The equation of a tangent to the parabola ${\mathrm{x}}^2=8\mathrm{y}$ which makes an angle $\mathrm{\theta }$ with the positive direction of x-axis is

$\begin{array}{rcl}{\mathrm{x}}^2& =& 8\mathrm{y} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2\mathrm{x}}{8}-\frac{\mathrm{x}}{4}=\tan \mathrm{\theta }\\ & \Rightarrow & \mathrm{x}=4\tan \mathrm{\theta }\\ \mathrm{y}& =& \frac{{\mathrm{x}}^2}{8}\\ & =& \frac{16{\tan }^2\mathrm{\theta }}{8}=2{\tan }^2\mathrm{\theta }\end{array}$

eq of tangent $\mathrm{y}-2{\tan }^2\mathrm{\theta }=\mathrm{tan\theta }(\mathrm{x}-4\mathrm{tan\theta })$

                $\mathrm{y}=\mathrm{x} \mathrm{tan\theta }-2{\tan }^2\mathrm{\theta }$