# Eccentricity of Hyperbola

The eccentricity of a hyperbola is defined as the ratio of the distance from any point on the hyperbola to its focus and the perpendicular distance from the same point to the nearest directrix.

## Formula

Mathematically, the eccentricity, denoted by e, in general form, is written as:

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${e=\dfrac{c}{a}}$

Here,

• c = distance from the center to the focus
• a = distance from the center to the vertex

The eccentricity of the hyperbola centered at the origin ${\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1}$, is given by:

${e=\dfrac{\sqrt{a^{2}+b^{2}}}{a}}$

Here,

• a = distance from the center of the hyperbola to a vertex along the transverse axis
• b = distance from the center to a point on the conjugate axis

The eccentricity of the hyperbola is always greater than 1 (e > 1), which indicates that the distance from any point on the hyperbola to its focus is always greater than the distance from that point to the corresponding directrix.

## Derivation

Let us consider a point P on the hyperbola with coordinates (x, y)

From the definition of hyperbola, we know the difference in the distances between a set of points present in a plane and two foci is always constant.

If the foci of the hyperbola are F (c, 0) and F’ (-c, 0), the difference in the distance from P to the foci F and F’ is 2a

PF’ – PF = 2a …..(i)

PF’ = ${\sqrt{\left( x+c\right) ^{2}+y^{2}}}$ and PF = ${\sqrt{\left( x-c\right) ^{2}+y^{2}}}$ …..(ii)

Now, substituting the values of (ii) in the equation (i), we get

${\sqrt{\left( x+c\right) ^{2}+y^{2}}-\sqrt{\left( x-c\right) ^{2}+y^{2}}=2a}$

⇒ ${\sqrt{\left( x+c\right) ^{2}+y^{2}}=2a+\sqrt{\left( x-c\right) ^{2}+y^{2}}}$

Squaring on both sides,

⇒ ${\left( x+c\right) ^{2}+y^{2}=4a^{2}+\left( x-c\right) ^{2}+y^{2}+4a\sqrt{\left( x-c\right) ^{2}+y^{2}}}$

⇒ ${x^{2}+c^{2}+2cx+y^{2}=4a^{2}+x^{2}+c^{2}-2cx+y^{2}+4a\sqrt{\left( x-c\right) ^{2}+y^{2}}}$

Simplifying,

⇒ ${2cx=4a^{2}-2cx+4a\sqrt{\left( x-c\right) ^{2}+y^{2}}}$

⇒ ${4cx-4a^{2}=4a\sqrt{\left( x-c\right) ^{2}+y^{2}}}$

⇒ ${cx-a^{2}=a\sqrt{\left( x-c\right) ^{2}+y^{2}}}$

Again, on squaring both sides,

⇒ ${\left( cx-a^{2}\right) ^{2}=a^{2}\left\{ \left( x-c\right) ^{2}+y^{2}\right\}}$

c2x2 + a4 – 2a2cx = a2(x2 + c2 – 2cx + y2)

Simplifying further,

c2x2 + a4 – 2a2cx = a2x2 + a2c2 – 2a2cx + a2y2

c2x2 + a4 = a2x2 + a2c2 + a2y2

c2x2 – a2x2 – a2y2 = a2c2 – a4

x2(c2 – a2) – a2y2 = a2(c2 – a2)

Dividing both sides by a2(c2 – a2),

⇒ ${\dfrac{x^{2}\left( c^{2}-a^{2}\right) }{a^{2}\left( c^{2}-a^{2}\right) }-\dfrac{a^{2}y^{2}}{a^{2}\left( c^{2}-a^{2}\right) }=\dfrac{a^{2}\left( c^{2}-a^{2}\right) }{a^{2}\left( c^{2}-a^{2}\right) }}$

⇒ ${\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{c^{2}-a^{2}}=1}$ …..(ii)

Since we know the standard equation of the hyperbola is ${\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1}$ …..(iii)

Now, on comparing the equations (ii) and (iii),

c2 – a2 = b2

c2 = a2 + b2

c = ${\sqrt{a^{2}+b^{2}}}$ …..(iv)

Now, from the definition of the eccentricity, we get ${e=\dfrac{c}{a}}$ …..(v)

Now, substituting the value of c from (iv) in the equation (iii), we get

${e=\dfrac{\sqrt{a^{2}+b^{2}}}{a}}$, which is the formula to calculate eccentricity in terms of lengths of its major axis and minor axis.

## Solved Examples

If the hyperbola is given by the equation ${\dfrac{x^{2}}{49}-\dfrac{y^{2}}{36}=1}$, find its eccentricity.

Solution:

Given, ${\dfrac{x^{2}}{49}-\dfrac{y^{2}}{36}=1}$
As we know, the standard equation of the hyperbola is ${\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1}$
On comparing both equations, we get
a = 7 and b = 6
Now, the eccentricity is ${e=\dfrac{\sqrt{a^{2}+b^{2}}}{a}}$
⇒ ${e=\dfrac{\sqrt{7^{2}+6^{2}}}{7}}$
⇒ ${e=\dfrac{\sqrt{49+36}}{7}}$
⇒ ${e=\dfrac{\sqrt{85}}{7}}$
Thus, the eccentricity of the hyperbola is ${e=\dfrac{\sqrt{85}}{7}}$

If the hyperbola is given by the equation ${\dfrac{y^{2}}{9}-\dfrac{x^{2}}{16}=1}$, find its eccentricity.

Solution:

Given, ${\dfrac{y^{2}}{9}-\dfrac{x^{2}}{16}=1}$
As we know, the standard equation of the hyperbola is ${\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1}$
On comparing both equations, we get
a = 3 and b = 4
Now, the eccentricity is ${e=\dfrac{\sqrt{a^{2}+b^{2}}}{a}}$
⇒ ${e=\dfrac{\sqrt{3^{2}+4^{2}}}{3}}$
⇒ ${e=\dfrac{\sqrt{9+16}}{3}}$
⇒ ${e=\dfrac{\sqrt{25}}{3}}$
⇒ ${e=\dfrac{5}{3}}$
Thus, the eccentricity of the hyperbola is ${e=\dfrac{5}{3}}$

If the eccentricity of a horizontal hyperbola centered at (0, 0) is 2, and the semi-major axis is 8 units, find the equation of the hyperbola.

Solution:

Given, e = 2 and a = 8
⇒ e = 2 and a2 = 64
As we know, ${e=\dfrac{\sqrt{a^{2}+b^{2}}}{a}}$
⇒ ${2=\dfrac{\sqrt{64+b^{2}}}{8}}$
⇒ ${2\times 8=\sqrt{64+b^{2}}}$
⇒ ${16=\sqrt{64+b^{2}}}$
⇒ ${\left( 16\right) ^{2}=\left( \sqrt{64+b^{2}}\right) ^{2}}$
⇒ ${256=64+b^{2}}$
⇒ ${b^{2}=256-64}$
⇒ ${b^{2}=192}$
Since the hyperbola is centered at (0, 0) and its transverse axis is on the x-axis.
Thus, the equation of the horizontal hyperbola is ${\dfrac{x^{2}}{64}-\dfrac{y^{2}}{192}=1}$

Problem – Finding the eccentricity of a CONJUGATE HYPERBOLA

If the eccentricity of a hyperbola is ${\dfrac{5}{3}}$, find the eccentricity of its conjugate hyperbola.

Solution:

The sum of the square of the reciprocal of eccentricities of a hyperbola and its conjugate is 1
Given, e = ${\dfrac{5}{3}}$
If the eccentricity of the conjugate hyperbola is e1, then
${\dfrac{1}{e^{2}}+\dfrac{1}{e_{1}^{2}}=1}$
⇒ ${\left( \dfrac{3}{5}\right) ^{2}+\dfrac{1}{e_{1}^{2}}=1}$
⇒ ${\dfrac{9}{25}+\dfrac{1}{e_{1}^{2}}=1}$
⇒ ${\dfrac{1}{e_{1}^{2}}=1-\dfrac{9}{25}}$
⇒ ${\dfrac{1}{e_{1}^{2}}=\dfrac{25-9}{25}}$
⇒ ${\dfrac{1}{e_{1}^{2}}=\dfrac{16}{25}}$
⇒ ${e_{1}^{2}=\dfrac{25}{16}}$
⇒ ${e=\dfrac{5}{4}}$
Thus, the eccentricity of the conjugate hyperbola is ${e=\dfrac{5}{4}}$