If the angle between the asymptotes of a hyperbola is 30° then its eccentricity is
If the angle between the asymptotes of a hyperbola is 30° then its eccentricity is
√5 - √2
√6 - √3
√5 - √3
√6 - √2
The Correct Option is D
Solution and Explanation
The problem provides us with a hyperbola whose angle between the asymptotes is \(30^\circ\). We are required to find the eccentricity of this hyperbola.
The equation of a hyperbola is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
The asymptotes for this hyperbola are:
\[ y = \pm \frac{b}{a} x \]
The angle between the asymptotes is:
\[ 2\theta = 2 \tan^{-1}\left(\frac{b}{a}\right) \Rightarrow \theta = \tan^{-1}\left(\frac{b}{a}\right) \]
Given angle between asymptotes is \(30^\circ\), so:
\[ \theta = 15^\circ \]
Thus,
\[ \frac{b}{a} = \tan 15^\circ = 2 - \sqrt{3} \]
The eccentricity is:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Now,
\[ \left(\frac{b}{a}\right)^2 = (2 - \sqrt{3})^2 = 7 - 4\sqrt{3} \]
So,
\[ e = \sqrt{1 + (7 - 4\sqrt{3})} = \sqrt{8 - 4\sqrt{3}} \]
Simplifying:
\[ \sqrt{8 - 4\sqrt{3}} = \sqrt{6} - \sqrt{2} \]
Thus, the correct answer is \( \sqrt{6} - \sqrt{2} \).
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Concepts Used:
Coordinate Geometry
The central idea in coordinate geometry is to assign numerical coordinates to points in a plane or space, which allows us to describe their positions and relationships using algebraic equations. The most common coordinate system is the Cartesian coordinate system, named after the French mathematician and philosopher René Descartes.