At the position $X$, we have
Time taken by an electron $t=\frac{X}{V}$ ...(i)
and displacement of an electron
$y=\frac{1}{2} a t^{2}=\frac{1}{2} \frac{E_{e}}{m} t^{2}$
From Eq. (i), $y=\frac{1}{2} \frac{E_{e}}{m} \frac{x^{2}}{v^{2}}$
$\Rightarrow y=a x^{2}$
[$\therefore $ acceleration, $a=\frac{E_{e}}{2 m v^{2}}$]
So, the trajectory represented as
$y=a x^{2}$ (Parabola)