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(C) The given equation of the hyperbola is $7x^2 - 9y^2 = 63$. Dividing by $63$,we get $\frac{x^2}{9} - \frac{y^2}{7} = 1$. Here,$a^2 = 9$ and $b^2 =

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$\frac{1}{8}$

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(C) The given equation of the hyperbola is $7x^2 - 9y^2 = 63$. Dividing by $63$,we get $\frac{x^2}{9} - \frac{y^2}{7} = 1$. Here,$a^2 = 9$ and $b^2 = 7$,so $a = 3$ and $b = \sqrt{7}$. The equations of the asymptotes are $y = \pm \frac{b}{a}x$,which are $y = \frac{\sqrt{7}}{3}x$ and $y = -\frac{\sqrt{7}}{3}x$. The slopes are $m_1 = \frac{\sqrt{7}}{3}$ and $m_2 = -\frac{\sqrt{7}}{3}$. The angle $	heta$ between the asymptotes is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$. Since the asymptotes are symmetric about the axes,the angle $2\alpha$ between them satisfies $	an \alpha = \frac{b}{a} = \frac{\sqrt{7}}{3}$. Then $\cos 2\alpha = \frac{1 - 	an^2 \alpha}{1 + 	an^2 \alpha} = \frac{1 - 7/9}{1 + 7/9} = \frac{2/9}{16/9} = \frac{2}{16} = \frac{1}{8}$. Thus,$\cos 	heta = \frac{1}{8}$.

If $\theta$ is the acute angle between the asymptotes of a hyperbola $7x^2 - 9y^2 = 63$,then $\cos \theta =$

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