# The Eccentricity of the Hyperbola X2 − 4y2 = 1 is - Mathematics

MCQ

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

#### Options

• $\frac{\sqrt{3}}{2}$

• $\frac{\sqrt{5}}{2}$

• $\frac{2}{\sqrt{3}}$

• $\frac{2}{\sqrt{5}}$

#### Solution

$\frac{\sqrt{5}}{2}$

The equation of the hyperbola is $x^2 - 4 y^2 = 1$.

This can be rewritten in the following way:

$\frac{x^2}{1} - \frac{y^2}{\frac{1}{4}} = 1$
This is the standard form of a hyperbola, where
$a^2 = 1 \text { and }b^2 = \frac{1}{4}$
The value of eccentricity is calculated in the following way:

$b^2 = a^2 ( e^2 - 1)$

$\Rightarrow \frac{1}{4} = ( e^2 - 1)$

$\Rightarrow e^2 = \frac{5}{4}$

$\Rightarrow e = \frac{\sqrt{5}}{2}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 27 Hyperbola
Q 8 | Page 19