If the line $\mathrm{x}-2\mathrm{y}+1=0$ makes an intercept of length $2\sqrt{30}$ on the parabola ${\mathrm{y}}^2=\mathrm{kx} (\mathrm{k}>0)$ then k = ...........

$$\begin{gathered} \mathrm{Given} \mathrm{line} \mathrm{x}-2\mathrm{y}+1=0 \\ {\mathrm{y}}^2=\mathrm{kx} \\ 4\mathrm{a}=\mathrm{k} \\ \mathrm{a}=\frac{\mathrm{k}}{4} \end{gathered}$$

Given length= $2\sqrt{30}$

$$\begin{gathered} \mathrm{x}=2\mathrm{y}-1 \\ {\mathrm{y}}^2=\mathrm{k}\left(2\mathrm{y}-1\right) \\ {\mathrm{y}}^2-2\mathrm{ky}+\mathrm{k}=0 \end{gathered}$$

let ${\mathrm{y}}_1,{\mathrm{y}}_2$ be two roots

${\mathrm{y}}_1+{\mathrm{y}}_2=2\mathrm{k} and {\mathrm{y}}_1{\mathrm{y}}_2=\mathrm{k}$

$$\begin{gathered} 4\left(30\right)={\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)}^2+{\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)}^2 \\ 4\left(30\right)={\left(\frac{{{\mathrm{y}}_1}^2}{\mathrm{k}}-\frac{{{\mathrm{y}}_2}^2}{\mathrm{k}}\right)}^2+{\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)}^2 \\ 120=\frac{{\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)}^2{\left({\mathrm{y}}_1+{\mathrm{y}}_2\right)}^2}{{\mathrm{k}}^2}+{\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)}^2 \\ 120={\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)}^2\left[\frac{4{\mathrm{k}}^2+{\mathrm{k}}^2}{{\mathrm{k}}^2}\right] \\ 120={\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)}^2\left(5\right) \\ {\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)}^2=24 \\ \Rightarrow {({\mathrm{y}}_1+{\mathrm{y}}_2)}^2-4{\mathrm{y}}_1{\mathrm{y}}_2=24 \\ \Rightarrow 4{\mathrm{k}}^2-4\mathrm{k}-24=0 \\ \Rightarrow (\mathrm{k}+2)(\mathrm{k}-3)=0 \\ \Rightarrow \mathrm{k}=3 \end{gathered}$$