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(A) Given parabola is $y^2=4x$,which is of the form $y^2=4ax$. Comparing,we get $a=1$.The equation of a tangent with slope $m$ to the parabola $y^2=4a

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$\frac{\pi}{3}$

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(A) Given parabola is $y^2=4x$,which is of the form $y^2=4ax$. Comparing,we get $a=1$.The equation of a tangent with slope $m$ to the parabola $y^2=4ax$ is $y=mx+\frac{a}{m}$.Since the tangent passes through $(1,4)$,we have $4=m(1)+\frac{1}{m}$.Multiplying by $m$,we get $m^2-4m+1=0$.Let $m_1$ and $m_2$ be the slopes of the two tangents. Then $m_1+m_2=4$ and $m_1m_2=1$.The angle $	heta$ between the tangents is given by $	an 	heta = \left| \frac{m_1-m_2}{1+m_1m_2} ight|$.Using the identity $|m_1-m_2| = \sqrt{(m_1+m_2)^2-4m_1m_2}$,we get $	an 	heta = \frac{\sqrt{4^2-4(1)}}{1+1} = \frac{\sqrt{12}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}$.Thus,$	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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