The Ellipse - Bradley Bradley

Report
The Ellipse
Analytic Geometry
Section 3.3
1
Definition of “ellipse”
• An ellipse is
the set of all
points in a
plane such
that the
distance from
two fixed
points (foci) on
the plane is a
constant.
2
Equation of the Ellipse
• The equation of an
ellipse with its center
at the origin has one
of two forms:
x2 y 2
 2 1
2
a
b
or
• The position of the a2
(under the x or y) tells
you whether the
horizontal or the
vertical axis is the
major axis of the
ellipse.
y2 a2
 2 1
2
a
b
3
Ellipse
x2 y 2

1
64 25
• This ellipse has a
horizontal major axis
that is 16 units long.
a  64, a  8
2
2a  16
4
Ellipse
x2 y 2

1
64 25
• The minor axis of this
ellipse is 10 units in
length.
b  25, b  5
2
2b  10
5
Foci
• The two foci for this
ellipse are the two
points lying on the
horizontal axis that
appear to be a little
over 6 units from the
origin. The origin is
the center of the
ellipse. The distance
from the center to a
focus is “c”.
6
• The segments drawn
from the two foci to
the point (0,5) on the
ellipse are each 8
units in length. Their
total length is 16
units. This total
length is also the
length of the major
axis.
7
• Two more segments
are added, drawn
from the foci to the
point (2,4.84) on the
ellipse. Their lengths
are 9.556 and 6.434.
The sum of these
lengths is again 16
units.
8
• The two latest
segments, drawn to
the point (7,-2.42) on
the ellipse, are 13.463
units and 2.537 units
in length, a sum of 16
units.
9
The Ellipse
(0,b)
P(x,y)
(-a,0)
(a,0)
(-c,0)
(c,0)
(0,-b)
x2 y 2
 2 1
2
a
b
• The ends of the major
axis are at (a,0) and
(-a,0).
• The ends of the minor
axis are at (0,b) and
(0,-b).
• The foci are at (c,0)
and (-c,0).
10
The Ellipse
(0,b)
• The sum of the
distances from point
P to the foci is 2a.
P(x,y)
(-a,0)
(a,0)
(-c,0)
(c,0)
(0,-b)
x2 y 2
 2 1
2
a
b
2
2
2
c

a

b
• Also,
or
b  a c
2
2
2
11
•
A chord through a focus and
perpendicular to the major axis is
called a latus rectum.
• The endpoints of the
two latus recti are
found using the
equivalence :
c  a b
2
2
2
providing the
endpoints

b 
  c, 
a 

2
12
Latus Rectum
• When the equation of
the ellipse is
x2 y2

1
64 25
then c 2  64  25
 39
and c  6.24
So the endpoints of the latus recti are:



6.24, 3.125

13
The Ellipse
y2  x2 1
169 25
• The major axis is the
vertical axis with
endpoints (0,13) and
(0,-13). The endpoints
of the major axis are
called the vertices.
The minor axis has
endpoints of (5,0) and
(-5,0).
14
y2  x2 1
169 25
• The foci are found
using
c 2  a 2  b2
c 2  169  25  144
• so the values of c are
12 and -12. The
coordinates of the foci
are (0,12) and (0,-12).
15
y2  x2 1
169 25
• The endpoints of the
latus recti are:

b2 
  c, 
a 

25 

  12,  
13 

16
Problem: Determining an equation
• Find the equation of
the ellipse with foci at
(8,0) and (-8,0) and a
vertex at (12,0).
• First, place these
points on axes.
• The F and F’ are the
foci.
F
F'
17
Finding the equation of the ellipse with foci at (8,0) and (-8,0) and a
vertex at (12,0).
• Since the vertex is on the horizontal axis, the
ellipse will be of the form:
x2 y 2
 2 1
2
a
b
• The values of a and b need to be determined.
18
Finding the equation of the ellipse with foci at (8,0) and
(-8,0) and a vertex at (12,0).
• If the foci are at 8 and -8, then c = 8. Since a
vertex is at (12,0), that means that a = 12.
Relating these values to the standard form for
an ellipse whose center is at the origin and
whose major axis is horizontal, ax  by  1 ,
and the equivalence c2  a 2  b2
2
2
2
applies. Solve for b2 to get b  a  c
In this case, b2  122  82  144  64  80
2
2
2
2
19
Finding the equation of the ellipse with foci at (8,0) and
(-8,0) and a vertex at (12,0).
•
•
•
•
Since b2  80, b  80  8.94
The value of a is 12, and a2 is 144.
The value of b is 80 and b2 is 80.
So the equation of the ellipse is:
2
2
x
y
x
y


1


1
or
144 80
a 2 b2
2
•
2
20
Ellipse with center at (h,k)
• The ellipses with their centers at the origin
are just special cases of the more general
ellipse with its center at the point (h,k).
This more general ellipse has a standard
formula of:
( x  h) 2 ( y  k ) 2

1
2
2
a
b
or
( y  k ) 2 ( x  h) 2

1
2
2
a
b
21
Problem: Write the equation in standard form
• The general form of the equation is:
4x  8 y  4x  24 y 13  0
2
2
• After writing this in standard form, also find
the coordinates of the center, the foci, the
ends of the major and minor axes, and the
ends of each latus rectum.
22
Write
4x2  8 y2  4x  24 y 13  0
in standard form:
• First, group the terms with x’s and the
terms with y’s, and move the constant to
the other side of the equation.
4x  4x  8 y  24 y  13
2
2
23
Write
4x  8 y  4x  24 y 13  0
2
2
in standard form:
• Now factor out the coefficient of each squared term.
4( x  x)  8( y  3 y)  13
2
2
• Then complete the square for each variable.
1  2
9
 2
1 9
4  x  x    8  y  3 y    13  4    8  
4 
4

 4  4
2
2
1
3


4  x    8  y    13  1  18
2
2


24
To finish the problem:
• Simplify on the right.
2
2
1
3


4  x    8  y    32
2
2


• Then divide each side by 32.
2
2
1
3


4 x   8 y  
32
2
2




32
32
32
2
2
1 
3

x   y 
2 
2


1
8
4
25
The Ellipse
2
2
1 
3

x  y 
2 
2


1
8
4
1 3
• This ellipse has a center at  2 , 2  .


• The major axis is 2 8 in length, and the minor
axis is 4 in length, so their endpoints are ( 8 ,0)
and (- 8 ,0), (0,2) and (0,-2).
• The foci are at (2,0) and (-2,0).
26
To finish
2
2
1 
3

x  y 
2 
2


1
8
4
• Since the foci are at (2,0) and (-2,0), the
endpoints of the latus recti are at

b2  
4 
 2,  2
 c,     2, 

a 
8



27
Site and Assignment
• There’s a neat website that you might
want to look at for more on the ellipse. It’s
at:
http://mathworld.wolfram.com/Ellipse.html
• Your assignment, due Monday, is:
3.3: 2, 3, 15, 16, 17, 22, 25, 44
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