11-5
In Problems 35–40, find the coordinates of any foci relative
to the original coordinate system:
30. An ellipse with foci (3, 0) and (3, 6) and vertices
(3, 2) and (3, 8).
31. A parabola with axis the y axis and passing through the
points (1, 0) and (2, 4).
35. Problem 15
36. Problem 16
37. Problem 17
38. Problem 18
39. Problem 21
40. Problem 22
In Problems 41–44, use a graphing utility to find the coordinates of all points of intersection to two decimal places.
32. A parabola with vertex at (6, 2), axis the line y 2, and
passing through the point (0, 7).
41. 3x2 5y2 7x 2y 11 0, 6x 4y 15
33. An ellipse with vertices (1, 1), and (5, 1) that passes
through the origin.
42. 8x2 3y2 14x 17y 39 0, 5x 11y 23
43. 7x2 8x 5y 25 0, x2 4y2 4x y 12 0
34. A hyperbola with vertices at (2, 3), and (2, 5) that passes
through the point (4, 0).
11-5
821
C
29. A hyperbola with foci (2, 1) and (6, 1) and vertices (3, 1)
and (5, 1).
SECTION
Parametric Equations
44. 4x2 y2 24x 2y 35 0, 2x2 6y2 3x 34 0
Parametric Equations
• Parametric Equations and Plane Curves
• Projectile Motion
• Cycloid
• Parametric
Equations and
Plane Curves
Consider the two equations
xt1
t y t 2 2t
(1)
Each value of t determines a value of x, a value of y, and hence, an ordered pair
(x, y). To graph the set of ordered pairs (x, y) determined by letting t assume all real
values, we construct Table 1 listing selected values of t and the corresponding values
of x and y. Then we plot the ordered pairs (x, y) and connect them with a continuous
curve, as shown in Figure 1. The variable t is called a parameter and does not appear
on the graph. Equations (1) are called parametric equations because both x and y
are expressed in terms of the parameter t. The graph of the ordered pairs (x, y) is
called a plane curve.
TABLE 1
y
10
5
x
FIGURE 1 Graph of x t 1,
y t 2 2t, t .
t
0
x
1
y
0
2
3
4
1
2
2
3
4
5
0
1
1
0
3
8
3
8
1
In some cases it is possible to eliminate the parameter by solving one of the equations for t and substituting into the other. In the example just considered, solving the
first equation for t in terms of x, we have
tx1
822
11
Additional Topics in Analytic Geometry
Then, substituting the result into the second equation, we obtain
y (x 1)2 2(x 1)
x2 4x 3
We recognize this as the equation of a parabola, as we would guess from Figure 1.
In other cases, it may not be easy or possible to eliminate the parameter to obtain
an equation in just x and y. For example, for
x t log t
t0
y t et
you will not find it possible to solve either equation for t in terms of functions we
have considered.
Is there more than one parametric representation for a plane curve? The answer
is yes. In fact, there is an unlimited number of parametric representations for the same
plane curve. The following are two additional representations of the parabola in Figure 1.
xt3
y t 2 2t
xt
y t 2 4t 3
t (2)
t (3)
The concepts introduced in the preceding discussion are summarized in Definition 1.
DEFINITION 1
Parametric Equations and Plane Curves
A plane curve is the set of points (x, y) determined by the parametric
equations
x f(t)
y g(t)
where the parameter t varies over an interval I and the functions f and g are
both defined on the interval I.
Why are we interested in parametric representations of plane curves? It turns out
that this approach is more general than using equations with two variables as we have
been doing. In addition, the approach generalizes to curves in three- and higher-dimensional spaces. Other important reasons for using parametric representations of plane
curves will be brought out in the discussion and examples that follow.
11-5
EXAMPLE 1
Parametric Equations
823
Graphing Parametric Equations and Eliminating the Parameter
Graph the plane curve given parametrically by
x 8 cos y 4 sin (4)
Identify the curve by eliminating the parameter .
Solution
Construct a table and graph:
0
/6
/3
/2
2/3
5/6
7/6
4/3
3/2
5/3
11/6
2
x
8
4
3
4
0
4
4
3
8
4
3
4
0
4
4
3
8
y
0
2
2
3
4
2
3
2
0
2
2
3
4
2
3
2
0
To eliminate the parameter , we solve the first equation in (4) for cos , the second
for sin , and substitute into the Pythagorean identity cos2 sin2 1:
y
cos 10
x
8
sin and
y
4
cos2 sin2 1
10
10
x
1
x
8
Matched Problem 1
EXPLORE-DISCUSS 1
y
4
2
x2
y2
1
64 16
10
FIGURE 2
Graph of x 8 cos ,
y 4 sin , .
2
The graph is an ellipse (Fig. 2).
Graph the plane curve given parametrically by x 4 cos , y 4 sin , 0. Identify the curve by eliminating the parameter .
Graph one period (0 2) of each of the three plane curves given parametrically by
x1 5 cos x2 2 cos x3 5 cos y1 5 sin y2 2 sin y3 2 sin Identify the curves by eliminating the parameter.
824
11
Additional Topics in Analytic Geometry
EXAMPLE 2
Parametric Equations for Conic Sections
Find parametric equations for the conic section with the given equation:
(A) 25x2 9y2 100x 54y 44 0
(B) x2 16y2 10x 32y 7 0
Solutions
(A) By completing the square in x and y we obtain the standard form
(x 2)2 (y 3)2
1. So the graph is an ellipse with center (2, 3) and
9
25
major axis on the line x 2. Since cos2 sin2 1, a parametric reprex2
y3
sentation with parameter is obtained by letting
cos ,
sin :
3
5
x 2 3 cos y 3 5 sin (B) By completing the square in x and y we obtain the standard form
(x 5)2
(y 1)2 1. So the graph is a hyperbola with center (5, 1) and
16
transverse axis on the line y 1. Since sec2 tan2 1, a parametric
x5
representation with parameter is obtained by letting
sec , y 1 4
tan :
4
2
12
x 5 4 sec y 1 tan 2
FIGURE 3
x 5 4 sec , y 1 tan .
Matched Problem 2
,
k, k an integer
2
Note that when the parametric equations are graphed using a graphing utility in
connected mode, the graph appears to show the asymptotes of the hyperbola (see
Fig. 3).
Find parametric equations for the conic section with the given equation:
(A) 36x2 16y2 504x 96y 1332 0
(B) 16y2 9x2 36x 128y 76 0
• Projectile Motion
Newton’s laws and advanced mathematics can be used to determine the path of a projectile. If v0 is the initial speed of the projectile at an angle with the horizontal (see
Fig. 4) and air resistance is neglected, then the path of the projectile is given by
x (v0 cos )t
y (v0 sin )t 4.9t 2
0tb
(5)
11-5
Parametric Equations
825
The parameter t represents time in seconds, and x and y are distances measured in
meters. Solving the first equation in (5) for t in terms of x, substituting into the second equation, and simplifying results in the following equation:
y (tan )x 4.9
x2
v cos2 2
0
(6)
You should verify this by supplying the omitted details.
FIGURE 4 Projectile motion.
y
v0
v 0 sin x
v 0 cos We recognize equation (6) as a parabola. This equation in x and y describes the
path the projectile follows but tells us little else about its flight. On the other hand,
the parametric equations (5) not only determine the path of the projectile but also tell
us where it is at any time t. Furthermore, using concepts from physics and calculus,
the parametric equations can be used to determine the velocity and acceleration of the
projectile at any time t. This illustrates another advantage of using parametric representations of plane curves.
The range of a projectile is the distance from the point of firing to the point of
impact. If we keep the initial speed v0 of the projectile constant and vary the angle in Figure 4, we obtain different parabolic paths followed by the projectile and different ranges. The maximum range is obtained when 45°. Furthermore, assuming that the projectile always stays in the same vertical plane, then there are points
in the air and on the ground that the projectile cannot reach, irrespective of the angle
used, 0° 180°. Using more advanced mathematics, it can be shown that the
reachable region is separated from the nonreachable region by a parabola called an
envelope of the other parabolas (see Fig. 5).
FIGURE 5 Reachable region of a
projectile.
Envelope
• Cycloid
We now consider an unusual curve called a cycloid, which has a fairly simple parametric representation and a very complicated representation in terms of x and y only.
The path traced by a point on the rim of a circle that rolls along a line is called a
cycloid. To derive parametric equations for a cycloid we roll a circle of radius a along
the x axis with the tracing point P on the rim starting at the origin (see Figure 6).
826
11
Additional Topics in Analytic Geometry
y
FIGURE 6 Cycloid.
P (x, y)
a
O R
C
Q
x
S
Since the circle rolls along the x axis without slipping (refer to Figure 6), we see
that
d(O, S) arc PS
a
in radians
(7)
where S is the point of contact between the circle and the x axis. Referring to triangle CPQ, we see that
d(P, Q) a sin 0 /2
(8)
d(Q, C) a cos 0 /2
(9)
Using these results, we have
x d(O, R)
d(O, S) d(R, S)
(arc PS) d(P, Q)
a a sin Use equations (7) and (8).
y d (R, P)
d(S, C ) d(Q, C)
a a cos Use equation (9) and the fact that d(S, C ) a.
Even though in equations (8) and (9) was restricted so that 0 /2, it can be
shown that the derived parametric equations generate the whole cycloid for . The graph specifies a periodic function with period 2a. Thus, in general, we
have Theorem 1.
Theorem 1
Parametric Equations for a Cycloid
For a circle of radius a rolled along the x axis, the resulting cycloid generated
by a point on the rim starting at the origin is given by
x a a sin y a a cos 11-5
Parametric Equations
827
The cycloid is a good example of a curve that is very difficult to represent without the use of a parameter. A cycloid has a very interesting physical property. An
object sliding without friction from a point P to a point Q lower than P, but not on
the same vertical line as P, will arrive at Q in a shorter time traveling along a cycloid
than on any other path (see Fig. 7).
P
Q
FIGURE 7 Cycloid path.
EXPLORE-DISCUSS 2
(A) Let Q be a point b units from the center of a wheel of radius a, where 0 b a. If the wheel rolls along the x axis with the tracing point Q starting at
(0, a b), explain why parametric equations for the path of Q are given by
x a b sin y a b cos (B) Use a graphing utility to graph the paths of a point on the rim of a wheel of
radius 1, and a point halfway between the rim and center, as the wheel makes
two complete revolutions rolling along the x axis.
Answers to Matched Problems
1. x2 y2 16; circle
y
5
5
5
x
5
2. (A) x 7 4 cos , y 3 6 sin , (B) x 2 4 tan , y 4 3 sec , , EXERCISE
k, k an integer
2
11-5
A
B
In Problems 1–10, plot each plane curve by use of a table
of values (see Example 1). Obtain an equation in x and y by
eliminating the parameter, and identify the curve. (In this
exercise set, the interval for the parameter is the whole real
line, unless stated to the contrary.)
In Problems 11–18, obtain an equation in x and y by eliminating the parameter. Use the simpler of the two forms to
plot the curve. Name the curve if it is a curve we have
identified.
11. x 3 sin , y 4 cos 1. x t, y 2t 2
2. x t, y t 1
12. x 3 sin , y 3 cos 3. x t 2, y 2t 2 2
4. x t 2, y t 2 1
13. x 2 2 sin , y 3 2 cos 5. x 3t, y 2t
6. x 2t, y t
14. x 3 4 sin , y 2 2 cos 7. x 14 t 2, y t
8. x 2t, y t 2
9. x 14 t 4, y t 2
10. x 2t 2, y t 4
15. x t 2, y 2
;t2
2t
828
11
Additional Topics in Analytic Geometry
16. x t 1, y 34. Repeat Problem 33 for x cot t, y (3t cot t)/t, t
, t 0.
2
;t1
t1
17. x t 1, y t; t 0
18. x t 3, y t 2 1
19. If A 0, C 0, and E 0, find parametric equations for
Ax2 Cy2 Dx Ey F 0. Identify the curve.
In Problems 35–38, find the standard form for each equation. Name the curve and find its center. Use parametric
equations to graph the curve on a graphing utility.
35. 25x2 200x 9y2 18y 616 0
20. If A 0, C 0, and D 0, find parametric equations for
Ax2 Cy2 Dx Ey F 0. Identify the curve.
36. 36x2 360x 4y2 8y 760 0
C
38. 16x2 32x 9y2 36y 164 0
37. 4x2 24x 49y2 392y 624 0
In Problems 21–26, obtain an equation in x and y by eliminating the parameter. Use the simpler of the two forms to
plot the curve. Name the curve if it is a curve we have
identified.
21. x t 2, y t 2; t 0
22. x e t, y et
APPLICATIONS
23. x cos 2, y 4 sin 24. x 3 sec2 , y 2 tan2 39. Plane Motion. An object follows a path as given by
25. x 8
4t
,y 2
t 4
t 4
2
26. x 4t 2
4t
,y 2
t 1
t 1
2
Graph, using a calculator, one period (0 2) of each
cycloid in Problems 27 and 28.
27. x sin , y 1 cos 28. x 2 2 sin , y 2 2 cos In Problems 29–32, use a graphing utility to graph the
parametric equations. Then eliminate the parameter and find
the standard equation for the curve. Name the curve and
find its center.
29. x 3 6 cos t, y 2 4 sin t, 0 t 2
30. x 1 3 sec t, y 2 2 tan t, 3
t ,t
2
2
2
31. x 3 2 tan t, y 1 5 sec t, 3
t ,t
2
2
2
32. x 4 5 cos t, y 1 8 sin t, 0 t 2
33. Find an equation of the form Ax2 Cy2 Dx Ey F 0
that has the same graph as the parametric equations
x 2 tan t, y 5 tan t, t .
2
2
x 5 sin 6t
y 5 cos 6t
t0
where t is time in seconds and x and y are distances in feet.
(A) What are the coordinates of the object when t 0.1
second? Compute answers to one decimal place.
(B) Eliminate the parameter and graph the resulting equation in x and y. Identify the path.
40. Plane Motion. Repeat Problem 39 for
x 4 sin t
y 2 cos t
t0
41. Projectile Motion. A projectile is fired with an initial speed
of 300 meters per second at an angle of 45° to the horizontal. Neglecting air resistance, find:
(A) The time of impact
(B) The horizontal distance covered (range) in meters and
kilometers at time of impact
(C) The maximum height in meters of the projectile
Compute all answers to three decimal places using a
calculator.
42. Projectile Motion. Repeat Problem 41 if the same projectile is fired at 40° to the horizontal instead of 45°.
CHAPTER 11 GROUP ACTIVITY Focal Chords
Many of the applications of the conic sections are based on their reflective or focal properties. One of the
interesting algebraic properties of the conic sections concerns their focal chords.
If a line through a focus F contains two points G and H of a conic section, then the line segment GH is
called a focal chord. Let G(x1, y1) and H(x2, y2) be points on the graph of x2 4ay such that GH is a focal
chord. Let u denote the length of GF and v the length of FH (see Fig. 1).
Chapter 11 Review
FIGURE 1 Focal chord GH of the
parabola x2 4ay.
829
y
G
F
u
v
H
(2a, a)
x
(A) Use the distance formula to show that u y1 a.
(B) Show that G and H lie on the line y a mx, where m (y2 y1)/(x2 x1).
(C) Solve y a mx for x and substitute in x2 4ay, obtaining a quadratic equation in y. Explain why
y1y2 a2.
1 1 1
(D) Show that .
u v a
(u 2a)2
(E) Show that u v 4a . Explain why this implies that u v 4a, with equality if and only
ua
if u v 2a.
(F) Which focal chord is the shortest? Is there a longest focal chord?
1 1
(G) Is a constant for focal chords of the ellipse? For focal chords of the hyperbola? Obtain evidence
u v
for your answers by considering specific examples.
(H) The conic section with focus at the origin, directrix the line x D 0, and eccentricity E 0 has the
DE
1 1 1
polar equation r . Explain how this polar equation makes it easy to show that 1 E cos u v a
1 1
for a parabola. Use the polar equation to determine the sum for a focal chord of an ellipse or
u v
hyperbola.
Chapter 11 Review
11-1
CONIC SECTIONS; PARABOLA
The plane curves obtained by intersecting a right circular cone with a plane are called conic sections. If the
plane cuts clear through one nappe, then the intersection curve is called a circle if the plane is perpendicular
to the axis and an ellipse if the plane is not perpendicular to the axis. If a plane cuts only one nappe, but does
not cut clear through, then the intersection curve is
called a parabola. If a plane cuts through both nappes,
but not through the vertex, the resulting intersection
curve is called a hyperbola. A plane passing through
the vertex of the cone produces a degenerate conic—a
point, a line, or a pair of lines. The figure illustrates the
four nondegenerate conics.
Circle
Ellipse
Parabola
Hyperbola