A tangent to the parabola $\;$ $y^2 = 16x$ $\;$ makes an angle of $\;$ $60^\circ$ $\;$ with the axis. Find the point of contact.
Equation of parabola: $\;$ $y^2 = 16x$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives
$4a = 16$ $\implies$ $a = 4$
Let slope of tangent $= m$
Since the tangent makes an angle of $60^\circ$ with the axis,
$\implies$ $m = \tan 60^\circ = \sqrt{3}$
Point of contact of the tangent with the parabola $= \left(\dfrac{a}{m^2}, \dfrac{2a}{m}\right) = \left(\dfrac{4}{\left(\sqrt{3}\right)^2}, \dfrac{2 \times 4}{\sqrt{3}}\right)$
$\therefore \;$ The point of contact $= \left(\dfrac{4}{3}, \dfrac{8}{\sqrt{3}}\right)$