\section*{Problem 3 - 121023}

In a right-angled Cartesian coordinate system, two parabolas are drawn. One is the graph of the function with the equation \(y=x^{2}\). The second is also symmetric with respect to the \(\mathrm{y}\)-axis; its vertex is \(S(0 ; 6)\).

Furthermore, it has the following property:

If perpendiculars are dropped from the intersection points \(A\) and \(B\) of both parabolas to the x-axis (the feet of the perpendiculars being \(A_{1}\) and \(B_{1}\) respectively), then the quadrilateral \(A_{1} B_{1} B A\) is a square.

Calculate the coordinates of the points where the second parabola intersects the x-axis!
}

Due to the symmetry about the y-axis and its vertex, the second parabola must have an equation of the form \(y=a x^{2}+6\) with a real \(a\).

The x-coordinates of the intersection points of \(y=x^{2}\) and \(y=a x^{2}+6\) are

\[
x_{1}=\frac{\sqrt{6}}{\sqrt{1-a}} \quad ; \quad x_{2}=-\frac{\sqrt{6}}{\sqrt{1-a}}
\]

For the quadrilateral \(A_{1} B_{1} B A\) to be a square, the ordinate of \(x_{1}\) must be equal to \(2 x_{1}\). From

\[
2 \cdot \frac{\sqrt{6}}{\sqrt{1-a}}=\left(\frac{\sqrt{6}}{\sqrt{1-a}}\right)^{2}
\]
we get \(a_{1}=-\frac{1}{2}\). The function \(y=-\frac{1}{2} x^{2}+6\) has its roots at

\[
x_{1}=2 \sqrt{3} \quad ; \quad x_{2}=-2 \sqrt{3}
\]

The sought points are thus \(S_{1}(2 \sqrt{3} ; 0)\) and \(S_{2}(-2 \sqrt{3} ; 0)\).