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(C) The equation of the parabola is $y^2 = 4x$,which is of the form $y^2 = 4ax$,where $a = 1$.Any tangent to the parabola $y^2 = 4ax$ is given by $y =

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$\frac{\pi}{3}$

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(C) The equation of the parabola is $y^2 = 4x$,which is of the form $y^2 = 4ax$,where $a = 1$.Any tangent to the parabola $y^2 = 4ax$ is given by $y = mx + \frac{a}{m}$.Substituting $a = 1$,we get $y = mx + \frac{1}{m}$.Since the tangent passes through the point $(1, 4)$,we have $4 = m(1) + \frac{1}{m}$.Multiplying by $m$,we get $4m = m^2 + 1$,or $m^2 - 4m + 1 = 0$.Let the slopes of the two tangents be $m_1$ and $m_2$. Then $m_1 + m_2 = 4$ and $m_1 m_2 = 1$.The angle $	heta$ between the tangents is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.We know that $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2 = 4^2 - 4(1) = 16 - 4 = 12$.Thus,$|m_1 - m_2| = \sqrt{12} = 2\sqrt{3}$.Substituting these values,$	an 	heta = |\frac{2\sqrt{3}}{1 + 1}| = \frac{2\sqrt{3}}{2} = \sqrt{3}$.Therefore,$	heta = 	an^{-1}(\sqrt{3}) = \frac{\pi}{3}$.

The angle between the tangents drawn from the point $(1,4)$ to the parabola $y^2=4x$ is

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