# If the normal to a parabola $y^2=4ax$ make an angle $\theta$ with the axis and $\phi$ is the angle it makes when it cut the parabola again find the relation between $\theta$ & $\phi$?

$\begin{array}{1 1}(a)\;\tan \phi=\large\frac{\tan \theta}{2}\\(b)\;\tan \phi=\large\frac{\tan \theta}{4}\\(c)\;\tan 2\phi=\large\frac{\tan \theta}{2}\\(d)\;4\tan \phi=\large\frac{\tan \theta}{2}\end{array}$

Let the normal at $P(at_1^2,2at_1)$ be $y=-t_1x+2at_1+at_1^3$
$\therefore \theta=-t_1$=slope of normal -------(1)
It meets the curve again at $\theta$ say $(at_2^2,2at_2)$
$t_2=-t_1-\large\frac{2}{t_1}$-------(2)
Now angle between the normal and parabola =Angle between the normal and tangent at $\theta$
(i.e) $t_2y=x+at_2^2$
If $\phi$ be the angle
$\tan \phi=\large\frac{m_1-m_2}{1+m_1m_2}$
$\Rightarrow \large\frac{-t_1-\Large\frac{1}{t_2}}{1+(-t_1)(\large\frac{1}{t_2})}$
$\Rightarrow \large\frac{-(t_1t_2+1)}{t_2-t_1}$
$\Rightarrow -\large\frac{(-t_1^2-1)}{-2((1+t_1^2)/t_1)}$
$\Rightarrow -\large\frac{-t_1}{2}$
$\tan \phi=\large\frac{\tan \theta}{2}$
Hence (a) is the correct answer.