Can we write a Lagrangian for the classical system with $H=kqp$?

Say we have the following Lagrangian:

$$H = kqp.$$

The equations of motion are easy to find:

$$\dot{q} = kq \\ \dot{p} = -kp,$$

and to solve:

$$q=q_0 e^{kt} \\ p=p_0 e^{-kt}.$$

I'm curious if we can write a meaningful Lagrangian for this system, and if not, why not?

$$H(q,p,t) = p\dot{q} - L(q,\dot{q},t)$$

so the Lagrangian should equal

$$p\dot{q} - kqp$$

To get $L(q,\dot{q},t)$, we must write $p$ in terms of $q$ and $\dot{q}$. Using the explicit solutions above, there are two ways we can do this:

$$p = p_0 q_0/q$$ or $$p = p_0 q_0 k/\dot{q}$$

I suppose the most general form is:

$$p = p_0 q_0(\alpha/q + \beta k/\dot{q})$$ with $\alpha+\beta=1$. Substituting $p$ with this expression, then removing constants and constant factors, we get:

$$L \sim \alpha \dot{q}/q - \beta k^2 q/\dot{q}.$$

Now:

$$\frac{\partial L}{\partial q} = -\frac{\beta k^2}{\dot{q}} - \frac{\alpha \dot{q}}{q^2} \\ \frac{d}{dt} \frac{\partial L}{\partial \dot{q}} = \frac{\beta k^2}{\dot{q}} - \frac{\alpha \dot{q}}{q^2} - \frac{2\beta k^2 q \ddot{q}}{\dot{q}^3}$$

So the Euler-Lagrange equation gives us:

$$\dot{q}^2 = q \ddot{q}$$ The solution here is $q=e^{wt}$ for all $w$, instead of just for $w=k$, disagreeing with the solution to Hamilton's equations.

Furthermore, if we had chosen $\beta=0$ above, we would have $L \sim \dot{q}/q$, which gives an Euler-Lagrange equation of $0=0$.

What is going on here? I feel like I probably did something illegitimate when deriving the Lagrangian from the Hamiltonian, but I don't know exactly what.

Is this particular Hamlitonian just illegitimate to start out with? Is it just not possible to use the Lagrangian formalism with this system? Is there any Lagrangian that will give equations of motion $\dot{q} = k q$ (or, more likely, $\ddot{q} = k \dot{q}$)?

Finally, does any of this matter? Is there any imaginable physical systems where we can define a coordinate $q$ such that $H\sim qp$ or $L\sim q/\dot{q}$ or $L\sim \dot{q}/q$?