Ellipse intersections

Choose $P$ with coordinates $(20, q)$ . Then $PA$ is a vertical line, and by symmetry $Q$ has coordinates $(20, -q)$ . From the equation of the ellipse, $A_1$ has coordinates $(-\sqrt{2020}, 0)$ and $A_2$ has coordinates $(\sqrt{2020}, 0)$ .

The equation of the line $A_1P$ is then $y = \frac{q}{\sqrt{2020} + 20}(x + \sqrt{2020})$ and the equation of the line $QA_2$ is $y = \frac{q}{\sqrt{2020} - 20}(x - \sqrt{2020})$ .

Since $S$ is on both $A_1P$ and $QA_2$ , $\frac{q}{\sqrt{2020} + 20}(\lambda + \sqrt{2020}) = \frac{q}{\sqrt{2020} - 20}(\lambda - \sqrt{2020})$ , which solves to $\lambda = \boxed{101}$ .

If the equation of the ellipse be $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ and the position coordinates of $A$ be $(p, 0)$ , then the locus of $S$ is $x=\dfrac{a^2}{p}$ . In this question, $a^2=2020, p=20$ . So the locus of $S$ is $x=\dfrac{2020}{20}=101$ . So $\lambda =\boxed {101}$ .