13.417 A bumblebee flew to an apple tree at a speed of $v_{1}$ m/min. At the same time, a bee flew to another apple tree at a speed of $v_{2}$ m/min. The bumblebee had to cover a distance of $2 a$ m, while the bee had to cover a distance of $2 b$ m. Suppose the trajectories of their flights are mutually perpendicular straight lines intersecting at a point that divides the path of the bumblebee and the path of the bee in half. Find the formula expressing the dependence of the distance $y$ between the bumblebee and the bee on the time $x$ of their flight. Determine the moment when the distance between the bumblebee and the bee reaches its minimum value during their flight.
Solution. Let $O$ be the point of intersection of the trajectories of the bumblebee and the bee. Then, after time $x$, the distances of the bumblebee and the bee from point $O$ will be respectively $\pm\left(x v_{1}-a\right)$ and $\pm\left(x v_{2}-b\right)$. From here, by the Pythagorean theorem, the distance between the bumblebee and the bee after time $x$ will be $f(x)=\sqrt{\left(x v_{1}-a\right)^{2}+\left(x v_{2}-b\right)^{2}}, f^{\prime}(x)=0 \Leftrightarrow$ $\Leftrightarrow v_{1}\left(x v_{1}-a\right)+v_{2}\left(x v_{2}-b\right)=0 \Rightarrow$ Answer: $x=\frac{v_{1} a+v_{2} b}{v_{1}^{2}+v_{2}^{2}}$ min (from the start of the flight).