Through every point $M$ of the plane, two hyperbolas can be drawn, whose equations, referred to the rectangular coordinate axes $O x, O y$, are

$$
a^{2} x y+a y+x=0
$$

where $a$ is a variable parameter; on which curve $(C)$ must the point $M$ lie so that the hyperbolas intersect at right angles at this point?

How many points does the curve $(C)$ intersect the hyperbolas with the equation $a^{2} x y+a y+x=0$. Only those points should be considered which are neither at infinity nor coincide with the origin of the coordinate system.

What relationship must hold between $a$ and $b$ so that the following two hyperbolas, whose equations are

$$
\begin{aligned}
& a^{2} x y+a y+x=0 \\
& b^{2} x y+b y+x=0
\end{aligned}
$$

intersect at right angles at a point different from the origin of the coordinate system. This relationship defines a curve $(C')$; the curve must be constructed. How many finite and non-origin points does the previously defined curve $(C)$ intersect the curve $(C')$?

(Licentiatus stipendium competition written paper, 1894.)
To express the condition that the curve $(a)$ passes through the point $M(\alpha, \beta)$, we obtain the following equation:

$$
a^{2} \alpha \beta + a \beta + \alpha = 0
$$

which provides two values of $a$, denoted as $a'$ and $a''$. To express the condition that the tangents to the hyperbolas at point $M$ are perpendicular to each other, we use the following equations for the tangents:

$$
\begin{gathered}
\left(a'^{2} \beta + 1\right)(x - \alpha) + \left(a'^{2} \alpha + a'\right)(y - \beta) = 0 \\
\left(a''^{2} \beta + 1\right)(x - \alpha) + \left(a''^{2} \alpha + a''\right)(y - \beta) = 0
\end{gathered}
$$

Thus, the condition for perpendicularity is:

$$
\left(a'^{2} \beta + 1\right)\left(a''^{2} \beta + 1\right) + \left(a'^{2} \alpha + a'\right)\left(a''^{2} \alpha + a''\right) = 0
$$

or expanded,

$$
\left(a'^{2} a''^{2} (\alpha^{2} + \beta^{2}) + (a'^{2} + a''^{2}) \beta + a' a'' (a' + a'') \alpha + a' a'' + 1 = 0\right.
$$

From the equation (1), it follows that:

$$
a' a'' = \frac{1}{\beta} \quad a' + a'' = -\frac{1}{\alpha}
$$

which gives

$$
a'^{2} + a''^{2} = (a' + a'')^{2} - 2 a' a'' = \frac{1}{\alpha^{2}} - \frac{2}{\beta}
$$

Substituting these values into (2), we get:

$$
\frac{\alpha^{2} + \beta^{2}}{\beta^{2}} + \frac{\beta}{\alpha^{2}} - 1 = 0
$$

or, after removing the denominators,

$$
\alpha^{4} + \beta^{3} = 0
$$

Thus, the curve $(C)$ is a fourth-order, parabolic curve. To determine the number of real and finite intersection points of the curves

$$
\begin{gathered}
x^{4} + y^{3} = 0 \\
a^{2} x y + a y + x = 0
\end{gathered}
$$

we find the number of lines passing through the origin of the coordinate system and one of the intersection points of the two curves. The equation of such a line is:

$$
y = u x
$$

From this equation and the two preceding ones, we eliminate $x$ and $y$. The number of real roots of the resulting equation, where $u$ is the unknown, will be the number of the sought points. After elimination, we obtain the following equation:

$$
a^{2} u^{4} - a u - 1 = 0
$$

This equation has only 2 real roots, one positive and one negative. If $\alpha$ and $\beta$ are the coordinates of a common point of the hyperbolas defined by the equations

$$
\begin{aligned}
& a^{2} x y + a y + x = 0 \\
& b^{2} x y + b y + x = 0
\end{aligned}
$$

the condition that these curves are perpendicular to each other at point $M(\alpha, \beta)$ is expressed by the following equation:

$$
\left(a^{2} \beta + 1\right)\left(b^{2} \beta + 1\right) + \left(a^{2} \alpha + a\right)\left(b^{2} \alpha + b\right) = 0
$$

Substituting the values of $\alpha$ and $\beta$ from the equations

$$
\begin{aligned}
& a^{2} \alpha \beta + a \beta + \alpha = 0 \\
& b^{2} \alpha \beta + b \beta + \alpha = 0
\end{aligned}
$$

we obtain a relation between $a$ and $b$ that defines the curve $(C')$. Eliminating $\alpha \beta$ from the preceding equations, we get:

$$
a b \beta + (a + b) \alpha = 0
$$

which, together with the following equations:

$$
\begin{gathered}
a^{2} \alpha \beta + a \beta + \alpha = 0 \\
\beta = \frac{1}{a b} \\
\alpha = -\frac{1}{a + b}
\end{gathered}
$$

gives the equation of $(C')$ as:

$$
\left(a^{2} \frac{1}{a b} + 1\right)\left(b^{2} \frac{1}{a b} + 1\right) + a b \left(1 - \frac{a}{a + b}\right)\left(1 - \frac{b}{a + b}\right) = 0
$$

or, after removing the denominators,

$$
(a + b)^{4} + a^{3} b^{3} = 0
$$

If $a$ and $b$ are considered as the coordinates of a point, this relation defines a curve whose equation is

$$
(x + y)^{4} + x^{3} y^{3} = 0
$$

The curve $(C')$ passes through the origin of the coordinate system. Let's find its tangent at this point.

The form of the tangent is:

$$
y - u x = 0
$$

We ask for which values of $u$ this line and the curve $(C')$ have at least two coinciding points. Eliminating $y$ from the two equations, we get:

$$
x^{2} = -\frac{(1 + u)^{4}}{u^{3}}
$$

and this equation has two equal roots when $u = -1$; thus, the equation of the sought tangent is

$$
x + y = 0
$$

or a line that forms a $135^{\circ}$ angle with the positive $x$-axis.

It is also evident from the equation of $(C')$ that the curve has real points only where the coordinates of the points have different signs; or from the equations

$$
\begin{gathered}
y = u x \\
x^{2} = -\frac{(1 + u)^{4}}{u^{3}}
\end{gathered}
$$

if $u < 0$.

The curve is centrally symmetric with respect to the origin of the coordinate system; it is sufficient to study the variations of $x^{2}$ as $u$ varies; $u$ can vary from $-\infty$ to 0.

Let's examine the variations of the function

$$
z = -\frac{(1 + u)^{4}}{u^{3}}
$$

The derivative of the function, or the expression

$$
\lim _{u_{1} \to u} \frac{z_{1} - z}{u_{1} - u} = z' = \frac{(1 + u)^{3}}{u^{4}} (3 - u)
$$

always has the same sign as $(1 + u)$; thus, the variations of $x^{2} = z$ are shown in the following table:

| $u$ | $-\infty$ | - | -1 | - | 0 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $z'$ | $-\infty$ | - | 0 | + | $+\infty$ |
| $x^{2}$ | $+\infty$ | decreases | 0 | increases | $+\infty$ |

Let's now find the intersection points of $(C)$ and $(C')$. Eliminating $x$ and $y$ from the equations

$$
\begin{aligned}
x^{3} y^{3} + (x + y)^{4} & = 0 \\
x^{4} + y^{3} & = 0 \\
y - u x & = 0
\end{aligned}
$$

we obtain the following result:

$$
u^{9} + (1 + u)^{4} = 0
$$

It also follows from the discussion in the next section that the above equation has only one real root, which lies between 1 and -1. Therefore, the two curves intersect at only one finite point different from the origin.

*) See the next section.

**) Omitted due to space constraints in this issue. It will be published in the next double issue.