Question

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 :sq. units.

Answer 1

The equation of the given hyperbola is $\frac{x^2}{9}-\frac{y^2}{4}=1.$
Thus, the equation of the asymptotes are
$\Rightarrow 2x+3y=0$ and $2x-3y=0$
The equation of any tangent to the hyperbola
$\frac{x^2}{9}-\frac{y^2}{4}=1$ is
$\frac{x}{3}\sec \varphi -\frac{y}{2}\tan \varphi =1$
Let the points of intersection of $2x+3y=0,2x-3y=0$
and $\frac{x}{3}\sec \varphi -\frac{y}{2}\tan \varphi =1$ are $O,P$ and $Q$ respectively.
Therefore, $O=(0,0),P=(\frac{3}{\sec \varphi +\tan \varphi },\frac{-2}{\sec \varphi +\tan \varphi })$ and $Q=(\frac{3}{\sec \varphi -\tan \varphi },\frac{2}{\sec \varphi -\tan \varphi })$

Area=$\frac{1}{2}\left[0(\frac{-2}{\sec \varphi +\tan \varphi }-\frac{2}{\sec \varphi -\tan \varphi })\right)+\left(\frac{3}{\sec \varphi +\tan \varphi }(\frac{2}{\sec \varphi -\tan \varphi }-0)\right)+\left(\frac{3}{\sec \varphi -\tan \varphi }(0-\frac{-2}{\sec \varphi +\tan \varphi })\right]$
$=\frac{12}{2}$
$=6$ sq. units