A conic section is given by its most general equation and two points: $M\left(x_{0}, y_{0}\right)$, and $N\left(x^{\prime}, y^{\prime}\right)$. Assuming that point $M$ is fixed, what geometric locus does $N$ describe if, in any position of $M N$, the conic section is intersected at two coincident points?
Let the most general equation of the conic section be:

$$
A x^{2}+2 B x y+C y^{2}+2 D x+2 E y+F=0
$$

The coordinates of a point $L$ on the arbitrary line $M N$ are:

$$
\frac{x_{0}+\lambda x^{\prime}}{1+\lambda}, \frac{y_{0}+\lambda y^{\prime}}{1+\lambda}
$$

and for $L$ to lie on the conic section, its coordinates must satisfy the equation (1) identically when substituted. After performing the substitution and arranging the equation in descending powers of $\lambda$, it takes the following form:

$$
P \lambda^{2}+2 Q \lambda+R=0
$$

where

$$
\begin{gathered}
P=A x^{\prime 2}+2 B x^{\prime} y^{\prime}+C y^{\prime 2}+2 D x^{\prime 2}+2 E y^{\prime}+F \\
Q=\left(A x_{0}+B y_{0}+D\right) x^{\prime}+\left(B x_{0}+C y_{0}+E\right) y^{\prime}+\left(D x_{0}+E y_{0}+F\right) \\
R=A x_{0}^{2}+B x_{0} y_{0}+C y_{0}^{2}+2 D x_{0}+2 E y_{0}+F
\end{gathered}
$$

For the line $M N$ to intersect the conic section at two coincident points, the equation (2) must have two equal roots. The condition for this is:

$$
Q^{2}-P R=0
$$

which is simultaneously the equation of the desired geometric locus.

If we rearrange this with respect to $x^{\prime}$ and $y^{\prime}$, we get:

$$
a x^{\prime 2}+2 b x^{\prime} y^{\prime}+c y^{\prime 2}+2 d x^{\prime}+2 e y^{\prime}+f=0
$$

where

$$
\begin{array}{lll}
a=\alpha^{2}-R A, & b=\alpha \beta-R B, & c=\beta^{2}-R C \\
d=\alpha \gamma-R D, & e=\beta \gamma-R E, & f=\gamma^{2}-R F
\end{array}
$$

when

$$
\begin{aligned}
\alpha & =A x_{0}+B y_{0}+D \\
\beta & =B x_{0}+C y_{0}+E \\
\gamma & =D x_{0}+E y_{0}+F
\end{aligned}
$$

The detailed form of the coefficients in equation (4) is:

$$
\begin{gathered}
a=\left(B^{2}-A C\right) y_{0}^{2}+2(B D-A D) y_{0}+\left(D^{2}-A F\right) \\
b=\left(A C-B^{2}\right) x_{0} y_{0}+(A E-B D) x_{0}+(C D-B E) y_{0}+(D E-B F) \\
c=\left(B^{2}-A C\right) x_{0}^{2}+2(B E-C D) x_{0}+\left(E^{2}-C F\right) \\
d=(A E-B D) x_{0} y_{0}+(B E-C D) y_{0}^{2}+\left(A F-D^{2}\right) x_{0}+(B F-D E) y_{0} \\
e=(B D-A E) x_{0}^{2}+(C D-B E) x_{0} y_{0}+(B F-D E) x_{0}+\left(C F-E^{2}\right) y_{0} \\
f=\left(D^{2}-A F\right) x_{0}^{2}+2(D E-B F) x_{0} y_{0}+\left(E^{2}-C F\right) y_{0}^{2}
\end{gathered}
$$

while this can also be written in the following form:

$$
\left(a x^{\prime}+b y^{\prime}+d\right) x^{\prime}+\left(b x^{\prime}+c y^{\prime}+e\right) y^{\prime}+\left(d x^{\prime}+e y^{\prime}+f\right)=0
$$

If in this equation $x^{\prime}$ and $y^{\prime}$ are replaced by

$$
x+x_{0} \text { and } y+y_{0}
$$

i.e., the coordinate system is translated parallel to itself so that its origin coincides with the point $M\left(x_{0}, y_{0}\right)$, equation (5) takes the following form:

$$
\begin{gathered}
\left(a x^{2}+2 b x y+c^{2}+2 a x_{0}+b y_{0}+d\right) x+2\left(b x_{0}+c y_{0}+c\right) y+ \\
+\left(a x_{0}+b y_{0}+d\right) x_{0}+\left(b x_{0}+c y_{0}+e\right) y_{0}+\left(d x_{0}+e y_{0}+f\right)=0
\end{gathered}
$$

Substituting the detailed values of $a, b, c, d, e$, and $f$, the following equations are identically satisfied:

$$
\begin{aligned}
a x_{0}+b y_{0}+d & =0 \\
b x_{0}+c y_{0}+e & =0 \\
d x_{0}+c y_{0}+f & =0
\end{aligned}
$$

and equation (6) finally takes the following simple form:

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

from which it is immediately clear that it is the equation of a pair of lines, as it does not contain a term independent of $x$ and $y$.