If the angle between the asymptotes of the hy | vedclass

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Question

(A) The given hyperbola is $\frac{x^2}{4a^2} - \frac{y^2}{12} = 1$. Comparing with $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$,we have $A^2 = 4a^2$ and $B

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

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(A) The given hyperbola is $\frac{x^2}{4a^2} - \frac{y^2}{12} = 1$. Comparing with $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$,we have $A^2 = 4a^2$ and $B^2 = 12$,so $B = 2\sqrt{3}$. The angle between the asymptotes is $2 	an^{-1}(\frac{B}{A}) = \frac{\pi}{3}$. Thus,$	an^{-1}(\frac{B}{A}) = \frac{\pi}{6}$,which implies $\frac{B}{A} = 	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$. Substituting $B = 2\sqrt{3}$,we get $\frac{2\sqrt{3}}{A} = \frac{1}{\sqrt{3}}$,so $A = 6$. Thus $A^2 = 36$. The hyperbola is $\frac{x^2}{36} - \frac{y^2}{12} = 1$. The conjugate hyperbola 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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