Let $k$ be the number of digits of $a$. Therefore, $10^{k-1} \le a < 10^k$. We have $\frac{a}{b} = b.a = b + \frac{a}{10^k}$ and consequently $\frac{a-b^2}{b} = \frac{a}{10^k}$. Since $(a,b) = 1$, $(a-b^2,b) = 1$; therefore, $\frac{a-b^2}{b}$ is the irreducible form of $\frac{a}{10^k}$. Thus, for some natural number $s$, $sb = 10^k$ and $s(a-b^2) = a$.

Now, $a-b^2|a$ and $(a-b^2,a) = 1$. Hence, $a-b^2 = \pm 1$, but $\frac{a-b^2}{b} = \frac{a}{10^k} > 0$, so $a-b^2 = 1$. Therefore, $a = b^2+1$ and $\frac{1}{b} = \frac{a}{10^k}$.

Since $a \ge 10^{k-1}$, we have $\frac{1}{b} = \frac{a}{10^k} \ge \frac{1}{10}$ which implies $b \le 10$. On the other hand, $ab = b(b^2+1) = 10^k$, so prime factors of $b$ and $b^2+1$ are only 2 or 5. Since $b \le 10$, we have $b \in \{1, 2, 4, 5, 8, 10\}$. A simple calculation shows that only for $b = 2$, $b(b^2+1)$ is a power of 10. For this case, $a = 5$ and $k = 1$. Therefore, $\frac{5}{2} = 2.5$, this is the only solution.