Consider the hyperbola determined by $$\Big(\frac{y-k}{b}\Big)^2-\Big(\frac{x}{a}\Big)^2=1.$$ The foci are at $(0,k \pm\sqrt{a^2+b^2}).$ Let's parametrize the upper half of the hyperbola by $$x=t,y=k+b\sqrt{1+\Big(\frac{t}{a}\Big)^2},$$
and define $r(t)=\langle x(t), y(t)\rangle.$ Under this parametrization I have $$r'(t)=\Bigg\langle 1, \frac{bt}{a^2\sqrt{1+\Big( \frac{t}{a}\Big)^2}}\Bigg \rangle$$ so that the speed is given by $$\vert r'(t) \vert=\sqrt{1+\frac{b^2t^2}{a^4\Bigg(1+\Big(\frac{t}{a}\Big)^2\Bigg)}}.$$
However, the speed is smallest when the point is near the foci (at time $t=0)$ and grows as $t\to \pm \infty$, counter to what I was expecting.
Is there a way to parametrize the hyperbola so that it reaches its maximum speed near the foci?
Yes. I will work with the simpler equation $y^2-x^2=1$ We note that the slope of this is monotonically increasing from $0$ at the vertex toward $1$ as $x \to \infty$ Let $y=1+t, x=\sqrt{y^2-1}=\sqrt{t^2+2t}$ Then $v=\sqrt{x'^2+y'^2}=\sqrt {\frac{(t+1)^2}{t^2+2t}+1}=\sqrt{2+\frac 1{t^2+2t}}$ is a decreasing function of $t$, so decreasing as we get away from the vertex.