Accelerated Precalculus
Ellipses
One Minute Question
• Find the diameter of:
• x2 + y2 + 6x - 14y + 9 = 0
One Minute Question
• Find the diameter of:
• x2 + y2 + 6x - 14y + 9 = 0
• x2 + 6x + 9 + y2 - 14y + 49= 49
• So the radius is 7 and the
diameter is 14.
Questions on Homework?
Check so far...
• Write the equation of the
parabola with focus at (4, 6)
and directrix at y = -2.
• (1/16)(x - 4)2 = (y - 2) or
• (x - 4)2 = 16(y - 2)
Check so far...
• Write the focus, vertex, and
directrix of: y2 = 4y + 2x + 8
• y2 - 4y + 4 = 2x + 8 + 4
• (y - 2)2 = 2(x + 6)
• ½(y - 2)2 = (x + 6)
Check so far...
• Write the focus, vertex, and
directrix of: ½(y - 2)2 = (x + 6)
• The vertex is (-6, 2).
• It opens to the right.
• So the focus is (-5.5, 2)
• and the directrix is x = -6.5
Check so far...
• Write the equation of a circle
whose diameter has endpoints
at (4, 5) and (-2, 9).
• The midpoint is (1, 7)
• The radius is 32 + 22 = 13
• So... (x - 1)2 + (y - 7)2 = 13
The Ellipse
• An ellipse is the set of all
points P in a plane, so that the
sum of the distances from P
to two fixed points in the
plane (called foci) is a
constant.
The Ellipse
• Place two pushpins on your
graph paper at horizontal or
vertical lattice points. These
will be your foci.
The Ellipse
• Cut a piece of string longer
than the distance between
your points to become your
constant sum of lengths.
The Ellipse
• Tie each end of the string to
the push pin and, using your
pencil to stretch the string as
far as you can, draw the
ellipse.
The Ellipse
• The center of the ellipse is
the midpoint of your foci.
• The segment joining two
points on the ellipse that
contains the foci is called the
major axis. Measure your
major axis.
The Ellipse
• The segment joining two
points on the ellipse that
is perpendicular to the
major axis is called the
minor axis. Find the length
of your minor axis.
The Ellipse
• We call the distance from the
center to a focus the focal length
(c), the distance from the center
to an endpoint of the major axis
the semimajor axis (a) and the
distance from the center to an
endpoint of the minor axis the
semiminor axis (b).
The Ellipse
b
c
a
The Ellipse
• Pick up your string and note
that its length is the length
of the major axis. Then put
the string back in its original
place and make it taut at the
endpoint of the minor axis.
The Ellipse
• See that b2 + c2 = a2
b
a
c
a
The Ellipse
• The algebra…
• This is basically a circle with
two different radii - one in
the x direction, and one in the
y direction. So, if
(x - h)2 + (y - k)2 = r2 for a
circle...
The Ellipse
( x  h)
r2
2

( y  k )2
r2
1
• if you divide both sides by r.
Make the two radii different
and... ( x  h) 2 ( y  k ) 2
r1 2

r2 2
1
The Ellipse
• But a and b are the radii, so
( x  h)
2
a2
• or
( x  h)
b
2

2
( y  k)

2
1
b2
( y  k)
a
2
2
1
The Ellipse
• If your ellipse is centered at the
origin, write its equation and
verify that b2 + c2 = a2.
The Ellipse
• Given:
( x  3)

25
2
( y  1) 2
9
1
• sketch it, find the center, foci,
endpoints of the major axis
(vertices) and endpoints of the
minor axis (co-vertices.)
The Ellipse
• Given:
( y  1)
( x  3)

1
25
9
2
2
• Center: (3, -1),
• Foci: (-1,-1) and (7,-1)
• Vertices: (-2, -1) and (8, -1)
• Co-vertices: (3, 2) and (3, -4)
The Ellipse
• Given: 4x2 + 9y2 + 6x - 8y = 11
• Write it in standard form,
sketch it, find the center, foci,
endpoints of the major axis
(vertices) and endpoints of the
minor axis (co-vertices.)
The Ellipse
• Given: 4x2 + 9y2 + 6x - 18y = 11
• 4(x2 + (3/2)x + (9/16)) +
9(y2 – 2y + 1) = 11 +9/4 + 9
4(x + ¾)2 + 9(y – 1)2 = 89/4
2
2
(x + ¾) + (y – 1) = 1
(89/16) (89/36)
The Ellipse
(x + ¾)2 + (y – 1)2 = 1
(89/16) (89/36)
The center is (-3/4, 1)
The vertices are ((3 ± √89)/4, 1)
The foci are ((9 ± √445)/12, 1)
The co-vertices are:
(3/4, (6 + √89)/6)
The Ellipse (Proof)
• Let the ellipse be centered at the
origin with foci located at (c, 0)
and (-c, 0). Let (x, y) be any point
on the ellipse. By the definition
of the ellipse, the sum of the
distances from (x, y) to (c, 0) and
(-c, 0) is the constant 2a.
The Ellipse (Proof)
 x  c
2
  y  0 
 x  c
2
  y  0   2a 
 x  c
   2a 
 x  c

 x  c
 x  c
2
2
 x  c
2
2
  y
2
2
  y   4a  4a
2
2
4a
2
 x  c
2
 x  c
2
2
  y   4a  4cx
2
  y  0   2a
2
2
2
2
  y  0
  y
2

2
2
  y   x  c   y 
2
2
2
2
x

c

y

x

2
cx

c

  
x  2cx  c  4a  4a
2
2
2
2
2
The Ellipse (Proof (cont.))
4a
2
x

c

y

4
a
 4cx

  
a
 x  c
2
2
2
  y
2
  a  cx 
2
2
2
a 2  x 2  2 xc  c 2  y 2   a 4  2a 2cx  c 2 x 2
a 2 x 2  2a 2 xc  a 2c 2  a 2 y 2  a 4  2a 2cx  c 2 x 2
a 2 x 2  a 2c 2  a 2 y 2  a 4  c 2 x 2
a 2 x 2  c 2 x 2  a 2 y 2  a 4  a 2c 2
2
2
2
2 2
2
2
2
a

c
x

a
y

a
a

c




2
2
2
a

c
x


a2 y2
 2 2
1
2
2
2
2
a a  c  a a  c 
The Ellipse (Proof (cont.))
• But since a2 – c2 = b2
a
2
c x
2
2
2
2
a y
 2 2 2 1
2
2
2
a a  c  a a  c 
2
2
2
2
b x
a y
 2 2 1
2 2
ab
ab
2
2
x
y
 2 1
2
a
b