Find focal length of electron for a parabolic motion

[Istiak]

Homework Statement

An electron of charge e is moving at a constant velocity u, along x-axis. It enters a region of constant electric field E, which is pointing perpendicular to x-axis. The electron moves in a parabola. Which of the following represents the focal length of the parabola? Neglect any effects due to gravity. (for a parabola of type $x^2=4ay$, a is the focal length)

Relevant Equations

$\int \vec F\cdot d\vec s = \frac{1}{2}mu^2$

Here I was going to use $\int \vec F \cdot d\vec l = \frac{1}{2}mu^2$

What I got that is $l=\frac{mu^2}{2eE}$. Here the question is what is $l$ (I took $x$ while doing the work but here I used $l$ instead of $x$)? I was assuming that it's $x$ since I am calculating work in the parabola. So my equation stands $a=\frac{m^2u^4}{16 y(eE)^2}$. But there's no option of it. But what I found for $l$ that satisfies. But my question is how $l$ is focal length?

 

[BvU]

$l$ is in the direction of $\vec F$, so perpendicular to your $x$. It plays the role of $y$, not $x$

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[Istiak]

$l$ is in the direction of $\vec F$, so perpendicular to your $x$. It plays the role of $y$, not $x$

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So if I take $y$ into account then I get...

$y=\frac{mu^2}{2eE}$
$\frac{x^2}{4a}=\frac{mu^2}{2eE}$
$a=\frac{x^2 eE}{2a mu^2}$

but...

 

[BvU]

No you don't:
$$\left . \begin {array} {ll} x &= ut \\y & = \displaystyle {eEt^2\over 2m} \end{array}\right \}\Rightarrow y =x^2 {eE\over 2mu^2}\Rightarrow x^2 = 4\left(mu^2\over 2eE\right ) y$$so that $$a=\displaystyle {mu^2\over 2eE}$$

[edit]sorry I had to fumble with the 2's a few times...

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