The condition that the line x = my + c may be a tangent of $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=-1$ is

$$\begin{gathered} Given hyperbola \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=-1 \\ \mathrm{Given} \mathrm{tangent} \mathrm{x}=\mathrm{my}+\mathrm{c} \Rightarrow \mathrm{x}-\mathrm{my}-c=0 — ① \\ \mathrm{let} \mathrm{P}({\mathrm{x}}_1, {\mathrm{y}}_1) \mathrm{be} \mathrm{point} \mathrm{of} \mathrm{contact}. \\ \mathrm{Now} \mathrm{tangent} \mathrm{at} \mathrm{P} \mathrm{is} {\mathrm{S}}_1=0 \\ \Rightarrow \frac{\mathrm{x} {\mathrm{x}}_1}{{\mathrm{a}}^2}-\frac{\mathrm{y} {\mathrm{y}}_1}{{\mathrm{b}}^2}+1=0 — ② \\ \mathrm{Since} ① \& ② \mathrm{represents} \mathrm{same} \mathrm{line} \mathrm{then} \\ \Rightarrow \frac{{\mathrm{x}}_1}{{\displaystyle \frac{{\mathrm{a}}^2}{1}}}=\frac{-{\mathrm{y}}_1}{{\displaystyle \frac{{\mathrm{b}}^2}{-m}}}=\frac{1}{-c} \\ \Rightarrow {\mathrm{x}}_1=\frac{-{\mathrm{a}}^2}{\mathrm{c}}, {\mathrm{y}}_1=\frac{-{\mathrm{b}}^2\mathrm{m}}{c} \\ \mathrm{sub} \mathrm{point} \mathrm{in} \mathrm{line} \mathrm{x}=\mathrm{my}+c \\ \Rightarrow \frac{-{\mathrm{a}}^2}{c}=\mathrm{m}\left(\frac{-{\mathrm{b}}^2\mathrm{m}}{c}\right)+c \\ \Rightarrow c^2={\mathrm{b}}^2{\mathrm{m}}^2-{\mathrm{a}}^2 \end{gathered}$$