The angle between the asymptotes of the hyper | vedclass

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Question

(A) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.The equations of the asymptotes are given by $\frac{x^2}{a^2} - \frac{y^2

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$2 	an^{-1} \left( \frac{b}{a} ight)$

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(A) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.The equations of the asymptotes are given by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0$,which implies $y = \pm \frac{b}{a} x$.Let the slopes of the asymptotes be $m_1 = \frac{b}{a}$ and $m_2 = -\frac{b}{a}$.The angle $	heta$ between the two lines is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the values,$	an 	heta = \left| \frac{\frac{b}{a} - (-\frac{b}{a})}{1 + (\frac{b}{a})(-\frac{b}{a})} ight| = \left| \frac{\frac{2b}{a}}{1 - \frac{b^2}{a^2}} ight| = \left| \frac{2ab}{a^2 - b^2} ight|$.Alternatively,if $2\alpha$ is the angle between the asymptotes,then $	an \alpha = \frac{b}{a}$,so $\alpha = 	an^{-1}(\frac{b}{a})$.The total angle between the asymptotes is $2\alpha = 2 	an^{-1}(\frac{b}{a})$.

The angle between the asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is equal to:

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