Consider the exponential family $f(x;\theta) = c(\theta)\exp\{b(\theta) T(x))\}h(x)$ and I would like to find a UMP test for
$$H_0: \theta \leq 0\quad \hbox{vs}\quad H_1: \theta > 0.$$
It is madotory in all books: Two steps. First, we fix $\theta_0 = 0$ and take $\theta_1 > \theta_0 = 0$. That is, first we have to find a test $\phi$ for
\begin{equation} H_0: \theta = 0\quad \hbox{vs}\quad H_1: \theta = \theta_1 \quad \quad (1) \end{equation}
and we have to show that $\phi$ does not depend on $\theta_1$. This the first step and I do not to describe the second step beacause I don't have problems with its procedure.
For the hypothesis test $(1)$, by Neyman–Pearson theorem, there is some $k>0$ such that
$$\phi(x) = \left\{ \begin{array}{lr} 1 & : f(x;0)k < f(x;\theta_1) \\ 0 & : f(x;0)k > f(x;\theta_1) \end{array} \right.$$
$$E[\phi(X)] = 0.01$$
is a UMP teste with size $0.01$. But, we can do
\begin{equation} \begin{split} f(x;0)k < f(x;\theta_1) & \Longleftrightarrow k< \frac{c(\theta_1)}{c(0)}\exp\{T(x) (b(\theta_1) - b(0))\}\\ & \Longleftrightarrow T(x)> \frac{\ln(k\frac{c(0)}{c(\theta_1)})}{(b(\theta_1) - b(0))}:= \tilde{k}(\theta_1,k)\\ & \Longleftrightarrow T(x)> \tilde{k}(\theta_1,k) \end{split} \end{equation}
And
$$E[\phi(X)] = 0.01 \implies \tilde{k}(\theta_1,k) = F^{-1}_{T}(0.99 | \theta = 0) = q_0(0.99)$$
That is $\tilde{k}(\theta_1,k)$ depends only on the null hypothesis $\theta_0 = 0$ and $0.99$. In other words, $\tilde{k}(\theta_1,k)$ is constan in the first stage. If I look at the test with this expression:
$$\phi(x) = \left\{ \begin{array}{lr} 1 & : T(x) > q_0 (\theta, 0.99) \\ 0 & : T(x) < q_0 (\theta, 0.99) \end{array} \right.$$
I can fully understand that this test does not depend on $\theta_1$, because $q_0 (\theta, 0.99)$ has nothing to do with $\theta_1$. But looking at this,
$$ \frac{\ln(k\frac{c(0)}{c(\theta_1)})}{(b(\theta_1) - b(0))} = q (\theta, 0.99) $$
I can not understand how I can keep $k$ fixed so that if I change $\theta_1$ by $\theta_1^{'}$ I still have exactly the same test $\phi(x) = \left\{ \begin{array}{lr} 1 & : f(x;0)k < f(x;\theta_1^{'}) \\ 0 & : f(x;0)k > f(x;\theta_1^{'}) \end{array} \right.$ In the sense of being exactly equal to $\phi(x) = \left\{ \begin{array}{lr} 1 & : f(x;0)k < f(x;\theta_1) \\ 0 & : f(x;0)k > f(x;\theta_1) \end{array} \right.$
Since if I really want to state that this test does not depend on $\theta_1> 0$, I have to keep $k$ fixed, in case I want to change to a $ \theta_{1}^{'}> 0 $. Am I confusing the true meaning of the expression "The test will not depend on $ \theta_1>0$?