The foci and vertices of the ellipse are located at (± c,0) and (± a,0) respectively.
B
b
The foci and vertices of the ellipse are located at (± c,0) and (± a,0) respectively.
C
c
Use the fact c^2=a^2-b^2 to determine the value of b and find the covertices of the ellipse in Part B.
D
d
Repeat the process from Part C to graph the ellipse in Part A.
A
a
0.9
B
b
0.1
C
c
The shape is close to a circle.
D
d
The shape is close to a line segment.
Practice makes perfect
a
Using the properties of a horizontal ellipse, let's determine the values of a and c.
Foci=(± c, 0)=(± 9, 0)
Vertices=(± a, 0)=(± 10, 0)
We have identified that c= 9 and a= 10. Let's substitute these values into the formula for eccentricity.
Eccentricity=c/a=9/10=0.9
b
Let's repeat the process from Part A to determine the eccentricity of an ellipse with foci ( ± 1,0).
Foci=(± c, 0)=(± 1, 0)
Vertices=(± a, 0)=(± 10, 0)
For this ellipse, c= 1 and a= 10. Let's substitute these values into the formula for eccentricity.
Eccentricity=c/a=1/10=0.1
c
We can see that the eccentricity of 0.1 in Part B is close to 0. Let's graph this ellipse to see its shape. In order to do this, we will need to find its co-vertices, (0, ± b). We can use the fact c^2=a^2-b^2 for an ellipse to determine the value of b.
We use the value of b to identify the covertices (0, ± 9.9). Let's plot the vertices and covertices to graph the ellipse from Part B.
We can see that if the eccentricity is close to 0, the ellipse looks like a circle.
d
We can see that the eccentricity of 0.9 in Part A is close to 1. Let's graph this ellipse to see its shape. We will repeat the process from Part C to find the co-vertices, (0, ± b).
