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APPENDIX E Rotation and the General Second-Degree Equation

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APPENDIX E Rotation and the General Second-Degree Equation 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 1 2 1 1 1 5 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 1 Bxy 1 Cy2 1 Dx 1 Ey 1 F 5 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 x9-axis and the y9-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 x9y9-plane will clockwise through an angle u, the rotated have the form axes are denoted as the x9-axis and y9-axis. A9sx9d2 1 C9sy9d2 1 D9x9 1 E9y9 1 F9 5 0. Equation in x9y9-plane Figure E.1 Because this equation has no x9y9-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 A9, C9, D9, E9, and F9. THEOREM A.1 Rotation of Axes The general equation of the conic Ax2 1 Bxy 1 Cy2 1 Dx 1 Ey 1 F 5 0, where B Þ 0, can be rewritten as A9sx9d2 1 C9sy9d2 1 D9x9 1 E9y9 1 F9 5 0 by rotating the coordinate axes through an angle u, where A 2 u 5 C cot 2 B . The coefficients of the new equation are obtained by making the substitutions x 5 x9 cos u 2 y9 sin u y 5 x9 sin u 1 y9 cos u. 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 x9y9-system, choose a point P 5 sx, yd in the original system and attempt to find its coordinates sx9, y9d 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, x9, and y9 are those given in Figure E.2. Using the formulas for the sine and cosine of the difference r of two angles, you obtain x9 5 r cossa 2 ud 5 rscos a cos u 1 sin a sin ud α 5 r cos a cos u 1 r sin a sin u 5 x cos u 1 y sin u x y9 5 r sin sa 2 ud 5 rssin a cos u 2 cos a sin ud 5 r sin a cos u 2 r cos a sin u 5 y cos u 2 x sin u. Original: x 5 r cos a y 5 r sin a Solving this system for x and y yields 5 9 u 2 9 u 5 9 u 1 9 u 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 9 5 A cos2 u 1 B cos u sin u 1 C sin2 u C 9 5 A sin2 u 2 B cos u sin u 1 C cos2 u r x′ α − θ D 9 5 D cos u 1 E sin u θ E 9 5 2D sin u 1 E cos u x F 9 5 F Now, in order to eliminate the x9y9-term, you must select u such that B9 5 0, as Rotated: x9 5 r cossa 2 ud follows. y9 5 r sinsa 2 ud B9 5 2sC 2 Ad sin u cos u 1 Bscos2 u 2 sin2 ud Figure E.2 5 sC 2 Ad sin 2u 1 B cos 2u C 2 A 5 Bssin 2ud1 1 cot 2u2 5 0, sin 2u Þ 0 B If B 5 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 B9 5 0 is to let A 2 C cot 2u 5 , 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 2 1 5 0 in standard form. Solution Because A 5 0, B 5 1, and C 5 0, you have sfor 0 < u < py2d A 2 C p p cot 2u 5 5 0 2u 5 u 5 . B 2 4 x) )2) y)2 − = 1 ) 2)2) 2)2 The equation in the x9y9-system is obtained by making the following substitutions. y p p !2 !2 x9 2 y9 x 5 x9 cos 2 y9 sin 5 x9 2 y9 5 4 4 1 2 2 1 2 2 !2 y2x p p !2 !2 x9 1 y9 1 y 5 x9 sin 1 y9 cos 5 x9 1 y9 5 4 4 1 2 2 1 2 2 !2 xSubstituting these expressions into the equation xy 2 1 5 0 produces −22−1 1 xy − 1 = 0 x9 2 y9 x9 1 y9 −1 1 21 2 2 1 5 0 !2 !2 sx9d2 2 sy9d2 2 1 5 0 2 sx9d2 sy9d2 2 5 1. Standard form Vertices: s!2d2 s!2d2 s!2, 0d, s2!2, 0d in x9y9-system x1, 1c, x21, 21c in xy-system This is the equation of a hyperbola centered at the origin with vertices at s±!2, 0d in Figure E.3 the x9y9-system, as shown in Figure E.3. EXAMPLE 2 Rotation of an Ellipse Sketch the graph of 7x2 2 6!3xy 1 13y2 2 16 5 0. y Solution Because A 5 7, B 5 26!3, and C 5 13, you have sfor 0 < u < py2d y x) )2 ) y )2 A 2 C 7 2 13 1 p + = 1 cot 2u 5 5 5 u 5 . 2) )2 ) 1)2 2 ! ! 2 B 6 3 3 6 x Therefore, the equation in the x9y9-system is derived by making the following substitutions. p p !3 1 !3x9 2 y9 x x 5 x9 cos 2 y9 sin 5 x9 2 y91 2 5 −2 2 6 6 1 2 2 2 2 p p 1 !3 x9 1 !3y9 y 5 x9 sin 1 y9 cos 5 x91 2 1 y9 5 6 6 2 1 2 2 2 −2 Substituting these expressions into the original equation eventually simplifies (after 7x2 − 6 3xy + 13y2 − 16 = 0 considerable algebra) to 4sx9d2 1 16sy9d2 5 16 Vertices: sx9d2 sy9d2 x±2, 0c, x0, ±1c in x9y9-system 1 5 1. Standard form s2d2 s1d2 1 !3 s±!3, ±1d, 1± , ± 2 in xy-system 2 2 This is the equation of an ellipse centered at the origin with vertices at s±2, 0d and Figure E.4 s0, ±1d in the x9y9-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 u would be one of the common angles 308, 458, and so forth. Of course, many second-degree equations do not yield such common solutions to the equation A 2 C cot 2u 5 . B Example 3 illustrates such a case. EXAMPLE 3 Rotation of a Parabola Sketch the graph of x2 2 4xy 1 4y2 1 5!5y 1 1 5 0. Solution Because A 5 1, B 5 24, and C 5 4, you have A 2 C 1 2 4 3 cot 2u 5 5 5 . B 24 4 5 The trigonometric identity cot 2u 5 scot2 u 2 1dys2 cot ud produces 1 3 cot2 u 2 1 cot 2u 5 5 θ 4 2 cot u 2 from which you can obtain the equation Figure E.5 6 cot u 5 4 cot2 u 2 4 4 cot2 u 2 6 cot u 2 4 5 0 s2 cot u 2 4ds2 cot u 1 1d 5 0. Considering 0 < u < py2, it follows that 2 cot u 5 4. Thus, cot u 5 2 u < 26.68. x2− 4xy + 4y2+ 5 5y + 1 = 0 From the triangle in Figure E.5, you can obtain sin u 5 1y!5 and cos u 5 2y!5. y Consequently, you can write the following. y′ 2 1 2x9 2 y9 2 x 5 x9 cos u 2 y9 sin u 5 x91 2 2 y91 2 5 !5 !5 !5 1 2 x9 1 2y9 1 x′ y 5 x9 sin u 1 y9 cos u 5 x91 2 1 y91 2 5 !5 !5 !5 θ ≈ 2 6 . 6 ° x Substituting these expressions into the original equation produces 2x9 2 y9 2 2x9 2 y9 x9 1 2y9 x9 1 2y9 2 −1 1 2 2 41 21 2 1 41 2 1 !5 !5 !5 !5 x9 1 2y9 −2 5!51 2 1 1 5 0 !5 which simplifies to 5sy9d2 1 5x9 1 10y9 1 1 5 0.
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