A normal to parabola, whose inclination is $30^o$, cuts it again at an angle of.

Equation of normal in terms of parameter

To parabola $y^2=4ax$ at $(at^2, 2at)$, the equation of normal is $y=-tx+2at+at^3$

To parabola $x^2=4ay$ at $(2at, at^2)$ the equation of normal is $x+ty=2at+at^3$

Let the normal $P(at_1,2at_1)$ by $y= -t_1x + 2at_1+at_1^3$

$\theta$ $=-t_1=$ slope of normal

It meets the curve at $\theta$ say $(at_2^2,2at_2)$

Now angle between the normal and parabola $=$ Angle the normal and tangent at $\theta$ (i.e) $t_2= x + at_2^2$

$\tan \phi$ =$\dfrac{m_1-m_2}{1+m_1m_2}$

$=\dfrac{-t_1-\dfrac{1}{t_2}}{1+\dfrac{(-t_1)}{(-t_2)}}$

$=\dfrac{-(t_1t_2+1)}{t_2-t_1}$

$=\dfrac{-(t_1^2+1)}{t_2-t_1}$

$=\dfrac{-(-t_1^2-1)}{\dfrac{-2(1+t_1^2)}{(t_1)}}$

$=\dfrac{-t_1}{2}$

${\tan}\phi=\dfrac{\tan\theta}{2}$

hence,

$\phi$=$\tan^{-1}\dfrac{\tan30^.}{2} = \tan^{-1}(\dfrac{1}{2\times3^{\tfrac{1}{2}}})$