If ax + by = 1 is tangent to the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\text{, then }{\mathrm{a}}^2-{\mathrm{b}}^2$ is equal to

Any tangent to the hyperbola is $\frac{\mathrm{x}}{\mathrm{a}}\sec \mathrm{\theta }-\frac{\mathrm{y}}{\mathrm{b}}\tan \mathrm{\theta }=1 ...\left(\mathrm{i}\right)$
The given tangent is ax + by = 1 Comparing (i) and (ii), we have ….(ii)
$$\begin{gathered} \sec \mathrm{\theta }={\mathrm{a}}^2\text{ and }\tan \mathrm{\theta }=-{\mathrm{b}}^2 \\ \text{Eliminating }\mathrm{\theta }\text{, we have } \\ {\mathrm{a}}^4-{\mathrm{b}}^4=1\text{ ⇒ }\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)=1 \end{gathered}$$
Also,
 $$\begin{gathered} {\mathrm{a}}^2+{\mathrm{b}}^2={\mathrm{a}}^2{\mathrm{e}}^2 \\ \text{⇒ }{\mathrm{a}}^2-{\mathrm{b}}^2=\frac{1}{{\mathrm{a}}^2{\mathrm{e}}^2} \end{gathered}$$