The parabolas given by the equations $y=-x^{2}+6 x-10$ and $2 y=x^{2}+6 x+13$ are centrally similar. Find the center of similarity.
Solution. The center of central similarity is the point where the line connecting any two corresponding points intersects. Since we know that the two parabolas are centrally similar, it is enough to choose two pairs of corresponding points and determine the intersection point of the lines passing through these points.

The coordinates of the foci and vertices of the parabolas can be easily read from the equations of the two lines.

From the equation $y=-x^{2}+6 x-10=-(x-3)^{2}-1$, we get $p_{1}=-\frac{1}{2}, C_{1}(3 ;-1), F_{1}\left(3 ;-\frac{5}{4}\right)$.

From the equation $2 y=x^{2}+6 x+13 ; y=\frac{1}{2}(x+3)^{2}+2$, we get $p_{2}=1, C_{2}(-3 ; 2), F_{2}\left(-3 ; \frac{5}{2}\right)$.

The equations of the lines passing through the points $C_{1}, C_{2}$ and $F_{1}, F_{2}$ are:

$$
y=-\frac{1}{2} x+\frac{1}{2}, \quad \text { and } \quad y=-\frac{5}{8} x+\frac{5}{8}
$$

The coordinates of the intersection point, which is the center of similarity, are: $(1 ; 0)$.

Based on the work of Péter Matkó (Debrecen, Landler J. Szki., III. o. t.)