Eccentricity of Hyperbola (original) (raw)

Last Updated : 23 Jul, 2025

Eccentricity of Hyperbola refers to the deviation of the conic section from being circular and closeness towards being oval in shape. In other words, it can be defined as the measure of how flattened or elongated a hyperbola is. It is calculated as the ratio of the distance of a point on the hyperbola from its focus and the directrix. This ratio for a hyperbola is always greater than one, which implies that the two branches of the hyperbola diverge away from each other when extended to infinity. It is denoted by the letter 'e'. It can be used to predict the shape of the hyperbola.

In this article, we will discuss, the eccentricity of a hyperbola, its formula, derivation, solved examples, practice problems and related frequently asked questions.

Table of Content

What is the Eccentricity of Hyperbola?

Eccentricity of a Hyperbola can be defined as the measure of flatness or elongation in its shape. Conic sections are locus of points which follow a certain relation between their distances from a fixed point called focus and a fixed line called directrix. Mathematically, eccentricity is the ratio of the distances from a point on hyperbola to the focus and the directrix. Its value ranges from 1 to infinity, i.e. it is always greater than 1. It is denoted by the symbol 'e'. If the distance between a point on hyperbola and the focus of hyperbola is 'c' and that between the same point and directrix of the hyperbola is 'a', eccentricity 'e' can be written as, e = c/a.

Formula of Eccentricity of Hyperbola

The formula to find eccentricity of a hyperbola is written as follows,

**e = c/a

where,

e = eccentricity of hyperbola,

c = distance between a point on hyperbola and its focus,

a = distance between the same point and directrix of the hyperbola.

For a hyperbola having the following equation,

x2/a2 - y2/b2 = 1

Formula for eccentricity is written as,

e = √(1 + b2/a2)

where,

e = eccentricity of the hyperbola,

a = length of major axis of the hyperbola,

b = length of minor axis of the hyperbola.

From the above formula, we see that eccentricity of a hyperbola depends on the length of major axis and that of minor axis of the hyperbola.

Diagram of Hyperbola

The curve represented by a hyperbola is shown as follows,

Eccentricity-of-Hyperbola-Page-4-(1)

Derivation of Eccentricity of Hyperbola

Let us see how the above mentioned formula for eccentricity can be derived. We know that, eccentricity of a hyperbola is the ratio of the distances from a point on it to the focus and the directrix. By using this definition, we would derive expression for eccentricity of hyperbola for following equation of the hyperbola,

x2/a2 - y2/b2 = 1

Let the coordinates for Focii for the hyperbola be F(c,0) and F'(-c,0). As per definition of hyperbola, for a point on hyperbola P(x,y).

PF' - PF = 2a

By using the formula for distance between two points, we get,

√((x+c)2+y2) - √((x-c)2+y2) = 2a

√((x+c)2+y2) = 2a + √((x-c)2+y2)

On squaring both sides, we get,

(x+c)2 + y2 = 4a2 + (x-c)2 + y2 + 4a√((x-c)2+y2)

On further simplification and rearrangement of terms, we get,

x2/a2 + y2/(c2-a2) = 1

On comparing with the equation of hyperbola, we get,

c2 - a2 = b2

Now, as per definition of hyperbola,

e = c/a

where,

c = distance of a point on hyperbola from focus,

a = distance of the point from the directrix.

If we take the point on hyperbola, as vertex of the hyperbola. Then, we get after substituting the required distances in above equation,

e = √(a2+b2)/a

or, e = √(1 + b2/a2)

Thus, we have derived the expression for eccentricity of the hyperbola in terms of lengths of its major axis and minor axis.

Solved Examples - Eccentricity of Hyperbola

**Example 1: Find the value for eccentricity of hyperbola represented by the following equation: x 2 /144 - y 2 /36 = 1.

**Solution:

Comparing the given equation with standard equation of hyperbola, we get,

a2 = 144 and b2 = 36, giving a = 12 and b = 6.

We know that, eccentricity (e) is given by,

e = √(1 + b2/a2)

⇒ e = √(1 + 36/144) = √(5/4) = √5/2

Thus, eccentricity of given hyperbola comes out to be 1.19 approximately.

**Example 2: Find the value for eccentricity of hyperbola represented as x 2 /16 - y 2 /64 = 1.

**Solution:

For the given hyperbola, we have, a2 = 16 and b2 = 64

Formula for eccentricity of the hyperbola, e = √(1 + b2/a2)

⇒ e = √(1 + 64/16)

⇒ e = √(1 + 4) = √5

⇒ e = √5

Hence, eccentricity of given hyperbola comes out to be √5 which equals to 2.24 approximately.

Practice Problems - Eccentricity of Hyperbola

**P1: What would eccentricity value for given hyperbola: x 2 - y 2 = 5.

**P2: Find the eccentricity for hyperbola represented as x 2 /81 - y 2 /121 = 1.

**P3: What is the value of eccentricity for hyperbola given by x 2 - y 2 = 1.

**P4: A hyperbola lies along X-axis. Length of its major and minor axis is 8 units and 2 units respectively. Find the value of eccentricity for this hyperbola.

**P5: The equation of a hyperbola is written as, x 2 /16 - y 2 /b 2 = 1. It has eccentricity value as √3. Find the value of b.