Eccentricity of Conic Sections

Introduction to eccentricity of conic sections:
A conic is the locus of a point in a plane such that the ratio of its distance from a fixed point in the plane to its distance from a fixed line in the plane is a constant. Here, the fixed point is called the focus, the fixed line is called the directrix and the constant ratio is called the eccentricity of the conic. It is denoted by ‘e’.

(i)     If e = 1, the conic is called the parabola.

(ii)    If e < 1, the conic is called the ellipse

(iii)   If e > 1, the conic is called the hyperbola

(iv)   If e = 0, the conic is called the circle.

Now, let us see few problems on this topic eccentricity of conic sections.

Example Problems on Eccentricity of Conic Sections

Ex 1. Find the eccentricity of conic  x2/ 36 + y 2 / 64 = 1

Soln: Given: `x^ 2 / 64` +` y ^2 / 36 ` = 1 ………… (1)

Here (1) represents an ellipse. To get the eccentricity, are follow this formula: e 2 = 1 – `b ^2 /a^ 2` .

Here a = `sqrt 64` , b = `sqrt 36`     [This we get from the standard form of an ellipse

`x^ 2 / a^ 2` +` y ^2 / b ^2` = 1        Here a >b]

a = 8, b = 6

Therefore, e 2 = 1 – `6 ^2 / 8^ 2`

= 1 – `36 / 64`    = 1 – `9 / 16`    = 16 – `9 / 16`    = `7 / 16`

Therefore e = `sqrt7 / 16`    [Eccentricity is always positive]

Therefore e = `sqrt7 / 4`

Ex 2: Find the eccentricity of the conic:` x ^2 / 49` +` y ^2 / 225` = 1

Sol: Given: ` x^2 / 49` +` y ^2 / 225` = 1

Here a = 7, b = 15,   = a < b

Therefore e 2 = 1 – `a^ 2 / 15^ 2` = 1 – ` 49 / 225`

= `(225 ** 49) / 225`   = `176 / 225`

Therefore  e = `sqrt (176 / 225)` = `sqrt ((16 * 11) / 225)`    =  `(4sqrt 11) / 15`

Ex 3: Find the eccentricity of the conic: ` x^ 2 / 625 **y ^2 / 144`   = 1

Soln: Given:` x^ 2 / 625 ** y ^2 / 144` = 1 ………… (1)

Here (1) represents a hyperbola

The eccentricity is given by e 2 = 1 + `(b^2/ a^2)`

= 1 + `144 / 625`      =` 769 / 625`

Therefore e = `sqrt 769 / 25` .

Ex 4: Find the eccentricity of the conic` x ^2/ 25 ** y ^2 / 36` = 1

Soln: Given:` x ^2 / 25** y ^2 / 36` = 1 ………….. (1)

This is a hyperbola.  a = 5, b = 6

Therefore e 2 = 1 + `b ^2 / a^ 2` = 1 + `36 / 25` = `61 / 25`

Therefore e = `(sqrt 61) / 5`

Note: Here the fraction  b 2 /  a 2 will change as whether a < b or a >b.

Practice Problems on Eccentricity of Conic Sections

Find the eccentricity of the following conics:
A) `x^2/16 + y^2/9` = 1

[Ans: e = `sqrt 7 / 4` ]

B)    ` x ^2 / 25 + y ^2 / 64` = 1

[Ans: e = `sqrt 39 / 8` ]

2. Find the eccentricity of the following conics:

A)    `x ^2 / 64 ** y ^2 / 36` = 1

[Ans: e = `5 / 2` ]

B)    ` x ^2 / 81** y ^2 / 144` = 1

[Ans: `5/3` ]