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
(a) $\sqrt{2}$
(b) $\sqrt{3}$
(c) $2\sqrt{3}$
(d) $3\sqrt{2}$

Answer 1

(a) $\sqrt{2}$
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:
$$\begin{gathered} \frac{4x^2}{a^2}=t^2+\frac{1}{t^2}+2 \\ \Rightarrow \frac{4x^2}{a^2}-2=t^2+\frac{1}{t^2}...\left(1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Also},\frac{4y^2}{a^2}=t^2+\frac{1}{t^2}-2 \\ \Rightarrow \frac{4y^2}{a^2}+2=t^2+\frac{1}{t^2}...\left(2\right) \end{gathered}$$

From (1) and (2), we get:
$$\begin{gathered} \frac{4x^2}{a^2}-\frac{4y^2}{a^2}=4 \\ \Rightarrow \frac{x^2}{a^2}-\frac{y^2}{a^2}=1 \end{gathered}$$

This is the standard equation of a hyperbola, where $a^2=b^2$.

Eccentricity of the hyperbola, $e=\frac{\sqrt{a^2+a^2}}{a}=\sqrt{2}$