Explanation

(c)
Equation of tangent to $y^{2} = 4x\ is\ y = mx + \frac{1}{m}$
Since, tangent passes through (1, 4)
$\therefore 4 = m + \frac{1}{m}\ \Rightarrow \ m^{2} - 4m + 1 = 0$
$\therefore\ m_{1} + m_{2} = 4\$and $m_{1}m_{2} = 1$
Now, $|m_{1} - m_{2}| = \sqrt{\left( m_{1} + m_{2} \right)^{2} - 4m_{1}m_{2}}$
$= \sqrt{16 - 4} = 2\sqrt{3}$
Thus, the angle between tangent
$\tan\begin{array}{r}
\theta = \left| \frac{m_{2} - m_{1}}{1 + m_{1}m_{2}} \right| = \left| \frac{2\sqrt{3}}{1 + 1} \right| = \sqrt{3} \\
\
\end{array}$
$\Rightarrow \ \theta = \frac{\pi}{3}$