Find the equation of tangent to the parabola y² = 16x inclined at an angle 60° with its axis and also find point of contact.

Given parabola y² =16x
$4\mathrm{a}=16\Rightarrow \mathrm{a}=4$
And slope of tangent $\mathrm{m}=\tan {60}^{\circ }=\sqrt{3}$
Now equation of tangents is $\mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\Rightarrow \mathrm{y}=\sqrt{3}\mathrm{x}+\frac{4}{\sqrt{3}}$
$3\mathrm{x}-\sqrt{3}\mathrm{y}+4=0$
and Point of contact $=\left(\frac{\mathrm{a}}{{\mathrm{m}}^2},\frac{2\mathrm{a}}{\mathrm{m}}\right)=\left(\frac{4}{3},\frac{2\left(4\right)}{\sqrt{3}}\right)=\left(\frac{4}{3},\frac{8}{\sqrt{3}}\right)$