tangents to parabola y2=4ax makes an angle of 60 degree with the axis find its point of contact

the slope of a line which makes an angle of 60 deg with x-axis is $m=\tan 60=\sqrt{3}$ the equation of the tangent is $y=mx+c i.e. y=\sqrt{3}.x+c ........\left(1\right)$ if the line (1) is tangent to the parabola $y^2=4ax$ , then $c=\frac{a}{m}=\frac{a}{\sqrt{3}}$ therefore the equation of the tangent is $y=\sqrt{3}.x+\frac{a}{\sqrt{3}}$  if this straight line is tangent to the given parabola,  $$\begin{gathered} {\left(\sqrt{3}x+\frac{a}{\sqrt{3}}\right)}^2=4.a.x \\ \Rightarrow 3x^2+\frac{a^2}{3}+2*\sqrt{3}.x*\frac{a}{\sqrt{3}}=4ax \\ 3x^2+\frac{a^2}{3}+2ax-4ax=0 \\ 9x^2+a^2-6ax=0 \\ \Rightarrow (3x-a{)}^2=0 \\ \Rightarrow 3x-a=0 \\ \Rightarrow x=\frac{a}{3} \end{gathered}$$
Therefore 
$$\begin{gathered} y=\sqrt{3}.x+\frac{a}{\sqrt{3}} \\ y=\sqrt{3}.\frac{a}{3}+\frac{a}{\sqrt{3}} \\ y=\frac{2a}{\sqrt{3}} \end{gathered}$$
thus the point of contact is $\left(\frac{a}{3},\frac{2a}{\sqrt{3}}\right)$ hope this helps you