Question

The eccentricity the hyperbola \[x = \frac{a}{2}\left( t + \frac{1}{t} \right), y = \frac{a}{2}\left( t - \frac{1}{t} \right)\] is

Options

• \[\sqrt{2}\]

• \[\sqrt{3}\]

• \[2\sqrt{3}\]

• \[3\sqrt{2}\]

Answer 1

\[\sqrt{2}\]

We eliminate t in \[x = \frac{a}{2}\left( t + \frac{1}{t} \right), y = \frac{a}{2}\left( t - \frac{1}{t} \right)\].

On squaring both the sides in the equations  \[x = \frac{a}{2}\left( t + \frac{1}{t} \right), y = \frac{a}{2}\left( t - \frac{1}{t} \right)\] , we get: 

\[\frac{4 x^2}{a^2} = t^2 + \frac{1}{t^2} + 2\]

\[ \Rightarrow \frac{4 x^2}{a^2} - 2 = t^2 + \frac{1}{t^2} . . . \left( 1 \right)\]

\[\text { Also }, \frac{4 y^2}{a^2} = t^2 + \frac{1}{t^2} - 2\]

\[ \Rightarrow \frac{4 y^2}{a^2} + 2 = t^2 + \frac{1}{t^2} . . . \left( 2 \right)\]

From (1) and (2), we get:

\[\frac{4 x^2}{a^2} - \frac{4 y^2}{a^2} = 4\]

\[ \Rightarrow \frac{x^2}{a^2} - \frac{y^2}{a^2} = 1\]

This is the standard equation of a hyperbola, where \[a^2 = b^2\].

Eccentricity of the hyperbola,

\[e = \frac{\sqrt{a^2 + a^2}}{a} = \sqrt{2}\]