Eccentricity of Hyperbola

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.

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.

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.

The curve represented by a hyperbola is shown as follows,

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.

Example 1: Find the value for eccentricity of hyperbola represented by the following equation: x²/144 – y²/36 = 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 x²/16 – y²/64 = 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.

P1: What would eccentricity value for given hyperbola: x² – y² = 5.

P2: Find the eccentricity for hyperbola represented as x²/81 – y²/121 = 1.

P3: What is the value of eccentricity for hyperbola given by x² – y² = 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, x²/16 – y²/b² = 1. It has eccentricity value as √3. Find the value of b.

Related Articles:

• Conic Sections
• Hyperbola | Equation, Definition & Properties
• Eccentricity Formula of Circle, Parabola, Ellipse, Hyperbola

What shape is defined by a Hyperbola?

The shape defined by a hyperbola can be visualized as a curve obtained by inclined intersection of a plane with two cones (double cone). It is one of the conic sections having eccentricity value greater than 1.

What is meant by eccentricity of a hyperbola?

Eccentricity of a hyperbola gives measure of how flatten or elongated are the branches of hyperbola. Its value is always greater than 1 which means that branches of a hyperbola tend to diverge from each other when extended till infinity.

What is the range of eccentricity of a hyperbola?

Eccentricity of a hyperbola is always greater than 1, i.e. e > 1. Thus, range of eccentricity of a hyperbola comes out to be 1 to infinity.

What are some practical applications of eccentricity of a hyperbola?

Eccentricity of hyperbola can be used to design various hyperbolic mirrors, satellite dishes, bridges, arcs, furnaces, chimneys, etc. due to their hyperbolic shape in nature.

What is formula for eccentricity of a hyperbola in terms of a and b?

Formula for eccentricity is: e = √(1+b²/a²)