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Rotation of Axes

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Rotation of Axes ROTATION OF AXES ■■ For a discussion of conic sections, see In precalculus or calculus you may have studied conic sections with equations of the form Review of Conic Sections. Ax2 ϩ Cy2 ϩ Dx ϩ Ey ϩ F ෇ 0 Here we show that the general second-degree equation 1 Ax2 ϩ Bxy ϩ Cy2 ϩ Dx ϩ Ey ϩ F ෇ 0 can be analyzed by rotating the axes so as to eliminate the term Bxy. In Figure 1 the x and y axes have been rotated about the origin through an acute angle ␪ to produce the X and Y axes. Thus, a given point P has coordinates ͑x, y͒ in the first coor- dinate system and ͑X, Y͒ in the new coordinate system. To see how X and Y are related to x and y we observe from Figure 2 that X ෇ r cos ␾ Y ෇ r sin ␾ x ෇ r cos͑␪ ϩ ␾͒ y ෇ r sin͑␪ ϩ ␾͒ y y Y P(x, y) Y P(X, Y) P X Y X r y ˙ ¨ ¨ 0 x 0 x X x FIGURE 1 FIGURE 2 The addition formula for the cosine function then gives x ෇ r cos͑␪ ϩ ␾͒ ෇ r͑cos ␪ cos ␾ Ϫ sin ␪ sin ␾͒ ෇ ͑r cos ␾͒ cos ␪ Ϫ ͑r sin ␾͒ sin ␪ ෇ ⌾ cos ␪ Ϫ Y sin ␪ A similar computation gives y in terms of X and Y and so we have the following formulas: 2 x ෇ X cos ␪ Ϫ Y sin ␪ y ෇ X sin ␪ ϩ Y cos ␪ By solving Equations 2 for X and Y we obtain 3 X ෇ x cos ␪ ϩ y sin ␪ Y ෇ Ϫx sin ␪ ϩ y cos ␪ EXAMPLE 1 If the axes are rotated through 60Њ, find the XY-coordinates of the point whose xy-coordinates are ͑2, 6͒. SOLUTION Using Equations 3 with x ෇ 2,y ෇ 6 , and ␪ ෇ 60Њ, we have X ෇ 2 cos 60Њϩ6 sin 60Њ ෇ 1 ϩ 3s3 Y ෇ Ϫ2 sin 60Њϩ6 cos 60Њ ෇ Ϫs3 ϩ 3 The XY-coordinates are (1 ϩ 3s3, 3 Ϫ s3). 1 Thomson Brooks-Cole copyright 2007 2 ■ ROTATION OF AXES Now let’s try to determine an angle ␪ such that the term Bxy in Equation 1 disappears when the axes are rotated through the angle ␪. If we substitute from Equations 2 in Equation 1, we get A͑X cos ␪ Ϫ Y sin ␪͒2 ϩ B͑X cos ␪ Ϫ Y sin ␪͒͑⌾ sin ␪ ϩ Y cos ␪͒ ϩ C͑X sin ␪ ϩ Y cos ␪͒2 ϩ D͑X cos ␪ Ϫ Y sin ␪͒ ϩ E͑X sin ␪ ϩ Y cos ␪͒ ϩ F ෇ 0 Expanding and collecting terms, we obtain an equation of the form 4 AЈX 2 ϩ BЈXY ϩ CЈY 2 ϩ DЈX ϩ EЈY ϩ F ෇ 0 where the coefficient BЈ of XY is BЈ ෇ 2͑C Ϫ A͒ sin ␪ cos ␪ ϩ B͑cos2␪ Ϫ sin2␪͒ ෇ ͑C Ϫ A͒ sin 2␪ ϩ B cos 2␪ To eliminate the XY term we choose ␪ so that BЈ ෇ 0, that is, ͑A Ϫ C͒ sin 2␪ ෇ 〉 cos 2␪ or A Ϫ C 5 cot 2␪ ෇ B EXAMPLE 2 Show that the graph of the equation xy ෇ 1 is a hyperbola. SOLUTION Notice that the equation xy ෇ 1 is in the form of Equation 1 where A ෇ 0, B ෇ 1, and C ෇ 0. According to Equation 5, the xy term will be eliminated if we choose ␪ so that A Ϫ C cot 2␪ ෇ ෇ 0 B This will be true if 2␪ ෇ ␲͞2, that is,␪ ෇ ␲͞4 . Then cos ␪ ෇ sin ␪ ෇ 1͞s2 and Equa- tions 2 become X@ Y@ xy=1 or - =1 2 2 y X Y X Y X x ෇ Ϫ y ෇ ϩ Y s2 s2 s2 s2 Substituting these expressions into the original equation gives π 4 x X Y X Y X 2 Y 2 0 ͩ Ϫ ͪͩ ϩ ͪ ෇ 1 or Ϫ ෇ 1 s2 s2 s2 s2 2 2 We recognize this as a hyperbola with vertices (Ϯs2, 0) in the XY-coordinate system. The asymptotes are Y ෇ ϮX in the XY-system, which correspond to the coordinate axes FIGURE 3 in the xy-system (see Figure 3). Thomson Brooks-Cole copyright 2007 ROTATION OF AXES ■ 3 EXAMPLE 3 Identify and sketch the curve 73x 2 ϩ 72xy ϩ 52y 2 ϩ 30x Ϫ 40y Ϫ 75 ෇ 0 SOLUTION This equation is in the form of Equation 1 with A ෇ 73,B ෇ 72 , and C ෇ 52. Thus A Ϫ C 73 Ϫ 52 7 cot 2␪ ෇ ෇ ෇ B 72 24 From the triangle in Figure 4 we see that ␪ ෇ 7 cos 2 25 25 ␪ ␪ 24 The values of cos and sin can then be computed from the half-angle formulas: 1 ϩ cos 2␪ 1 ϩ 7 4 2¨ cos ␪ ෇ ͱ ෇ ͱ 25 ෇ 2 2 5 7 FIGURE 4 1 Ϫ cos 2␪ 1 Ϫ 7 3 sin ␪ ෇ ͱ ෇ ͱ 25 ෇ 2 2 5 The rotation equations (2) become ෇ 4 Ϫ 3 ෇ 3 ϩ 4 x 5 X 5 Y y 5 X 5 Y Substituting into the given equation, we have 4 Ϫ 3 2 ϩ 4 Ϫ 3 3 ϩ 4 ϩ 3 ϩ 4 2 73(5 X 5Y) 72(5 X 5Y)(5 X 5Y) 52(5 X 5Y) ϩ 4 Ϫ 3 Ϫ 3 ϩ 4 Ϫ ෇ 30(5 X 5Y) 40(5 X 5Y) 75 0 which simplifies to 4X 2 ϩ Y 2 Ϫ 2Y ෇ 3 Completing the square gives ͑Y Ϫ 1͒2 4X 2 ϩ ͑Y Ϫ 1͒2 ෇ 4 or X 2 ϩ ෇ 1 4 and we recognize this as being an ellipse whose center is ͑0, 1͒ in XY-coordinates. ␪ ෇ Ϫ1 4 Ϸ Њ Since cos (5) 37 , we can sketch the graph in Figure 5. y X Y (0, 1) ¨Å37° 0 x 73≈+72xy+52¥+30x-40y-75=0 or FIGURE 5 4X@+(Y-1)@=4 Thomson Brooks-Cole copyright 2007 4 ■ ROTATION OF AXES EXERCISES (c) Find an equation of the directrix in the xy-coordinate A Click here for answers. S Click here for solutions. system. 14. (a) Use rotation of axes to show that the equation 1–4 Find the XY-coordinates of the given point if the axes are rotated through the specified angle. 2x 2 Ϫ 72xy ϩ 23y 2 Ϫ 80x Ϫ 60y ෇ 125 1.͑1, 4͒, 30Њ 2. ͑4, 3͒, 45Њ represents a hyperbola. 3.͑Ϫ2, 4͒, 60Њ 4. ͑1, 1͒, 15Њ (b) Find the XY-coordinates of the foci. Then find the ■■■■■■■■■■■■■ xy-coordinates of the foci. (c) Find the xy-coordinates of the vertices. 5–12 Use rotation of axes to identify and sketch the curve. (d) Find the equations of the asymptotes in the xy-coordinate 5. x 2 Ϫ 2xy ϩ y 2 Ϫ x Ϫ y ෇ 0 system. (e) Find the eccentricity of the hyperbola. 6. x 2 Ϫ xy ϩ y 2 ෇ 1 15. Suppose that a rotation changes Equation 1 into Equation 4. x 2 ϩ xy ϩ y 2 ෇ 1 7. Show that s ϩ 2 ෇ 8. 3xy y 1 AЈϩCЈ ෇ A ϩ C 9. 97x 2 ϩ 192xy ϩ 153y 2 ෇ 225 16. Suppose that a rotation changes Equation 1 into Equation 4. 2 Ϫ ϩ 2 ϩ ෇ 10. 3x 12s5xy 6y 9 0 Show that 11. 2s3xy Ϫ 2y 2 Ϫ s3x Ϫ y ෇ 0 ͑BЈ͒2 Ϫ 4AЈCЈ ෇ B 2 Ϫ 4AC 12. 16x 2 Ϫ 8s2xy ϩ 2y 2 ϩ (8s2 Ϫ 3)x Ϫ (6s2 ϩ 4)y ෇ 7 17. Use Exercise 16 to show that Equation 1 represents (a) a ■■■■■■■■■■■■■ parabola if B 2 Ϫ 4AC ෇ 0, (b) an ellipse if B 2 Ϫ 4AC Ͻ 0, and (c) a hyperbola if B 2 Ϫ 4AC Ͼ 0,except in degenerate 13. (a) Use rotation of axes to show that the equation cases when it reduces to a point, a line, a pair of lines, or no 36x 2 ϩ 96xy ϩ 64y 2 ϩ 20x Ϫ 15y ϩ 25 ෇ 0 graph at all. represents a parabola. 18. Use Exercise 17 to determine the type of curve in (b) Find the XY-coordinates of the focus. Then find the Exercises 9–12. xy-coordinates of the focus. Thomson Brooks-Cole copyright 2007 ROTATION OF AXES ■ 5 ANSWERS 9. X 2 ϩ ͑Y 2͞9͒ ෇ 1, ellipse S Click here for solutions. y X s ϩ ͞ s Ϫ ͞ 1. (( 3 4) 2, (4 3 1) 2) Y 3. (2s3 Ϫ 1, s3 ϩ 2) 4 sin–!” 5 ’ 2 5. X ෇ s2Y , parabola 0 x y Y X 11. ͑X Ϫ 1͒2 Ϫ 3Y 2 ෇ 1, hyperbola 1 1 y 0 x Y X 7. 3X 2 ϩ Y 2 ෇ 2, ellipse 0 x y Y X 0 x Ϫ ෇ 2 17 Ϫ17 51 13. (a)Y 1 4X (b) (0, 16 ), ( 20 , 80 ) (c) 64x Ϫ 48y ϩ 75 ෇ 0 Thomson Brooks-Cole copyright 2007 6 ■ ROTATION OF AXES SOLUTIONS √3 1 1. X =1 cos 30◦ +4sin30◦ =2+ , Y = 1 sin 30◦ +4cos30◦ =2√3 . · 2 − · − 2 3. X = 2cos60◦ +4sin60◦ = 1+2√3, Y =2sin60◦ +4cos60◦ = √3+2. − − A C 5. cot 2θ = − =0 2θ = π θ = π [by Equations 2] B ⇒ 2 ⇔ 4 ⇒ X Y X + Y x = − and y = . Substituting these into the curve equation √2 √2 X gives 0=(x y)2 (x + y)=2Y 2 √2X or Y 2 = . − − − √2 [Parabola, vertex (0, 0), directrix X = 1/ 4 √2 , focus 1/ 4 √2 , 0 ]. − A C 7. cot 2θ = − =0 2θ = π θ = π [by B ⇒ 2 ⇔ 4 ⇒ X Y X + Y Equations 2] x = − and y = . Substituting these into the √2 √2 curve equation gives X2 2XY + Y 2 X2 Y 2 X2 +2XY + Y 2 1= − + − + 2 2 2 ⇒ X2 Y 2 3X2 + Y 2 =2 + =1.
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