## Task A-4.6.

Determine points $A$ and $B$ on the parabola $y^{2}=x$ such that the point $(2,1)$ lies on the line segment $\overline{A B}$, and the midpoint of the line segment $\overline{A B}$ is as close as possible to the $y$-axis.
## Solution.

The equation of the line passing through the point $(2,1)$ is $y=a(x-2)+1$, for some real parameter $a$. Therefore, the coordinates of the endpoints of the segment in question are solutions to the system

$$
\begin{aligned}
y^{2} & =x \\
y & =a(x-2)+1
\end{aligned}
$$

The ordinates of these points are the solutions to the quadratic equation

$$
a y^{2}-y+(1-2 a)=0
$$

Notice that the abscissa of the midpoint of the segment in question is always a positive number, so we need to minimize this abscissa.

![](https://cdn.mathpix.com/cropped/2024_05_30_a374c05c2f1967f14be6g-24.jpg?height=620&width=851&top_left_y=181&top_left_x=525)

If points $A$ and $B$ have coordinates $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, then the value we want to minimize is

$$
\begin{aligned}
\frac{x_{1}+x_{2}}{2} & =\frac{y_{1}^{2}+y_{2}^{2}}{2}=\frac{\left(y_{1}+y_{2}\right)^{2}-2 y_{1} y_{2}}{2} \\
& =\frac{\left(\frac{1}{a}\right)^{2}-2 \cdot \frac{1-2 a}{a}}{2} \\
& =\frac{1}{2 a^{2}}-\frac{1}{a}+2
\end{aligned}
$$

In the first line, we used the fact that $y_{1}$ and $y_{2}$ are solutions to the derived quadratic equation, and applied Viète's formulas.

Let $t=\frac{1}{a}$, the quadratic equation $\frac{1}{2} t^{2}-t+2$ achieves its minimum at the point $t=1$.

Thus, the desired minimum distance is achieved for the real parameter $a=1$.

This minimum distance is $\frac{3}{2}$, and the points at which it is achieved are

$$
\left(\frac{3+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right) \quad \text { and } \quad\left(\frac{3-\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2}\right)
$$