Four. (20 points) Given a positive integer $n(n \geqslant 5)$, let the side lengths of rectangle $ABCD$ be $AB=2a, BC=2b (a>b)$. Points $P_{1}, P_{2}, \cdots, P_{n}=C$, $Q_{1}, Q_{2}, \cdots, Q_{n}=D$, $R_{0}=D, R_{1}, R_{2}, \cdots, R_{n}=C$ satisfy
$$
\begin{array}{c}
BP_{1}=P_{1}P_{2}=\cdots=P_{n-1}P_{n}=\frac{2b}{n}, \\
AQ_{1}=Q_{1}Q_{2}=\cdots=Q_{n-1}Q_{n}=\frac{2b}{n}, \\
R_{0}R_{1}=R_{1}R_{2}=\cdots=R_{n-1}R_{n}=\frac{2a}{n}.
\end{array}
$$
(1) For $i=1,2, \cdots, n$, find the equation of the conic section $\Gamma_{1}$ on which the intersection points of $AP_{i}$ and $BR_{n-i}$, $AR_{i}$ and $BQ_{i}$ lie;
(2) If points $S_{1}, S_{2}, \cdots, S_{n}$ on the extension of $DC$ satisfy $CS_{1}=S_{1}S_{2}=\cdots=S_{n-1}S_{n}=\frac{2a}{n}$, for $i=1,2, \cdots, n$, find the equation of the conic section $\Gamma_{2}$ on which the intersection points of $AP_{i}$ and $BS_{i}$ lie;
(3) Let $M$ be the point on the conic section $\Gamma_{1}$ that is farthest from $AB$. Find the minimum distance between $M$ and a point on the conic section $\Gamma_{2}$.
(1) Establish a Cartesian coordinate system with $AB$ as the $x$-axis and the perpendicular bisector of $AB$ as the $y$-axis. Let $A P_{i}$ and $B R_{n-i}$ intersect at point $E$, and let the projections of $E$ and $R_{n-i}$ on $AB$ be $F$ and $G$, respectively, with $E(x, y)$.
From $\triangle A E F \backsim \triangle A P_{i} B$,
$\triangle B E F \backsim \triangle B R_{n-i} G$,

we have $\frac{y}{a+x}=\frac{\frac{2 i b}{n}}{2 a}=\frac{i b}{n a}, \frac{y}{a-x}=\frac{2 b}{\frac{2 i a}{n}}=\frac{n b}{i a}$.
Multiplying these two equations, we get the equation of the conic section $\Gamma_{1}$ that point $E$ satisfies: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
(2) Similarly, the intersection point of $A R_{i}$ and $B Q_{i}$ also lies on the ellipse $\frac{x^{2}}{a^{2}}+$ $\frac{y^{2}}{b^{2}}=1$. Let $A P_{i}$ and $B S_{i}$ intersect at point $H$, and let the projections of $H$ and $S_{i}$ on the line $AB$ be $K$ and $L$, respectively, with $H(x, y)$.
From $\triangle A B P_{i} \backsim \triangle A K H$,
$\triangle B H K \backsim \triangle B S_{i} L$,

we have $\frac{y}{x+a}=\frac{\frac{2 i b}{n}}{2 a}=\frac{i b}{n a}, \frac{y}{x-a}=\frac{2 b}{\frac{2 i a}{n}}=\frac{n b}{i a}$.
Multiplying these two equations, we get the equation of the conic section $\Gamma_{2}$ that point $H$ satisfies: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.
(3) Let $M(0, b)$. Consider a circle with center $M$ and radius $r$, $\odot M$. The radius when $\odot M$ is tangent to $\Gamma_{2}$ is the minimum value we seek.

Let the equation of $\odot M$ be $x^{2}+(y-b)^{2}=r^{2}$, and the equation of $\Gamma_{2}$ be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Solving these equations simultaneously and eliminating $x$, we get $\left(a^{2}+b^{2}\right) y^{2}-2 b^{3} y+a^{2} b^{2}+b^{4}-b^{2} r^{2}=0$.
By setting $\Delta=0$, we get $r^{2}=a^{2}+\frac{a^{2} b^{2}}{a^{2}+b^{2}}$.
Therefore, the minimum value we seek is
$$
\sqrt{a^{2}+\frac{a^{2} b^{2}}{a^{2}+b^{2}}}=a \sqrt{\frac{a^{2}+2 b^{2}}{a^{2}+b^{2}}} \text {. }
$$