Tightly Bounded Area

Let the tangent touches $f(x)$ at $\left(a, \dfrac 1a\right)$ .

The gradient of the tangent is $\dfrac d{dx} f(x) \bigg|_{x=a} =$ $\dfrac d{dx} \left(\dfrac 1x\right) \bigg|_{x=a} =$ $- \dfrac 1{a^2}$ .

The equation of the tangent is given by:

$\begin{aligned} \frac {y - \frac 1a}{x-a} & = - \frac 1{a^2} \\ & = \frac {a-x}{a^2} \\ \implies y & = \frac {2a-x}{a^2} \end{aligned}$

Let $x-$ and $y$ -axis intercepts be $x_0$ and $y_0$ respectively. Then we have:

$\begin{aligned} 0 & = \frac {2a-x_0}{a^2} \\ \implies x_0 & = 2a \\ y_0 & = \frac {2a-0}{a^2} \\ \implies y_0 & = \frac 2a \end{aligned}$

Then we have $X = \dfrac 12 x_0 y_0 = \dfrac 12 \cdot 2a \cdot \dfrac 2a = \boxed{2}$ , which is independent of $a$ .

Cute problem indeed.

Here's a walkthrough from an MIT Single Variable Calculus lecture:

Lec 1 | MIT 18.01 Single Variable Calculus, Fall 2007