If the angle between the asymptotes of the hy | vedclass

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(C) The given hyperbola is $\frac{x^2}{4a^2} - \frac{y^2}{12} = 1$.Here,$A^2 = 4a^2$ and $B^2 = 12$,so $A = 2a$ and $B = 2\sqrt{3}$.The angle $	heta$

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$\frac{y^2}{12} - \frac{x^2}{36} = 1$

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(C) The given hyperbola is $\frac{x^2}{4a^2} - \frac{y^2}{12} = 1$.Here,$A^2 = 4a^2$ and $B^2 = 12$,so $A = 2a$ and $B = 2\sqrt{3}$.The angle $	heta$ between the asymptotes is given by $2 	an^{-1}(\frac{B}{A}) = \frac{\pi}{3}$.Thus,$	an^{-1}(\frac{B}{A}) = \frac{\pi}{6}$,which implies $\frac{B}{A} = 	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$.Substituting the values,$\frac{2\sqrt{3}}{2a} = \frac{1}{\sqrt{3}}$ $\Rightarrow \frac{\sqrt{3}}{a} = \frac{1}{\sqrt{3}}$ $\Rightarrow a = 3$.Then $A^2 = 4(3)^2 = 36$.The equation of the hyperbola is $\frac{x^2}{36} - \frac{y^2}{12} = 1$.The conjugate hyperbola is $\frac{x^2}{36} - \frac{y^2}{12} = -1$,which is $\frac{y^2}{12} - \frac{x^2}{36} = 1$.

If the angle between the asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{3} = 4$ is $\frac{\pi}{3}$,then its conjugate hyperbola is:

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