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(C) The equation of the parabola is $y^2 = 4ax$,where $a = 1$. The point is $(x_1, y_1) = (1, 4)$. The locus of the point of intersection of perpendic

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$\frac{\pi}{3}$

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(C) The equation of the parabola is $y^2 = 4ax$,where $a = 1$. The point is $(x_1, y_1) = (1, 4)$. The locus of the point of intersection of perpendicular tangents to a parabola is its directrix. The directrix of the parabola $y^2 = 4x$ is $x = -a$,which is $x = -1$. Since the point $(1, 4)$ does not lie on the directrix $x = -1$,the tangents are not perpendicular. The angle $	heta$ between the tangents from a point $(x_1, y_1)$ to the parabola $y^2 = 4ax$ is given by $	an 	heta = \left| \frac{\sqrt{y_1^2 - 4ax_1}}{x_1 + a} ight|$. Substituting the values: $a = 1, x_1 = 1, y_1 = 4$. $	an 	heta = \left| \frac{\sqrt{4^2 - 4(1)(1)}}{1 + 1} ight| = \left| \frac{\sqrt{16 - 4}}{2} ight| = \frac{\sqrt{12}}{2} = \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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