Eccentricity of Hyperbola
The eccentricity of hyperbola is greater than 1. The eccentricity of hyperbola helps us to understand how closely in circular shape, it is related to a circle. Eccentricity also measures the ovalness of the Hyperbola and eccentricity close to one refers to high degree of ovalness.
Eccentricity is the ratio of the distance of a point on the hyperbola from the focus, and from the directrix. Let us learn more about the definition, formula, and derivation of the eccentricity of hyperbola.
What Is Eccentricity of Hyperbola?
The two important terms to refer to before we talk about eccentricity is the focus and the directrix of the hyperbola. For a conic section, the locus of any point on it is such that the ratio of its distance from the fixed point  focus, and its distance from the fixed line  directrix is a constant value, which is called the eccentricity.
Eccentricity: (e < 1). The ratio of the distance of the focus from the center of the Hyperbola, and the distance of one vertex of the hyperbola from the center of the hyperbola. If the distance of the focus from the center of the hyperbola is 'c' and the distance of the vertex of the hyperbola from the center is 'a', then eccentricity e = c/a.
Formula of Eccentricity of Hyperbola
The eccentricity of a hyperbola is always greater than 1. i.e. e > 1. The eccentricity of a hyperbola can be taken as the ratio of the distance of the point on the hyperbole, from the focus, and its distance from the directrix.
Eccentricity = Distance from Focus/Distance from Directrix
e = c/a
Substituting the value of c we have the following value of eccentricity.
\(e = \sqrt {1 + \dfrac{b^2}{a^2}}\)
Here a is the length of the semimajor axis and b is a constant value.
Derivation of Eccentricity of Hyperbola
The eccentricity of the hyperbola can be derived from the equation of the hyperbola. Let us consider the basic definition of Hyperbola. A hyperbola represents a locus of a point such that the difference of its distances from the two fixed points is a constant value. Let P(x, y) be a point on the hyperbola and the coordinates of the two foci are F(c, 0), and F' (c, 0). From the definition of hyperbola, we have the following expression.
PF'  PF= 2a
\(\sqrt {(x + c)^2 + y^2}  \sqrt{(x  c)^2 + y^2} = 2a \)
\(\sqrt {(x + c)^2 + y^2} =2a + \sqrt{(x  c)^2 + y^2} \)
Squaring this above expression on both sides, we have the below expression.
\((x + c)^2 + y^2 = 4a^2 + (x  c)^2 + y^2 + 4a\sqrt {(x  c)^2 + y^2} \)
This on further simplification we have the following equation of hyperbola.
\(\dfrac{x^2}{a^2}  \dfrac{y^2}{c^2  a^2} = 1 \)
Here we replace \(c^2  a^2 = b^2\) to obtain the equation of hyperbola as \(\dfrac{x^2}{a^2}  \dfrac{y^2}{b^2} = 1\). Further we simplify it to c^{2} = a^{2} + b^{2} , and to c = \(\sqrt {a^2 + b^2}\). This is now substituted in the expression e = c/a, to obtain the below formula of eccentricity of hyperbola.
\(e = \dfrac{\sqrt {a^2 + b^2}}{a}\)
\(e = \sqrt {1 + \dfrac{b^2}{a^2}}\)
Related Topics
The following topics are helpful for a better understanding of eccentricity of hyperbola.
Solved Examples on Eccentricity of Hyperbola

Example 1: Find the eccentricity of hyperbola having the equation x^{2}/49  y^{2}/25 = 1.
Solution:
The given equation of the hyperbola is x^{2}/49  y^{2}/25 = 1. Comparing this with the equation of the hyperbola x^{2}/a^{2}  y^{2}/b^{2} = 1, we have a^{2} = 49, and b^{2 } = 25.
The formula for eccentricity of a hyperbola is as follows.
\(e = \sqrt {1 + \dfrac{b^2}{a^2}}\)
\(e = \sqrt {1 + \dfrac{25}{49}}\)
\(e = \sqrt {\dfrac{49 +25}{49}}\)
\(e = \sqrt {\dfrac{74}{49}}\)
\(e = \dfrac{8.6}{7}\)
e = 1.2Answer: Therefore the eccentricity of hyperbola is 1.2.

Example 2: The eccentricity of a hyperbola is 1.2, and the value of a = 10. Find the equation of the hyperbola.
Solution:
Given e = 1.2, and a = 10. Applying this in the eccentricity formula we have the following expression.
\(e = \sqrt {1 + \dfrac{b^2}{a^2}}\)
\(1.2 = \sqrt {1 + \dfrac{b^2}{10^2}}\)
\(\dfrac{12}{10} = \sqrt {\dfrac{100 + b^2}{100}}\)
\((\dfrac{12}{10})^2 = \dfrac{100 + b^2}{100}\)
\(\dfrac{144}{100} = \dfrac{100 + b^2}{100}\)
144 = 100 + b^{2}
b^{2} = 144  100
b^{2 } = 44
Hence the required equation of the hyperbola is as follows.x^{2}/a^{2}  y^{2}/b^{2} = 1
x^{2}/100  y^{2}/44 = 1Answer: Therefore the required equation of the ellipse is x^{2}/100  y^{2}/44 = 1
FAQs on Eccentricity of Ellipse
What Is Eccentricity of Hyperbola?
The eccentricity of hyperbola is greater than 1. The eccentricity of hyperbola helps us understand how circular it is with reference to a circle. Eccentricity is basically the ratio of the distances of a point on the hyperbola from the focus, and the directrix. If the distance of the focus from the center of the hyperbola is 'c' and the distance of the vertex of the hyperbola from the center is 'a', then eccentricity of hyperbola e = c/a.
Another formula to find the eccentricity of hyperbola is \(e = \sqrt {1  \dfrac{b^2}{a^2}}\).
Why Is Eccentricity of Hyperbola Greater than 1?
The eccentricity of the hyperbola is greater than 1 because it has a shape beyond a circle and an oval shape. The circle has an eccentricity of 0, and an oval has an eccentricity of 1. The eccentricity of a hyperbola is always greater than 1. (e >1).
How to Find the Eccentricity of Hyperbola?
The eccentricity of hyperbola can be found from the formula \(e = \sqrt {1 + \dfrac{b^2}{a^2}}\). For this formula, the values a, and b are the lengths of semimajor axes and semiminor axes of the hyperbola. And these values can be calculated from the equation of the hyperbola. x^{2}/a^{2}  y^{2}/b^{2} = 1
What Is the Use of Eccentricity of Hyperbola?
The eccentricity of hyperbola is used to give a relationship between the semimajor axis and the semiminor axis of the hyperbola. This major axis of the hyperbola is of length 2a units, and the minor axis of the hyperbola is of length 2b units. And the semimajor axis and the semiminor axis are of lengths a units and b units respectively. This can be understood from the formula of the eccentricity of hyperbola. \(e = \sqrt {1 + \dfrac{b^2}{a^2}}\).
visual curriculum