APPENDIX E Rotation and the General Second-Degree Equation
Rotation of Axes • Invariants Under Rotation
y y′ Rotation of Axes In Section 9.1, you learned that equations of conics with axes parallel to one of the coordinate axes can be written in the general form 2 ϩ 2 ϩ ϩ ϩ ϭ x′ Ax Cy Dx Ey F 0. Horizontal or vertical axes
θ Here you will study the equations of conics whose axes are rotated so that they are not x parallel to the x-axis or the y-axis. The general equation for such conics contains an xy-term.
Ax2 ϩ Bxy ϩ Cy2 ϩ Dx ϩ Ey ϩ F ϭ 0 Equation in xy-plane To eliminate this xy-term, you can use a procedure called rotation of axes. You want to rotate the x- and y-axes until they are parallel to the axes of the conic. (The rotated axes are denoted as the xЈ -axis and the yЈ -axis, as shown in Figure E.1.) After the After rotation of the x- and y-axes counter- rotation has been accomplished, the equation of the conic in the new xЈyЈ -plane will clockwise through an angle , the rotated have the form axes are denoted as the xЈ -axis and yЈ -axis. AЈ͑xЈ͒2 ϩ CЈ͑yЈ͒2 ϩ DЈxЈ ϩ EЈyЈ ϩ FЈ ϭ 0. Equation in xЈyЈ -plane Figure E.1 Because this equation has no xЈyЈ -term, you can obtain a standard form by complet- ing the square. The following theorem identifies how much to rotate the axes to eliminate an xy-term and also the equations for determining the new coefficients AЈ, CЈ, DЈ, EЈ, and FЈ.
THEOREM A.1 Rotation of Axes The general equation of the conic Ax2 ϩ Bxy ϩ Cy2 ϩ Dx ϩ Ey ϩ F ϭ 0, where B Þ 0, can be rewritten as AЈ͑xЈ͒2 ϩ CЈ͑yЈ͒2 ϩ DЈxЈ ϩ EЈyЈ ϩ FЈ ϭ 0 by rotating the coordinate axes through an angle , where 〈 Ϫ ϭ C cot 2 〉 .
The coefficients of the new equation are obtained by making the substitutions x ϭ xЈ cos Ϫ yЈ sin y ϭ xЈ sin ϩ yЈ cos .
E1 E2 APPENDIX E Rotation and the General Second-Degree Equation
y Proof To discover how the coordinates in the xy-system are related to the coordinates in the xЈyЈ-system, choose a point P ϭ ͑x, y͒ in the original system and attempt to find its coordinates ͑xЈ, yЈ͒ in the rotated system. In either system, the distance r between P = (x, y) the point P and the origin is the same, and thus the equations for x, y, xЈ, and yЈ are those given in Figure E.2. Using the formulas for the sine and cosine of the difference r of two angles, you obtain xЈ ϭ r cos͑␣ Ϫ ͒ ϭ r͑cos ␣ cos ϩ sin ␣ sin ͒ α ϭ r cos ␣ cos ϩ r sin ␣ sin ϭ x cos ϩ y sin x yЈ ϭ r sin ͑␣ Ϫ ͒ ϭ r͑sin ␣ cos Ϫ cos ␣ sin ͒ ϭ r sin ␣ cos Ϫ r cos ␣ sin ϭ y cos Ϫ x sin . Original: x ϭ r cos ␣ y ϭ r sin ␣ Solving this system for x and y yields ϭ Ј Ϫ Ј ϭ Ј ϩ Ј y x x cos y sin and y x sin y cos . y′ Finally, by substituting these values for x and y into the original equation and collect- ing terms, you obtain the following. P = (x′, y′) A Ј ϭ A cos2 ϩ B cos sin ϩ C sin2 C Ј ϭ A sin2 Ϫ B cos sin ϩ C cos2 r x′ α − θ D Ј ϭ D cos ϩ E sin
θ E Ј ϭ ϪD sin ϩ E cos x FЈ ϭ F Now, in order to eliminate the xЈyЈ-term, you must select such that BЈ ϭ 0, as Rotated: xЈ ϭ r cos͑␣ Ϫ ͒ follows. yЈ ϭ r sin͑␣ Ϫ ͒ BЈ ϭ 2͑C Ϫ A͒ sin cos ϩ B͑cos2 Ϫ sin2 ͒ Figure E.2 ϭ ͑C Ϫ A͒ sin 2 ϩ B cos 2 C Ϫ A ϭ B͑sin 2͒ ϩ cot 2 ϭ 0, sin 2 Þ 0 B If B ϭ 0, no rotation is necessary, because the xy-term is not present in the original equation. If B Þ 0, the only way to make BЈ ϭ 0 is to let A Ϫ C cot 2 ϭ , B Þ 0. B Thus, you have established the desired results. APPENDIX E Rotation and the General Second-Degree Equation E3
EXAMPLE 1 Rotation of a Hyperbola
Write the equation xy Ϫ 1 ϭ 0 in standard form.
Solution Because A ϭ 0, B ϭ 1, and C ϭ 0, you have ͑for 0 < < ͞2͒ A Ϫ C cot 2 ϭ ϭ 0 2 ϭ ϭ . B 2 4 x) )2) y)2 − = 1 ) 2)2) 2)2 The equation in the xЈyЈ-system is obtained by making the following substitutions. y Ί2 Ί2 xЈ Ϫ yЈ x ϭ xЈ cos Ϫ yЈ sin ϭ xЈ Ϫ yЈ ϭ 4 4 2 2 Ί2 y2x Ί2 Ί2 xЈ ϩ yЈ 1 y ϭ xЈ sin ϩ yЈ cos ϭ xЈ ϩ yЈ ϭ 4 4 2 2 Ί2
xSubstituting these expressions into the equation xy Ϫ 1 ϭ 0 produces −22−1 1 xy − 1 = 0 xЈ Ϫ yЈ xЈ ϩ yЈ −1 Ϫ 1 ϭ 0 Ί2 Ί2 ͑xЈ͒2 Ϫ ͑yЈ͒2 Ϫ 1 ϭ 0 2 ͑xЈ͒2 ͑yЈ͒2 Ϫ ϭ 1. Standard form Vertices: ͑Ί2͒2 ͑Ί2͒2 Ί2, 0͒, ͑؊Ί2, 0͒ in xy-system͑ ؊1, ؊1ͨ in xy-system This is the equation of a hyperbola centered at the origin with vertices at ͑±Ί2, 0͒ inͧ ,1ͨ ,1ͧ Figure E.3 the xЈyЈ-system, as shown in Figure E.3.
EXAMPLE 2 Rotation of an Ellipse
Sketch the graph of 7x2 Ϫ 6Ί3xy ϩ 13y2 Ϫ 16 ϭ 0.
y Solution Because A ϭ 7, B ϭ Ϫ6Ί3, and C ϭ 13, you have ͑for 0 < < ͞2͒ y x) )2 ) y )2 A Ϫ C 7 Ϫ 13 1 + = 1 cot 2 ϭ ϭ ϭ ϭ . 2) )2 ) 1)2 Ϫ Ί Ί 2 B 6 3 3 6 x Therefore, the equation in the xЈyЈ-system is derived by making the following substitutions. Ί3 1 Ί3xЈ Ϫ yЈ x x ϭ xЈ cos Ϫ yЈ sin ϭ xЈ Ϫ yЈ ϭ −2 2 6 6 2 2 2 1 Ί3 xЈ ϩ Ί3yЈ y ϭ xЈ sin ϩ yЈ cos ϭ xЈ ϩ yЈ ϭ 6 6 2 2 2 −2 Substituting these expressions into the original equation eventually simplifies (after 7x2 − 6 3xy + 13y2 − 16 = 0 considerable algebra) to 4͑xЈ͒2 ϩ 16͑yЈ͒2 ϭ 16 Vertices: ͑xЈ͒2 ͑yЈ͒2 in xy-system ϩ ϭ 1. Standard form ±1ͨ ,0ͧ ,0ͨ ,±2ͧ ͑2͒2 ͑1͒2 1 Ί3 ͑±Ί3, ±1͒, ± , ± in xy-system 2 2 This is the equation of an ellipse centered at the origin with vertices at ͑±2, 0͒ and Figure E.4 ͑0, ±1͒ in the xЈyЈ-system, as shown in Figure E.4. E4 APPENDIX E Rotation and the General Second-Degree Equation
In writing Examples 1 and 2, we chose the equations such that would be one of the common angles 30Њ, 45Њ, and so forth. Of course, many second-degree equations do not yield such common solutions to the equation A Ϫ C cot 2 ϭ . B Example 3 illustrates such a case.
EXAMPLE 3 Rotation of a Parabola
Sketch the graph of x2 Ϫ 4xy ϩ 4y2 ϩ 5Ί5y ϩ 1 ϭ 0.
Solution Because A ϭ 1, B ϭ Ϫ4, and C ϭ 4, you have A Ϫ C 1 Ϫ 4 3 cot 2 ϭ ϭ ϭ . B Ϫ4 4 5 The trigonometric identity cot 2 ϭ ͑cot2 Ϫ 1͒͑͞2 cot ͒ produces 1 3 cot2 Ϫ 1 cot 2 ϭ ϭ θ 4 2 cot 2 from which you can obtain the equation Figure E.5 6 cot ϭ 4 cot2 Ϫ 4 4 cot2 Ϫ 6 cot Ϫ 4 ϭ 0 ͑2 cot Ϫ 4͒͑2 cot ϩ 1͒ ϭ 0. Considering 0 < < ͞2, it follows that 2 cot ϭ 4. Thus, cot ϭ 2 Ϸ 26.6Њ. x2− 4xy + 4y2+ 5 5y + 1 = 0 From the triangle in Figure E.5, you can obtain sin ϭ 1͞Ί5 and cos ϭ 2͞Ί5. y Consequently, you can write the following. y′ 2 1 2xЈ Ϫ yЈ 2 x ϭ xЈ cos Ϫ yЈ sin ϭ xЈ Ϫ yЈ ϭ Ί5 Ί5 Ί5 1 2 xЈ ϩ 2yЈ 1 x′ y ϭ xЈ sin ϩ yЈ cos ϭ xЈ ϩ yЈ ϭ Ί5 Ί5 Ί5 θ ≈ 2 6 . 6 ° x Substituting these expressions into the original equation produces
2xЈ Ϫ yЈ 2 2xЈ Ϫ yЈ xЈ ϩ 2yЈ xЈ ϩ 2yЈ 2 −1 Ϫ 4 ϩ 4 ϩ Ί5 Ί5 Ί5 Ί5 xЈ ϩ 2yЈ −2 5Ί5 ϩ 1 ϭ 0 Ί5 which simplifies to 5͑yЈ͒2 ϩ 5xЈ ϩ 10yЈ ϩ 1 ϭ 0. (y′ + 1)2 = 4 − 1 x′− 4 ( 4 ( 5 ( By completing the square, you can obtain the standard form 5͑yЈ ϩ 1͒2 ϭ Ϫ5xЈ ϩ 4 4 Vertex: , ؊1 in xy-system 1 4 5 ͑yЈ ϩ 1͒2 ϭ 4Ϫ xЈ Ϫ . Standard form 13 6 4 5 , ؊ in xy-system 5Ί5 5Ί5 ͑4 Ϫ ͒ The graph of the equation is a parabola with its vertex at 5 , 1 and its axis parallel Figure E.6 to the xЈ-axis in the xЈyЈ-system, as shown in Figure E.6. APPENDIX E Rotation and the General Second-Degree Equation E5
Invariants Under Rotation In Theorem A.1, note that the constant term FЈ ϭ F is the same in both equations. Because of this, F is said to be invariant under rotation. Theorem A.2 lists some other rotation invariants. The proof of this theorem is left as an exercise (see Exer- cise 34).
THEOREM A.2 Rotation Invariants The rotation of coordinate axes through an angle that transforms the equation Ax2 ϩ Bxy ϩ Cy2 ϩ Dx ϩ Ey ϩ F ϭ 0 into the form AЈ͑xЈ͒2 ϩ CЈ͑yЈ͒2 ϩ DЈxЈ ϩ EЈyЈ ϩ FЈ ϭ 0 has the following rotation invariants. 1. F ϭ FЈ 2. A ϩ C ϭ AЈ ϩ CЈ 3. B2 Ϫ 4AC ϭ ͑BЈ͒2 Ϫ 4AЈCЈ
You can use this theorem to classify the graph of a second-degree equation with an xy-term in much the same way you do for a second-degree equation without an xy- term. Note that because BЈ ϭ 0, the invariant B2 Ϫ 4AC reduces to
B2 Ϫ 4AC ϭ Ϫ4AЈCЈ Discriminant which is called the discriminant of the equation Ax2 ϩ Bxy ϩ Cy2 ϩ Dx ϩ Ey ϩ F ϭ 0. Because the sign of AЈCЈ determines the type of graph for the equation AЈ͑xЈ͒2 ϩ CЈ͑yЈ͒2 ϩ DЈxЈ ϩ EЈyЈ ϩ FЈ ϭ 0 the sign of B2 Ϫ 4AC must determine the type of graph for the original equation. This result is stated in Theorem A.3.
THEOREM A.3 Classification of Conics by the Discriminant The graph of the equation Ax2 ϩ Bxy ϩ Cy2 ϩ Dx ϩ Ey ϩ F ϭ 0 is, except in degenerate cases, determined by its discriminant as follows. 1. Ellipse or circle B2 Ϫ 4AC < 0 2. Parabola B2 Ϫ 4AC ϭ 0 3. Hyperbola B2 Ϫ 4AC > 0 E6 APPENDIX E Rotation and the General Second-Degree Equation
EXAMPLE 4 Using the Discriminant
Classify the graph of each of the following equations. a. 4xy Ϫ 9 ϭ 0 b. 2x2 Ϫ 3xy ϩ 2y2 Ϫ 2x ϭ 0 c. x2 Ϫ 6xy ϩ 9y2 Ϫ 2y ϩ 1 ϭ 0 d. 3x2 ϩ 8xy ϩ 4y2 Ϫ 7 ϭ 0 Solution a. The graph is a hyperbola because B2 Ϫ 4AC ϭ 16 Ϫ 0 > 0. b. The graph is a circle or an ellipse because B2 Ϫ 4AC ϭ 9 Ϫ 16 < 0. c. The graph is a parabola because B2 Ϫ 4AC ϭ 36 Ϫ 36 ϭ 0. d. The graph is a hyperbola because B2 Ϫ 4AC ϭ 64 Ϫ 48 > 0.
E X E R C I S E S F O R APPENDIX E
In Exercises 1–12, rotate the axes to eliminate the xy-term. Give In Exercises 19–26, use the discriminant to determine whether the resulting equation and sketch its graph showing both sets the graph of the equation is a parabola, an ellipse, or a hyper- of axes. bola. 1. xy ϩ 1 ϭ 0 19. 16x2 Ϫ 24xy ϩ 9y2 Ϫ 30x Ϫ 40y ϭ 0 2. xy Ϫ 4 ϭ 0 20. x2 Ϫ 4xy Ϫ 2y2 Ϫ 6 ϭ 0 3. x2 Ϫ 10xy ϩ y2 ϩ 1 ϭ 0 21. 13x2 Ϫ 8xy ϩ 7y2 Ϫ 45 ϭ 0 4. xy ϩ x Ϫ 2y ϩ 3 ϭ 0 22. 2x2 ϩ 4xy ϩ 5y2 ϩ 3x Ϫ 4y Ϫ 20 ϭ 0 5. xy Ϫ 2y Ϫ 4x ϭ 0 23. x2 Ϫ 6xy Ϫ 5y2 ϩ 4x Ϫ 22 ϭ 0 6. 13x2 ϩ 6Ί3xy ϩ 7y2 Ϫ 16 ϭ 0 24. 36x2 Ϫ 60xy ϩ 25y2 ϩ 9y ϭ 0 7. 5x2 Ϫ 2xy ϩ 5y2 Ϫ 12 ϭ 0 25. x2 ϩ 4xy ϩ 4y2 Ϫ 5x Ϫ y Ϫ 3 ϭ 0 8. 2x2 Ϫ 3xy Ϫ 2y2 ϩ 10 ϭ 0 26. x2 ϩ xy ϩ 4y2 ϩ x ϩ y Ϫ 4 ϭ 0 9. 3x2 Ϫ 2Ί3xy ϩ y2 ϩ 2x ϩ 2Ί3y ϭ 0 10. 16x2 Ϫ 24xy ϩ 9y2 Ϫ 60x Ϫ 80y ϩ 100 ϭ 0 In Exercises 27–32, sketch the graph (if possible) of the degen- erate conic. 11. 9x2 ϩ 24xy ϩ 16y2 ϩ 90x Ϫ 130y ϭ 0 27. y2 Ϫ 4x2 ϭ 0 12. 9x2 ϩ 24xy ϩ 16y2 ϩ 80x Ϫ 60y ϭ 0 28. x2 ϩ y2 Ϫ 2x ϩ 6y ϩ 10 ϭ 0 In Exercises 13–18, use a graphing utility to graph the conic. 29. x2 ϩ 2xy ϩ y2 Ϫ 1 ϭ 0 Determine the angle through which the axes are rotated. 30. x2 Ϫ 10xy ϩ y2 ϭ 0 Explain how you used the utility to obtain the graph. 31. ͑x Ϫ 2y ϩ 1͒͑x ϩ 2y Ϫ 3͒ ϭ 0 13. x2 ϩ xy ϩ y2 ϭ 10 32. ͑2x ϩ y Ϫ 3͒2 ϭ 0 14. x2 Ϫ 4xy ϩ 2y2 ϭ 6 15. 17x2 ϩ 32xy Ϫ 7y2 ϭ 75 33. Show that the equation x2 ϩ y2 ϭ r2 is invariant under rotation 16. 40x2 ϩ 36xy ϩ 25y2 ϭ 52 of axes. 17. 32x2 ϩ 50xy ϩ 7y2 ϭ 52 34. Prove Theorem A.2. 18. 4x2 Ϫ 12xy ϩ 9y2 ϩ ͑4Ί13 ϩ 12͒x Ϫ ͑6Ί13 ϩ 8͒y ϭ 91
The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system. The solutions of other exercises may also be facilitated by use of appropriate technology.