Eccentricity of Conic Sections

The eccentricity of a conic section is a numerical value that measures its shape and how much it deviates from being circular. Denoted by "e" or "ε (epsilon)", it describes how stretched or flattened the curve appears.

Eccentricity of an Ellipse

Because every ellipse satisfies c < a, its eccentricity is always strictly less than 1, falling in the range (0 ≤ e < 1).

• When the eccentricity is close to 0, the ellipse closely resembles a circle. When e = 0, the ellipse becomes a circle. In this case, both foci coincide at the center (c = 0), which also means both semi-axes are equal.
• When the eccentricity approaches 1, the foci move farther from the center, causing the ellipse to become longer and more flattened.

The eccentricity can never equal 1; this would mean the foci coincide with the vertices, which cannot happen in an ellipse. However, the closer e is to 1, the flatter the ellipse appears, eventually resembling a straight line segment.

Worked Examples

Below are step-by-step examples on how to find the eccentricity of an ellipse.

Example 1

Find the eccentricity of the following ellipse: \(\dfrac{x^2}{64}+\dfrac{y^2}{9}=1\)

Solution: We can find the eccentricity either by calculating the focal distance first or by using the formula with only the semi-axes. We will use the first approach.

First, from the standard equation we see that \(a^2=64 → a=8\) and \(b^2=9,\) meaning this is a horizontal ellipse. Now we find the focal distance using the relation:

$$c=\sqrt{a^2-b^2}$$

$$c=\sqrt{64-9}$$

$$c=\sqrt{55}$$

$$c≈7.42$$

Next, we apply the eccentricity formula:

$$e=\dfrac{c}{a}=\dfrac{\sqrt{55}}{8}$$

$$e≈0.93$$

Example 2

Determine the eccentricity of the following ellipse: \(\dfrac{y^2}{25}+\dfrac{x^2}{16}=1\)

Solution: From the equation, this is a vertical ellipse with a² = 25 and b² = 16. We will compute the eccentricity using the alternate formula:

$$e= \sqrt{1-\dfrac{b^2}{a^2}}$$

$$e=\sqrt{1-\dfrac{16}{25}}$$

$$e=\sqrt{\dfrac{9}{25}}$$

$$e=\dfrac{3}{5}=0.6$$

Example 3

Find the eccentricity of the ellipse \(\dfrac{(x-1)^2}{15}+\dfrac{y^2}{14}=1\)

Solution: This is a horizontal ellipse with a² = 15 and b² = 14. We use the second formula since it requires fewer steps:

$$e=\sqrt{1-\dfrac{b^2}{a^2}}$$

$$e=\sqrt{1-\dfrac{14}{15}}$$

$$e=\sqrt{\dfrac{1}{15}}$$

$$e≈0.26$$

Example 4

Calculate the eccentricity of \(2x^2+4x+7y^2-28y-40=0\)

Solution: The equation is given in general form rather than standard form. We must first manipulate it algebraically to find the values of a and b.

By completing the square, we can rewrite the equation in standard form:

$$\dfrac{(x+1)^2}{35}+\dfrac{(y-2)^2}{10}=1$$

From this equation, we identify \(a^2=35\) and \(b^2=10,\) confirming that this is an ellipse with a horizontal major axis. We now find its eccentricity:

$$e=\sqrt{1-\dfrac{b^2}{a^2}}$$

$$e=\sqrt{1-\dfrac{10}{35}}$$

$$e=\sqrt{\dfrac{5}{7}}$$

$$e≈0.85$$

Eccentricity of a Parabola

Every parabola has an eccentricity equal to 1, regardless of its shape or orientation.

Worked Examples

Determine the eccentricity of the following parabolas:

1. \(3y+4=(x-2)^2-5\)
2. \(x+5=-2(y-1)^2+3\)
3. \(y=-\dfrac{1}{2}(x+3)^2+4\)

Solution: Every parabola has an eccentricity of e = 1.

Eccentricity of a Hyperbola

Just as with the ellipse, it can also be calculated using the distance from the center to a focus (c) and the distance from the center to a vertex (a):

e = c / a

For any hyperbola, the eccentricity is always strictly greater than 1, because the foci are farther from the center than the vertices (c > a). As the eccentricity increases, the branches open wider and flatten out. As e approaches 1, the branches become narrower and tightly curved around the vertices.

Worked Examples

Find the eccentricity of the following hyperbolas.

1. \((x-3)^2-\dfrac{(y+2)^2}{4}=1\)
2. \(\dfrac{(y-1)^2}{9}-\dfrac{(x+5)^2}{16}=1\)
3. \(\dfrac{x^2}{7}-\dfrac{(y-4)^2}{8}=1\)

Solutions

1) From \((x-3)^2-\dfrac{(y+2)^2}{4}=1,\) we have \(a^2=1\) and \(b^2=4,\) since the positive term determines the value of a.

Using these values, we calculate the eccentricity:

$$e=\sqrt{1+\dfrac{b^2}{a^2}}=\sqrt{1+\dfrac{4}{1}}=\sqrt{5}≈2.24$$

2) For \(\dfrac{(y-1)^2}{9}-\dfrac{(x+5)^2}{16}=1,\) the positive term contains the y variable, so \(a^2=9\) and \(b^2=16.\)

We compute the eccentricity:

$$e=\sqrt{1+\dfrac{b^2}{a^2}}=\sqrt{1+\dfrac{16}{9}}=\sqrt{\dfrac{25}{9}}=\dfrac{5}{3}\approx 1.67$$

3) For \(\dfrac{x^2}{7}-\dfrac{(y-4)^2}{8}=1,\) the positive term is \(\dfrac{x^2}{7},\) giving \(a^2=7\) and \(b^2=8.\)

Now we evaluate the eccentricity:

$$e=\sqrt{1+\dfrac{b^2}{a^2}}=\sqrt{1+\dfrac{8}{7}}=\sqrt{\dfrac{15}{7}}\approx 1.46$$

As expected for hyperbolas, all calculated eccentricities are strictly greater than 1.

Summary of Formulas

Conic Section	Eccentricity	Formulas
Circle	e = 0	-
Ellipse	0 ≤ e < 1	e = c / a \(e=\sqrt{1-\dfrac{b^2}{a^2}}\)
Parabola	e = 1	-
Hyperbola	e > 1	e = c / a \(e=\sqrt{1+\dfrac{b^2}{a^2}}\)