114. Find the geometric locus of points $M$ on the coordinate plane such that two tangents drawn from $M$ to the parabola $y=x^{2}$ are perpendicular to each other.
114. Find the geometric locus of points \(M\) on the coordinate plane such that two tangents drawn from \(M\) to the parabola \(y = x^2\) are perpendicular to each other.

Answer: the line \(y = -\frac{1}{4}\). Hint: Let \((a; b)\) be the coordinates of point \(M\), and \(M_{1}(x_{1}; x_{1}^{2}), M_{2}(x_{2}; x_{2}^{2})\) be the points of tangency with the parabola of the two tangents from point \(M\). The equation of the tangent at point \(M_{1}\) is \(y - x_{1}^{2} = 2 x_{1}(x - x_{1})\) (here we consider that the slope of the tangent is \((x^2)' = 2x\)). Substituting the coordinates of point \(M\), we get \(x_{1}^{2} - 2 x_{1} a + b = 0\). We will get the same equation with \(x_{1}\) replaced by \(x_{2}\) for point \(M_{2}\), i.e., \(x_{1}, x_{2}\) are the roots of the equation \(x^{2} - 2 a x + b = 0\), and by Vieta's theorem \(b = x_{1} x_{2}\). From the condition of perpendicularity of the two tangents, we have \((2 x_{2}) = -\frac{1}{2 x_{1}} \Leftrightarrow x_{1} x_{2} = -\frac{1}{4}\). Therefore, \(b = x_{1} x_{2} = -\frac{1}{4}\).