Any tangent to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is

$y=m x \pm \sqrt{a^{2} m^{2}-b^{2}}$

or $y=m x+c$

$\text { where } c=\pm \sqrt{a^{2} m^{2}-b^{2}}$

This will touch the hyperbola $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$

if the equation $\frac{(m x+c)^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$ has equal roots or $\mathrm{x}^{2}\left(\mathrm{b}^{2} \mathrm{m}^{2}-\mathrm{a}^{2}\right)+2 \mathrm{b}^{2} \mathrm{mc} \mathrm{x}+\left(\mathrm{c}^{2}-\mathrm{a}^{2}\right) \mathrm{b}^{2}=0$ is an quadratic eqaution have equal roots 

then $4 b^{4} m^{2} c^{2}=4\left(b^{2} m^{2}-a^{2}\right)\left(c^{2}-a^{2}\right) b^{2}$

or $c^{2}=a^{2}-b^{2} m^{2}$

$\quad a^{2} m^{2}-b^{2}=a^{2}-b^{2} m^{2} \quad$ put $\left(c^{2}=a^{2} m^{2}-b^{2}\right)$

$m^{2}\left(a^{2}+b^{2}\right)=a^{2}+b^{2} \quad \Rightarrow \quad m=\pm 1$