What is the parameter of the parabola $y^{2}=2 p x$ if the parabola touches the line $a x+b y+c^{2}=0$?
I. solution: To exclude trivial cases and ensure solvability, we assume that $a, b$ and $c \neq 0$ and, of course, $p \neq 0$.

The line touches the parabola if and only if the two intersection points coincide, i.e., if the system of equations

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

and

$$
-b y-c^{2}=a x
$$

has only one root pair.

Substituting the value of $x$ from (2) into (1), we get

$$
y^{2}=-\frac{2 p\left(b y+c^{2}\right)}{a}
$$

Rearranging,

$$
y^{2}+\frac{2 p b}{a} y+\frac{2 p c^{2}}{a}=0
$$

The roots of this equation coincide if and only if its discriminant is 0, i.e.,

$$
\frac{4 p^{2} b^{2}}{a^{2}}-\frac{8 p c^{2}}{a}=0
$$

From which, given $(p \neq 0)$,

$$
p=\frac{2 a c^{2}}{b^{2}}
$$

János Fülöp (Bp. V., Piarist g. III. o. t.)

II. solution: We use the fact that the equation of the tangent to the parabola at the point $\left(x_{1} y_{1}\right)$ is

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

or

$$
y_{1} y-p x-p x_{1}=0
$$

The equation

$$
b y+a x+c^{2}=0
$$

is also the equation of the tangent. Therefore, if we multiply (4) by an appropriate number, the coefficients will match those in (3).

Multiplying (4) by $\frac{y_{1}}{b}$, we get the equation

$$
y_{1} y+\frac{a y_{1}}{b} x+\frac{c^{2} y_{1}}{b}=0
$$

where the coefficient of $y$ is the same as the coefficient of $y$ in (3), so the other two coefficients must also be the same:

$$
-p=\frac{a y_{1}}{b}
$$

and

$$
-p x_{1}=\frac{c^{2} y_{1}}{b}
$$

From the last two relationships, we calculate the coordinates of the point of tangency:

$$
x_{1}=\frac{c^{2}}{a}, \quad y_{1}=-\frac{b p}{a}
$$

The point of tangency lies on the parabola, so its coordinates satisfy the equation of the parabola:

$$
\frac{b^{2} p^{2}}{a^{2}}=2 p \frac{c^{2}}{a}
$$

from which

$$
p=\frac{2 a c^{2}}{b^{2}}
$$

Tivadar Molnár (Kisvárda, Bessenyei g. IV. o. t.)

III. solution: We use the theorem that the tangent at the vertex of the parabola bisects any other tangent between the point of tangency and the axis of the parabola.

According to this, the abscissa of the point of tangency is the opposite of the $x$-intercept of the tangent, i.e.,

$$
x_{1}=\frac{c^{2}}{a}
$$

The ordinate of the point of tangency is twice the $y$-intercept, i.e.,

$$
y_{1}=-\frac{2 c^{2}}{b}
$$

The point of tangency lies on the parabola, etc., as in the second solution.

Béla Tokaji (Debrecen, Ép. ip. techn. IV. o. t.)

IV. solution: We use the theorem that the perpendicular dropped from the intersection of the tangent and the vertex tangent (the $y$-axis) passes through the focus. This means that the $y$-intercept of the tangent, as the height of a right triangle, is the geometric mean of $\frac{p}{2}$ and the absolute value of the $x$-intercept of the tangent.

Thus,

$$
\frac{c^{4}}{b^{2}}=\frac{p}{2} \cdot \frac{c^{2}}{a}
$$

from which

$$
p=\frac{2 a c^{2}}{b^{2}}
$$