For the parabola $y=x^{2}$, from which point of the line $y=x$ can two perpendicular tangents be drawn?
The axis of the parabola given by the equation $y=x^{2}$ is the $y$-axis, and its vertex is the origin. We are looking for the point $P\left(x_{0} ; x_{0}\right)$ on the line $y=x$ from which two perpendicular tangents can be drawn to the parabola.

Since the parabola does not have a tangent parallel to the axis, the tangents passing through $P$ have a slope, and their equations can be written as

$$
y-x_{0}=m\left(x-x_{0}\right)
$$

If a line of this form is tangent to the parabola, then it has exactly one common point with the parabola. The coordinates of the common point of the parabola and the line satisfy both the equation of the parabola and the equation of the line. The two curves have exactly one common point if the discriminant of the resulting quadratic equation is 0, that is, in the equation

$$
\begin{gathered}
x^{2}-m x+(m-1) x_{0}=0 \quad \text { we have } \\
m^{2}-4 x_{0} m+4 x_{0}=0,
\end{gathered}
$$

from which, using the relationship between the roots and coefficients, we get $m_{1} m_{2}=4 x_{0}$.

We also know that the product of the slopes of perpendicular lines is -1, so $4 x_{0}=-1$, from which $x_{0}=-\frac{1}{4}$. Therefore, the point we are looking for is $P\left(-\frac{1}{4} ;-\frac{1}{4}\right)$.

Zalán Péter (Budapesti Evangélikus Gimn., 11. o.t.)

Note. Besenyei Ádám (Tatabánya, Árpád Gimn., 12. o.t.) showed that for the parabola given by the equation $y=a x^{2}$, the set of points from which two perpendicular tangents can be drawn to the parabola is a line, the directrix of the parabola. The equation of this line is $y=-\frac{1}{4 a}$. This immediately yields the solution to our problem.

![](https://cdn.mathpix.com/cropped/2024_05_02_8daed038cf7ba6ef8b41g-1.jpg?height=623&width=399&top_left_y=1044&top_left_x=863)