$2 a y+b^{2}=0$ is externally tangent to $(a, b \in \mathbf{N}, a \neq 0)$, and the moving circle $C_{1}$ is tangent to the $x$-axis. Find:
(1) The equation of the locus $L$ of the center of the moving circle $C_{1}$;
(2) If the line $4(\sqrt{7}-1) a b x-4 a y+b^{2}+a^{2}-6958 a=0$ intersects the curve $L$ at exactly one point, find the values of $a$ and $b$.

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(1) Method 1: From $4 a^{2} x^{2}+4 a^{2} y^{2}-4 a b x-2 a y+b^{2}=0$, we can get $\left(x-\frac{b}{2 a}\right)^{2} +\left(y-\frac{1}{4 a}\right)^{2}=\left(\frac{1}{4 a}\right)^{2}$. Given $a, b \in \mathbf{N}$, and the two circles are on the same side of the $x$-axis, it is known that the center of the moving circle is above the $x$-axis. Let the coordinates of the center of the moving circle be $(x, y)$, then $\sqrt{\left(x-\frac{b}{2 a}\right)^{2}+\left(y-\frac{1}{4 a}\right)^{2}}=y+\frac{1}{4 a}$. After rearranging, we get the trajectory equation of the center of the moving circle as $y=a x^{2}-b x+\frac{b^{2}}{4 a} \quad\left(x \neq \frac{b}{2 a}\right)$.

Method 2: From the given, the trajectory of the center of the moving circle is a parabola with focus at $\left(\frac{b}{2 a}, \frac{1}{4 a}\right)$, directrix at $y=-\frac{1}{4 a}$, and vertex at $\left(\frac{b}{2 a}, 0\right)$ (excluding this point). The trajectory equation is $\left(x-\frac{b}{2 a}\right)^{2}=\frac{1}{a} y$, which is $y=a x^{2}-b x+\frac{b^{2}}{4 a} \quad\left(x \neq \frac{b}{2 a}\right)$.

(2) Solving the system of equations $\left\{\begin{array}{l}y=a x^{2}-b x+\frac{b^{2}}{4 a}\left(x \neq \frac{b}{2 a}\right), \\ 4(\sqrt{7}-1) a b x-4 a y+b^{2}+a^{2}-6958 a=0,\end{array}\right.$

Eliminating $y$,

we get $4 a^{2} x^{2}-4 \sqrt{7} a b x-\left(a^{2}-6958 a\right)=0$. From $\Delta=16 \times 7 a^{2} b^{2}+16 a^{2}\left(a^{2}-6958 a\right) =0$, rearranging gives $7 b^{2}+a^{2}=6958 a$.

(3) From (3), we know $7\left|a^{2} \Rightarrow 7\right| a$, so let $a=7 a_{1}$. Substituting into (3) gives $b^{2}+7 a_{1}^{2}=6958 a_{1} \Rightarrow 7\left|b^{2} \Rightarrow 7\right| b$. Let $b=7 b_{1}$, substituting into the above equation gives $7 b_{1}^{2}+a_{1}^{2} =994 a_{1}$. Similarly, we get $7\left|a_{1}, 7\right| b_{1}$, so let $a=49 n, b=49 m$, substituting into (3) gives $7 m^{2} + n^{2}=142 n$ (4). Completing the square for (4), we get $(n-71)^{2}+7 m^{2}=71^{2}$. Analyzing the parity of this equation, we know $m, n$ are both even, so $7 m^{2}=71^{2}-(n-71)^{2}$ is a multiple of 8, so $4 \mid m$. Let $m=4 r$, then $112 r^{2} \leqslant 71^{2} \Rightarrow r^{2} \leqslant 45$. Therefore, $|r|=0,1,2,3,4,5,6$, $\left\{\begin{array}{l}a=6272, \\ b=784,\end{array}\left\{\begin{array}{l}a=686 \\ b=784 .\end{array}\right.\right.$