Let's first examine the initial shape immediately when the water exits the pipe:
Initially, the stream possesses a well-defined radius of curvature $R$, which we can find by applying Euler's equation normal to a streamline:
$$ \frac{V^2}{R} = -\frac{1}{\rho} \frac{\partial P}{\partial n} - g\frac{\partial y}{\partial n}$$
where $n$ is a radial coordinate normal to a streamline in the jet, referenced from the center of curvature.
Let's say the pipe diameter $D$ is small ($R>>D$), then $\frac{\partial P}{\partial n} \sim 0$ since there is no appreciable pressure gradient across the streamtube.
Also, $\frac{\partial y}{\partial n} = -1$ from inspecting the coordinate system.
This shows that the local radius of curvature, right when the stream exits the pipe, is given by:
$$ R = \frac{V^2}{g} $$
This defines a circle given by (look at the coordinate system):
$$x^2 + (R-y)^2= R^2$$
$$ x^2 + y^2 = 2Ry$$
Extremely close to the origin, $x^2>>y^2$, so ignore $y^2$ and solve for $y$:
$$ y = \frac{x^2}{2R} = g\frac{x^2}{2V^2}$$
where $V$ is the exit velocity at the jet, and I substituted our previous expression for $R$.
This result makes sense; consider a jet with large exit velocity and note how we will get a parabola with a large focal length, as expected.
EDIT: How does the shape depend on pressure P and pipe diameter D?
What is the shape and size of this arc as a function of D, the inner
diameter of the pipe and, P, the pressure of the water?
Both $D$ and $P$ affect the velocity $V$ in our parabolic equation, but the specific equation $V=V(D,P)$ depends on what's happening before the pipe exit.
To see the effect of $D$, you could use mass conservation $A_cV = constant$ within the pipe where $A_c$ is the cross-section area of the pipe. Larger $D$ will therefore yield smaller $V$ and vice versa.
To see the effect of $P$, you need to clarify what your system includes before the pipe exit. For example, if our pipe is a small hole in a tank, $V=\sqrt{2 g h}$ via Bernoulli's equation, where $h$ is the distance of the hole from the bottom. You can see that a larger hydrostatic pressure increases $V$. If our pipe is connected to a pump of efficiency $\eta$, with power $\dot{W} = \eta Q \Delta P=\eta A_c V \Delta P$, you can see how pressure influences $V$ again. Generally, you can see that a larger pressure at the pipe exit will result in larger $V$, thus giving our parabola a larger focal length. Vice versa for smaller pressure.
The specific answer to how $V$ depends on $D$ and $P$ depends on what your system includes before the pipe exit.
