If the line 2 x-1=0 is the directrix of the parabola y²-k x+6=0 ,...

Given equation of parabola is $y²-k x+6=0$ (\Rightarrow y²=k x-6 \Rightarrow y²$=k\l$\left(x-\frac{6}{k}\right)$Now, directrix, \$x-\frac{6}{k}=-\frac{k}{4}$ (\Rightarrow $x=\frac{6}{k}-\frac{k}{4}$...(i) But directrix is given (\Rightarrow $x=\frac{1}{2}$...(ii) $$\begin{aligned}& \therefore \frac{6}{k}-\frac{k}{4}=\frac{1}{2} \Rightarrow24-k²=2 k \\ & \Rightarrow k²+2 k-24=0 \\ & \Rightarrow(k+6)(k-4)=0 \quad \Rightarrow k=-6, k=4 \end{aligned}$$)