If line 2x-1=0 is directrix of parabola y^2 - kx + 6 = 0, value of k is?  [4 or -6]

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$$\begin{gathered} 2\mathrm{x}=1 \mathrm{is} \mathrm{directrix} \\ \mathrm{i}.\mathrm{e}. \mathrm{x}=\frac{1}{2}...\left(\mathrm{i}\right) \\ \mathrm{We} \mathrm{have} \\ {\mathrm{y}}^2-\mathrm{kx}+6=0 \\ \Rightarrow {\mathrm{y}}^2=\mathrm{kx}-6 \\ \Rightarrow {\mathrm{y}}^2=\mathrm{k}\left(\mathrm{x}-\frac{6}{\mathrm{k}}\right) \\ \mathrm{Compare} \mathrm{with} {\mathrm{Y}}^2=4\mathrm{aX} \\ 4\mathrm{a}=\mathrm{k} \\ \mathrm{a}=\frac{\mathrm{k}}{4} \\ \mathrm{X}=-\mathrm{a} \\ \mathrm{x}-\frac{6}{\mathrm{k}}=-\frac{\mathrm{k}}{4} \\ \Rightarrow \mathrm{x}=-\frac{\mathrm{k}}{4}+\frac{6}{\mathrm{k}} \\ \Rightarrow -\frac{\mathrm{k}}{4}+\frac{6}{\mathrm{k}}=\frac{1}{2} \\ \Rightarrow -{\mathrm{k}}^2+24=2\mathrm{k} \\ \Rightarrow {\mathrm{k}}^2+2\mathrm{k}-24=0 \\ \Rightarrow {\mathrm{k}}^2+6\mathrm{k}-4\mathrm{k}-24=0 \\ \Rightarrow \mathrm{k}\left(\mathrm{k}+6\right)-4\left(\mathrm{k}+6\right)=0 \\ \Rightarrow \left(\mathrm{k}-4\right)\left(\mathrm{k}+6\right)=0 \\ \mathbf{k}\mathbf{=}\mathbf{4}\mathbf{ }\mathbf{or}\mathbf{ }\mathbf{k}\mathbf{=}\mathbf{-}\mathbf{6}\mathbf{ }\left[\mathbf{Answer}\right] \end{gathered}$$

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