# Eccentricity of Hyperbola – Formula and Examples

Here you will learn what is the eccentricity of hyperbola formula and how to find eccentricity with examples.

Let’s begin –

## Eccentricity of Hyperbola Formula

#### (i) For the hyperbola $$x^2\over a^2$$ – $$y^2\over b^2$$ = 1

Eccentricity (e) = $$\sqrt{1 + {b^2\over a^2}}$$

#### (ii) For the hyperbola -$$x^2\over a^2$$ + $$y^2\over b^2$$ = 1

Eccentricity (e) = $$\sqrt{1 + {a^2\over b^2}}$$

Example : For the given ellipses, find the eccentricity.

(i)  $$16x^2 – 9y^2$$ = 144

(ii)  $$9x^2 – 16y^2 – 18x + 32y – 151$$ = 0

Solution :

(i)  We have,

$$16x^2 – 9y^2$$ = 144 $$\implies$$ $$x^2\over 9$$ – $$y^2\over 16$$ = 1,

This is of the form $$x^2\over a^2$$ – $$y^2\over b^2$$ = 1

where $$a^2$$ = 9 and $$b^2$$ = 16 i.e. a = 3 and b = 4

Eccentricity (e) = $$\sqrt{1 + {b^2\over a^2}}$$

$$\implies$$ e = $$\sqrt{1 + 16/9}$$ = $$5\over 3$$

(ii) We have,

$$9x^2 – 16y^2 – 18x + 32y – 151$$ = 0

$$\implies$$ $$9(x^2 – 2x)$$ – $$16(y^2 – 2y)$$ = 151

$$\implies$$  $$9(x^2 – 2x + 1)$$ – $$16(y^2 – 2y + 1)$$ = 144

$$\implies$$  $$9(x – 1)^2$$ – $$16(y – 1)^2$$ = 144

$$\implies$$  $$(x – 1)^2\over 16$$ – $$(y – 1)^2\over 9$$ = 1

Here, a = 4 and b = 3

Eccentricity (e) = $$\sqrt{1 + {b^2\over a^2}}$$

$$\implies$$ e = $$\sqrt{1 + 9/16}$$ = $$5\over 4$$