Given the parabola $y^{2}=2 p x$ and the point $P(3 p, 0)$. Determine the point $Q$ on the parabola such that the distance $P Q$ is minimal.
I. solution: If $P Q$ is minimal, then for every point $Q^{\prime}$ in the vicinity of $Q$, $P Q^{\prime}>P Q$, which means the circle centered at $P$ with radius $P Q$ touches the parabola at $Q$, and thus the point $P$ lies on the normal to the parabola at $Q$.

Circles centered at $P$ generally intersect the parabola at four points. We seek among these circles the one that touches the parabola at two points. Since the $x$-axis is the axis of symmetry, the abscissas of the two points of tangency are the same.

The equation of the parabola is

$$
y^{2}=2 p x
$$

The equation of the circle centered at $P(3 p, 0)$ with radius $r$ is

$$
(x-3 p)^{2}+y^{2}=r^{2}
$$

For the points of intersection, equations (2) and (1) hold, from which

$$
(x-3 p)^{2}=r^{2}-2 p x
$$

follows, or rearranged,

$$
x^{2}-4 p x+9 p^{2}-r^{2}=0
$$

The abscissas of the points of intersection are equal if the discriminant $D$ of the quadratic equation (3) is $D=0$, that is,

$$
D=16 p^{2}-4\left(9 p^{2}-r^{2}\right)=0
$$

from which

$$
r^{2}=5 p^{2}
$$

Since in this case the coefficient of the linear term is the negative of the square root of -2, we have

$$
x_{1,2}=2 p, \quad \text { and thus from (1) } \quad y= \pm 2 p
$$

Thus, two points satisfy the requirements of the problem: $Q_{1}(2 p, 2 p)$ and $Q_{2}(2 p,-2 p)$.

Krix Eleonóra (Baja, III. Béla g. III. o. t.)

II. solution: The equation of the tangent at the point $\left(x_{1}, y_{1}\right)$ of the parabola is

$$
y_{1} y=p\left(x+x_{1}\right), \quad \text { or } \quad y=\frac{p}{y_{1}}\left(x+x_{1}\right)
$$

The equation of the normal, which passes through $\left(x_{1}, y_{1}\right)$ and is perpendicular to the tangent, is therefore

$$
y-y_{1}=-\frac{y_{1}}{p}\left(x-x_{1}\right)
$$

The normal to the sought point $Q\left(x_{1}, y_{1}\right)$, as seen in the first solution, passes through the point $P(3 p, 0)$, and thus

$$
0-y_{1}=-\frac{y_{1}}{p}\left(3 p-x_{1}\right), \text { or rearranged } y_{1}\left(2 p-x_{1}\right)=0
$$

This equation can hold either if $y_{1}=0$, or if $2 p-x_{1}=0$. The point on the parabola with $y_{1}=0$ is the vertex of the parabola, which is not the minimal distance from $P$ but rather the maximal distance (points of the curve near the vertex are closer to $P$ than the vertex). The condition $2 p-x_{1}=0$ holds for the point on the parabola with $x_{1}=2 p$, which agrees with the result obtained in the first solution.

Fillinger László (Bp., I., Toldy F. g. III. o. t.)

III. solution: Let $Q(x, y)$ be any point on the parabola. Its distance from $P$ is

$$
P Q=\sqrt{(3 p-x)^{2}+y^{2}}=\sqrt{(3 p-x)^{2}+2 p x}
$$

where we can substitute $2 p x$ for $y^{2}$ under the square root, since the point $Q$ lies on the parabola, and thus its coordinates satisfy the equation of the parabola. By rearranging the expression under the square root,

$$
P Q=\sqrt{(x-2 p)^{2}+5 p^{2}}
$$

On the right side, both terms under the square root are squares, hence positive, and $5 p^{2}$ is constant. Thus, the value of the expression is minimal when $(x-2 p)^{2}$ is as small as possible. The smallest possible value of $(x-2 p)^{2}$ is 0, which the expression takes at $x=2 p$. This result agrees with the previous ones. The minimal value of $P Q$ is $p \sqrt{5}$.