$$\begin{gathered} {\text{The length of the sub-tangent to the hyperbola x}}^2-4y^2=4 corresponding \\ to the normal having slope unity is \frac{1}{\sqrt{k}}, then k is equal to \end{gathered}$$

$\begin{array}{l}Slope of normal is unity\text{}\Rightarrow -\frac{dx}{dy}=1, \frac{dy}{dx}=-1\text{}\\ from curve \frac{dy}{dx}=\frac{x}{4y}=-1\\ \text{}\therefore x=-4y\text{\hspace{0.17em} sub in curve }16y^2-4y^2=4\text{\hspace{0.17em}}\\ y=\pm \frac{1}{\sqrt{3 }}\therefore x= \mp \frac{4}{\sqrt{3}}\\ \text{}\therefore sub-\tan gent=\left|y.\frac{dy}{dx}\right| =\frac{1}{\sqrt{3}}\\ \text{}\therefore k=3\end{array}$