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CHAPTER 10
Extension
What you should learn
Find the eccentricity
of a conic section.
GOAL
Why you should learn it
Eccentricity of Conic Sections
major
axis
Some ellipses are more oval than
others. In an ellipse that is nearly
circular, the ratio c:a is close to 0. In
a more oval ellipse, c:a is close to 1.
This ratio is called the eccentricity
of the ellipse. Every conic has an
eccentricity e associated with it.
To write equations for
real-life conics, such as the
moon’s orbit in Example 3.
CONCEPT
SUMMARY
c
c
a
a
minor
axis
ECCENTRICITY OF CONIC SECTIONS
Let c be the distance from each focus to the center of the conic section, and let a
be the distance from each vertex to the center.
c
a
•
The eccentricity of an ellipse is e = , and 0 < e < 1.
•
The eccentricity of a hyperbola is e = , and e > 1.
•
•
The eccentricity of a parabola is e = 1.
c
a
The eccentricity of a circle is e = 0.
EXAMPLE 1
Earth, as seen from
the moon
Finding Eccentricity
Find the eccentricity of the conic section described by the equation.
a. (x + 2)2 = 4(y º 1)
b. 25(x + 2)2 º 36(y º 1)2 = 900
SOLUTION
a. This equation describes a parabola. By definition, the eccentricity is e = 1.
b. This equation describes a hyperbola with a = 3
6 = 6, b = 25 = 5, and
c
a
61
6
c = a2+
b2 = 61. The eccentricity is e = = ≈ 1.302.
EXAMPLE 2
Using Eccentricity to Write an Equation
Find an equation of the hyperbola with center (3, º5), vertex (9, º5), and e = 2.
SOLUTION
(x º h)2
(y º k)2
a
b
c
c
center, so a = 6. Because e = = 2, you know that = 2, or c = 12. Therefore,
a
6
(x º 3)2
( y + 5)2
b2 = c2 º a2 = 144 º 36 = 108. The equation is º = 1.
36
108
Use the form º
= 1. The vertex lies 9 º 3 = 6 units from the
2
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Chapter 10 Extension
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EXAMPLE 3
Using Eccentricity to Write a Model
The moon orbits Earth in an elliptical path with the center of Earth at one focus.
The eccentricity of the orbit is e = 0.055 and the length of the major axis is about
768,800 kilometers. Find an equation of the moon’s orbit.
SOLUTION
Let the major axis of the ellipse be horizontal. The equation of the orbit has the form
y2
x2
2 + 2 = 1. Using the length of the major axis, you know that 2a = 768,800,
a
b
c
c
or a ≈ 384,400. Because e = , you know that 0.055 = , or c ≈ 21,142
a
384,400
c2 = 384,4
002º
21,1
422 = 1.4
7
1011
≈ 383,800. The
and b = a2º
2
y2
383,800
equation of the moon’s orbit is x2 + 2 = 1 where x and y are
384,400
measured in kilometers.
EXERCISES
Find the eccentricity of the conic section.
1. 3x2 º 5x + y + 20 = 0
2. 25(x º 3)2 + 9(y + 6)2 = 225
3. x2 + 16( y º 4)2 = 16
(x º 3)2
(y º 5)2
4. + = 8
8
8
(x + 6)2 ( y º 6)2
5. º = 1
25
100
(x + 2)2
( y + 2)2
6. + = 1
49
16
7. 4(x + 1)2 º 8(y º 2)2 = 16
8. (x º 4)2 º (y º 3)2 = 1
Write an equation of the conic section.
9. Ellipse with vertices at (º5, º1) and (5, º1), and e = 0.6
10. Ellipse with foci at (2, º4) and (2, 4), and e = 0.5
11. Ellipse with center at (2, 0), focus at (2, 2), and e = 0.25
12. Ellipse with center at (0, 6), vertex at (3, 6), and e = 0.1
13. Hyperbola with foci at (3, º7) and (3, 9), and e = 3
14. Hyperbola with vertices at (º10, 4) and (º2, 4), and e = 2.4
15. Hyperbola with center at (3, 2), vertex at (3, 5), and e = 1.9
16. Hyperbola with center at (º1, 2), focus at (4, 2), and e = 5
17.
ASTRONOMY Mercury orbits the sun in an elliptical path with the center
of the sun at one focus. The eccentricity of Mercury’s orbit is e = 0.2056. The
length of the major axis of the orbit is 72 million miles. Find an equation of
Mercury’s orbit.
18.
ASTRONOMY Mars orbits the sun in an elliptical path with the center of the
sun at one focus. The eccentricity of Mars’ orbit is e = 0.0932. The perihelion of
Mars’ orbit is the point where the planet is closest to the sun. At the perihelion,
Mars’ distance from the sun is 128.4 million miles. Find an equation of Mars’ orbit.
19.
Writing Explain why the definition of eccentricity for ellipses and hyperbolas
implies that 0 < e < 1 for an ellipse and e > 1 for a hyperbola.
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Chapter 10 Quadratic Relations and Conic Sections