The line $AB = 2d$ is perpendicularly bisected by the line $e$; on this, the segment $RS = d$ slides. The lines $AR$ and $BS$ intersect at a point $P$. What is the geometric locus of the point $P$?
In our rectangular coordinate system, the $X$-axis is the line $AB$, and the $Y$-axis is the line $e$. The coordinates of point $A$ are $(-d, 0)$, and those of point $B$ are $(d, 0)$. The coordinates of the variable point $R$ are $(0, \lambda)$, and those of point $S$ are $(0, \lambda+d)$.

![](https://cdn.mathpix.com/cropped/2024_05_02_c9d26b7ff8e887975d00g-1.jpg?height=504&width=423&top_left_y=263&top_left_x=840)

The equation of the line $AR$ is:

$$
y=\frac{\lambda}{d}(x+d) \ldots
$$

The equation of the line $BS$ is:

$$
y=-\frac{\lambda+d}{d}(x-d) \ldots
$$

The coordinates of the intersection point of the two lines are provided by solving a system of two equations, as functions of the parameter $\lambda$. If we eliminate $\lambda$ from the two equations, we obtain a relationship between the coordinates of the intersection point; this will be the equation of the desired geometric locus.

From (1), $\lambda=\frac{d y}{x+d}$. Substituting this into (2):

$$
y=-\frac{1}{d}\left(\frac{d y}{x+d}+d\right)(x-d) \quad \text { or } \quad(x+d) y=-(y+x+d)(x-d) \ldots
$$

After performing the indicated operations and simplifying, we get:

$$
x^{2}+2 x y=d^{2} \ldots
$$

It is evident that we are dealing with a hyperbola, whose center is the origin, passing through points $A$ and $B$, and whose asymptotes are the pair of lines given by:

$$
x^{2}+2 x y \equiv x(x+2 y=0) \ldots
$$

Thus, one asymptote of the hyperbola is the line $x=0$, which is the $Y$-axis, and the other asymptote is the line $x+2 y=0$. (Its slope is $-\frac{1}{2}!$)

If $R$ is at point $O$ $(\lambda=0)$, then $AR$ intersects $BS$ at point $B$ $(P \equiv B)$.

If $S$ is at point $O$ $(\lambda=-d)$, then the intersection point of $AR$ and $BS$ is $P \equiv A$.

If $R$ moves to infinity along the $Y$-axis, then $AR$ and $BS$ are parallel to the $Y$-axis: $P$ is the point at infinity on the $Y$-axis.

If $\lambda=-\frac{d}{2}$, i.e., $O$ bisects the distance $RS$, then $AR$ and $BS$ are parallel, and their slope is $-\frac{1}{2}$. The intersection point of $AR$ and $BS$ is at infinity on a line with slope $-\frac{1}{2}$.

Let the positions of points $R$ and $S$ in the latter case be $R_{0}\left(0,-\frac{d}{2}\right)$ and $S_{0}\left(0, \frac{d}{2}\right)$. If $R$ traces the part of the $Y$-axis above $R_{0}$ $\left(\lambda>-\frac{d}{2}\right)$, then point $P$ traces the branch of the hyperbola that lies on the positive side of the $X$-axis; if $R$ moves from $R_{0}$ towards $-\infty$ on the $Y$-axis, then point $P$ traces the other branch of the hyperbola, which lies on the negative side of the $X$-axis.

Géza Freud (Berzsenyi Dániel g. VII. o. Bp. V.)[^0]

Note. a) If we solve the system of equations (1) and (2) for $x$ and $y$:

$$
x=\frac{d^{2}}{2 \lambda+d}, \quad y=2 \lambda \frac{\lambda+d}{2 \lambda+d}
$$

From this parametric representation, the observations we made about the position of the curve can be read off.

b) Knowing the positions of the asymptotes, we can construct the transverse axes.

c) If the upper endpoint of the segment $RS$ is $R$, then point $P$ traces the curve:

$$
x^{2}-2 x y=d^{2}
$$

The asymptotes of this curve are the lines $x=0$ and $x-2 y=0$.

[^0]:    ${ }^{1}(-d, 0)$ and $(d, 0)$ satisfy equation (4)!

    ${ }^{2} \mathrm{If} \lambda=-\frac{d}{2}$, then equation (1) becomes: $y=-\frac{1}{2}(x+d)$, and equation (2) becomes: $y=-\frac{1}{2}(x-d)$.