Let $y=\frac{1}{x}$ be a hyperbola, and let $P$ be the point on this hyperbola with an abscissa $a \geq 1$. Denote by $e$ the tangent at $P$, and let $t(a)$ represent the area enclosed by this tangent $e$ and the coordinate axes; let $T(a)$ represent the area of the triangle bounded by the tangent $e$, the perpendicular line raised at $P$, and the $x$-axis. Does the ratio $\frac{t(a)}{T(a)}$ have a maximum and a minimum value?
Let's write down the equation of the tangent line passing through the point $P\left(a ; \frac{1}{a}\right)$. The derivative of the function $\frac{1}{x}$ is $-\frac{1}{x^{2}}$, so the slope of the tangent at $P\left(a, \frac{1}{a}\right)$ is $-\frac{1}{a^{2}}$. Therefore, the equation of the tangent is $y=-\frac{1}{a^{2}} x+\frac{2}{a}$. According to this, the tangent intersects the axes at the points $(2 a ; 0)$ and $\left(0, \frac{2}{a}\right)$. From this, $t(a)=\frac{1}{2} \cdot 2 a \cdot \frac{2}{a}=2$. (This result is also well-known from geometry.)

The above calculation shows that the point of tangency $P$ bisects the hypotenuse of the right triangle with vertices $(0,0),(2 a, 0),\left(0, \frac{2}{a}\right)$. The perpendicular raised at $P$ cuts off a similar triangle, and the ratio of similarity is $2 a: \overline{(2 a, 0) P}$. The square of this ratio is precisely the ratio of the areas $t(a) / T(a)$, so by the Pythagorean theorem,

$$
t(a) / T(a)=\left(\frac{2 a}{\frac{1}{2} \sqrt{(2 a)^{2}+\left(\frac{2}{a}\right)^{2}}}\right)^{2}=\frac{4}{1+\frac{1}{a^{4}}}
$$

Thus, $\frac{t(a)}{T(a)}$ is a strictly increasing function of $a$, which takes the minimal value of 2 at $a=1$ and has no maximum (although it is bounded from above), with a limit of 4 as $a$ approaches $+\infty$.