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Copyright © 2011 Pearson Education, Inc. Slide 6.3-1
Copyright © 2011 Pearson Education, Inc. Slide 6.3-2
Chapter 6: Analytic Geometry
6.1 Circles and Parabolas
6.2 Ellipses and Hyperbolas
6.3 Summary of the Conic Sections
6.4 Parametric Equations
Copyright © 2011 Pearson Education, Inc. Slide 6.3-3
• Conic sections presented in this chapter are of the form
where either A or C must be nonzero.
6.3 Summary of the Conic Sections
,022 FExCyDxAx
Conic Section Characteristic Example
Parabola
Circle
Ellipse
Hyperbola
Either A = 0 or C = 0, but not both.
y = x2
x = 3y2 + 2y – 4
0CA 1622 yx0, ACCA
0AC
125
2
16
2 yx
122 yx
Copyright © 2011 Pearson Education, Inc. Slide 6.3-4
6.3 Summary of the Conic Sections
Copyright © 2011 Pearson Education, Inc. Slide 6.3-5
6.3 Summary of the Conic Sections
Last 3 rows of the Table on pg 6-52
Copyright © 2011 Pearson Education, Inc. Slide 6.3-6
6.3 Determining Type of Conic Section
• To recognize the type of conic section, we may need to transform the equation.
Example Decide on the type of conic section represented by each equation.
(a)
(b)
(c)
(d)
22 525 yx 61549164 22 yyxx
41108 22 yyxx
07862 yxx
Copyright © 2011 Pearson Education, Inc. Slide 6.3-7
6.3 Determining Type of Conic Section
Solution (a)
This equation represents a hyperbola centered at the origin.
1525
255525
22
22
22
yxyx
yx
Copyright © 2011 Pearson Education, Inc. Slide 6.3-8
6.3 Determining Type of Conic Section
(b) Coefficients of the x2- and y2-terms are unequal and both positive. This equation may be an ellipse. Complete the square on x and y.
This is an ellipse centered at (2, –3).
14
)3(9
)2(
36)3(9)2(4
6181)96(916)44(4
61)996(9)444(4
61)6(9)4(4
22
22
22
22
22
yx
yx
yyxx
yyxx
yyxx
Copyright © 2011 Pearson Education, Inc. Slide 6.3-9
6.3 Determining Type of Conic Section
(c) Coefficients of the x2- and y2-terms are both 1. This equation may be a circle. Complete the square on x and y.
This equation is a circle with radius 0; that is, the point (4, –5).
0)5()4(
251641)2510()168(
41108
22
22
22
yx
yyxx
yyxx
Copyright © 2011 Pearson Education, Inc. Slide 6.3-10
6.3 Determining Type of Conic Section
(d) Since only one variable is squared, x2, the equation represents a parabola. Solve for y (the variable that is not squared) and complete the square on x (the squared variable).
The parabola has vertex (3, 2) and opens downward.
2)3(81
16)3(8
97)96(8
768
0786
2
2
2
2
2
xy
xy
xxy
xxy
yxx
Copyright © 2011 Pearson Education, Inc. Slide 6.3-11
6.3 Eccentricity
• The constant ratio is called the eccentricity of the conic, written e.
A conic is the set of all points P(x, y) in a plane such that the ratio of the distance from P to a fixed point and the distance from P to a fixed line is constant.
Copyright © 2011 Pearson Education, Inc. Slide 6.3-12
6.3 Eccentricity of Ellipses and Hyperbolas
• Ellipses and hyperbolas have eccentricity
where c is the distance from the center to a focus.
• For ellipses, a2 > b2 and
• Note that ellipses with eccentricity close to 0 have a circular shape because b a as c 0.
,ac
e
So, .22 bac
)hyperbolasfor 1(.10
10
0
eeac
ac
Copyright © 2011 Pearson Education, Inc. Slide 6.3-13
6.3 Finding Eccentricity of an Ellipse
Example Find the eccentricity of
Solution Since 16 > 9, let a2 = 16, giving a = 4.
.1169
22
yx
7916
22
bac
66.47
ac
e
Copyright © 2011 Pearson Education, Inc. Slide 6.3-14
6.3 Figures Comparing Different Eccentricities of Ellipses and Hyperbolas
Copyright © 2011 Pearson Education, Inc. Slide 6.3-15
6.3 Finding Equations of Conics using Eccentricity
Example Find an equation for each conic with center at the origin.
(a) Focus at (3, 0) and eccentricity 2(b) Vertex at (0, –8) and e =
Solution(a) e = 2 > 1, this conic is a hyperbola with c = 3.
21
Copyright © 2011 Pearson Education, Inc. Slide 6.3-16
6.3 Finding Equations of Conics using Eccentricity
The focus is on the x-axis, so the x2-term is positive. The equation is
427
49
923
3
. find to Use
23
32
222
2222
b
bacb
a
a
ac
e
.1274
94
or1
427
49
2222
yxyx
Copyright © 2011 Pearson Education, Inc. Slide 6.3-17
6.3 Finding Equations of Conics using Eccentricity
(b) Since the conic is an ellipse. The vertex at (0, –8) indicates that the vertices lie on the y-axis and a = 8.
Since the equation is
,121 e
c
cac
e
482
1
,481664222 cab
.14864
22
xy
Copyright © 2011 Pearson Education, Inc. Slide 6.3-18
6.3 Applying an Ellipse to the Orbit of a Planet
Example The orbit of the planet Mars is an ellipse with the sun at one focus. The eccentricity of the ellipse is 0.0935, and the closest Mars comes to the sun is 128.5 million miles. Find the maximum distance of Mars from the sun.
SolutionUsing the given figure, Mars is
- closest to the sun on the right - farthest from the sun on the left
Therefore, - smallest distance is a – c - greatest distance is a + c
Copyright © 2011 Pearson Education, Inc. Slide 6.3-19
6.3 Applying an Ellipse to the Orbit of a Planet
Since a – c = 128.5, c = a – 128.5. Using e = 0.0935, we find a.
The maximum distance of Mars from the sun is about 155.1 million miles.
128.50.0935
0.0935 128.5141.8
ce
aa
aa aa
1.1553.138.141 3.135.1288.141
cac
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