Answer: B
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Let $P(t^2,2t)$ be the parametric point on the parabola. Then the slope of the parabola at $P$ is given by
$$\frac{dy}{dx}=\frac{1}{t}$$
Now, the slope of the tangent at $P$ passing through the point $P$ and $(1,4)$ is given by $$\frac{2t-4}{t^2-1}$$.
Hence, equating the slopes at point $P$, we have the following $$\frac{1}{t}=\frac{2t-4}{t^2-1}$$
The roots of the equation are $t_1$ and $t_2$.
The angle between the two tangents is given by
$$\tan\alpha =\left|\frac{t_2-t_1}{1+t_1t_2}\right|$$
On solving, we will get $\alpha=\frac{\pi}{3}$.