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(A) The equation of the hyperbola is $4x^2 - 5y^2 = 20$,which can be written as $\frac{x^2}{5} - \frac{y^2}{4} = 1$. Here,$a^2 = 5$ and $b^2 = 4$. The

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$2\sqrt{21}$

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(A) The equation of the hyperbola is $4x^2 - 5y^2 = 20$,which can be written as $\frac{x^2}{5} - \frac{y^2}{4} = 1$. Here,$a^2 = 5$ and $b^2 = 4$. The point $(1,1)$ lies outside the hyperbola since $\frac{1^2}{5} - \frac{1^2}{4} = 0.2 - 0.25 = -0.05 The equation of a tangent to the hyperbola with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2} = mx \pm \sqrt{5m^2 - 4}$. Since the tangent passes through $(1,1)$,we have $1 = m \pm \sqrt{5m^2 - 4}$,so $(1 - m)^2 = 5m^2 - 4$. Expanding this gives $1 - 2m + m^2 = 5m^2 - 4$,which simplifies to $4m^2 + 2m - 5 = 0$. Let the slopes of the two tangents be $m_1$ and $m_2$. Then $m_1 + m_2 = -\frac{2}{4} = -\frac{1}{2}$ and $m_1m_2 = -\frac{5}{4}$. The angle $	heta$ between the tangents is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1m_2}|$. We know $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1m_2 = (-\frac{1}{2})^2 - 4(-\frac{5}{4}) = \frac{1}{4} + 5 = \frac{21}{4}$. Thus,$|m_1 - m_2| = \frac{\sqrt{21}}{2}$. Substituting these values,$	an 	heta = |\frac{\sqrt{21}/2}{1 - 5/4}| = |\frac{\sqrt{21}/2}{-1/4}| = |\frac{\sqrt{21}}{2} 	imes (-4)| = 2\sqrt{21}$.

If $\theta$ is the acute angle between the tangents drawn from the point $(1,1)$ to the hyperbola $4x^2 - 5y^2 = 20$,then $\tan \theta = $

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