In the parabola ${\mathrm{y}}^2-2\mathrm{y}+8\mathrm{x}-23=0$, the length of double ordinate at a distance of 4 units from its vertex is

$$\begin{gathered} {\mathrm{y}}^2-2\mathrm{y}=-8\mathrm{x}+23 \\ \Rightarrow {\mathrm{y}}^2-2\left(1\right)\mathrm{y}+1=-8\mathrm{x}+24 \\ \Rightarrow {\left(\mathrm{y}-1\right)}^2=-8\left(\mathrm{x}-3\right) \\ where 4\mathrm{a}=8 \\ \Rightarrow \mathrm{a}=2 \end{gathered}$$

$$\begin{gathered} \mathrm{vertex}=\left(\mathrm{h}-\mathrm{a},\mathrm{k}\right) \\ \mathrm{vertex}=\left(3-2,1\right)=\left(1,1\right) \end{gathered}$$

Length of double ordinate $=4\sqrt{ad}=4\sqrt{2\times 4}=8\sqrt{2}$