If $\mathrm{lx}+\mathrm{my}=1$ is a normal to the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ then ${\mathrm{a}}^2{\mathrm{m}}^2–{\mathrm{b}}^2{\mathrm{l}}^2=$

$$\begin{gathered} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \mathrm{Equation} \mathrm{of} \mathrm{normal} \mathrm{at} \mathrm{P}\left(\mathrm{\theta }\right) \mathrm{is} \\ \frac{\mathrm{ax}}{\mathrm{sec\theta }}+\frac{\mathrm{by}}{\mathrm{tan\theta }}={\mathrm{a}}^2+{\mathrm{b}}^2 — ① \\ \mathrm{Given} \mathrm{normal} \mathrm{is} lx+\mathrm{my}=1 — ② \\ \mathrm{since} ① \& ② \mathrm{represents} \mathrm{same} \mathrm{line} \mathrm{then} \\ \frac{\mathrm{l}}{{\displaystyle \frac{\mathrm{a}}{\mathrm{sec\theta }}}}=\frac{\mathrm{m}}{{\displaystyle \frac{\mathrm{b}}{\mathrm{tan\theta }}}}=\frac{1}{{\mathrm{a}}^2+{\mathrm{b}}^2} \\ \Rightarrow \frac{\mathrm{l} \mathrm{sec\theta }}{\mathrm{a}}=\frac{1}{{\mathrm{a}}^2+{\mathrm{b}}^2}, \frac{\mathrm{m} \mathrm{tan\theta }}{\mathrm{b}}=\frac{1}{{\mathrm{a}}^2+{\mathrm{b}}^2} \\ \Rightarrow \mathrm{sec\theta }=\frac{\mathrm{a}}{\mathrm{l}({\mathrm{a}}^2+{\mathrm{b}}^2)}, \mathrm{tan\theta }=\frac{\mathrm{b}}{\mathrm{m}({\mathrm{a}}^2+{\mathrm{b}}^2)} \\ \mathrm{we} \mathrm{K}.\mathrm{T} {\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }=1 \\ \Rightarrow {\left(\frac{\mathrm{a}}{\mathrm{l}({\mathrm{a}}^2+{\mathrm{b}}^2)}\right)}^2 -{\left(\frac{\mathrm{b}}{\mathrm{m}({\mathrm{a}}^2+{\mathrm{b}}^2)}\right)}^2=1` \\ \Rightarrow \left(\frac{{\mathrm{a}}^2}{{\mathrm{l}}^2}-\frac{{\mathrm{b}}^2}{{\mathrm{m}}^2}\right)={\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)}^2 \\ \Rightarrow {\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2{\mathrm{l}}^2={\mathrm{l}}^2{\mathrm{m}}^2{\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)}^2 \end{gathered}$$