Is this language with prefix regular?
Let $k=2009$ and $L = \{w : |u_a| \ge k\cdot |u_b|$}. I think it is easier to prove using Myhill-Neorde's Theorem.
Consider the strings $a^k, a^{2k}, a^{3k} \ldots$, we claim that all of these have a $L$-distinguishable suffix. Consider $a^{lk}$ and $a^{mk}$ with $l &lt; m$. Let $x = b^m$. We have $a^{lk}b^m \notin L$ but  $a^{mk}b^m \in L$ (Why?). Therefore, there are infinitely many $L$-equivalence classes, which proves that, by Myhill-Nerode's theorem, that $L$ is non-regular.

Your argument is almost correct:
Assume, for the sake of contradiction, that $L$ is regular. Therefore, by closure property of regular languages, $L' = L(a^*b^*) \cap L$ will also be regular. Notice that all the words in $L'$ will be of the form $a^{n_1}b^{n_2}$ s.t. $n_1 \ge kn_2$.
Now, consider a word $w = a^{kn}b^{n}$ in $L'$. Using pumping lemma, for a sufficiently large $n$, there should exist a partition of this string $w = xvy$ such that $v \ne \epsilon$. Now, three case arises:

If $v$ lies in the part $a^{kn}$: pumping down will lead to a contradiction
If $v$ overlaps both the $a$'s and $b$'s parts: $v$ should be of the form $a^{m_1}b^{m_2}$. Pumping up will result in a string of the form $a^{kn}b^{m_2}a^{m_2}b^{m_2 + n}$, which doesn't belong to $L'$ (why?)
If $v$ lies in the part $b^n$: pumping up will lead to a contradiction