The angle between the asymptotes of the hyper | vedclass

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(C) The given equation of the hyperbola is $x^2-3y^2=3$. Dividing by $3$,we get $\frac{x^2}{3}-\frac{y^2}{1}=1$.Here,$a^2=3$ and $b^2=1$,so $a=\sqrt{3

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$\frac{\pi}{3}$

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(C) The given equation of the hyperbola is $x^2-3y^2=3$. Dividing by $3$,we get $\frac{x^2}{3}-\frac{y^2}{1}=1$.Here,$a^2=3$ and $b^2=1$,so $a=\sqrt{3}$ and $b=1$.The equations of the asymptotes are $y = \pm \frac{b}{a}x$,which gives $y = \frac{1}{\sqrt{3}}x$ and $y = -\frac{1}{\sqrt{3}}x$.Let the slopes be $m_1 = \frac{1}{\sqrt{3}}$ and $m_2 = -\frac{1}{\sqrt{3}}$.The angle $	heta$ between the asymptotes 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{1}{\sqrt{3}} - (-\frac{1}{\sqrt{3}})}{1 + (\frac{1}{\sqrt{3}})(-\frac{1}{\sqrt{3}})} ight| = \left| \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} ight| = \left| \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} ight| = \sqrt{3}$.Since $	an 	heta = \sqrt{3}$,we have $	heta = \frac{\pi}{3}$.

The angle between the asymptotes of the hyperbola $x^2-3y^2=3$ is

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