# Answer: $\left(\frac{1}{8} ; \frac{1}{4}\right)$.

Solution. By plotting the parabolas with the given equations, it is easy to notice that they have 4 common points. Therefore, there can be at most one point equidistant from them. Rewrite the equations of the parabolas as $x^{2}-\frac{1}{2} y-\frac{1}{2}=0$ and $y^{2}-\frac{1}{4} x-\frac{1}{2}=0$ and add them. We get the equation

$$
x^{2}+y^{2}-\frac{1}{4} x-\frac{1}{2} y-1=0 \Longleftrightarrow\left(x-\frac{1}{8}\right)^{2}+\left(y-\frac{1}{4}\right)^{2}=\frac{69}{64}
$$

This is the equation of a circle with center $\left(\frac{1}{8} ; \frac{1}{4}\right)$ and radius $\frac{\sqrt{69}}{8}$, and it is satisfied by the coordinates of all the intersection points of the parabolas, while the center of the circle is equidistant from these intersection points.
Problem 6. Find the coordinates of all points in the plane that are equidistant from all points of intersection of the parabolas given in the Cartesian coordinate system on the plane by the equations $y=2 x^{2}-1$ and $x=4 y^{2}-2$.

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