Applying the appropriate simplifications to the Navier-Stokes equation yields the following:
$$(\partial_x^2+\partial_y^2)v_z=\frac1\mu \partial_zp$$
And assuming that the velocity profile is going to be an elliptic paraboloid we can easily find $v_z(x,y)$ (keeping in mind the no-slip condition). We can then deduce the volumetric flow rate: $$Q_e=-\frac \pi {4\mu}\frac{a^3b^3}{a^2+b^2}\partial_zp$$
Now, what I'd like to do, is deduce the Poiseuille law for two parallel plates: $$\frac {Q_p} l=-\frac 1 {12\mu}c^3\partial_zp$$ from $Q_e$. Dividing $Q_e$ by the $2a$ then computing the limit as $a$ approaches $\infty$ doesn't work since the quantity $Q_e /2a$ is not equivalent to $Q_p/l$. I then tried to compute the mean velocity: $$\bar{V_e}=\frac {Q_e}{\pi a b}$$ then its limit, but it didn't work either. I then imagined a parabolic function $f$ such that: $$\int_{-b}^bf(x)2adx=Q_e$$
which yielded: $$f(x)=\frac{3Q_e}{8ab}\left(1-\frac{x^2}{b^2}\right)$$
finding its limit yields a velocity distribution that doesn't correspond to the case of two parallel plates. I'm stumped at this point! Is there a way to directly deduce $Q_e$ from $Q_p/l$ without having to go through $v_z(x,y)$? In other words, without finding the limit of $v_z$ as $a$ approaches $\infty$ then finding $Q/l$. Suppose we're only given $Q_e$!
Let $v_{z}^{(e)}(x,y)$ denote the axial velocity in the elliptical tube (with semi-major and semi-minor axis lengths $a$ and $b$, respectively), and let $v_{z}^{(p)}(y)$ denote the velocity between infinite parallel plates separated by a distance $c = 2b$. Let $G = -p_z$ denote the (constant) pressure gradient.
The exact solutions to the Navier-Stokes equations for unidirectional, laminar flow are
$$v_z^{(p)}(y) = \frac{Gb^2}{2 \mu}\left(1 - \frac{y^2}{b^2}\right), \\ v_z^{(e)}(x,y) = \frac{Gb^2}{2 \mu}\frac{a^2}{a^2 + b^2 }\left(1 - \frac{y^2}{b^2}- \frac{x^2}{a^2 + b^2}\right), $$
and it is true that with fixed coordinates $(x,y)$
$$\tag{*} \lim_{a \to \infty}v_z^{(e)}(x,y) = v_z^{(p)}(y).$$
However, it would not be possible to deduce the precise equality given by (*) simply by examining the volumetric flow rates, which are definite integrals of the velocity fields over different cross-sectional areas. Furthermore, there is no easy way to "discover" $Q_p$ given $Q_e$.
To elaborate, note that on a per unit area basis the volumetric flow rate between parallel plates is
$$\frac{Q_p}{2bl} = \frac{1}{3}\frac{G}{\mu} b^2.$$
This represents the average of $u_z^{(p)}(y)$ over a rectangle with dimensions $2b \times l$.
The volumetric flow rate in the elliptical tube is
$$Q_e = \frac{\pi ab}{4} \frac{G}{\mu} \frac{a^2b^2}{a^2 + b^2}.$$
As $a \to \infty$ with $b$ fixed, we have $Q_e \to \infty$ since the cross-sectional area becomes infinite and the average velocity does not tend to zero if the pressure gradient is maintained (requiring, of course, infinite energy).
Considering instead the volumetric flow rate per unit area (i.e., the average velocity over the elliptical cross-section), we find
$$\lim_{a \to \infty} \frac{Q_e}{\pi a b} = \lim_{a \to \infty} \frac{1}{4} \frac{G}{\mu} \frac{a^2b^2}{a^2 + b^2} = \frac{1}{4} \frac{G}{\mu} b^2 = \frac{3}{4} \frac{Q_p}{2bl}.$$
Thus, we essentially recover the result for flow between plates (on a per unit area basis) up to a scaling factor of $3/4$. This factor represents the impact of averaging one velocity field over an elliptical area and averaging the other over a rectangle where the ellipse is inscribed in the rectangle. This persists in the limit as $a \to \infty$, even though the differences in velocity converge to zero.
Addendum
The $3/4$ factor is an artifact of computing the average velocity over an elliptical rather than rectangular cross section. The limit of the velocity profile for the elliptical tube is the parallel-plane velocity $v_z^{(p)}(y)$, which independent of the $x-$coordinate. The average over an elliptical cross-section $E(a,b)$ is
$$\bar{v}_z^{(p)} = \frac{1}{\pi a b} \int \int_{E(a,b)}v_z^{(p)}(y) \, dx\,dy = \frac{1}{\pi a b} \frac{G b^2}{2 \mu} \int \int_{E(a,b)}(1 - y^2/b^2)\,dx\,dy. $$
Changing variables with $x = ar \cos \theta $ and $y = b r \sin \theta$ we get
$$\begin{align}\bar{v}_z^{(p)} &= \frac{1}{\pi a b} \frac{G b^2}{2 \mu} \int_0^{2\pi} \int_0^1\left(1 - \frac{b^2r^2 \sin^2 \theta}{b^2}\right)\,a b \,r \, dr \, d\theta \\ &= \frac{1}{\pi a b} \frac{G b^2}{2 \mu} \frac{ab}{2}\int_0^{2\pi} \left(1 - \frac{\sin^2 \theta}{2}\right) \, d\theta \\ &= \frac{G b^2}{2 \mu}\frac{1}{2\pi} \left( 2 \pi - \frac{1}{2}\int_0^{2\pi}\sin^2 \theta \, d\theta \right) \\ &= \frac{G b^2}{2 \mu}\frac{1}{2\pi} \left( 2 \pi - \frac{\pi}{2}\right) \\ &= \frac{3}{4} \frac{G b^2}{2 \mu}\end{align},$$
and it is apparent how the $3/4$ factor arises.