Choose any point of this hyperbola

Let $y = \frac{2018}{x}$ be our hyperbola in question whose derivative equals $y' = -\frac{2018}{x^2}$ . Let $P(x_{0}, \frac{2018}{x_{0}})$ , where $x_{0} > 0,$ be any point on the hyperbola where the tangent line is expressible as:

$y - \frac{2018}{x_{0}} = -\frac{2018}{x_{0}^{2}} (x - x_{0}) \Rightarrow y = -\frac{2018}{x_{0}^{2}} x + \frac{4036}{x_{0}}$ (i).

Solving for the x and y-intercepts of this tangent line gives $(x,y) = (2x_{0}, 0 ); (0, \frac{4036}{x_{0}})$ , and the area of the right triangle contained between the coordinate axes + the tangent line equals:

$A = \frac{1}{2} \cdot (2x_{0})(\frac{4036}{x_{0}}) = \boxed{4036}.$