1. Determine the geometric locus of points $M$ in the Cartesian coordinate plane such that the tangents drawn from point $M$ to the parabola $y=x^{2}$ are mutually perpendicular.
1. The equation of the tangent passing through the point ($x_{0}, y_{0}$) can be written in the form

$$
y=k\left(x-x_{0}\right)+y_{0}
$$

where $k$ is the slope of the tangent. Since it touches the parabola $y=x^{2}$, we have

$$
\begin{gathered}
x^{2}=k\left(x-x_{0}\right)+y_{0} \\
x^{2}-k x+k x_{0}-y_{0}=0
\end{gathered}
$$

Since the tangent touches the parabola at one and only one point, the discriminant of this quadratic equation must be zero, i.e.,

$$
k^{2}-4 x_{0} k+4 y_{0}=0 . \quad 5 \text { points }
$$

The solutions of this quadratic equation are the slopes $k_{1}$ and $k_{2}$ of the tangents from the points of the desired set to the parabola. Since they are perpendicular to each other, we must have $k_{1} k_{2}=-1$. From the quadratic equation (1), we get

$$
k_{1} k_{2}=4 y_{0}, \quad \text { i.e., } \quad 4 y_{0}=-1 . \quad 5 \text { points }
$$

We see that the condition of perpendicularity does not depend on $x_{0}$. Therefore, the desired set of points is the line $y=-\frac{1}{4}$.