Ellipse: Eccentricity
If an ellipse is close to circular it has an eccentricity close to zero. If an ellipse has an eccentricity close to one it has a high degree of ovalness.
Figure 1 shows a picture of two ellipses one of which is nearly circular with an eccentricity close to zero and the other with a higher degree of eccentricity.
The formal definition of eccentricity is:
ECCENTRICITY OF AN ELLIPSE:
The eccentricity (e) of an ellipse is the ratio of the distance from the center to the foci (c) and the distance from the center to the vertices (a).
$e=\frac{c}{a}$
As the distance between the center and the foci (c) approaches zero, the ratio of $\frac{c}{a}$ approaches zero and the shape approaches a circle. A circle has eccentricity equal to zero.
As the distance between the center and the foci (c) approaches the distance between the center and the vertices (a), the ratio of $\frac{c}{a}$ approaches one. An ellipse with a high degree of ovalness has an eccentricity approaching one.
Let's use this concept in some examples:
Step 1: Determine the values for the distance between the center and the foci (c) and the distance between the center and the vertices (a). 
Length of a: The given equation for the ellipse is written in standard form. Since the major axis is 2a and the smaller minor axis is 2b, then a^{2} > b^{2}, therefore a^{2} = 16. ${a}^{2}=16\to {a}{=}{4}$ Length of c: To find c the equation c^{2} = a^{2} + b^{2} can be used but the value of b must be determined. From our discussion above, b^{2} = 9. Find b and solve for c. ${b}^{2}=9\to b=3$ ${c}^{2}={4}^{2}{3}^{2}\to {c}^{2}=7\to {c}{=}\sqrt{7}$ 
Step 2: Substitute the values for c and a into the equation for eccentricity. 
$e=\frac{c}{a}$ $e=\frac{\sqrt{7}}{4}\to {e}{\approx}{0.66}$ 
Step 1: Determine the following:
➢ the orientation of the major axis.
➢ the coordinates of the center (h, k).
➢ the length of half the major axis (a).
➢ the distance of half the minor axis (b).

Orientation of major axis: Since the two vertices fall on the horizontal line y = 2, the major axis is horizontal. Center: Since the vertices are equidistant from the center of the ellipse the center can be determine by finding the midpoint of the vertices. $\left(h,k\right)=\left(\frac{4+\left(6\right)}{2},\frac{2+2}{2}\right)=\left(\frac{2}{2},\frac{4}{2}\right)={\left(}1,2{\right)}$ Length of a: the length of a is the distance between the center and the vertices. To find a take one of the vertices and determine the distance from the center. Vertex (4, 2): $c=\left4\left(1\right)\right=\left5\right=5$ Vertex (6, 2): $c=\left6\left(1\right)\right=\left5\right=5$ a = 5 Length of b: To find b the equation c^{2} = a^{2}  b^{2}can be used but the value of c must be determined. Since the eccentricity is $\frac{4}{5}$ the length of c can be found using the value for a. Then solve for b. $e=\frac{4}{5}=\frac{c}{a}$ $\frac{4}{5}=\frac{c}{5}\to 20=5c\to c=4$ ${c}^{2}={a}^{2}{b}^{2}\to b=\sqrt{{a}^{2}{c}^{2}}$ $b=\sqrt{{5}^{2}{4}^{2}}\to b=\sqrt{9}\to {b}{=}{3}$ 
Step 2: Substitute the values for h, k, a and b into the equation for an ellipse with a horizontal major axis. 
Horizontal major axis equation: $\frac{{\left(xh\right)}^{2}}{{a}^{2}}+\frac{{\left(yk\right)}^{2}}{{b}^{2}}$ Substitute values: $\frac{{\left[x\left({}{1}\right)\right]}^{2}}{{{5}}^{2}}+\frac{{\left(y{2}\right)}^{2}}{{{3}}^{2}}=1$ Simplify: $\frac{{\left(x+1\right)}^{2}}{{5}^{2}}+\frac{{\left(y1\right)}^{2}}{{3}^{2}}=1$ 
Related Links: Math algebra Hyperbola: Standard Equation Hyperbola: Asymptotes Pre Calculus 
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