The area of the triangle formed by any tangent to the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ and its asymptotes is

The equation of the given hyperbola is  $\frac{x^2}{9}-\frac{y^2}{4}=1$.

Thus, the equations of the asymptotes are 

$\left(\frac{x}{3}+\frac{y}{2}\right)\left(\frac{x}{3}-\frac{y}{2}\right)=0$

$\Rightarrow \left(\frac{x}{3}+\frac{y}{2}\right)=0$ and $\left(\frac{x}{3}-\frac{y}{2}\right)=0$

The equation of any tangent to the hyperbola 

$\frac{x^2}{9}-\frac{y^2}{4}=1$ is

$\frac{x}{3}\sec \phi -\frac{y}{2}\tan \phi =1$

Let the points of intersection of $2x+3y=0,2x-3y=0$

and $\frac{x}{3}\sec \phi -\frac{y}{2}\tan \phi =1$ are $O$,$P$ and $Q$ respectively.

Therefore, $O=(0,0)$,

$P=\left(\frac{3}{\sec \phi +\tan \phi },-\frac{2}{\sec \phi +\tan \phi }\right)$

and $Q=\left(\frac{3}{\sec \phi -\tan \phi },\frac{2}{\sec \phi -\tan \phi }\right)$

Hence, the area of $\mathrm{△}OPQ$

$\begin{array}{l}=\frac{1}{2}\left|\begin{array}{cc}0& 0\\ \frac{3}{\sec \phi +\tan \phi }& -\frac{2}{\sec \phi +\tan \phi }\\ \frac{3}{\sec \phi -\tan \phi }& \frac{2}{\sec \phi -\tan \phi }\\ 0& 0\end{array}\right|\\ \end{array}$

$=\frac{1}{2}(6+6)=6\mathrm{sq}.\mathrm{u}$.