Question

The eccentricity of the hyperbola x² − 4y² = 1 is

(a) $\frac{\sqrt{3}}{2}$

(b) $\frac{\sqrt{5}}{2}$

(c) $\frac{2}{\sqrt{3}}$

(d) $\frac{2}{\sqrt{5}}$

Answer 1

(b) $\frac{\sqrt{5}}{2}$

The equation of the hyperbola is $x^2-4y^2=1$.
This can be rewritten in the following way:
$\frac{x^2}{1}-\frac{y^2}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}=1$
This is the standard form of a hyperbola, where $a^2=1\mathrm{and}{b}^2=\frac{1}{4}$.
The value of eccentricity is calculated in the following way:
$$\begin{gathered} b^2=a^2(e^2-1) \\ \Rightarrow \frac{1}{4}=(e^2-1) \\ \Rightarrow e^2=\frac{5}{4} \\ \Rightarrow e=\frac{\sqrt{5}}{2} \end{gathered}$$