The eccentricity of the hyperbola 16x² – 9y² = 1 is

Concept:

Hyperbola: The locus of a point which moves such that its distance from a fixed point is greater than its distance from a fixed straight line. (Eccentricity = e > 1)

Calculation:

Given:

16x² – 9y² = 1

\( \Rightarrow \frac{{{\rm{\;}}{{\rm{x}}^2}}}{{\frac{1}{{16}}}} - \frac{{{{\rm{y}}^2}}}{{\frac{1}{9}}} = 1\)

Compare with \(\frac{{{\rm{\;}}{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} - \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1\)

∴ a² = 1/16 and b² = 1/9

Eccentricity = \(\sqrt {1 + {\rm{\;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} = \;\sqrt {1 + \;\frac{{\left( {\frac{1}{9}} \right)}}{{\left( {\frac{1}{{16}}} \right)}}} = \;\sqrt {1 + \;\frac{{16}}{9}} = \;\sqrt {\frac{{25}}{9}} = \;\frac{5}{3}\)