Section 6-3: Ellipses
 
Definition of an Ellipse
If F1(c, 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
PF1 + 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!!

 Problems
 
1)  Sketch the graph and find the vertices, end points of the minor axis and foci points for:  x2 + 4y2 = 16
            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)
                 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)

 
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.
                 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 = 1

Translation of the Ellipse
 
 
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 = 3
                     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 
 
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