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

Also Read : Equation of the Hyperbola | Graph of a Hyperbola

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 = \(\sqrt{1 + 9/16}\) = \(5\over 4\)

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