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 Calculus, Early Transcendentals, Sixth Edition, Section 10.5. Ax2 Cy2 Dx Ey F 0

Here we show that the general second-degree equation

1 Ax2 Bxy Cy2 Dx Ey 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, y in the first coor- dinate system and X, Y 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 cos y r sin

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 cos rcos cos sin sin

r cos cos r sin sin cos Y sin

A similar computation gives y in terms of X and Y and so we have the following formulas:

2 x X cos Y sin y X sin Y cos

By solving Equations 2 for X and Y we obtain

3 X x cos y sin Y x sin y cos

EXAMPLE 1 If the axes are rotated through 60 , find the XY -coordinates of the point whose xy-coordinates are 2, 6 .

SOLUTION Using Equations 3 with x 2 ,y 6 , and 60 , we have X 2 cos 606 sin 60 1 3s3

Sixth Edition. ISBN: 0495011606. © 2008 Brooks/Cole. All rights reserved. Y 2 sin 606 cos 60 s3 3

Calculus, The XY-coordinates are (1 3s3, 3 s3) . Stewart: 1 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 Y sin ͒2 B͑X cos Y sin ͒͑ sin Y cos ͒

C͑X sin Y cos ͒2 D͑X cos Y sin ͒

E͑X sin Y cos ͒ F ෇ 0

Expanding and collecting terms, we obtain an equation of the form

4 AX 2 BXY CY 2 DX EY F ෇ 0

where the coefficient B of XY is

B ෇ 2͑C A͒ sin cos B͑cos2 sin2͒

෇ ͑C A͒ sin 2 B cos 2

To eliminate the XY term we choose so that B ෇ 0 , that is,

͑A C͒ sin 2 ෇ cos 2

or

A 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 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 X Y X Y y ෇ ෇ X x y 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 ͩ ͪͩ ͪ ෇ 1 or ෇ 1 s2 s2 s2 s2 2 2

We recognize this as a hyperbola with vertices (s2, 0) in the XY -coordinate system. The asymptotes are Y ෇ X in the XY -system, which correspond to the coordinate axes

FIGURE 3 in the xy -system (see Figure 3). Sixth Edition. ISBN: 0495011606. © 2008 Brooks/Cole. All rights reserved.

Calculus, Stewart: ROTATION OF AXES ■ 3

EXAMPLE 3 Identify and sketch the curve

73x 2 72xy 52y 2 30x 40y 75 ෇ 0

SOLUTION This equation is in the form of Equation 1 with A ෇ 73 ,B ෇ 72 , and C ෇ 52 . Thus A C 73 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: 7 1 cos 2 1 25 4 2¨ cos ෇ ͱ ෇ ͱ ෇ 2 2 5 7

FIGURE 4 1 cos 2 1 7 3 sin ෇ ͱ ෇ ͱ 25 ෇ 2 2 5

The rotation equations (2) become ෇ 4 3 ෇ 3 4 x 5 X 5 Y y 5 X 5 Y Substituting into the given equation, we have

4 3 2 4 3 3 4 3 4 2 73(5 X 5Y) 72(5 X 5Y)(5 X 5Y) 52(5 X 5Y)

4 3 3 4 ෇ 30(5 X 5Y) 40(5 X 5Y) 75 0

which simplifies to 4X 2 Y 2 2Y ෇ 3 Completing the square gives

͑Y 1͒2 4X 2 ͑Y 1͒2 ෇ 4 or X 2 ෇ 1 4

and we recognize this as being an ellipse whose center is ͑0, 1͒ in XY -coordinates. ෇ 1 4 Ϸ 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 Sixth Edition. ISBN: 0495011606. © 2008 Brooks/Cole. All rights reserved.

Calculus, Stewart: 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 72xy 23y 2 80x 60y ෇ 125 1.͑1, 4͒, 30 2. ͑4, 3͒, 45 represents a hyperbola. 3.͑2, 4͒, 60 4. ͑1, 1͒, 15 (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 2xy y 2 x y ෇ 0 system. (e) Find the eccentricity of the hyperbola. 6. x 2 xy y 2 ෇ 1 15. Suppose that a rotation changes Equation 1 into Equation 4. x 2 xy y 2 ෇ 1 7. Show that s 2 ෇ 8. 3xy y 1 AC ෇ A C 9. 97x 2 192xy 153y 2 ෇ 225 16. Suppose that a rotation changes Equation 1 into Equation 4. 2 2 ෇ 10. 3x 12s5xy 6y 9 0 Show that 11. 2s3xy 2y 2 s3x y ෇ 0 ͑B͒2 4AC ෇ B 2 4AC 12. 16x 2 8s2xy 2y 2 (8s2 3)x (6s2 4)y ෇ 7 17. Use Exercise 16 to show that Equation 1 represents (a) a ■■■■■■■■■■■■■ parabola if B 2 4AC ෇ 0 , (b) an ellipse if B 2 4AC 0, and (c) a hyperbola if B 2 4AC 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 96xy 64y 2 20x 15y 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. Sixth Edition. ISBN: 0495011606. © 2008 Brooks/Cole. All rights reserved.

Calculus, Stewart: ROTATION OF AXES ■ 5

ANSWERS

9. X 2 Y 29 1, ellipse S Click here for solutions. y X s s 1. (( 3 4) 2, (4 3 1) 2) Y

3. (2s3 1, s3 2) 4 sin–!” 5 ’ 2 5. X s2Y , parabola 0 x

y

Y X

11. X 12 3Y 2 1, hyperbola 1 1 y 0 x Y

X

7. 3X 2 Y 2 2, ellipse 0 x y

Y X

0 x 2 17 17 51 13. (a)Y 1 4X (b) (0, 16 ), ( 20 , 80 ) (c) 64x 48y 75 0 Sixth Edition. ISBN: 0495011606. © 2008 Brooks/Cole. All rights reserved.

Calculus, Stewart: 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. [An ellipse, center (0, 0),focion ⇒ 2/3 2 Y -axis with a = √2, b = √6/3, c =2√3/3.] 97 153 7 9. cot 2θ = − = − tan 2θ = 24 π < 2θ<π 192 24 ⇒ − 7 ⇒ 2 7 π π 3 4 and cos 2θ = − <θ< , cos θ = , sin θ = 25 ⇒ 4 2 5 5 ⇒ 3X 4Y x = X cos θ Y sin θ = − and − 5 4X +3Y y = X sin θ + Y cos θ = . Substituting, we get 5 97 (3X 4Y )2 + 192 (3X 4Y )(4X +3Y )+ 153 (4X +3Y )2 =225, 25 − 25 − 25 Y 2 which simplifies to X2 + =1(an ellipse with foci on Y-axis, centered 9 at origin, a =3, b =1). A C 1 π √3 X Y 11. cot 2θ = − = θ = x = − , B √3 ⇒ 6 ⇒ 2 X + √3 Y y = . Substituting into the curve equation and simplifying gives 2 4X2 12Y 2 8x =0 (X 1)2 3Y 2 =1[a hyperbola with foci − − ⇒ − − on X-axis, centered at (1, 0), a =1,b=1/√3, c =2/√3 ]. Sixth Edition. ISBN: 0495011606. © 2008 Brooks/Cole. All rights reserved.

Calculus, Stewart: ROTATION OF AXES ■ 7

A C 7 3X 4Y 4X +3Y 13. (a) cot 2θ = − = − so, as in Exercise 9, x = − and y = . B 24 5 5 Substituting and simplifying we get 100X2 25Y +25=0 4X2 = Y 1, which is a parabola. − ⇒ − 1 17 (b) The vertex is (0, 1) and p = 16 ,sotheXY -coordinates of the focus are 0, 16 ,andthexy-coordinates are 0 3 17 4 17 0 4 17 3 51 x = · = and y = · + = .   5 − 16 5 − 20 5 16 5 80 (c) The directrix is Y = 15 ,so x 4 + y 3 =15   64x 48y +75=0. 16 − · 5 · 5 16 ⇒ − 15. A rotation through θ changes Equation 1 to A(X cos θ Y sin θ)2 + B(X cos θ Y sin θ)(X sin θ + Y cos θ)+ − − C(X sin θ + Y cos θ)2 + D(X cos θ Y sin θ)+E(X sin θ + Y cos θ)+F =0. − 2 2 2 2 ComparingthistoEquation4,weseethatA0 + C0 = A(cos θ +sin θ)+C(sin θ +cos θ)=A + C.

2 2 17. Choose θ so that B0 =0.ThenB 4AC =(B0) 4A0C0 = 4A0C0.ButA0C0 will be 0 for a parabola, − − − negative for a hyperbola (where the X2 and Y 2 coefficients are of opposite sign), and positive for an ellipse (same

sign for X2 and Y 2 coefficients). So :

B2 4AC =0for a parabola, B2 4AC > 0 for a hyperbola, B2 4AC < 0 for an ellipse. − − −

2 2 Note that the transformed equation takes the form A0X + C0Y + D0X + E0Y + F =0, or by completing the

2 2 square (assuming A0C0 =0), A0(X0) + C0(Y 0) = F 0,sothatifF 0 =0, the graph is either a pair of intersecting 6

lines or a point, depending on the signs of A0 and C0.IfF 0 =0and A0C0 > 0, then the graph is either an ellipse, a 6

point, or nothing, and if A0C0 < 0, the graph is a hyperbola. If A0 or C0 is 0, we cannot complete the square, so we

2 2 get A0(X0) + E0Y + F =0or C0(Y 0) + D0X + F 0 =0. This is a parabola, a straight line (if only the

second-degree coefficient is nonzero), a pair of parallel lines (if the first-degree coefficient is zero and the other two have opposite signs), or an empty graph (if the first-degree coefficient is zero and the other two have the same sign). Sixth Edition. ISBN: 0495011606. © 2008 Brooks/Cole. All rights reserved.

Calculus, Stewart: