4A. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, where $a>b>0$. What does the geometric locus of points represent that are midpoints of all chords of the ellipse passing through the right focus $F_{2}(e, 0)$?
Solution. The equation of an arbitrary line through the right focus is:

$$
y=k(x-e) .
$$

By substituting into the equation of the ellipse, we get:

$$
\left(b^{2}+a^{2} k^{2}\right) x^{2}-2 a^{2} k^{2} e x+a^{2} k^{2} e^{2}-a^{2} b^{2}=0
$$

If we denote the solutions of the quadratic equation (2) by $x_{1}$ and $x_{2}$, which are actually the abscissas of the endpoints of the chord defined by the line (1), the abscissa of the midpoint of the corresponding chord, according to Vieta's formulas, is:

$$
x=\frac{x_{1}+x_{2}}{2}=\frac{a^{2} k^{2} e}{b^{2}+a^{2} k^{2}}
$$

The corresponding ordinate is:

$$
y=k(x-e)=-\frac{b^{2} k e}{b^{2}+a^{2} k^{2}}
$$

From (3) and (4) it follows:

$$
\frac{x}{y}=-\frac{a^{2} k}{b^{2}}
$$

By eliminating $k$ from (1) and (5), we obtain the equation of the desired geometric locus of points:

$$
\begin{aligned}
& b^{2}\left(x-\frac{e}{2}\right)^{2}+a^{2} y^{2}=\left(\frac{e b}{2}\right)^{2} \\
& \frac{\left(x-\frac{e}{2}\right)^{2}}{\left(\frac{e}{2}\right)^{2}}+\frac{y^{2}}{\left(\frac{e b}{2 a}\right)^{2}}=1
\end{aligned}
$$

Thus, the desired geometric locus of points is an ellipse.