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 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 coordinate 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 Y 2 sin 606 cos 60 s3 3

The XY-coordinates are (1 3s3, 3 s3) .

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

AX cos Y sin 2 BX cos Y sin sin Y cos

CX sin Y cos 2 DX cos Y sin

EX 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 2C A sin cos Bcos2 sin2

C A sin 2 B cos 2

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

A C sin 2 cos 2

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 1s2 and Equa- tions 2 become X@ Y@ xy=1 or - =1 2 2 y X Y X 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). Thomson Brooks-Cole copyright 2007 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:

1 cos 2 1 7 4 2¨ cos 25 2 2 5 7

FIGURE 4 1 cos 2 1 7 3 sin 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 12 4X 2 Y 12 4 or X 2 1 4

and we recognize this as being an ellipse whose center is 0, 1 in XY -coordinates. 1 4 Since cos (5) 37 , we can sketch the graph in Figure 5.

y X Y

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, 4, 30 2. 4, 3, 45 represents a hyperbola. 3.2, 4, 60 4. 1, 1, 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 B2 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. Thomson Brooks-Cole copyright 2007 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 X

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

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

Y X

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 ]. Thomson Brooks-Cole copyright 2007 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). Thomson Brooks-Cole copyright 2007