If the line lx + my + n = 0 is tangent to the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ then show that ${\mathrm{a}}^2{\mathrm{l}}^2-{\mathrm{b}}^2{\mathrm{m}}^2={\mathrm{n}}^2$

Given hyperbola $\text{s=}\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$
Given tangent $\mathrm{lx}+\mathrm{my}+\mathrm{n}=0\to \left(1\right)$
Let p(x_1,y₁) be a point on the hyperbola
Equation of tangent at p(x_1,y₁) w.r.t to the hyperbola S = 0 is S₁ = 0
$S_1=\frac{{\mathrm{xx}}_1}{{\mathrm{a}}^2}-\frac{{\mathrm{yy}}_1}{{\mathrm{b}}^2}-1=0\to \left(2\right)$
Equations (1) and (2) represents same line then
$\begin{array}{l}\frac{\mathrm{l}}{\frac{{\mathrm{x}}_1}{{\mathrm{a}}^2}}=\frac{\mathrm{m}}{\frac{-{\mathrm{y}}_1}{{\mathrm{b}}^2}}=\frac{\mathrm{n}}{-1}\Rightarrow \frac{{\mathrm{a}}^2\mathrm{l}}{{\mathrm{x}}_1}=-\mathrm{n},\frac{-{\mathrm{b}}^2\mathrm{m}}{{\mathrm{y}}_1}=-\mathrm{n}\\ {\mathrm{x}}_1=\frac{-{\mathrm{a}}^2\mathrm{l}}{\mathrm{n}},{\mathrm{y}}_1=\frac{{\mathrm{b}}^2\mathrm{m}}{\mathrm{n}}\end{array}$
$\begin{array}{l}\therefore \mathrm{p}\left({\mathrm{x}}_1{\mathrm{y}}_1\right)=\left(\frac{-{\mathrm{a}}^2\mathrm{l}}{\mathrm{n}},\frac{{\mathrm{b}}^2\mathrm{m}}{\mathrm{n}}\right)\\ \mathrm{Since} \mathrm{p}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\text{ lies on equation (}1) then\\ \mathrm{l}\left(\frac{-{\mathrm{a}}^2\mathrm{l}}{\mathrm{n}}\right)+\mathrm{m}\left(\frac{{\mathrm{b}}^2\mathrm{m}}{\mathrm{n}}\right)+\mathrm{n}=0\\ \frac{-{\mathrm{a}}^2{\mathrm{l}}^2+{\mathrm{b}}^2{\mathrm{m}}^2+{\mathrm{n}}^2}{\mathrm{n}}=0\\ -\left({\mathrm{a}}^2{\mathrm{l}}^2-{\mathrm{b}}^2{\mathrm{m}}^2-{\mathrm{n}}^2\right)=0\\ \therefore {\mathrm{a}}^2{\mathrm{l}}^2-{\mathrm{b}}^2{\mathrm{m}}^2={\mathrm{n}}^2\end{array}$