ROTATION OF AXES
■ ■ For a discussion of conic sections, see
Review of Conic Sections.
In precalculus or calculus you may have studied conic sections with equations of the form
Ax 2 Cy 2 Dx Ey F 0
Here we show that the general second-degree equation
Ax 2 Bxy Cy 2 Dx Ey F 0
1
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
y
Y
X r cos Y r sin x r cos y r sin P(x, y)
P(X, Y)
y
Y
P
X
Y
¨
0
FIGURE 1
x
0
r
y
˙
¨
x
X
X
x
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
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X 2 cos 60 6 sin 60 1 3s3
Y 2 sin 60 6 cos 60 s3 3
The XY -coordinates are (1 3s3, 3 s3 ).
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
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
A
X 2 B
XY C
Y 2 D
X E
Y F 0
4
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
or
cot 2 5
AC
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
cot 2 xy=1 or
This will be true if 2 2, that is, 4. Then cos sin 1s2 and Equations 2 become
X@ Y @
- =1
2
2
y
x
X
Y
AC
0
B
X
Y
s2
s2
y
X
Y
s2
s2
Substituting these expressions into the original equation gives
π
4
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0
FIGURE 3
x
X
Y
s2
s2
X
Y
s2
s2
1
or
X2
Y2
1
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
in the xy-system (see Figure 3).
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
cot 2 AC
73 52
7
B
72
24
From the triangle in Figure 4 we see that
cos 2 257
25
The values of cos and sin can then be computed from the half-angle formulas:
24
cos 2¨
7
FIGURE 4
sin 1 cos 2
2
1 cos 2
2
1 257
4
2
5
1 257
3
2
5
The rotation equations (2) become
x 45 X 35 Y
y 35 X 45 Y
Substituting into the given equation, we have
73( 45 X 35 Y) 72( 45 X 35 Y)( 35 X 45 Y) 52( 35 X 45 Y)
2
2
30( 45 X 35 Y) 40( 35 X 45 Y) 75 0
4X 2 Y 2 2Y 3
which simplifies to
Completing the square gives
4X 2 Y 12 4
or
X2 Y 12
1
4
and we recognize this as being an ellipse whose center is 0, 1 in XY -coordinates.
Since cos1( 45 ) 37, we can sketch the graph in Figure 5.
X
y
Y
(0, 1)
¨Å37°
0
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FIGURE 5
x
73≈+72xy+52¥+30x-40y-75=0
or
4X @+(Y-1)@=4
4 ■ ROTATION OF AXES
EXERCISES
A Click here for answers.
S
(c) Find an equation of the directrix in the xy-coordinate
system.
Click here for solutions.
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.
1. 1, 4, 30
2. 4, 3, 45
3. 2, 4, 60
4. 1, 1, 15
■
■
5–12
■
■
■
■
■
■
■
■
■
■
2x 2 72xy 23y 2 80x 60y 125
■
Use rotation of axes to identify and sketch the curve.
5. x 2xy y 2 x y 0
2
6. x 2 xy y 2 1
represents a hyperbola.
(b) Find the XY -coordinates of the foci. Then find the
xy-coordinates of the foci.
(c) Find the xy-coordinates of the vertices.
(d) Find the equations of the asymptotes in the xy-coordinate
system.
(e) Find the eccentricity of the hyperbola.
15. Suppose that a rotation changes Equation 1 into Equation 4.
7. x 2 xy y 2 1
Show that
8. s3 xy y 2 1
A
C A C
9. 97x 2 192xy 153y 2 225
16. Suppose that a rotation changes Equation 1 into Equation 4.
10. 3x 2 12s5 xy 6y 2 9 0
Show that
11. 2s3 xy 2y 2 s3 x y 0
B
2 4A
C B 2 4AC
12. 16x 8s2 xy 2y (8s2 3) x (6s2 4) y 7
2
■
■
2
■
■
■
■
■
■
■
■
■
13. (a) Use rotation of axes to show that the equation
36x 2 96xy 64y 2 20x 15y 25 0
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represents a parabola.
(b) Find the XY -coordinates of the focus. Then find the
xy-coordinates of the focus.
■
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
cases when it reduces to a point, a line, a pair of lines, or no
graph at all.
18. Use Exercise 17 to determine the type of curve in
Exercises 9–12.
ROTATION OF AXES ■ 5
ANSWERS
S
9. X 2 Y 29 1, ellipse
Click here for solutions.
y
1.
((s3 4)2, (4s3 1)2)
X
Y
3. (2s3 1, s3 2)
4
sin–! ” 5 ’
5. X s2 Y 2, parabola
0
x
y
X
Y
1
11. X 12 3Y 2 1, hyperbola
1
y
x
0
Y
X
7. 3X Y 2, ellipse
2
2
0
y
x
X
Y
0
x
17
17 51
(b) (0, 16 ), ( 20 , 80 )
(c) 64x 48y 75 0
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13. (a) Y 1 4X 2
6 ■ ROTATION OF AXES
SOLUTIONS
√
Y = −1 · sin 30◦ + 4 cos 30◦ = 2 3 − 12 .
√
√
3. X = −2 cos 60◦ + 4 sin 60◦ = −1 + 2 3, Y = 2 sin 60◦ + 4 cos 60◦ = 3 + 2.
A−C
= 0 ⇒ 2θ = π2 ⇔ θ = π4 ⇒ [by Equations 2]
5. cot 2θ =
B
1. X = 1 · cos 30◦ + 4 sin 30◦ = 2 +
x=
√
3
,
2
X +Y
X −Y
√
and y = √ . Substituting these into the curve equation
2
2
√
X
2X or Y 2 = √ .
2
√ √ [Parabola, vertex (0, 0), directrix X = −1/ 4 2 , focus 1/ 4 2 , 0 ].
gives 0 = (x − y)2 − (x + y) = 2Y 2 −
7. cot 2θ =
A−C
= 0 ⇒ 2θ =
B
Equations 2] x =
π
2
⇔
θ=
π
4
⇒ [by
X −Y
X +Y
√
and y = √ . Substituting these into the
2
2
curve equation gives
1=
X2 − Y 2
X 2 + 2XY + Y 2
X 2 − 2XY + Y 2
+
+
2
2
2
⇒
Y2
X2
+
= 1. [An ellipse, center (0, 0), foci on
2/3
2
√
√
√
Y -axis with a = 2, b = 6/3, c = 2 3/3.]
3X 2 + Y 2 = 2 ⇒
9. cot 2θ =
−7
97 − 153
=
192
24
and cos 2θ =
−7
25
⇒
π
4
⇒ tan 2θ = − 24
7
<θ<
π
,
2
⇒
π
2
cos θ = 35 , sin θ =
x = X cos θ − Y sin θ =
3X − 4Y
and
5
y = X sin θ + Y cos θ =
4X + 3Y
. Substituting, we get
5
97
(3X
25
− 4Y )2 +
192
(3X
25
which simplifies to X 2 +
− 4Y )(4X + 3Y ) +
153
(4X
25
< 2θ < π
4
5
⇒
+ 3Y )2 = 225,
Y2
= 1 (an ellipse with foci on Y-axis, centered
9
at origin, a = 3, b = 1).
√
1
A−C
3X − Y
π
= √
⇒ x=
,
⇒ θ=
B
6
2
3
√
X + 3Y
y=
. Substituting into the curve equation and simplifying gives
2
11. cot 2θ =
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4X 2 − 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 ].
ROTATION OF AXES ■ 7
13. (a) cot 2θ =
−7
3X − 4Y
4X + 3Y
A−C
=
so, as in Exercise 9, x =
and y =
.
B
24
5
5
Substituting and simplifying we get 100X 2 − 25Y + 25 = 0 ⇒ 4X 2 = Y − 1, which is a parabola.
1
, and the xy-coordinates are
(b) The vertex is (0, 1) and p = 16
, so the XY -coordinates of the focus are 0, 17
16
4
3
x = 0 5· 3 − 17
= − 17
= 51
and y = 0 5· 4 + 17
.
16
5
20
16
5
80
(c) The directrix is Y =
15
,
16
so −x ·
4
5
+y·
3
5
=
15
16
⇒ 64x − 48y + 75 = 0.
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.
Comparing this to Equation 4, we see that A0 + C 0 = A(cos2 θ + sin2 θ) + C(sin2 θ + cos2 θ) = A + C.
17. Choose θ so that B 0 = 0. Then B 2 − 4AC = (B 0 )2 − 4A0 C 0 = −4A0 C 0 . But A0 C 0 will be 0 for a parabola,
negative for a hyperbola (where the X 2 and Y 2 coefficients are of opposite sign), and positive for an ellipse (same
sign for X 2 and Y 2 coefficients). So :
B 2 − 4AC = 0 for a parabola,
B 2 − 4AC > 0 for a hyperbola,
B 2 − 4AC < 0 for an ellipse.
Note that the transformed equation takes the form A0 X 2 + C 0 Y 2 + D0 X + E 0 Y + F = 0, or by completing the
square (assuming A0 C 0 6= 0), A0 (X 0 )2 + C 0 (Y 0 )2 = F 0 , so that if F 0 = 0, the graph is either a pair of intersecting
lines or a point, depending on the signs of A0 and C 0 . If F 0 6= 0 and A0 C 0 > 0, then the graph is either an ellipse, a
point, or nothing, and if A0 C 0 < 0, the graph is a hyperbola. If A0 or C 0 is 0, we cannot complete the square, so we
get A0 (X 0 )2 + E 0 Y + F = 0 or C 0 (Y 0 )2 + D0 X + 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
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have opposite signs), or an empty graph (if the first-degree coefficient is zero and the other two have the same sign).