Normal AO, A1, A2 drawn to the parabola y^2=8x from the point A(h,0) if triangle OA1A2 is equilateral triangle then possible values of h is

Dear Student,
Given that,
Equation of Parabola, 
$y^2=8x$
a =2​​​ 

$O{A}_1{A}_2 is an equilateral triangle$
Therefore,
$\angle A_{1}OA=\frac{\mathrm{\pi }}{6}$
Slope of $O{A}_1= \tan \frac{\mathrm{\pi }}{6}= \frac{1}{\sqrt{3}}$
That is $\frac{y}{x}$= $\frac{2at}{a{t}^2}=\frac{1}{\sqrt{3}}$ (Parametric Form)
        
Hence, t =  $2\sqrt{3}$
Now, we have to find the equation of the normal from point $A_1$
The equation of a normal in parametric form is;
$y+tx = a{t}^3 + 2at$
Also, point (h,0) lies on this eqation,
Therefore, by putting values of a=2, t=$2\sqrt{3}$,  x = h and  y = 0
We get 0 + h = 2*4*3 + 2*2
h = 24+4
h=28
  ​​​​​​​​​​Regards.