\section*{Exercise 2 - 121012}

In a right-angled Cartesian coordinate system, a parabola is given by the equation \(y=x^{2}\).

Provide an equation of a line that does not run parallel to the \(y\)-axis and intersects the parabola at exactly one point \(P\) with the abscissa 3!
}

The condition of not being parallel to the y-axis is precisely met by a line if it has an equation of the form \(y = mx + b\). Every line with such an equation has common points with the parabola \(y = x^2\) if and only if the equation

\[
x^2 - mx - b = 0
\]

has real solutions, and the solutions are then the x-coordinates of the common points. Therefore, \(m\) and \(b\) satisfy the conditions of the problem if and only if the equation (1) has exactly the solution \(x = 3\), which means that (1) must have the form \((x - 3)^2 = 0\), i.e., \(m = 6\) and \(b = -9\). Thus, the line \(y = 6x - 9\) is the only one that satisfies the conditions of the problem.