The area of the triangle that a tangent at a point on the hyperbola $\frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{9}=1,$ makes with it’s asymptotes is $\mathrm{\_\_\_\_\_}$ square units

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{are}a \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{formed} \mathrm{by} \mathrm{any} \mathrm{tangent} \mathrm{to} \mathrm{the} \mathrm{hyperbola} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \mathrm{and} \mathrm{its} \mathrm{asy}m\mathrm{ptotes} \mathrm{is} \text{'}\mathrm{ab}\text{'} \mathrm{square} \mathrm{units} \\ \therefore \mathrm{Required} \mathrm{area} =\mathrm{ab}=4\times 3=12 \\ \left(\mathrm{or}\right) \\ \frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{9}=1 \\ \mathrm{equations} \mathrm{of} \mathrm{asymptotes} \mathrm{are} 3\mathrm{x}+4\mathrm{y}=0 — ① \\ 3\mathrm{x}-4\mathrm{y}=0 — ② \\ \mathrm{Equation} \mathrm{of} \mathrm{tangent} \mathrm{at} \mathrm{P}\left(\mathrm{\theta }\right) \mathrm{is} \frac{\mathrm{x}}{4}\mathrm{sec\theta }-\frac{\mathrm{y}}{3}\mathrm{tan\theta }=1 — ③ \\ \therefore \mathrm{vertices} \mathrm{of} \mathrm{triangle} \mathrm{are} \end{gathered}$$

$$\begin{gathered} 0=(0, 0), \mathrm{A}=\left(\frac{4}{\mathrm{sec\theta }+\mathrm{tan\theta }}, \frac{-3}{\mathrm{sec\theta }+\mathrm{tan\theta }}\right) \\ \mathrm{and} \mathrm{B}=\left(\frac{4}{\mathrm{sec\theta }-\mathrm{tan\theta }}, \frac{3}{\mathrm{sec\theta }-\mathrm{tan\theta }}\right) \\ \therefore \mathrm{Area}=\frac{1}{2}\left|\left(\frac{4}{\mathrm{sec\theta }+\mathrm{tan\theta }}\right) \left(\frac{3}{\mathrm{sec\theta }-\mathrm{tan\theta }}\right)-\left(\frac{4}{\mathrm{sec\theta }-\mathrm{tan\theta }}\right) \left(\frac{-3}{\mathrm{sec\theta }+\mathrm{tan\theta }}\right)\right| \\ =\frac{1}{2}\left|\frac{12}{{\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }}+\frac{12}{{\sec }^2{\theta }^{ }-{\tan }^2\mathrm{\theta }}\right| \\ =\frac{1}{2}\left|24\right| \\ =12 \end{gathered}$$