The trajectory as y(x)-function
Here also the equations for the motions in x- ans y-direction must be plugged in each other. There we solve the equation$$$$x(t)= v_0\cdot t$$$$ for t and plug it in$$$$y(t)=\frac{1}{2}a_y\cdot t^2$$$$ This yields the searched equation:$$$$y(x)= \frac{1}{2}\cdot a_y \cdot \frac{x^2}{{v_0}^2}.$$$$ Here $a_y$ can be replaced by the acceleration of the electrons in the electric field:$$$$y(x)= -\frac{1}{2}\cdot \frac{V_{\text p}\cdot e}{\text{d}\cdot m_e \cdot{v_0}^2}\cdot x^2.$$$$ With the equation for $v_0$ determined here, it yields:$$$$y(x)= -\frac{1}{4\cdot \text{d}}\cdot \frac{V_{\text p}}{V_{\text a}}\cdot x^2.$$$$
$v_0=$ initial speed in horizontal direction,
$V_{\text{p}}=$ potential between capacitor plates,
$V_{\text{a}}=$ acceleration voltage,
d = distance between capacitor plates,
$e=$ Elementary charge,
$m_e=$ Mass of an electron