How I can explain this result geometrically for the single real $a$? The mean values theorem says that there exists a $c∈(u,v)$ such that $$f(v)−f(u)=f'(c)(v−u)$$
My question is: Assume that $f$ is a real analytic function. Fix the value of $v$ as $v=a$. Then we conclude the existence of a $c∈(u,a)$ for all $u≠a$ such that
$$f'(c)=(f(a)-f(u))/(a-u)$$
How I can explain this result geometrically for the single real $a$? I know about the general setting: the slope of the line joining two points equal to the slope of the tangent to the curve at the point $(c,f(c))$.
 A: First, your reformulation seems a bit missleading. You write "Then we conclude the existence of a $c \in (u,a)$ for all $u \neq a$ such that...". That kind of makes it sound as if there was one single $c$ for all possible $u \neq a$, which is wrong. For each $u$ you can find a $c$, but its value will generally depend on $u$. 
Why not start from the original statement, i.e. for every $(u,v)$ there's a $c \in (u,v)$ with $$
  f(v) - f(u) = f'(c)(v-u) \text{.}
$$
To simplify things, you may require wlog that $u < v$. That removes the case $u = v$ and allows you to rewrite this like you did as $$
  \frac{f(v) - f(u)}{v-u} = f'(c) \text{.}
$$
You can then interpret $\frac{f(v) - f(u)}{v-u}$ as an approximation of the slope of $f$ between $u$ and $v$. If you reduce the distance between $u$ and $v$, that approximation will get better, and in the limit it will converge to the value of $f'$ at $u$ (or $v$, in the limit both are the same). The theorem now tells you that even if you don't do the limit, there's still at least one point between $u$ and $v$ for which the approximation is the true value of $f'$ at the point. With a few additional constraints, that is quite plausible.
Assume that there were not such point $c$, and that $f'$ is continous. Then you either have $f'(x) < c$ for all $x \in (u,v)$, or $f'(x) > c$ for all $x \in (u,v)$. In other words, the curve either has to have a smaller or a larger slope than $c$ over the whole interval $(u,v)$. But if the slope is smaler than $c$ on the whole interval, the function grows to slowly to attain $f(v)$ at $v$. And if it's larger, it grows to fast and will have grown beyond $f(v)$ at $v$.
