= Solution
For the <electric-dominated crossed-field orbit>, choose a boost $V=c^2B/E$ along $x$. Since $|E|>c|B|$, it is subluminal. The <Lorentz transformation of electromagnetic fields> gives $B'_z=0$ and $E'_y=E/\Gamma$, where $\Gamma=(1-V^2/c^2)^{-1/2}$. The initially stationary <electron> has $U'(0)=(\Gamma c,-\Gamma V,0,0)$, not zero <velocity> in this frame.
Let $q=-e$, $a=qE'_y/m_e$ and $s=a\tau/c$. Neglect radiation reaction, as in the prescribed external-field model. Integrating the <relativistic Lorentz force> in this purely electric frame gives the exact trajectory
$$
\boxed{t'=\frac{\Gamma c}{a}\sinh s,\qquad
x'=-\Gamma V\tau,\qquad
y'=\frac{\Gamma c^2}{a}(\cosh s-1),\qquad z'=0.}
$$
Indeed $U'^0=\Gamma c\cosh s$, $U'^x=-\Gamma V$, and $U'^y=\Gamma c\sinh s$ satisfy both the force equation and $U'^2=c^2$. The constant transverse <momentum> must be retained.
The inverse boost gives a useful exact laboratory parametrization as well:
$$
\boxed{t=\Gamma^2\left[\frac ca\sinh s-\frac{V^2}{c^2}\tau\right],\qquad
x=\Gamma^2V\left[\frac ca\sinh s-\tau\right],\qquad y=y'.}
$$
For $E,B>0$, the lab <electron> starts accelerating toward negative $y$ and bends toward positive $x$, the direction of $\mathbf E\times\mathbf B$. It never acquires a cyclotron orbit in this electric-dominated case. At late times, $v_x\to V$ and $v_y\to\operatorname{sgn}(qE')c/\Gamma$, so its speed approaches $c$ along an oblique asymptote. Reversing field signs reverses the appropriate directions. The zero-$B$ limit is the ordinary electric-field <hyperbolic motion>.
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