6-2 ellipses (presentation)
TRANSCRIPT
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6-2 Ellipses
Unit 6 Conic Sections
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Concepts and Objectives
Ellipses (Obj. 20)
Identify the equation of an ellipse Find the center,x-radius, andy-radius of an ellipse
Find the major and minor axes
Find the foci and focal length of an ellipse Write the equation of an ellipse
Solve problems involving ellipses
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Ellipses
An ellipse, geometrically speaking, is a set of points in a
plane such that for each point, the sum of its distances,d1 + d2, from two fixed points F1 and F2, is constant.
What does this mean?
.
Place the two pins at least 3" apart.
Tie your piece of string in a loop that will fit around
the pins without going off the edge.
Put the loop around the pins, pull it taut, and trace
around the loop.
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Parts of an Ellipse
Parts of an ellipse:
Center Vertices (ea. vertex)
Major axis
Minor axis Foci (ea. focus)
What we have done with the string is kept the distancebetween the foci and the points on the ellipse constant
(i.e. the definition).
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Parts of an Ellipse
Other important parts:
The semi-majorand semi-minoraxes are half thelength of the major and minor axes.
The distance from the center to the ellipse in thex-
- . ,
in they-direction is called they-radius.
The distance between the foci is called thefocal
length. The distance between the center and a focus
is called thefocal radius.
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Ellipses
Generally speaking,
a is the length of the semi-major axis b is the length of the semi-minor axis
c is the length of the focal radius.
Therefore,
The length of the major axis is 2a
The length of the minor axis is 2b The sum of the distances from a point(x,y) to a point
on the ellipse is 2a
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Focal Radius
Not only does the dotted
line trace the outline of theellipse, but it is also the
length of the major axis. As
ou can see from theb
a
picture, half of that length(a) is the hypotenuse of the
triangle formed by the semi-
minor axis and the focal
radius. This gives us the
formula:= 2 2 2c a b
c
(careful!)
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Standard Form
The standard form of an ellipse centered at(h, k) is
+ =
22
1
x y
x h y k
r r
where rxis thex-radius and ryis they-radius. To graph an ellipse from the standard form, plot the
center, mark thex- andy-radii, and sketch in the curve.
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Standard Form
Example: Sketch the graph of
+ + =
2 24 11
3 5
x y
The center is at(4, 1)Thex-radius is 3
They-radius is 5
Sketch in the curves
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Standard Form of an Ellipse
Example: Sketch the graph of
To sketch the graph, we have to rewrite the equation.+ + + =
2 24 9 16 90 205 0 x y x y
+ + = 2 24 16 9 90 205 x x y y
( ) ( ) ( ) ( ) + + = + + + +2 2 222 24 4 9 10 2052 5 4 2 9 5 x x y y
( ) ( ) ++ =
2 2
4 2 9 5 36
36 36 36
x y
+ + =
2 2
2 51
3 2
x y( ) ( ) ++ =
2 2
2 51
9 4
x y
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Standard Form of an Ellipse
Example (cont.):
+ + =
2 2
2 51
3 2
x y
The center is at(2, 5)Thex-radius is 3
They-radius is 2
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Focal Length
Example: Write the equation of the ellipse having center
at the origin, foci at(5, 0) and (5, 0), and major axis oflength 18 units.
, , .
The distance between the center and a focus, c, is 5.
Therefore, we can find b using the formula:
= 2 2 2c a b
= 2 2 2
5 9 b= =2 2 29 5 56b
= 56b
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Focal Length
Example (cont.):
The foci lie along the major axis, so we know thatrx= a.
Putting it all together, we have:
+ =
22
19 56
x y
+ =2 2
1
81 56
x y
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Eccentricity The eccentricity of an ellipse is a measure of its
roundness, and it is the ratio of the focal length to themajor axis.
This ratio is written as
=e a
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Eccentricity Example: The orbit of Jupiter is an ellipse with the sun
at one focus (mostly). The eccentricity of the ellipse is0.0489, and the maximum distance of Jupiter from the
sun is 507.4 million miles. Find the closest distance that
u iter comes to the sun.
Jupiter
Sun
a + c
a c= 0.0489ca
= 0.0489c a+ = 507.4
a c+ =0.0489 507.4a a
= 507.4
483.741.0489
a
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Eccentricity Example (cont.):
( )= 0.0489 483.74c
= 23.65c
Jupiter
Sun
a + c
a c
= . .a c= 460.1 million miles
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Homework Algebra & Trigonometry(green book)
Page 476: 3-18 (3s) Turn in: 9, 15
College Algebra (brown book) Page 968: 15-39 (3s), 45, 47a, 48
Turn in: 18, 24, 47, 48