The central equation for a hyperbola is
x^{2} y^{2}
   = 1 (1)
a^{2} b^{2}
where 2a is the distance between the vertices and
y = ±bx/a (2)
are the equations of the asymptotes.
Differentiating equation (1) gives
b^{2}x
y' =  (3)
a^{2}y
Because of the symmetry of the hyperbola, we only
need to consider the point of tangency (x_{t},y_{t}) in
the first quadrant. The tangent line
y = y'_{t}x + k
b^{2}x_{t}
=  x + k
a^{2}y_{t}
b^{2}x_{t}^{2}
y_{t} =  + k
a^{2}y_{t}
b^{2}x_{t} b^{2}x_{t}^{2}
y =  x + y_{t}  
a^{2}y_{t} a^{2}y_{t}
a^{2}y_{t}y = b^{2}x_{t}x + a^{2}y_{t}^{2}  b^{2}x_{t}^{2}
a^{2}y_{t}y = b^{2}x_{t}x  a^{2}b^{2}
b^{2}x_{t} b^{2}
y =  x   (4)
a^{2}y_{t} y_{t}
Equation (4) and y = bx/a intersect at point
L = (x_{l},y_{l}) = ( a^{2}b/[bx_{t}  ay_{t}] , ab^{2}/[bx_{t}  ay_{t}] ) (5)
which lies above the xaxis.
Equation (4) and y = bx/a intersect at point
N = (x_{n},y_{n}) = ( a^{2}b/[bx_{t} + ay_{t}] , ab^{2}/[bx_{t} + ay_{t}] ) (6)
which lies below the xaxis.
Equation (4) and xaxis intersect at point
I = (x_{i},y_{i}) = ( a^{2}/x_{t},0 ) (7)
We can now calculate the area.
Area(ΔLMN) = Area(ΔLMI) + Area(ΔNMI)
= (a^{2}/2x_{t})(ab^{2}/[bx_{t}  ay_{t}] ) + (a^{2}/2x_{t})(ab^{2}/[bx_{t} + ay_{t}] )
= (a^{3}b^{2}/2x_{t})([bx_{t} + ay_{t} + bx_{t}  ay_{t}] /[b^{2}x_{t}^{2}  a^{2}y_{t}^{2} ])
= (a^{3}b^{2}/2x_{t})(2bx_{t}/a^{2}b^{2})
= ab (8)
Note:
The above argument does not hold if the point of tangency is the
vertex (a,0) because of equation (3). But, the area for this case is
easily calculated.
Area(ΔLMN) = (1/2)(base)(altitude)
= (1/2)LNMI
= ab (9)
Equations (8) and (9) agree on the constant ab.
QED
Note:
See Ady's post for how simple the problem is for a rectangular
hyperbola.
