Example 4 Given the ellipse $C:(x-2)^{2}+\frac{y^{2}}{4}=1$, the line $l: x-y+1=0$, and the point $R(1,4)$.
( I ) Find a point $M$ inside the ellipse $C$ that satisfies the following condition: If a line through $M$ intersects line $l$ at point $Q$ and intersects the ellipse $C$ at two points $A, B$, then it must be true that $\frac{|A M|}{|M B|}=\frac{|A Q|}{|Q B|}$. Does such a point $M$ exist?
(II) Draw a line through point $R$ that intersects the ellipse at points $E, F$. If $|R E|=|E F|$, find the equation of the line $E F$.
(I) Let the coordinates of point $Q$ be $(a, a+1)$ (where $a$ is a parameter), the equation of the ellipse $C$ can be transformed into $4 x^{2}+y^{2}-16 x+12=0$. Using Property 8, in this equation, substitute $a x$ for $x^{2}$, $(a+1) y$ for $y^{2}$, and $\frac{x+a}{2}$ for $x$, then we get $4 a x+(a+1) y-16 \cdot \frac{x+a}{2}+12=0$, which can be simplified to $y-8 x+12+a(4 x+y-8)=0$. This represents a system of lines passing through the intersection point $\left(\frac{5}{3}, \frac{4}{3}\right)$ of the lines $y-8 x+12=0$ and $4 x+y-8=0$.
Point $\left(\frac{5}{3}, \frac{4}{3}\right)$ is inside the ellipse $C$, and it is the required point $M$.
(II) Let $P$ be a point on the line segment $E F$ and satisfy $\frac{|E P|}{|P F|}=\frac{|E R|}{|R F|}$.
In the equation of the ellipse $C$, $4 x^{2}+y^{2}-16 x^{2}+12=0$, using Property 8, substitute $x$ for $x^{2}$, $4 y$ for $y^{2}$, and $\frac{x+1}{2}$ for $x$, we get $4 x+4 y-16 \cdot \frac{x+1}{2}+12=0$, which simplifies to $x-y-1=0$.

Since point $P$ satisfies $\frac{|E P|}{|P F|}=\frac{|E R|}{|R F|}$, it must lie on the line $x-y-1=0$, so let its coordinates be $(b, b-1)$.

Since $R$ is a point outside the ellipse $C$, $P$ is a point inside the ellipse $C$, and $\frac{|E P|}{|P F|}=\frac{|E R|}{|R F|}=\frac{1}{2}$, point $E$ is the internal division point of the directed line segment $\overrightarrow{R P}$, and we can find that $\frac{R E}{E P}=3$.

Let the coordinates of point $E$ be $(x, y)$, then $x=\frac{1+3 b}{1+3}=\frac{3 b+1}{4}, y=\frac{4+3(b-1)}{1+3}=\frac{3 b+1}{4}$, substituting these into the equation of $C$ $(x-2)^{2}+\frac{y^{2}}{4}=1$, and simplifying, we get $45 b^{2}-162 b+133=0$, from which we find $b=\frac{7}{3}$ or $b=\frac{19}{15}$, so the coordinates of point $P$ are $\left(\frac{7}{3}, \frac{4}{3}\right)$ or $\left(\frac{19}{15}, \frac{4}{15}\right)$.
From the coordinates of points $R$ and $P$, we can find that the equation of the line $E F$ is $2 x+y-6=0$ or $14 x+y-18=0$.