Section 6-3: Ellipses
Definition of an Ellipse
0) and F2(-c, 0) are
two fixed points in the plane and a is a constant, 0< c < a, then
the set of all points P in the plane such that
+ PF2 = 2a
is an ellipse.
F1 and F2
are the foci of the ellipse.
Reflective Property of Ellipses (Manipula Math)
Notice the two fixed
points in the graph, (-4, 0) and (4, 0). These are the foci points
for the graph. By the definition, the distance from these points
to a point on the ellipse is a constant. In this case the constant
is 10. No matter what point you take on the ellipse, the distance
to the fixed points will always equal 10. As one distance gets longer
the other will get shorter so that the sum will always be 10.
Equation of the ellipse
This ellipse has the
major axis parallel to the x-axis making it open longer across. The
length is 2a. The minor axis is parallel to the y-axis and has length
2b. The foci points are 2c units apart. The center of the ellipse
until we translate it will remain at (0, 0). The vertex points are
at the end points of the major axis. Look at the equation. The a
value is always the biggest number!!
Notice the major axis
and the minor axis have reversed. The longer axis is now vertical.
What causes this to happen? Look at the equation closely. The
a value is now under the y value rather than the x value in the previous
equation. We now have an easy method to tell which way the ellipse
opens!! Look to see whether the larger value is under the x value
or the y value!!
1) Sketch the graph and
find the vertices, end points of the minor axis and foci points for:
x2 + 4y2
Solution: First put the equation in
the correct form by dividing by16.
x2/16 + y2/4
= 1 The larger value is a2
= 16 and b2 = 4.
Since the larger value is under x, the ellipse has the major axis horizontal.
The values are a = 4, b = 2. To find c, subtract 16 - 4 and take
the square root. Thus c = 3.5 (rounded to tenths)
2) An ellipse has its center
at the origin. Find an equation of the ellipse with Vertex (8, 0)
and minor axis 4 units long.
Center at (0, 0)
Vertices: (4, 0) and (-4, 0)
End points of minor axis: (0, 2) and (0, -2)
Foci: (3.5, 0) and (-3.5, 0)
Solution: a = 8 and b = 2 the
minor axis is 2b = 4, so b = 2.
The equation is: x2/64
+ y2/4 = 1 The
vertex is on the x-axis.
3) An ellipse has its
center at the origin. Find an equation of the ellipse with vertex
(0, -12) and focus ( 0, -4).
Solution: a = 12 and c = 4. Both
are on the y-axis, so the major axis is vertical. To find b2,
square a and c and subtract. b2
= 144 - 16 = 128
Thus the equation is:
x2/128 + y2/144
Translation of the
The center is now at (h, k).
All values are now calculated from this point rather than from (0, 0)
1) Sketch the graph
and find the vertices, end points of the minor axis and foci for:
Solution: The larger of the values is
under x. The major axis is horizontal. a = 5, b = 4 and c =
Center: (2, -1)
Vertices: (7, -1) and (-3, -1) add/subtract 5 from the x-value!
End points of minor: (2, 3) and (2, -5) add/subt 4 from y-value!
Foci points: (5, -1) and (-1, -1) add/subt 3 from x-value!
Notice this ellipse is
almost circular. The reason is because a and b are close in value.
In truth, a circle is a special case of an ellipse with a = b!!
2) Find the equation
of the ellipse with center (2,5), one focus (5,5) and one vertex (7, 5).
Solution: The focus and vertex are on
the same horizontal line. (y values are the same!) a is the distance
from the center to a vertex, so x = 5
c is the distance from the
center to a focus point, so c = 3. This make b = 4.
Thus the equation is:
On to Hyperbolas
Back up to circles