Electrified parabola

Let's just do this for $x^{2} = 4y$

Consider a small element $dl$ along the curve, at $(x,y)$

The line joining the focus to the point makes an angle $\theta$ with positive x-axis.
The electric field by the element at focus,
$dE_{1} = \dfrac{2k\lambda dl}{(y+1)^{2}}$
$dl = \sqrt{y+1}\dfrac{dy}{\sqrt{y}}$

Consider a similar element on the other branch of the parabola, making the same angle $\theta$ with negative x-axis.
The electric field by these two elements cancel in the horizontal direction , but add up in the vertical direction.

The vertical electric field $dE$ in downward direction is given by,
$dE = 2dE_{1}\sin(\theta)$

$\sin(\theta) = \dfrac{y-1}{y+1}$
$\therefore dE = 2k\lambda\dfrac{y-1}{(y+1)^{\frac{5}{2}}\sqrt{y}}dy$

$E = \displaystyle 2k\lambda\int_{0}^{\infty}\dfrac{y-1}{(y+1)^{\frac{5}{2}}\sqrt{y}}dy$

Substitue $y = \tan^{2}(u)$ , the integral gets converted to,

$E = 4k\lambda \displaystyle \int_{0}^{\frac{\pi}{2}}\dfrac{\tan^{2}(u)-1}{\sec^{3}(u)}du = 4k\lambda \displaystyle \int_{0}^{\frac{\pi}{2}}\cos(u)(\sin^{2}(u) - \cos^{2}(u))du$
$E = -4k\lambda \displaystyle \int_{0}^{\frac{\pi}{2}} \cos(u)\cos(2u)du$
$E = -2k \lambda \displaystyle \int_{0}^{\frac{\pi}{2}}\cos(u) + \cos(3u)du$
$E = -2k\lambda \left[\sin(u) + \dfrac{\sin(3u)}{3}\right]_{0}^{\frac{\pi}{2}} = -2k\lambda \dfrac{2}{3}$
Substitute, $k = \dfrac{1}{4\pi \epsilon_{0}}$
$E = -\dfrac{\lambda}{3\pi\epsilon_{0}}$
$\therefore \vec{E} =( -\dfrac{\lambda}{3\pi \epsilon_{0}}) (- \hat{j}) = \dfrac{\lambda}{3\pi\epsilon_{0}}(\hat{j})$

I used the following properties :
Distance of a point on parabola from focus = Distance from directrix.

$dl = \sqrt{1+\left(\dfrac{dx}{dy}\right)^{2}}dy$
$\cos(A)+\cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$

you can easily solvethis que

points needed to be kept in mind

only one component of force will be nonzero

distance between elemnt and focus can be foundd by knowing that distance betwwen focus and point of parabola is equalt to the perpendicular to the vertex