Consider the curve family characterized by the equation

$$
y=(p-1) x^{2}+2 p x+4
$$

1. Determine the parameter $p$ so that the parabola $g_{p}$ is tangent to the $x$-axis. (Denote the point of tangency by $A$.)
2. Determine $p$ so that the vertex of the parabola $g_{p}$ (denoted by $B$) lies on the $y$-axis.
3. Show that the two curves mentioned above are centrally symmetric with respect to the midpoint of the segment $AB$.
4. Does there exist a point through which every curve in the family passes?
If $p=1$, then the equation represents a straight line. If $p \neq 1$, then by transforming the curve equation, we can write it as:

$$
y=(p-1)\left(x+\frac{p}{p-1}\right)^{2}-\frac{p^{2}}{p-1}+4
$$

which shows that the curve is a parabola whose axis is parallel to the $y$-axis, and the coordinates of its vertex are:

$$
u=-\frac{p}{p-1} \quad v=4-\frac{p^{2}}{p-1}
$$

1. If $g_{p}$ is tangent to the $x$-axis, then $v=0$,

$$
\frac{p^{2}}{p-1}=4, \quad \text { from which } \quad p^{2}-4 p+4=(p-2)^{2}=0
$$

thus

$$
p=2
$$

and the equation of $g_{p}$ is:

$$
y=x^{2}+4 x+4=(x+2)^{2}
$$

The vertex in this case is $\mathbf{A}(-\mathbf{2} ; \mathbf{0})$.

2. In the second case, the axis of symmetry of the parabola is the $y$-axis, which means that if we replace $x$ with $-x$, we get the same $y$ value, so the first-degree term containing $x$ must be missing from the equation of $g_{p}$, i.e.,

$$
p=0
$$

and the equation of $g_{p}$ is then:

$$
y=-x^{2}+4=-\left(x^{2}-4\right)
$$

The vertex is: $\mathbf{B}(\mathbf{0} ; \mathbf{4})$.

3. Both parabolas can be derived from the parabola $y=x^{2}$, specifically, the one with vertex $A$ by translating the parabola $y=x^{2}$ 2 units to the left, and the one with vertex $B$ by reflecting it over the point $(0 ; 2)$. Therefore, the two parabolas are congruent. If we reflect the parabola with vertex $B$ over the midpoint of $A B$, then $B$ will move to $A$, and the axes will also coincide, so the two congruent parabolas will cover each other. This proves the third statement.
4. If there exists a point through which every curve in the curve series passes, then its coordinates cannot depend on $p$. Rearrange the given equation of $g_{p}$ in terms of $p$:

$$
y=p\left(x^{2}+2 x\right)-x^{2}+4
$$

The coefficient of $p$ is:

$$
x^{2}+2 x=x(x+2)
$$

which disappears if

$$
x=0, \quad \text { in which case } \quad y=4
$$

or if

$$
x=-2, \quad \text { and then } \quad y=0
$$

These points are $A$ and $B$, so these two points lie on every curve in the series defined by $g_{p}$.

Puruczky Éva (Makó, József A. g. IV. o. t.)