A tangent of ${\mathrm{y}}^2=12\mathrm{x}$ makes an angle 60° with the line $\mathrm{y}=5\mathrm{x}+4$. then its equation

Let Equation  of tangent  is $y=mx+\frac{3}{m}$
$Now \tan {60}^0=\left|\frac{\mathrm{m}-5}{1+5\mathrm{m}}\right|,$
$\Rightarrow \mathrm{m}\equiv \frac{5+\sqrt{3}}{1-5\sqrt{3}},\frac{5-\sqrt{3}}{1+5\sqrt{3}},$
tangents are $\mathrm{y}=\left(\frac{5-\sqrt{3}}{1+5\sqrt{3}}\right)\mathrm{x}+3\left(\frac{1+5\sqrt{3}}{5-\sqrt{3}}\right)$ and 
$\Rightarrow \mathrm{y}=\left(\frac{5+\sqrt{3}}{1-5\sqrt{3}}\right)\mathrm{x}+3\left(\frac{1-5\sqrt{3}}{5+\sqrt{3}}\right)$