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# Angle between the asymptotes of a hyperbola is $30^\circ$ then $e =$A. $\sqrt 6$B. $\sqrt 2$C. $\sqrt 6 - \sqrt 2$D. $\sqrt 6 - \sqrt 3$

Last updated date: 13th Jun 2024
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Hint: Here, we will use the given angles and the asymptotes of a hyperbola to form a quadratic equation and solve it further to find the value of the variable. Then use this value we will find the required value of eccentricity of the given hyperbola. A hyperbola is an open curve with two branches, the intersection of a plane with both halves of a double cone.

Formula Used:
1. Equation of a hyperbola is $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$
2. Quadratic formula can be written as $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
3. Eccentricity of hyperbola, $e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}}$

Equation of a hyperbola is $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$
The asymptotes of a hyperbola are $y = \pm \dfrac{b}{a}x$
According to the question, angle between the asymptotes of a hyperbola is $30^\circ$
Hence, we get,
$\tan 30^\circ = \dfrac{{\dfrac{b}{a} + \dfrac{b}{a}}}{{1 - \dfrac{{{b^2}}}{{{a^2}}}}} = \dfrac{{\dfrac{{2b}}{a}}}{{\dfrac{{{a^2} - {b^2}}}{{{a^2}}}}} = \dfrac{{2ab}}{{{a^2} - {b^2}}}$
Using the trigonometric table, we know that $\tan 30^\circ = \dfrac{1}{{\sqrt 3 }}$
$\Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{{2ab}}{{{a^2} - {b^2}}}$
Dividing numerator and denominator by ${a^2}$, we get,
$\Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{{\dfrac{{2ab}}{{{a^2}}}}}{{1 - \dfrac{{{b^2}}}{{{a^2}}}}} = \dfrac{{2\left( {\dfrac{b}{a}} \right)}}{{1 - {{\left( {\dfrac{b}{a}} \right)}^2}}}$
Now, substitute $\left( {\dfrac{b}{a}} \right) = x$
$\Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{{2x}}{{1 - {x^2}}}$
Now, cross multiplying, we get
$\Rightarrow 1 - {x^2} = 2\sqrt 3 x$
This can be written as:
$\Rightarrow {x^2} + 2\sqrt 3 x - 1 = 0$
Comparing this quadratic equation with $a{x^2} + bx + c = 0$, we have $a = 1$, $b = 2\sqrt 3$and $c = - 1$
Hence, substituting these in quadratic formula, we get,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$\Rightarrow x = \dfrac{{ - 2\sqrt 3 \pm \sqrt {{{\left( {2\sqrt 3 } \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}}$
Solving further, we get,
$\Rightarrow x = \dfrac{{ - 2\sqrt 3 \pm \sqrt {12 + 4} }}{2} = \dfrac{{ - 2\sqrt 3 \pm \sqrt {16} }}{2} = \dfrac{{ - 2\sqrt 3 \pm 4}}{2}$
Dividing the numerator and denominator by 2, we get
$\Rightarrow x = - \sqrt 3 \pm 2$
But, $x$can’t be negative
Therefore, we get,
$\Rightarrow x = 2 - \sqrt 3$
Hence eccentricity, $e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} = \sqrt {1 + {x^2}}$
Using the value $x = 2 - \sqrt 3$, we get,
$e = \sqrt {1 + {{\left( {2 - \sqrt 3 } \right)}^2}}$
Now, using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
$\Rightarrow e = \sqrt {1 + 4 - 4\sqrt 3 + 3} = \sqrt {8 - 4\sqrt 3 }$
$\Rightarrow e = \sqrt {1 + 4 - 4\sqrt 3 + 3} = 2\sqrt {2 - \sqrt 3 }$

Hence, we get,
$\Rightarrow e = \sqrt 6 - \sqrt 2$
Hence, option C is the correct answer.

Note:
The plane does not have to be parallel to the axis of the cone for the hyperbola to be symmetrical. A hyperbola will be symmetrical in any case and every hyperbola has two asymptotes. A hyperbola with a horizontal transverse axis and center at $\left( {h,k} \right)$ has one asymptote with equation $y = k + \left( {x - h} \right)$and the other with equation $y = k - \left( {x - h} \right)$. Also, the eccentricity of a circle is zero. The eccentricity of an ellipse which is not a circle is greater than zero but less than 1. The eccentricity of a parabola is 1 and the eccentricity of a hyperbola is greater than 1.