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Q.
A normal, to parabola $y^{2}=4 a x$ whose inclination is $30^{\circ}$, to a parabola cuts it again at an angle of
Conic Sections
Solution:
The normal at $P\left(a t_{1}^{2}, 2 a t_{1}\right)$ is
$y+x t_{1}=2 a t_{1}+a t_{1}^{3}$
with slope say $\tan \alpha=-t_{1}=\frac{1}{\sqrt{3}}$
If it meets curve at $Q\left(a t_{2}^{2}, 2 a t_{2}\right)$
then $t_{2}=-t_{1}-\frac{2}{t_{1}}=\frac{7}{\sqrt{3}}$.
Then angle $\theta$ between parabola (tangent at $Q$ ) and normal at $P$ is given by
$\tan \theta=\frac{-t_{1}-\frac{1}{t_{2}}}{1-\frac{t_{1}}{t_{2}}}=\frac{1}{2 \sqrt{3}}$ $\Rightarrow \theta=\tan ^{-1}\left(\frac{1}{2 \sqrt{3}}\right)$