It is known - and can be easily read from the central equation of the ellipse - that points of an ellipse defined by its two axes can be constructed by the following procedure: over both axes, as if they were diameters, we draw circles, then we draw perpendiculars to the appropriate axes through the intersection points of a ray originating from the center with the circles, the intersection of these is a point of the ellipse. Let us give a procedure for constructing points of a hyperbola based on its central equation.
The quoted construction can also be considered a procedure for constructing the ordinate of a point on the ellipse given by the equation $(x / a)^{2}+(y / b)^{2}=1$, if the abscissa of the point is given, or for constructing the abscissa if the ordinate is given. Indeed, if we mark the point on the semicircle with radii $a$ and $b$, we have selected the abscissa or the ordinate of the point to be constructed. Similarly, with two different starting points, we can interpret the task of constructing points on the hyperbola given by the equation $(x / a)^{2}-(y / b)^{2}=1$.

The procedure to be demonstrated is similar to the quoted one in that the compass is used only in the first point construction, and the construction of further points can be done using a pair of right-angled rulers (assuming that drawing parallel and perpendicular lines by sliding and rotating is allowed, as is customary in the quoted procedure). The hyperbola is symmetric with respect to both axes, so we only consider points in the first quadrant, without regard to the direction of the hyperbola's axes as coordinate axes.

![](https://cdn.mathpix.com/cropped/2024_05_02_eec0f77664febb907a99g-1.jpg?height=561&width=542&top_left_y=603&top_left_x=781)

As a preparatory step, we draw the circle $k$ with radius $a$ centered at the origin $O$ (the focal circle of our hyperbola) and the line $e$ perpendicular to the $X$-axis and intersecting it at a distance $b$ from $O$ (the segment $b$ can be constructed from the segments $a$ and $O F_{1}=c$ using $b=\sqrt{c^{2}-a^{2}}$). Given the ordinate $y_{1}$ of the sought point $P$ on the hyperbola, we can obtain the point itself through the following steps: 1. Measure $y_{1}$ on $e$ from the point where it intersects the $X$-axis, and draw a perpendicular $q$ to $e$ through the endpoint $Q$; draw: 2. the ray $O Q$, let its intersection with $k$ be $R$; 3. the tangent to $k$ at $R$, let its intersection with the $X$-axis be $S$, and finally 4. the line through $S$ perpendicular to the $X$-axis; its intersection with $q$ is the sought point $P$.

We will show that the abscissa of $P$ at $O S$ is equal to the value

$$
x_{1}=\frac{a}{b} \sqrt{b^{2}+y_{1}^{2}}
$$

calculated from the ordinate $y_{1}$ based on the equation of the hyperbola. Let the positive half of the $X$-axis intersect $e$ at $T$, $k$ at $U$, and the tangent to $k$ at $U$ intersect $O Q$ at $V$. Then $\triangle O R S$ and $\triangle O U V$ are congruent triangles, and $\triangle O T Q$ is similar to them, so

$$
O S=O V=\frac{O U}{O T} \cdot O Q=\frac{a}{b} \sqrt{O T^{2}+T Q^{2}}=\frac{a}{b} \sqrt{b^{2}+y_{1}^{2}}=x_{1}
$$

which is what we wanted to prove.

Conversely, if the abscissa of $P$ is given, we measure it on the $X$-axis to get $S$, draw a perpendicular $n$ to the $X$-axis through $S$, draw a tangent to $k$ from $S$, find the intersection of the line through the tangent point with $O$ and $e$, and finally draw a line parallel to the $X$-axis through this intersection to find $P$ from the intersection with $n$.

We can consider our procedures as the construction of the intersection points of the hyperbola (or the ellipse) with a line parallel to one of its axes.

László Lovász (Budapest, Fazekas M. G.)

Remarks. 1. The correctness of the procedure can also be easily verified by expressing $x$ and $y$ in terms of the angle $\varphi = \angle Q O T$.

Róbert Kersner (Ajka, Bródy I. g. III. o. t.)

2. Several papers transferred the segment $O V$ to $O S$ or vice versa using a compass. In this case, the line $U V$ at a distance $a$ from the $Y$-axis should be drawn instead of $k$. The above procedure skillfully transfers $O V$ to $O S$ without using a compass, but in the reverse procedure, the drawing of the tangent by simply aligning the ruler can lead to errors in practice.