Problem 305

(1) Substitute the point $P(\sqrt{2},1)$ into the equation of the ellipse $\frac{x^{2}}{a^{2}}+ \frac{y^{2}}{b^{2}}=1$, which simplifies to: $\frac{2}{a^{2}}+ \frac{1}{b^{2}}=1$.

Given that the eccentricity $e = \frac{c}{a} = \frac{\sqrt{a^{2}-b^{2}}}{a} = \frac{\sqrt{2}}{2}$, solving for $a$ and $b$ yields: $a^{2} = 4$ and $b^{2} = 2$.

Hence, the equation of ellipse $C$ is $\frac{x^{2}}{4}+ \frac{y^{2}}{2}=1$.

(2) Proof: From (1), we have $A_{1}(-2,0)$ and $A_{2}(2,0)$. Let $M(2,y_{0})$ and $R(x_{1},y_{1})$.

The equation of line $MA_{1}$ is given by: $y = \frac{y_{0}}{4}x + \frac{y_{0}}{2}$. Substitute this expression for $y$ into the equation of the ellipse, we obtain: $(1+ \frac{y_{0}^{2}}{8})x^{2} + \frac{y_{0}^{2}}{2}x + \frac{y_{0}^{2}}{2} - 4 = 0$.

Since $x_{1}$ satisfies this quadratic equation and is not equal to $-2$ (the x-coordinate of $A_{1}$), we have: $(-2)x_{1} = \frac{4(y_{0}^{2} - 8)}{y_{0}^{2} + 8}$, which leads to $x_{1} = \frac{-2(y_{0}^{2} - 8)}{y_{0}^{2} + 8}$. Consequently, $y_{1} = \frac{8y_{0}}{y_{0}^{2} + 8}$.

Calculating the dot product $\overrightarrow{OR} \cdot \overrightarrow{OM}$: $(\frac{-2(y_{0}^{2} - 8)}{y_{0}^{2} + 8}, \frac{8y_{0}}{y_{0}^{2} + 8}) \cdot (2,y_{0}) = \frac{-4(y_{0}^{2} - 8)}{y_{0}^{2} + 8} + \frac{8y_{0}^{2}}{y_{0}^{2} + 8} = 4$.

Therefore, $\boxed{\overrightarrow{OR} \cdot \overrightarrow{OM} = 4}$, which is a constant value.