Question

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

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 $Q=(\frac{3}{\sec \varphi -\tan \varphi },\frac{2}{\sec \varphi -\tan \varphi })$

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

Alternate solution:
Given: hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1\Rightarrow a=3,b=2$


Tangent at $D(a,0)$ is parallel to $y-$axis.
$\therefore$ Co-ordinates of $A,B$ and $C$ are $(0,0),(a,b)$ and $(a,-b)$ respectively.
$\Rightarrow BC=2b,AD=a$
$\Rightarrow \text{Area of}△ABC=\frac{1}{2}\times BC\times AD$
$\Rightarrow \frac{1}{2}\times 2b\times a=ab=6\text{sq.units}$

Note: The area of the triangle formed by two asymptotes and a tangent to the hyperbola $\frac{x^2}{a}-\frac{y^2}{b}=1$ is always constant and its equal to $ab.$