Transcript 6.2 Ellipses and Hyperbolas

Copyright © 2011 Pearson Education, Inc.
Slide 6.2-1
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
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Slide 6.2-2
6.2 Ellipses and Hyperbolas
An ellipse is the set of all points in a plane, the sum
of whose distances from two fixed points is constant.
Each fixed point is called a focus of the ellipse.
• The graph of an ellipse
is not that of a function.
• The foci lie on the major
axis – the line from V to
V.
• The minor axis – B to B
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Slide 6.2-3
6.2 The Equation of an Ellipse
• Let the foci of an ellipse be at the points (c, 0).
The sum of the distances from the foci to a point
(x, y) on the ellipse is 2a.
d ( P, F )  ( x  c) 2  y 2
d ( P, F ' )  ( x  (c))2  y 2  ( x  c) 2  y 2
So, we rewrite the following equation.
( x  c ) 2  y 2  ( x  c ) 2  y 2  2a
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Slide 6.2-4
6.2 The Equation of an Ellipse
( x  c ) 2  y 2  ( x  c ) 2  y 2  2a
( x  c ) 2  y 2  2a  ( x  c ) 2  y 2
( x  c ) 2  y 2  4a 2  4a ( x  c ) 2  y 2  ( x  c ) 2  y 2
x 2  2cx  c 2  y 2  4a 2  4a ( x  c) 2  y 2  x 2  2cx  c 2  y 2
4a ( x  c) 2  y 2  4a 2  4cx
a ( x  c) 2  y 2  a 2  cx
a 2 (( x  c) 2  y 2 )  a 4  2ca 2 x  c 2 x 2
a 2 x 2  2ca 2 x  a 2 c 2  a 2 y 2  a 4  2ca 2 x  c 2 x 2
a 2 x 2  c 2 x 2  a 2 y 2  a 4  a 2c 2
x 2 (a 2  c 2 )  a 2 y 2  a 2 (a 2  c 2 )
x2
y2
 2
1
2
2
a (a  c )
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Slide 6.2-5
6.2 The Equation of an Ellipse
• Replacing a2 – c2 with b2 gives the standard
equation of an ellipse with the foci on the x-axis.
2
2
x
y
 2 2 1
2
a (a  c )
2
2
x y
 2 1
2
a b
• Similarly, if the foci were on the y-axis, we get
x2 y 2
 2  1.
2
b a
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Slide 6.2-6
6.2 The Equation of an Ellipse
The ellipse with center at the origin and equation
x2 y 2
 2  1  a  b  0
2
a b
has vertices (a, 0), endpoints of the minor axis
(0, b), and foci (c, 0), where c2 = a2 – b2.
The ellipse with center at the origin and equation
x2 y 2
 2  1  a  b  0
2
b
a
has vertices (0, a), endpoints of the minor axis
(b, 0), and foci (0, c), where c2 = a2 – b2.
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Slide 6.2-7
6.2 Graphing an Ellipse Centered at the
Origin
Example
Graph 4 x 2  9 y 2  36.
Solution
Divide both sides by 36.
x2 y2
 1
9 4
This ellipse, centered at the
origin, has x-intercepts 3 and
–3, and y-intercepts 2 and –2.
The domain is [–3, 3]. The
range is [–2, 2].
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Slide 6.2-8
6.2 Finding Foci of an Ellipse
Example Find the coordinates of the foci of the
equation 4 x 2  9 y 2  36.
Solution
From the previous example, the2
2
y
equation of the ellipse in standard form is x   1.
9
4
Since 9 > 4, a2 = 9 and b2 = 4.
2
2
2
c  a b
c2  9  4  5
c 5
The major axis is along the x-axis, so the foci have
coordinates ( 5, 0) and ( 5, 0).
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Slide 6.2-9
6.2 Finding the Equation of an Ellipse
Example Find the equation of the ellipse having
center at the origin, foci at (0, ±3), and major axis of
length 8 units. Give the domain and range.
Solution
2a = 8, so a = 4.
a 2  b2  c2
42  b 2  32
b 7
2
Foci lie on the y-axis, so the larger intercept, a, is used
to2 find2 the denominator for y2. The standard form is
y
 x  1, with domain [ 7, 7 ] and range [–4, 4].
16
7
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Slide 6.2-10
6.2 Ellipse Centered at (h, k)
An ellipse centered at (h, k) and either a horizontal or vertical
major axis satisfies one of the following equations, where
2
2
2
a > b > 0, and c  a  b with c > 0:
( x  h) 2 ( y  k ) 2

1
2
2
a
b
( x  h) 2 ( y  k ) 2

1
2
2
b
a
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Slide 6.2-11
6.2 Ellipse Centered at (h, k)
( x  2) 2 ( y  1) 2

 1.
Example Graph
9
16
Analytic Solution
Center at (2, –1).
Since a > b, a = 4 is associated with
the y2 term, so the vertices are on the
vertical line through (2, –1).
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Slide 6.2-12
6.2 Ellipse Centered at (h, k)
Graphical Solution
Solving for y in the equation yields
( x  2) 2
y  1  4 1 
.
9
The + sign indicates the upper half of the ellipse, while the
– sign yields the bottom half.
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Slide 6.2-13
6.2 Finding the Standard Form of an Ellipse
Example Write the equation in standard form.
4 x 2  16 x  9 y 2  54 y  61  0
Solution
4 x 2  16 x  9 y 2  54 y  61  0
4  x 2  4 x  4   9  y 2  6 y  9   61  4  4   9  9 
4  x  2   9  y  3  36
2
 x  2
9
2
2
y  3


4
2
1
Center (2, –3); Vertices (2±3, –3) = (5, –3), (–1, –3)
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Slide 6.2-14
6.2 Hyperbolas
A hyperbola is the set of all points in a plane such that the
absolute value of the difference of the distances from two
fixed points is constant. The two fixed points are called the
foci of the hyperbola.
• If the center is at the origin, the
foci are at (±c, 0).
• The midpoint of the line segment
FF is the center of the hyperbola.
• The vertices are at (±a, 0).
• The line segment VV is called
the transverse axis.
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Slide 6.2-15
6.2 Standard Forms of Equations for
Hyperbolas
The hyperbola with center at the origin and equation
x2 y2
 2 1
2
a
b
has vertices (±a, 0), asymptotes y = ±b/ax, and foci (±c, 0),
where c2 = a2 + b2.
The hyperbola with center at the origin and equation
y2 x2
 2 1
2
a
b
has vertices (0, ±a), asymptotes y = ±a/bx, and foci (0, ±c),
where c2 = a2 + b2.
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Slide 6.2-16
6.2 Standard Forms of Equations for
Hyperbolas
Solving for y in the first equation gives y   ba x 2  a 2 .
If |x| is large, the difference x  a approaches x2.
Thus, the hyperbola has asymptotes y   ba x.
2
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2
Slide 6.2-17
6.2 Using Asymptotes to Graph a
Hyperbola
Example
Sketch the asymptotes and graph the hyperbola
x2 y 2
  1.
25 49
Solution
a = 5 and b = 7
b
7
y x x
a
5
Choosing x = 5 (or –5) gives y = ±7.
These four points: (5, 7), (–5, 7),
(5, –7), and (–5, –7), are the corners
of the fundamental rectangle shown.
The x-intercepts are ±5.
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Slide 6.2-18
6.2 Graphing a Hyperbola with the
Graphing Calculator
Example
Graph 25 y 2  4 x 2  9.
Solution
Solve the given equation for y.
25 y 2  4 x 2  9
25 y 2  9  4 x 2
5 y   9  4x2
1
y   9  4x2
5
1
Let y1 
9  4 x 2 , and
5
1
y2   9  4 x 2 .
5
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Slide 6.2-19
6.2 Graphing a Hyperbola Translated
from the Origin
Example
( y  2) 2 ( x  3) 2

 1.
Graph
9
4
Solution
This hyperbola has the same graph as
y2 x2
  1, except that it is centered at (–3, –2).
9 4
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Slide 6.2-20
6.2 Finding the Standard Form for a Hyperbola
Example Write the equation in standard form.
9 x 2  18 x  4 y 2  16 y  43
Solution
9 x 2  18 x  4 y 2  16 y  43
9  x 2  2 x  1  4  y 2  4 y  4   43  9 1  4  4 
9  x  1  4  y  2   36
2
 x  1
4
2
2
y  2


9
2
1
Center (1, –2); Vertices (1±2, –2) = (3, –2), (–1, –2)
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Slide 6.2-21