In a Cartesian coordinate system, the line $x-d=0$ is given. Consider all conic sections whose focus is the origin and the corresponding directrix is the line $x-d=0$. Draw tangents to these conic sections that are parallel to the line $y=x$. What is the geometric locus of the points of tangency?
We start from the definition of conic sections, which states that a conic section is the geometric locus of points for which the ratio of the distances from a fixed point and a fixed line is constant. In the given case, the fixed point is the origin, one of the foci of the conic section, and the fixed line is the corresponding directrix, whose equation is $x-d=0$.

For a point $P(x, y)$, the distance from the origin is: $\left(x^{2}+y^{2}\right)^{\frac{1}{2}}$.

The distance from the point $P$ to the given line is $|x-d|$.

The point $P$ must satisfy the following equation:

$$
\frac{\left(x^{2}+y^{2}\right)^{\frac{1}{2}}}{|x-d|}=k \quad \text { or } \quad x^{2}+y^{2}-\lambda(x-d)^{2}=0 \ldots
$$

where $\lambda=k^{2}$ is an arbitrary positive number (or zero), a parameter. The 1) series of conic sections includes ellipses, a parabola, and hyperbolas.

Let's now examine the points on these conic sections for which the tangents are parallel to the line $y=x$, i.e., the direction coefficient of these tangents is 1.

Differentiating equation (1) with respect to $x$:

$$
2 x+2 y y^{\prime}-2 \lambda(x-d)=0, \quad \text { or } \quad x+y y^{\prime}-\lambda(x-d)=0 \ldots
$$

If $y^{\prime}$, the direction coefficient of the tangent, is $=1$, then the following relationship holds between the coordinates of the points of tangency:

$$
x+y-\lambda(x-d)=0 \ldots
$$

Solving the system of equations (1) and (3) for $x$ and $y$ as functions of $\lambda$, we obtain the parametric equations of the geometric locus of the points of tangency. If we eliminate $\lambda$ from (1) and (3), we get the relationship between the $x, y$ coordinates of the points of tangency, which is the equation of the geometric locus in question.

From (3), $\lambda=\frac{x+y}{x-d}$. Substituting this into (1), we get:

$$
\begin{gathered}
x^{2}+y^{2}-(x+y)(x-d)=0 \\
\text { or } \quad y^{2}-x y+d(x+y)=0 \ldots
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_0adb2b30d826f90d39ccg-1.jpg?height=523&width=724&top_left_y=1468&top_left_x=690)

Equation (4) is clearly the equation of a hyperbola; the quadratic terms are:

$y^{2}-x y=y(y-x)$. From this, we can see that the asymptotes of the hyperbola are parallel to the lines $y=0$ and $y-x=0$. The center of the hyperbola is determined by the system of equations:

$$
\begin{gathered}
-y+d=0 \ldots \\
2 y-x+d=0 \ldots
\end{gathered}
$$

The coordinates of the center are: $3 d, d$. Therefore, the equations of the asymptotes are:

$$
\begin{gathered}
y=d \ldots \\
y=x-2 d \ldots
\end{gathered}
$$

From (4), it follows that the line $x+y=0$ is tangent to the hyperbola at the origin. (The line $y=-x$ has two coincident points of intersection with the hyperbola (4) at the origin!)

In the case $\lambda=0$, we obtain the circle of zero radius, $x^{2}+y^{2}=0$.