate the -termand also the equations for determining the new coefficients and F�.A�, C�, D�, E�,
xy
x�y�
x�y�A� �x��2 � C� �y��2 � D�x� � E�y� � F� � 0.
x�y�y�x�
yxxy
xyAx2 � Bxy � Cy2 � Dx � Ey � F � 0
xyyx
Ax2 � Cy2 � Dx � Ey � F � 0.
x
y
x ′
y ′
θ
After rotation of the - and -axescounterclockwise through an angle the rotated axes are denoted as the
-axis and -axis.Figure D.1
y�x�
�,yx
THEOREM D.1 Rotation of Axes
The general second-degree equation
where can be rewritten as
by rotating the coordinate axes through an angle where
The coefficients of the new equation are obtained by making the substitutions
y � x� sin � � y� cos �.
x � x� cos � � y� sin �
cot 2� �� � C
�.
�,
A� �x��2 � C� �y��2 � D�x� � E�y� � F� � 0
B 0,
Ax2 � Bxy � Cy2 � Dx � Ey � F � 0
9781285057095_AppD.qxp 2/18/13 8:24 AM Page D1
Proof To discover how the coordinates in the -system are related to the coordinatesin the -system, choose a point in the original system and attempt to find itscoordinates in the rotated system. In either system, the distance between thepoint and the origin is the same, and so the equations for and are those givenin Figure D.2. Using the formulas for the sine and cosine of the difference of twoangles, you obtain
and
Solving this system for and yields
and
Finally, by substituting these values for and into the original equation and collectingterms, you obtain the following.
Now, in order to eliminate the -term, you must select such that as follows.
When no rotation is necessary, because the -term is not present in the original equation. When the only way to make is to let
So, you have established the desired results.
B 0.cot 2� �A � C
B,
B� � 0B 0, xyB � 0,
� 0, sin 2� 0
� B�sin 2���C � AB
� cot 2�� � �C � A� sin 2� � B cos 2�
B� � 2�C � A� sin � cos � � B�cos2 � � sin2 ��
B� � 0,�x�y�
F� � F
E� � �D sin � � E cos �
D� � D cos � � E sin �
C� � A sin2 � � B cos � sin � � C cos2 �
A� � A cos2 � � B cos � sin � � C sin2 �
yx
y � x� sin � � y� cos �.x � x� cos � � y� sin �
yx
� y cos � � x sin �.
� r sin cos � � r cos sin �
� r�sin cos � � cos sin �� y� � r sin � � ��
� x cos � � y sin �
� r cos cos � � r sin sin �
� r�cos cos � � sin sin �� x� � r cos� � ��
y�x�,y,x,r�x�, y��
�x, y�x�y�xy
D2 Appendix D Rotation and the General Second-Degree Equation
x
(x, y)
r
α
y
Original:y � r sin x � r cos
x
x ′
y ′y
r
θ
θα −
(x ′, y ′)
Rotated:
Figure D.2y� � r sin� � ��x� � r cos� � ��
9781285057095_AppD.qxp 2/18/13 8:24 AM Page D2
Rotation of Axes for a Hyperbola
Write the equation in standard form.
Solution Because and you have
The equation in the -system is obtained by making the following substitutions.
Substituting these expressions into the equation produces
Write in standard form.
This is the equation of a hyperbola centered at the origin with vertices at inthe -system, as shown in Figure D.3.
Rotation of Axes for an Ellipse
Sketch the graph of
Solution Because and you have
The equation in the -system is derived by making the following substitutions.
Substituting these expressions into the original equation eventually simplifies (afterconsiderable algebra) to
Write in standard form.
This is the equation of an ellipse centered at the origin with vertices at andin the -system, as shown in Figure D.4.x�y��0, ±1�
�±2, 0�
�x��2
22 ��y��2
12 � 1.
4�x��2 � 16�y��2 � 16
y � x� sin �
6� y� cos
�
6� x��1
2� � y���32 � �
x� � �3y�
2
x � x� cos �
6� y� sin
�
6� x���3
2 � � y��12� �
�3x� � y�
2
x�y�
� ��
6.cot 2� �
A � CB
�7 � 13
�6�3�
1�3
�for 0 < � < ��2�C � 13,B � �6�3,A � 7,
7x2 � 6�3xy � 13y2 � 16 � 0.
x�y��±�2, 0�
�x��2
��2 �2 ��y��2
��2 �2 � 1.
�x��2 � �y��2
2� 1 � 0
�x� � y�
�2 ��x� � y�
�2 � � 1 � 0
xy � 1 � 0
y � x� sin �
4� y� cos
�
4� x���2
2 � � y���22 � �
x� � y�
�2
x � x� cos �
4� y� sin
�
4� x���2
2 � � y���22 � �
x� � y�
�2
x�y�
� ��
4.2� �
�
2cot 2� �
A � CB
� 0
�for 0 < � < ��2�C � 0,B � 1,A � 0,
xy � 1 � 0
Appendix D Rotation and the General Second-Degree Equation D3
1
1−1−2 2
−1
2
y2( )2
x
− = 1(x ′)2
2( )2
(y ′)2
x ′y ′
xy − 1 = 0
Vertices:in -system
in -systemFigure D.3
xy�1, 1�, ��1, �1�x�y���2, 0�, ���2, 0�
y
2
−2
2−2x
7x2 − 6 3xy + 13y2 − 16 = 0
22+ = 1
(x ′)2
12
(y ′)2
x ′
y ′
Vertices:in -system
in -system
Figure D.4
xy�±�3, ±1�, �±12
, ±�32 �
x�y��±2, 0�, �0, ±1�
9781285057095_AppD.qxp 2/18/13 8:24 AM Page D3
In Examples 1 and 2, the values of were the common angles and respectively. Of course, many second-degree equations do not yield such common solutions to the equation
Example 3 illustrates such a case.
Rotation of Axes for a Parabola
Sketch the graph of
Solution Because and you have
The trigonometric identity produces
from which you obtain the equation
Considering it follows that So,
From the triangle in Figure D.5, you obtain and Consequently, you can write the following.
Substituting these expressions into the original equation produces
which simplifies to
By completing the square, you obtain the standard form
Write in standard form.
The graph of the equation is a parabola with its vertex at and its axis parallel tothe -axis in the -system, as shown in Figure D.6.x�y�x�
�45, �1�
�y� � 1�2 � ��1��x� �45�.
5�y� � 1�2 � �5x� � 4
5�y��2 � 5x� � 10y� � 1 � 0.
�2x� � y�
�5 �2
� 4�2x� � y�
�5 ��x� � 2y�
�5 � � 4�x� � 2y�
�5 �2
� 5�5�x� � 2y��5 � � 1 � 0
y � x� sin � � y� cos � � x�� 1�5� � y�� 2
�5� �x� � 2y�
�5
x � x� cos � � y� sin � � x�� 2�5� � y�� 1
�5� �2x� � y�
�5
cos � � 2��5.sin � � 1��5
� � 26.6�.cot � � 2
2 cot � � 4.0 < � < ��2,
0 � �2 cot � � 4��2 cot � � 1�. 0 � 4 cot2 � � 6 cot � � 4
6 cot � � 4 cot2 � � 4
cot 2� �34
�cot2 � � 1
2 cot �
cot 2� � �cot2 � � 1���2 cot ��
cot 2� �A � C
B�
1 � 4�4
�34
.
C � 4,B � �4,A � 1,
x2 � 4xy � 4y2 � 5�5y � 1 � 0.
cot 2� �A � C
B.
30�,45��
D4 Appendix D Rotation and the General Second-Degree Equation
2
15
θ
Figure D.5
(y + 1)2 = (−1) x − 45
x2 − 4xy + 4y2 + 5 5y + 1 = 0
x
y
−1
−2
1
2
θ ≈ 26.6°
x ′
y ′
( )
Vertex:
in -system
in -system
Figure D.6
xy� 13
5�5, �
6
5�5�x�y��4
5, �1�
9781285057095_AppD.qxp 2/18/13 8:24 AM Page D4
Invariants Under RotationIn Theorem D.1, note that the constant term is the same in both equations—that is,
Because of this, is said to be invariant under rotation. Theorem D.2 listssome other rotation invariants. The proof of this theorem is left as an exercise (seeExercise 34).
You can use this theorem to classify the graph of a second-degree equation with an-term in much the same way you do for a second-degree equation without an -term. Note that because the invariant reduces to
Discriminant
which is called the discriminant of the equation
Because the sign of determines the type of graph for the equation
the sign of must determine the type of graph for the original equation. Thisresult is stated in Theorem D.3.
B2 � 4AC
A��x��2 � C��y��2 � D�x� � E�y� � F� � 0
A�C�
Ax2 � Bxy � Cy2 � Dx � Ey � F � 0.
B2 � 4AC � �4A�C�
B2 � 4ACB� � 0,xyxy
FF� � F.
Appendix D Rotation and the General Second-Degree Equation D5
THEOREM D.2 Rotation Invariants
The rotation of coordinate axes through an angle that transforms the equationinto the form
has the following rotation invariants.
1.
2.
3. B2 � 4AC � �B��2 � 4A�C�
A � C � A� � C�
F � F�
A��x��2 � C��y��2 � D�x� � E�y� � F� � 0
Ax2 � Bxy � Cy2 � Dx � Ey � F � 0�
THEOREM D.3 Classification of Conics by the Discriminant
The graph of the equation
is, except in degenerate cases, determined by its discriminant as follows.
1. Ellipse or circle:
2. Parabola:
3. Hyperbola: B2 � 4AC > 0
B2 � 4AC � 0
B2 � 4AC < 0
Ax2 � Bxy � Cy2 � Dx � Ey � F � 0
9781285057095_AppD.qxp 2/18/13 8:24 AM Page D5
D6 Appendix D Rotation and the General Second-Degree Equation
Rotation of Axes In Exercises 1–12, rotate the axes to eliminate the xy-term in the equation. Write the resulting equation in standard form and sketch its graph showing bothsets of axes.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
Graphing a Conic In Exercises 13–18, use a graphing utilityto graph the conic. Determine the angle through which theaxes are rotated. Explain how you used the graphing utility toobtain the graph.
13.
14.
15.
16.
17.
18.
Using the Discriminant In Exercises 19–26, use the discriminant to determine whether the graph of the equation isa parabola, an ellipse, or a hyperbola.
19.
20.
21.
22.
23.
24.
25.
26.
Degenerate Conic In Exercises 27–32, sketch the graph (ifpossible) of the degenerate conic.
27.
28.
29.
30.
31.
32.
33. Invariant Under Rotation Show that the equationis invariant under rotation of axes.
34. Proof Prove Theorem D.2.
x2 � y2 � r2
�2x � y � 3�2 � 0
�x � 2y � 1��x � 2y � 3� � 0
x2 � 10xy � y2 � 0
x2 � 2xy � y2 � 1 � 0
x2 � y2 � 2x � 6y � 10 � 0
y2 � 4x2 � 0
x2 � xy � 4y2 � x � y � 4 � 0
x2 � 4xy � 4y2 � 5x � y � 3 � 0
36x2 � 60xy � 25y2 � 9y � 0
x2 � 6xy � 5y2 � 4x � 22 � 0
2x2 � 4xy � 5y2 � 3x � 4y � 20 � 0
13x2 � 8xy � 7y2 � 45 � 0
x2 � 4xy � 2y2 � 6 � 0
16x2 � 24xy � 9y2 � 30x � 40y � 0
� �6�13 � 8�y � 914x2 � 12xy � 9y2 � �4�13 � 12�x32x2 � 50xy � 7y2 � 52
40x2 � 36xy � 25y2 � 52
17x2 � 32xy � 7y2 � 75
x2 � 4xy � 2y2 � 6
x2 � xy � y2 � 10
�
9x2 � 24xy � 16y2 � 80x � 60y � 0
9x2 � 24xy � 16y2 � 90x � 130y � 0
16x2 � 24xy � 9y2 � 60x � 80y � 100 � 0
3x2 � 2�3xy � y2 � 2x � 2�3y � 0
2x2 � 3xy � 2y2 � 10 � 0
5x2 � 2xy � 5y2 � 12 � 0
13x2 � 6�3xy � 7y2 � 16 � 0
xy � 2y � 4x � 0
xy � x � 2y � 3 � 0
x2 � 10xy � y2 � 1 � 0
xy � 4 � 0
xy � 1 � 0
D Exercises
Using the Discriminant
Classify the graph of each equation.
a. b.
c. d.
Solution
a. The graph is a hyperbola because
b. The graph is a circle or an ellipse because
c. The graph is a parabola because
d. The graph is a hyperbola because
B2 � 4AC � 64 � 48 > 0.
B2 � 4AC � 36 � 36 � 0.
B2 � 4AC � 9 � 16 < 0.
B2 � 4AC � 16 � 0 > 0.
3x2 � 8xy � 4y2 � 7 � 0x2 � 6xy � 9y2 � 2y � 1 � 0
2x2 � 3xy � 2y2 � 2x � 04xy � 9 � 0
9781285057095_AppD.qxp 2/18/13 8:24 AM Page D6