632
Chapter 22
◆
Analytic Geometry
22–3 The Parabola
Directrix
L
9.0 cm
Definition
A parabola has the following definition:
P2
Definition
of a
Parabola
9.0 cm
V
F
A parabola is the set of points in a plane, each of
which is equidistant from a fixed point, the focus,
and a fixed line, the directrix.
307
6.0 cm
Example 27: Use Definition 307 above to construct a parabola for which the distance from
focus to directrix is 6.0 cm.
◆◆◆
P1
Solution: We draw a line to represent the directrix as shown in Fig. 22–42, and we indicate a
focus 6.0 cm from that line. Then we draw a line L parallel to the directrix at some arbitrary
distance, say, 9.0 cm. With that same 9.0-cm distance as radius and F as centre, we use a comFIGURE 22–42 Construction pass to draw arcs intersecting L at P1 and P2. Each of these points is now at the same distance
of a parabola.
(9.0 cm) from F and from the directrix, and is hence a point on the parabola. Repeat the con◆◆◆
struction with distances other than 9.0 cm to get more points on the parabola.
Figure 22–43 shows the typical shape of a parabola. The parabola has an axis of symmetry
which intersects it at the vertex. The distance p from directrix to vertex is equal to the directed
distance from the vertex to the focus.
Directrix
Vertex
p
Axis of symmetry
p
Focus
FIGURE 22–43 Parabola.
◆◆◆
Standard Equation of a Parabola with Vertex at the Origin
Let us place the parabola on coordinate axes with the vertex at the origin and with the axis of
symmetry along the x axis, as shown in Fig. 22–44. Choose any point P on the parabola. Then,
by the definition of a parabola, FP AP. But in right triangle FBP,
FP (x p)2 y2
and
AP p x
But, since FP AP,
(x p)2 y2 p x
Squaring both sides yields
(x p)2 y2 p2 2px x2
x2 2px p2 y2 p2 2px x2
Section 22–3
◆
633
The Parabola
y
P(x, y)
A
y
0
p
V
p
F
x−p
B
x
Directrix
x
FIGURE 22–44
Collecting terms, we get the standard equation of a parabola with vertex at the origin.
Standard Equation
of a Parabola;
Vertex at Origin,
Axis Horizontal
y
V
0 p
y
F
x
y2 4px
308
2
F(−0.875, 0)
−2
We have defined p as the directed distance VF from the vertex V to the focus F. Thus if p is
positive, the focus must lie to the right of the vertex, and hence the parabola opens to the right.
Conversely, if p is negative, the parabola opens to the left.
◆◆◆
Example 28: Find the coordinates of the focus of the parabola 2y2 7x 0.
Solution: Subtracting 7x from both sides and dividing by 2 gives us
y2 3.5x
x
1
−2
FIGURE 22–45
We have graphed the parabola
before. Glance back at Sec. 5–2.
Thus 4p 3.5 and p 0.875. Since p is negative, the parabola opens to the left. The focus is
◆◆◆
thus 0.875 unit to the left of the vertex (Fig. 22–45) and has the coordinates (0.875, 0).
Example 29: Find the coordinates of the focus and write the equation of a parabola that
has its vertex at the origin, that has a horizontal axis of symmetry, and that passes through the
point (5, 6).
◆◆◆
Solution: Our given point must satisfy the equation y2 4px. Substituting 5 for x and 6 for
y gives
y
(6)2 4p(5)
36
9
So 4p and p . The equation of the parabola is then y2 36x5, or
5
5
5y2 36x
Since the axis is horizontal and p is positive, the focus must be on the x axis and is a distance
9
9
p to the right of the origin. The coordinates of F are thus p , 0 q, as shown in Fig. 22–46.
◆◆◆
5
5
V
0
( )
F 95 , 0
x
(5, −6)
FIGURE 22–46
634
Chapter 22
◆
Analytic Geometry
Standard Equation of a Parabola with Vertical Axis
The standard equation for a parabola having a vertical axis is obtained by switching the positions
of x and y in Eq. 308.
Note that only one variable is
squared in any parabola
equation. This gives us the best
way to recognize such equations.
Standard Equation
of a Parabola;
Vertex at Origin,
Axis Vertical
y
F p
0 V
309
x2 4py
x
When p is positive, the parabola opens upward; when p is negative, it opens downward.
Example 30: A parabola has its vertex at the origin and passes through the points (3, 2)
and (3, 2). Write its equation and find the focus.
◆◆◆
Solution: The sketch of Fig. 22–47 shows that the axis must be vertical, so our equation is of
the form x2 4py. Substituting 3 for x and 2 for y gives
32 4p(2)
9
9
from which 4p and p . Our equation is then x2 9y2 or 2x2 9y. The focus is on the
2
8
9q
9 from the origin, so its coordinates are p 0, y axis at a distance p 8
8
◆◆◆
y
y
A(p, h)
( )
F 0, 98
(−3, 2)
h L
V
0
(3, 2)
0 V
9
L= 2
p
F
x
x
B
FIGURE 22–48
FIGURE 22–47
Focal Width of a Parabola
The latus rectum of a parabola is a line through the focus which is perpendicular to the axis of
symmetry, such as line AB in Fig. 22–48. The length of the latus rectum is also called the focal
width. We will find the focal width, or length L of the latus rectum, by substituting the coordinates (p, h) of point A into Eq. 308.
h2 4p(p) 4p2
h 2p
The focal width is twice h, so we have the following equation:
The focal width is useful for
making a quick sketch of the
parabola.
Focal Width
of a Parabola
L 4p 110
The focal width (length of the latus rectum) of a parabola is four times the distance
from vertex to focus.
9
◆◆◆
Example 31: The focal width for the parabola in Fig. 22–47 is , or 4.5 units.
2
Shift of Axes
As with the circle, when the vertex of the parabola is not at the origin but at (h, k), our equations will
be similar to Eqs. 308 and 309, except that x is replaced with x h and y is replaced with y k.
◆◆◆
Section 22–3
◆
635
The Parabola
y
k
Standard Axis Horizontal
Equation of
a Parabola;
Vertex at
(h, k)
Axis Vertical
0
V
h
y
F
k
V
h
0
F
p
(y k)2 4p(x h)
310
(x h)2 4p(y k)
311
x
p
x
◆◆◆ Example 32: Find the vertex, focus, focal width, and equation of the axis for the parabola
(y 3)2 8(x 2).
Solution: The given equation is of the same form as Eq. 310, so the axis is horizontal. Also,
h 2, k 3, and 4p 8. So the vertex is at V(2, 3) (Fig. 22–49). Since p 84 2, the
focus is 2 units to the right of the vertex, at F(0, 3). The focal width is 4p, so L 8. The
◆◆◆
axis is horizontal and 3 units from the x axis, so its equation is y 3.
Example 33: Write the equation of a parabola that opens upward, with vertex at
(1, 2), and that passes through the point (1, 3) (Fig. 22–50). Find the focus and the
focal width.
y
◆◆◆
L=8
V(−2, 3)
Solution: We substitute h 1 and k 2 into Eq. 311.
y=3
F(0, 3)
0
x
(x 1) 4p(y 2)
2
Now, since (1, 3) is on the parabola, these coordinates must satisfy our equation.
Substituting yields
(1 1)2 4p(3 2)
FIGURE 22–49
y
Solving for p, we obtain 22 4p, or p 1. So the equation is
(1, 3)
(x 1)2 4(y 2)
V(−1, 2)
The focus is p units above the vertex, at (1, 3). The focal width is, by Eq. 313,
0
L 4p 4(1) 4 units
◆◆◆
x
FIGURE 22–50
General Equation of a Parabola
We get the general equation of the parabola by expanding the standard equation (Eq. 310) as
follows:
(y k)2 4p(x h)
y2 2ky k2 4px 4ph
or
y2 4px 2ky (k2 4ph) 0
which is of the following general form (C, D, E, and F are constants):
General Equation
of a Parabola with
Horizontal Axis
Compare this with the general
second-degree equation.
Cy2 Dx Ey F 0
312
636
Chapter 22
y
V
2
−6
−8
0
F
2
x
−2
−4
FIGURE 22–51
◆
Analytic Geometry
We see that the equation of a parabola having a horizontal axis of symmetry has a y2
term but no x2 term. Conversely, the equation for a parabola with vertical axis has an x2 term
but not a y2 term. The parabola is the only conic for which there is only one variable squared.
If the coefficient B of the xy term in the general second-degree equation (Eq. 297)
were not zero, it would indicate that the axis of symmetry was rotated by some amount and
was no longer parallel to a coordinate axis. The presence of an xy term indicates rotation
of the ellipse and hyperbola as well.
Completing the Square
As with the circle, we go from general to standard form by completing the square.
◆◆◆
Example 34: Find the vertex, focus, and focal width for the parabola
x2 6x 8y 1 0
Solution: Separating the x and y terms, we have
x2 6x 8y 1
Completing the square by adding 9 to both sides, we obtain
x2 6x 9 8y 1 9
Factoring yields
(x 3)2 8(y 1)
which is the form of Eq. 311, with h 3, k 1, and p 2.
The vertex is (3, 1). Since the parabola opens downward (Fig. 22–51), the focus is 2 units
◆◆◆
below the vertex, at (3, 1). The focal width is 4 p , or 8 units.
Exercise 3
◆
The Parabola
Parabola with Vertex at Origin
Find the focus and focal width of each parabola.
1. y2 8x
2. x2 16y
3. 7x2 12y 0
4. 3y2 5x 0
Write the equation of each parabola.
5. focus at (0, 2)
6. passes through (25, 20); axis horizontal
7. passes through (3, 2) and (3, 2)
8. passes through (3, 4) and (3, 4)
Parabola with Vertex Not at Origin
Find the vertex, focus, focal width, and equation of the axis for each parabola.
9.
11.
13.
15.
(y 5)2 12(x 3)
(x 3)2 24(y 1)
3x 2y2 4y 4 0
y 3x x2 1 0
10.
12.
14.
16.
(x 2)2 16(y 6)
(y 3)2 4(x 5)
y2 8y 4 6x 0
x2 4x y 6 0
Write the equation, in general form, for each parabola.
17. vertex at (1, 2); L 8; axis is y 2; opens to the right
18. axis is y 3; passes through (6, 1) and (3, 1)
Section 22–3
◆
637
The Parabola
19. passes through (3, 3), (6, 5), and (11, 7); axis horizontal
20. vertex (0, 2); axis is x 0; passes through (4, 2)
21. axis is y 1; passes through (4, 2) and (2, 1)
Trajectories
22. A ball thrown into the air will, neglecting air resistance, follow a parabolic path, as shown
in Fig. 22–52. Write the equation of the path, taking axes as shown. Use your equation to
find the height of the ball when it is at a horizontal distance of 30.0 m from O.
y
26.0 m
x
O
21.0 m
FIGURE 22–52
Ball thrown into the air.
23. Some comets follow a parabolic orbit with the sun at the focal point (Fig. 22–53). Taking
axes as shown, write the equation of the path if the distance p is 75 million kilometres.
y
y
3520 m
Sun
0
p
x
Water
x
0
Comet
FIGURE 22–53
Path of a comet.
2150 m
FIGURE 22–54 Object dropped from an aircraft.
24. An object dropped from a moving aircraft (Fig. 22–54) will follow a parabolic path if air
resistance is negligible. A weather instrument released at a height of 3520 m is observed
to strike the water at a distance of 2150 m from the point of release. Write the equation
of the path, taking axes as shown. Find the height of the instrument when x is 1000 m.
638
Chapter 22
◆
Analytic Geometry
Parabolic Arch
25. A 3.0-m-high truck passes under a parabolic arch, as shown in Fig. 22–55. Find the maximum distance x that the side of the truck can be from the centre of the road.
x
6.0 m
3.0 m
12 m
FIGURE 22–55
Parabolic arch.
26. Assuming the bridge cable AB of Fig. 22–56 to be a parabola, write its equation, taking
axes as shown.
y
A cable will hang in the shape of
a parabola if the vertical load per
horizontal foot is constant.
A
B
60.0 m
x
300 m
FIGURE 22–56
Parabolic bridge cable.
27. A parabolic arch supports a roadway as shown in Fig. 22–57. Write the equation of the
arch, taking axes as shown. Use your equation to find the vertical distance from the roadway to the arch at a horizontal distance of 50.0 m from the centre.
y
Roadway
0
115 m
224 m
FIGURE 22–57
Parabolic arch.
x
Section 22–3
◆
639
The Parabola
Parabolic Reflector
28. A certain solar collector consists of a long panel of polished steel bent into a parabolic
shape (Fig. 22–58), which focuses sunlight onto a pipe P at the focal point of the parabola.
At what distance x should the pipe be placed?
29. A parabolic collector for receiving television signals from a satellite is shown in
Fig. 22–59. The receiver R is at the focus, 1.00 m from the vertex. Find the depth d of
the collector.
3.00 m
d
10.0 m
R
P
x
1.00 m
3.00 m
FIGURE 22–58
Parabolic solar collector.
FIGURE 22–59 Parabolic antenna.
Vertical Highway Curves
30. A parabolic curve is to be used at a dip in a highway. The road
dips 32.0 m in a horizontal distance of 125 m and then rises to
its previous height in another 125 m. Write the equation of the
curve of the roadway, taking the origin at the bottom of the
dip and the y axis vertical.
31. Write the equation of the vertical highway curve in Fig. 22–60,
taking axes as shown.
y
75 m
P
0
600 m
FIGURE 22–60 Road over a hill.
Beams
32. A simply supported beam with a concentrated load at its midspan will deflect in the shape
of a parabola, as shown in Fig. 22–61. If the deflection at the midspan is 1.00 in., write the
equation of the parabola (called the elastic curve), taking axes as shown.
32.0 ft.
Load
x
y
FIGURE 22–61
Deflection of a beam.
33. Using the equation found in problem 32, find the deflection of the beam in Fig. 22–61 at a
distance of 10.0 ft. from the left end.
Q
x
Chapter 22
22–4
◆
Analytic Geometry
The Ellipse
Definition
The ellipse is defined in the following way:
Definition of An ellipse is the set of all points in a plane such that the
sum of the distances from each point on the ellipse to
an Ellipse two fixed points (called the foci) is constant.
315
One way to draw an ellipse that is true to its definition is to pass a loop of string around two
tacks, pull the string taut with a pencil, and trace the curve (Fig. 22–62).
FIGURE 22–62 Construction of an ellipse.
Figure 22–63 shows the typical shape of the ellipse. An ellipse has two axes of symmetry:
the major axis and the minor axis, which intersect at the centre of the ellipse. A vertex is a point
where the ellipse crosses the major axis.
P
Vertex V′
Minor axis
640
b
Focus F′
Latus rectum
Focus F
Centre
Vertex V
Major axis
c
a
c
a
FIGURE 22–65 Ellipse
It is often convenient to speak of half the lengths of the major and minor axes, and these are
called the semimajor and semiminor axes, whose lengths we label a and b, respectively. The
distance from either focus to the centre is labelled c.