Notes 10-3: Ellipses
I.
Ellipse- Definition and Vocab
An ellipse is the set of points P(x, y) in
a plane such that the sum of the
distances from any point P on the
ellipse to two fixed points F1 and F2,
called the foci (singular: focus), is the
constant sum d = PF1 + PF2. This
distance d can be represented by the
length of a piece of string connecting two pushpins located at the foci. The center is the
midpoint of the foci. To help visualize this: https://www.youtube.com/watch?v=7UD8hOs-vaI
Instead of a single radius, an ellipse has two axes. The longer the axis of an ellipse is the major
axis and passes through both foci. The endpoints of the major axis are the vertices of the ellipse.
The shorter axis of an ellipse is the minor axis. The endpoints of the minor axis are the covertices of the ellipse. The major axis and
minor
axis are perpendicular and intersect at the
center
of the ellipse. The center is also the midpoint
of the
two foci.
So, b2 + c2 = a2, where b is the length of the semi-minor axis, a is the length of the semi-major
axis, and c is the distance from the center to one of the foci.
Ex 1: SPACE The graph models the elliptical path
of a space probe around two moons of a planet.
The foci of the path are the centers of the
moons. Find the coordinates of the foci.
The center of the ellipse is the origin.
1
a = 2(314) or 157
1
b = 2(110) or 55
The distance from the origin to the foci is c km.
c2 + b2 = a2
c2 + 552 = 1572
c  147
The foci are at (147, 0) and (-147, 0).
II.
Standard Form of an Ellipse
Remember that for b2 + c2 = a2, b is the length of the semi-minor axis, a is the length of the
semi-major axis, and c is the distance from the center to one of the foci.
Example 2
Consider the ellipse graphed at the right.
a. Write the equation of the ellipse in standard
form.
The center of the graph is at (-2, 1). Therefore, h
= -2 and k = 1.
Since the ellipse’s vertical axis is longer than its horizontal axis, a is the distance between
points at (-2, 1) and (-2, -3) or 4. The value of b is the distance between points at (-2, 1) and
(1, 1) or 3.
Therefore, the standard form of the equation is
(y - 1)2 (x - (-2))2
(y - 1)2 (x + 2)2
= 1 or 16 + 9
= 1.
42 +
32
b. Find the coordinates of the foci.
Using the equation c = a2 - b2, we find that c = 7. The foci are located on the vertical axis,
7 units from the center of the ellipse. Therefore, the foci have coordinates (-2, 1 - 7) and (-2,
1 + 7).
Example 3
(y - 2)2 (x + 1)2
For the equation 100 + 64 = 1, find the coordinates of the center, foci, and vertices of
the ellipse. Then graph the equation.
Determine the values of a, b, c, h, and k.
Since a2 > b2, a2 = 100 and b2 = 64. Therefore, a = 10 and b = 8.
c = a2 - b2
c = 100 - 64 or 6
x–h
=x+1y–k =y-2
h = -1
k =2
Since a2 is the denominator of the y term, the major axis is parallel to the y-axis.
center: (-1, 2)
(h, k)
foci: (-1, 8) and (-1, -4)
(h, k  c)
major axis vertices: (-1, 12) and (-1, -8)
(h, k
 a)
minor axis vertices: (-9, 2) and (7, 2) (h  b, k)
Graph these ordered pairs. Other points on the
ellipse can be found by substituting values for x
and y. Complete the ellipse.
III.
General Form of an Ellipse
Ax2 + Cx2 + Dx + Ey + F = 0
; A and C have the same sign.
Example 4
Find the coordinates of the center, the foci, and the vertices of the ellipse with the equation 9x2
+ 16y2 + 54x – 32y – 47 = 0. Then graph the equation.
First, write the equation in standard form.
9x2 + 16y2 + 54x – 32y – 47
=0
9(x2 + 6x + ?) + 16(y2 – 2y + ?)
= 47 + ? + ?
9(x2 + 6x + 9) + 16(y2 – 2y + 1)
= 47 + 9(9) + 16(1) CTS
9(x + 3)2 + 16(y – 1)2 = 144
Factor.
(x + 3)2 (y - 1)2
16 + 9 = 1
Divide each side by 144.
Since a2 > b2, a2 = 16 and b2 = 9. Thus, a = 4 and b = 3.
Since c2 = a2 – b2, c = 7.
Since a2 is the denominator of the x term, the
major axis is parallel to the x-axis.
center: (-3, 1)
(h, k)
foci: (-3  7, 1)
(h  c, k)
major axis vertices: (-7, 1), (1, 1) (h  a, k)
minor axis vertices: (-3, 4), (-3, -2) (h, k  b)
Sketch the ellipse.
IV. Eccentricity
The eccentricity of an ellipse, denoted by e, is a measure that describes the shape of an ellipse.
It is defined as
. e is always between 0 and 1.
Sometimes you may need to know the value of b when you know the values of a and e. In any
ellipse, c2 = a2 – b2 , so b2 = c2 – a2. This means that b2 = a2(1 – e2).
Example 5
ASTRONOMY The eccentricity of Uranus is 0.047. Its orbit is about 18.3 AU (astronomical
units) from the sun at its closest point to the sun. The length of the semi-major axis of the orbit
is about 19.21 AU.
Sketch a diagram.
Find the length of the semi-minor axis of the orbit.
We are looking for b.
b2 = a2(1 – e2)
b2 = (19.21)2(1 – (.047)2)
b2 =368.20
b = 19.19 AU
Find the distance of Uranus from the sun at its farthest point.
Aphelion = length of major axis – sun to perihelion
d = 2(19.19) – 18.3
d = 20
Uranus is about 20 AU from the sun at its aphelion.