First slide

Hyperbola in conic sections

Question

$If e is the eccentricity of the hyperbola \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 and θ is the angle between the asymptotes, then \cos \frac{θ}{2} is equal to$

Moderate

Solution

$\begin{array}{l} θ = 2 \tan^{- 1} \frac{b}{a} \Rightarrow \tan \frac{θ}{2} = \frac{b}{a} \\ \Rightarrow \cos \frac{θ}{2} = \frac{a}{\sqrt{a^{2} + b^{2}}} = \frac{1}{\sqrt{1 + \frac{b^{2}}{a^{2}}}} = \frac{1}{e} \end{array}$

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