China Mathematical Competition

As shown in Fig. 11.1, we find $F_1(-1, 0), Q_2(1, 0)$.
Denote $P(x_0, y_0)$. By the condition, it follows that $x_0, y_0 > 0$, $y_1 < 0$, $y_2 < 0$.

Fig. 11.1

The equation of line $PF_1$ is $x = \frac{(x_0 + 1)y}{y_0} - 1$. Substituting it into $\frac{x^2}{2} + y^2 = 1$ and organizing it yields
$$\left( \frac{(x_0 + 1)^2}{2y_0^2} + 1 \right) y^2 - \frac{x_0 + 1}{y_0} y - \frac{1}{2} = 0.$$
Multiplying both sides by $2y_0^2$ and noting that $x_0^2 + 2y_0^2 = 2$, we get
$$(3 + 2x_0)y^2 - 2(x_0 + 1)y_0y - y_0^2 = 0.$$
The two roots of this equation are $y_0, y_1$. By Vieta's formulas, we get $y_0y_1 = -\frac{y_0^2}{3+2x_0}$. Thus,
$$y_1 = -\frac{y_0}{3 + 2x_0}.$$
Similarly, we can get $y_2 = -\frac{y_0}{3 - 2x_0}$. Therefore,
$$y_1 - y_2 = \frac{y_0}{3 - 2x_0} - \frac{y_0}{3 + 2x_0} = \frac{4x_0y_0}{9 - 4x_0^2}.$$
Since $9 - 4x_0^2 = \frac{1}{2}x_0^2 + 9y_0^2 \ge 2\sqrt{\frac{1}{2}x_0^2 \cdot 9y_0^2} = 3\sqrt{2}x_0y_0$, it follows that
$$y_1 - y_2 \le \frac{4x_0y_0}{3\sqrt{2}x_0y_0} = \frac{2\sqrt{2}}{3},$$
where the equal sign holds when $\frac{1}{2}x_0^2 = 9y_0^2$ is required, and accordingly $x_0 = \frac{3\sqrt{5}}{5}, y_0 = \frac{\sqrt{10}}{10}$.
Therefore, the maximum of $y_1 - y_2$ is $\frac{2\sqrt{2}}{3}$.