Answer
$2\sqrt 6$
Work Step by Step
$y$ = $\frac{3}{x}$
$y'$ = $-\frac{3}{x^2}$
so an equation of the tangent line at the point $\left(a,\frac{3}{a}\right)$ is
$y-\frac{3}{a}$ = $-\frac{3}{a^2}(x-a)$
$y$ = $-\frac{3}{a^2}(x-a)+\frac{3}{a}$
The $y$-intercept $x$ = $0$ is $\frac{6}{a}$
The $x$-intercept $y$ = $0$ is $2a$
The distance $d$ of the line segment that has endpoints at the intercept is
$d$ = $\sqrt {(2a-0)^2+\left(0-\frac{6}{a}\right)^2}$
Let
$S$ = $d^2$
so
$S$ = $4a^2+\frac{36}{a^2}$
$S'$ = $8a-\frac{72}{a^3}$
$8a-\frac{72}{a^3}$ = $0$
$8a$ = $\frac{72}{a^3}$
$a^4$ = $9$
$a$ = $\sqrt 3$
$S''$ = $8+\frac{216}{a^4}$
$8+\frac{216}{a^4}$ $\gt$ $0$
so there is an absolute minimum at $a$ = $\sqrt 3$
Thus
$S$ = $4(3)+\frac{36}{3}$ = $24$
and
$d$ = $\sqrt {24}$ = $2\sqrt 6$
