Question

Eccentricity of hyperbola \(\frac{{{{\rm{x}}^2}}}{{16}} - \frac{{{{\rm{y}}^2}}}{9} = 1\)
1. 5/3
2. 3/5
3. 3/4
4. 5/4
5. None of these

Answer 1

Correct Answer - Option 4 : 5/4

Concept:

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

• Equation of hyperbola: \(\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} - \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1{\rm{\;}},{\rm{\;a}} > {\rm{b}}\)
• Eccentricity of hyperbola = \(\sqrt {1 + \left( {\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} \right)} \)

Calculation:

Given: Equation of hyperbola \(\frac{{{{\rm{x}}^2}}}{{16}} - \frac{{{{\rm{y}}^2}}}{9} = 1\)

Compare given equation with standard equation of ellipse \(\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} - \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1{\rm{\;}},{\rm{\;a}} > {\rm{b}}\)

We get,

a² = 16 and b² = 9

Eccentricity of hyperbola = \(\sqrt {1 + \left( {\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} \right)} \Rightarrow \sqrt {\frac{{({{\rm{a}}^2} + {{\rm{b}}^2})}}{{{{\rm{a}}^2}}}} \)

\( = \sqrt {\frac{{16 + 9}}{{16}}} \)

\(= \sqrt {\frac{{25}}{{16}}} \)

\(= \frac{5}{4}\)

Hence, option (4) is correct.