Determine the geometric place of those points from which the parabola $y^{2}-2 p x=0$ is seen at an angle of $45^{\circ}$.
1. solution: We write down the equation of the tangents that can be drawn from an arbitrary point $P(u, v)$ to the parabola, and then the condition that the two tangents form a $45^{\circ}$ angle with each other. Thus, we obtain a relationship between the coordinates of the points satisfying the condition, which is the equation of the sought geometric locus.

The line passing through the point $P(u, v)$, whose equation is

$$
y-v=m(x-u)
$$

will be tangent to the parabola

$$
y^{2}=2 p x
$$

if the two intersection points of the line with the parabola coincide. Express $x$ from (1) and substitute the obtained value into (2). After rearranging, we get the following equation:

$$
m y^{2}-2 p y+(2 p v-2 m p u)=0
$$

We will get coinciding roots for $y$ if the discriminant of the equation is

$$
4 p^{2}-8 m p v+8 m^{2} p u=0
$$

simplified and rearranged, this is

$$
2 u m^{2}-2 v m+p=0
$$

Solving this equation for $m$, we get the slopes of the two tangents that can be drawn from the point $P(u, v)$ to the parabola.

$$
m_{1}=\frac{v+\sqrt{v^{2}-2 p u}}{2 u} \quad \text { and } m_{2}=\frac{v-\sqrt{v^{2}-2 p u}}{2 u}
$$

The condition that the two tangents form a $45^{\circ}$ angle with each other is

$$
\operatorname{tg} 45^{\circ}=1=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}, \quad \text { hence } \quad 1+m_{1} m_{2}=m_{1}-m_{2}
$$

Substituting the values of $m_{1}$ and $m_{2}$ from (3) into this last equation, we get the equation of the sought geometric locus:

$$
1+\frac{v+\sqrt{v^{2}-2 p u}}{2 u} \cdot \frac{v-\sqrt{v-2 p u}}{2 u}=\frac{v+\sqrt{v^{2}-2 p u}}{2 u}-\frac{v-\sqrt{v^{2}-2 p u}}{2 u}
$$

simplified, this is

$$
1+\frac{p}{2 u}=\frac{\sqrt{v^{2}-2 p u}}{u}
$$

We will transform this equation to determine what kind of second-order curve it represents and what are the characteristic data of this curve. Multiply through by $u$, square both sides, and complete the square for the terms involving $u$:

$$
\left(u+\frac{3}{2} p\right)^{2}-v^{2}=2 p^{2}, \quad \text { or } \quad \frac{\left(u+\frac{3}{2} p\right)^{2}}{2 p^{2}}-\frac{v^{2}}{2 p^{2}}=1
$$

This is the equation of a hyperbola with equal axes. The hyperbola's major axis coincides with the axis of the parabola, its center is at $O\left(-\frac{3 p}{2}, 0\right)$, and the lengths of its semi-axes are $a=b=p \sqrt{2}$. The linear eccentricity is $c=\sqrt{a^{2}-b^{2}}=a \sqrt{2}=2 p$, so one of the foci of the hyperbola, $F_{2}\left(\frac{p}{2}, 0\right)$, coincides with the focus $F$ of the parabola (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_1974a359c6352896c2b7g-1.jpg?height=501&width=854&top_left_y=2205&top_left_x=625)

The sought geometric locus is not the entire hyperbola, but only the branch that is farther from the parabola. The other branch of the hyperbola consists of points from which the parabola is seen at an angle of $135^{\circ}$. Indeed, during the transformation, we squared and thus combined the equation

$$
1=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}} \quad \text { with the equation } \quad-1=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}
$$

The latter equation, however, expresses that the two tangents form a $135^{\circ}$ angle with each other.

$$
\text { László Tarlacz (Szombathely, Nagy Lajos g. III. o. t.) }
$$

Second solution: Let $P(x, y)$ be a point on the sought geometric locus, then the tangents that can be drawn from this point to the parabola form a $45^{\circ}$ angle. The equation of the tangent to the parabola at some point $P\left(x_{1}, y_{1}\right)$ is

$$
y_{1} y-p\left(x_{1}+x\right)=0
$$

Since the point of tangency lies on the parabola, we have $y_{1}^{2}-2 p x_{1}=0$, and thus the equation of the tangent can also be written as

$$
y_{1}^{2}-2 y y_{1}+2 p x=0
$$

Solving this equation for $y_{1}$, we get

$$
y_{1}= \pm \sqrt{y^{2}-2 p x}
$$

Thus, the ordinates of the points of tangency of the tangents that can be drawn from the point $P(x, y)$ to the parabola are

$$
y_{1}=y+\sqrt{y^{2}-2 p x}, \quad y_{2}=y-\sqrt{y^{2}-2 p x}
$$

The slopes of the tangents are

$$
m_{4}=\frac{p}{y_{1}}, \quad \text { and } \quad m_{2}=\frac{p}{y_{2}}
$$

Thus, using the formula for the tangent of the angle between two lines, we get

$$
1=\frac{p y_{1}-p y_{2}}{y_{1} y_{2}+p^{2}}=\frac{2 \sqrt{y^{2}-2 p x}}{2 x+p}
$$

From this, the equation of the sought geometric locus is

$$
4 x^{2}-4 y^{2}+12 p x+p^{2}=0
$$

This can also be written as:

$$
\left(x+\frac{3 p}{2}\right)^{2}-y^{2}-2 p^{2}=0
$$

which agrees with the result obtained in the first solution.

Pál Bártfai (Bp. I. Petőfi g. III. o. t.)