If the locus of point of intersection of tangents to ${\mathrm{y}}^2=8\mathrm{x}$ which makes angles ${\mathrm{\theta }}_1$ and ${\mathrm{\theta }}_2$ with axis so that 

$cot{\theta }_1+cot{\theta }_2=k$ is $\mathrm{y}=8$ then $\mathrm{k}$ is

$$\begin{gathered} \mathrm{Given} \mathrm{parabola} {\mathrm{y}}^2 = 8\mathrm{x} \\ \Rightarrow 4\mathrm{a} = 8 \\ \Rightarrow \mathrm{a}=2 \\ \mathrm{Eq}uation \mathrm{of} \mathrm{tangent} \mathrm{to} \mathrm{Parabola} \mathrm{is} \end{gathered}$$

$$\begin{gathered} \mathrm{y}=\mathrm{mx} +\frac{2}{\mathrm{m}} \\ \Rightarrow {\mathrm{m}}^2 \mathrm{x} -\mathrm{my} +2=0 \\ let m_1,m_2 be roots and {\mathrm{m}}_1= {\mathrm{tan\theta }}_1,{\mathrm{m}}_2={\mathrm{tan\theta }}_2 \\ Now {\mathrm{tan\theta }}_1 +{\mathrm{tan\theta }}_2=\raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right. \end{gathered}$$

$$\begin{gathered} {\mathrm{tan\theta }}_1 . \tan {\mathrm{\theta }}_2=\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right. \\ \mathrm{Cot} {\mathrm{\theta }}_1 + \mathrm{Cot} {\mathrm{\theta }}_2 =\mathrm{k} \\ \frac{\tan {\mathrm{\theta }}_1 + \tan {\mathrm{\theta }}_2}{\tan {\mathrm{\theta }}_1. \tan {\mathrm{\theta }}_2} =\mathrm{k} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.}}{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.}} = \mathrm{k} \\ \Rightarrow \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$2 $}\right.=\mathrm{k} \\ \mathrm{y}=2\mathrm{k} and given \mathrm{y} =8 \\ \Rightarrow 2\mathrm{k}=8 \\ \Rightarrow \mathrm{k}=4 \end{gathered}$$