The Brachistochrone problem is usually presented with the having a ball dropped into the slide with initially zero velocity and at position $(x, y)=(0, 0)$.
I would like to know the more general solution to this problem. What happens when we start with initial velocity $\vec{v}_0 = (v_x, v_y)$?
There is not necessarily a path that extremizes the travel time and also satisfies the given initial conditions. The Euler-Lagrange equation is an ODE that is a necessary condition for a given functional to be extremized; but this does not guarantee us that there exists a solution to the Euler-Lagrange equation with all of the properties we want it to have.
Let's assume that we want to find a path $y(x)$ between $(0,0)$ and $(x_f, y_f)$ which extremizes the travel time for a particle under gravity, and for which the particle is also is moving at a speed $\vec{v} = (v_x,v_y)$ initially. If the particle in the brachistochrone problem has an initial speed $v_0 = \sqrt{v_x^2 + v_y^2}$, then it is not too hard to show that the speed of the particle along its subsequent trajectory will be $v = \sqrt{2g}\sqrt{y + y_0}$, where $y_0 \equiv v_0^2/(2g)$. (We have defined $y$ to increase in the direction of gravity.) The time taken along the path, following the usual argument, will be $$ \tau = \frac{1}{\sqrt{2g}}\int_0^{x_f} \sqrt{\frac{1+(y')^2}{y+y_0}} \, dx $$ Defining $w = y + y_0$, this reduces to the standard integral for the brachistochrone problem: $$ \tau = \frac{1}{\sqrt{2g}} \int_0^{x_f} \sqrt{\frac{1+(w')^2}{w}} \, dx $$ This implies that the function $w(x)$ also satisfies the brachistochrone problem, under the boundary conditions $w(0) = y_0$ and $w(x_f) = y_f + y_0$.
So far so good. The problem arises when we also try to satisfy the initial direction conditions. In particular, Beltrami's identity implies that if $w(x)$ extremizes the transit time, we will have $$ w(1 + (w')^2) = 2a $$ where $a$ is a constant. (Physically, $a$ corresponds to the radius of the "rolling circle" that generates the cycloid.) Defining $\alpha = v_y/v_x = w'(0)$, we see that we must have $$ y_0 (1 + \alpha^2) = 2 a. $$ In other words, $a$ is completely determined by the initial conditions. If the final point $(x_f, y_0 + y_f)$ is not on the cycloid of this particular size passing through $(0, y_0)$, then there cannot be a (smooth) path between these two points which extremizes the travel time.
To understand why this might be the case, consider a curve that starts out in the given initial direction for a distance $\epsilon$, and then immediately turns smoothly (if quickly) and follows a cycloid path to the specified final location. By making $\epsilon$ arbitrarily small, we can make the transit time arbitrarily close to the transit time of the standard cycloid problem, which we know has the minimum transit time among all paths. But among the paths with the given initial direction, there is no path that actually attains this minimum; so the extremum is not attained among the class of paths under consideration.
(Note, however, that if you just specify the initial speed $v_0$ rather than the initial velocity, then $a$ is not determined by the boundary conditions at $x = 0$, and can instead be determined by the condition that $w(x_f) = y_0 + y_f$.)
