If the tangents to the hyperbola ${\mathrm{x}}^2-9{\mathrm{y}}^2=9$ ate drawn from point (3, 2), then

Given hyperbola is ${\mathrm{x}}^2-9{\mathrm{y}}^2=9$

or $\frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{1}=1$

Equation of tangent is 

$\mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2}$

This tangent passes through (3,2). 

$\begin{array}{l}\mathrm{so},2=3\mathrm{m}\pm \sqrt{9{\mathrm{m}}^2-1}\\ \Rightarrow 4+9{\mathrm{m}}^2-12\mathrm{m}=9{\mathrm{m}}^2=1\\ \therefore {\mathrm{m}}_1=5/12\text{ and }{\mathrm{m}}_2=\mathrm{\infty }\end{array}$

So, equations of tangents are

$\begin{array}{l}\mathrm{y}-2=(5/12)(\mathrm{x}-3)\\ \Rightarrow 5\mathrm{x}-12\mathrm{y}+9=0 \left(1\right)\\ \mathrm{and} \mathrm{x}=3 \left(2\right)\end{array}$

Now, equation of chord of contact w.r.t. point P(3,2) is

$\begin{array}{l}\mathrm{T}=0\\ \Rightarrow 3\mathrm{x}-18\mathrm{y}=9\\ \Rightarrow \mathrm{x}-6\mathrm{y}=3 \left(3\right)\end{array}$

Solving (1) and (3), we get $\mathrm{x}=-5,\mathrm{y}=-\frac{4}{3}$.

Solving (2) and (3), we get $\mathrm{x}=3,\mathrm{y}=0$

Thus, vertices of triangle are  $(3,2),(3,0),(-5,-4/3)$

$\begin{array}{l}\mathrm{area} \mathrm{of} \mathrm{triangle}=\frac{1}{2}\left|\begin{array}{ccc}3& 2& 1\\ 3& 0& 1\\ -5& -\frac{4}{3}& 1\end{array}\right|\\ =\frac{1}{2}\times \left|3\left(\frac{4}{3}\right)-2(3+5)+1(-4)\right|\\ =\frac{1}{2}|4-16-4|\\ =8\text{ sq. units }\end{array}$