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

MCQ

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

Options

  • \[\frac{\sqrt{3}}{2}\]

  • \[\frac{\sqrt{5}}{2}\]

  • \[\frac{2}{\sqrt{3}}\]

  • \[\frac{2}{\sqrt{5}}\]

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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?
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APPEARS IN

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