APPENDIX D Rotation and the General Second-Degree Equation Rotation of Aes Invariants Under Rotation After rotation of the - and -aes counterclockwise through an angle, the rotated aes are denoted as the -ais and -ais. Figure D. θ Rotation of Aes Previousl, ou learned that equations of conics with aes parallel to one of the coordinate aes can be written in the general form A C D E F 0. Horizontal or vertical aes Here ou will stud the equations of conics whose aes are rotated so that the are not parallel to either the -ais or the -ais. The general equation for such conics contains an -term. A C D E F 0 Equation in -plane To eliminate this -term, ou can use a procedure called rotation of aes. You want to rotate the - and -aes until the are parallel to the aes of the conic. (The rotated aes are denoted as the -ais and the -ais, as shown in Figure D.. After the rotation has been accomplished, the equation of the conic in the new -plane will have the form A C D E F 0. Equation in -plane ecause this equation has no -term, ou can obtain a standard form b completing the square. The following theorem identifies how much to rotate the aes to eliminate an -term and also the equations for determining the new coefficients A, C, D, E, and F. THEOREM D. Rotation of Aes The general second-degree equation of A C D E F 0, where 0, can be rewritten as A C D E F 0 b rotating the coordinate aes through an angle, where cot C The coefficients of the new equation are obtained b making the substitutions cos sin sin cos.. D
D APPENDIX D Rotation and the General Second-Degree Equation Original: Rotated: Figure D. r α r cos r sin r P = (, P = (, α θ θ cos r sin r Proof To discover how the coordinates in the -sstem are related to the coordinates in the -sstem, choose a point P, in the original sstem and attempt to find its coordinates, in the rotated sstem. In either sstem, the distance r between the point P and the origin is the same, and so the equations for,,, and are those given in Figure D.. Using the formulas for the sine and cosine of the difference of two angles, ou obtain r cos rcos cos sin sin r cos cos r sin sin cos sin r sin rsin cos cos sin r sin cos r cos sin cos sin. Solving this sstem for and ields cos sin and sin cos. Finall, b substituting these values for and into the original equation and collecting terms, ou obtain the following. A A cos cos sin C sin C A sin cos sin C cos D D cos E sin E D sin E cos F F Now, in order to eliminate the -term, ou must select such that as follows. C A sin cos cos sin C A sin cos sin C A 0, sin 0 If 0, no rotation is necessar, because the -term is not present in the original equation. If 0, the onl wa to make is to let cot A C, 0. cot So, ou have established the desired results. 0 0,
APPENDIX D Rotation and the General Second-Degree Equation D3 EXAMPLE Rotation of a Hperbola Write the equation 0 in standard form. = 0 Vertices:, 0,, 0 in -sstem,,,, in -sstem Figure D.3 = Solution ecause A 0,, and C 0, ou have for 0 < cot A C The equation in the -sstem is obtained b making the following substitutions. cos 4 sin 4 sin 4 cos 4 Substituting these epressions into the equation 0 produces 0 Write in standard form. This is the equation of a hperbola centered at the origin with vertices at ±, 0 in the -sstem, as shown in Figure D.3. 0 0. 4. < EXAMPLE Rotation of an Ellipse + = 7 6 3 + 3 6 = 0 Vertices: ±, 0, 0, ± in -sstem ±3, ±, ± in -sstem, ±3 Figure D.4 Sketch the graph of 7 63 3 6 0. Solution ecause A 7, 63, and C 3, ou have for 0 < cot A C Therefore, the equation in the substitutions. cos 6 sin 6 sin 6 cos -sstem is derived b making the following Substituting these epressions into the original equation eventuall simplifies (after considerable algebra to 4 6 6. Write in standard form. This is the equation of an ellipse centered at the origin with vertices at ±, 0 and 0, ± in the -sstem, as shown in Figure D.4. 7 3 63 3 3 3 6 6. 3 3 <
D4 APPENDIX D Rotation and the General Second-Degree Equation In Eamples and, the values of were the common angles 45 and 30, respectivel. Of course, man second-degree equations do not ield such common solutions to the equation cot A C. Eample 3 illustrates such a case. EXAMPLE 3 Rotation of a Parabola θ Figure D.5 4 + 4 + 5 5 + = 0 Verte: 4 5, ( + = 4 ( ( 4 4 5 in -sstem 3 in -sstem 55, 55 6 Figure D.6 5 ( ( θ 6.6 Sketch the graph of 4 4 55 0. Solution ecause A, 4, and C 4, ou have cot A C The trigonometric identit cot cot cot produces cot 3 4 cot cot from which ou obtain the equation 6 cot 4 cot 4 Considering 0 < it follows that cot 4. So, cot From the triangle in Figure D.5, ou obtain sin 5 Consequentl, ou can write the following. cos sin 5 5 5 sin cos 5 5 5 Substituting these epressions into the original equation produces 5 4 5 5 4 55 5 0 which simplifies to 5 5 0 0. completing the square, ou obtain the standard form 5 5 4 4 4 4 5. and Write in standard form. The graph of the equation is a parabola with its verte at 4 5, and its ais parallel to the -ais in the -sstem, as shown in Figure D.6. 4 4 3 4. <, 6.6. 4 cot 6 cot 4 0 cot 4 cot 0. 5 cos 5.
APPENDIX D Rotation and the General Second-Degree Equation D5 Invariants Under Rotation In Theorem D., note that the constant term is the same in both equations that is, ecause of this, F is said to be invariant under rotation. Theorem D. lists some other rotation invariants. The proof of this theorem is left as an eercise (see Eercise 34. F F. THEOREM D. Rotation Invariants The rotation of coordinate aes through an angle that transforms the equation A C D E F 0 into the form A C D E F 0 has the following rotation invariants.. F F. A C A C 3. 4AC 4AC You can use this theorem to classif the graph of a second-degree equation with an -term in much the same wa ou do for a second-degree equation without an -term. Note that because the invariant 4AC reduces to 0, 4AC 4AC Discriminant which is called the discriminant of the equation A C D E F 0. ecause the sign of AC determines the tpe of graph for the equation A C D E F 0 the sign of 4AC must determine the tpe of graph for the original equation. This result is stated in Theorem D.3. THEOREM D.3 Classification of Conics b the Discriminant The graph of the equation A C D E F 0 is, ecept in degenerate cases, determined b its discriminant as follows.. Ellipse or circle 4AC < 0. Parabola 4AC 0 3. Hperbola 4AC > 0
D6 APPENDIX D Rotation and the General Second-Degree Equation EXAMPLE 4 Using the Discriminant Classif the graph of each equation. a. 4 9 0 b. c. 6 9 0 d. 3 0 3 8 4 7 0 Solution a. The graph is a hperbola because 4AC 6 0 > 0. b. The graph is a circle or an ellipse because 4AC 9 6 < 0. c. The graph is a parabola because 4AC 36 36 0. d. The graph is a hperbola because 4AC 64 48 > 0. EXERCISES FOR APPENDIX D In Eercises, rotate the aes to eliminate the -term in the equation. Write the resulting equation in standard form and sketch its graph showing both sets of aes.. 0. 4 0 3. 0 0 4. 3 0 5. 4 0 6. 3 63 7 6 0 7. 5 5 0 8. 3 0 0 9. 3 3 3 0 0. 6 4 9 60 80 00 0. 9 4 6 90 30 0. 9 4 6 80 60 0 In Eercises 3 8, use a graphing utilit to graph the conic. Determine the angle through which the aes are rotated. Eplain how ou used the graphing utilit to obtain the graph. 3. 0 4. 4 6 5. 7 3 7 75 6. 40 36 5 5 7. 3 50 7 5 8. 4 9 43 63 8 9 In Eercises 9 6, use the discriminant to determine whether the graph of the equation is a parabola, an ellipse, or a hperbola. 9. 6 4 9 30 40 0 0. 4 6 0. 3 8 7 45 0. 4 5 3 4 0 0 3. 6 5 4 0 4. 36 60 5 9 0 5. 4 4 5 3 0 6. 4 4 0 In Eercises 7 3, sketch the graph (if possible of the degenerate conic. 7. 4 0 8. 6 0 0 9. 0 30. 0 0 3. 3 0 3. 3 0 33. Show that the equation r is invariant under rotation of aes. 34. Prove Theorem D..