Eccentricity Solved Examples

Here are some solved examples on the eccentricity of different conic sections.

Example 1: Calculate the Eccentricity for an Ellipse with a semi-major axis of 8 units and a distance from the centre to a focus of 5 units.

Solution:

Formula to calculate Eccentricity (e) for an Ellipse is:

e = c/a

Given values are:

Semi-major axis (a) is 8 units

Distance from the centre to a focus (c) is 5 units

Eccentricity formula is:

e = 5/8

Now, calculate the value of

e = 0.625

So, Eccentricity of the Ellipse is = 0.625

Example 2: Find the Eccentricity of the Ellipse for the given equation 16x² + 25y² = 400

Solution:

Given equation is: 16x² + 25y² = 400

General equation of Ellipse is

x² /a² +y²/b²= 1

To make it in standard form, divide both sides by 400

x² /5² +y²/4²= 1

So, value of semi-major axis length a = 5 and semi-minor axis length b = 4

From the formula of the Eccentricity of an Ellipse,

e = √(a²-b²)/a²

⇒ e = √(5²-4²)/5²

⇒ e = √9/25

⇒ e = 3/5

Therefore, for given equation, the Eccentricity is 3/5.

Example 3: Find the Eccentricity of the conic section (x²/25) + (y²/16) = 1.

Solution:

Given equation is: (x²/25) + (y²/16) = 1

It can also be written as (x²/5²) + (y²/4²) =1

Given conic section is an Ellipse in the form of (x²/a²) + (y²/b²) = 1

Here, a = 5 and b = 4.

We know that c² = a²-b²

⇒ c² = 5² – 4²

⇒ c² = 25 -16 = 9

Hence, c = √9 = 3.

Formula for Eccentricity is:

e = c/a

Now, put the values of c and a, we get

e = 3/5

Hence, the Eccentricity of the given conic section (x²/25) + (y²/16) = 1 is 3/5.

Example 4: Find the Eccentricity of the hyperbola (x²/36) – (y²/9) = 1.

Solution:

Given equation is: (x²/36) – (y²/9) = 1

Given Hyperbola can be written as

(x²/6²) – (y²/3²) = 1

Given conic section is an Hyperbola in the form of (x²/a²) – (y²/b²) = 1

So, axis of Hyperbola is x-axis

Now, by comparing the equation, we get a = 6 and b = 3

Eccentricity formula for Hyperbola is e = √[1+(b²/a²)]

Now, put the values in the formula, we get

e = √[1+(3²/6²)]

⇒ e = √[1+(9/36)]

⇒ e = √(45/16)

⇒ e = 3/4√5

Therefore, 3/4√5 is the Eccentricity of the Hyperbolic equation (x²/36) – (y²/9) = 1.

Eccentricity is a non-negative real number that describes the shape of a conic section. It measures how much a conic section deviates from being circular. Generally, eccentricity measures the degree to which a conic section differs from a uniform circular shape.

Let’s discuss Eccentricity formula for circle, parabola, ellipse, and hyperbola, along with examples.