Question

If a normal to a parabola make an angle $\varphi$ with the axis, show that it will cut the curve again at an angle ${\tan }^{-1}\left(\frac{1}{2}\tan \varphi \right).$

Answer 1

Let normal is drawn at $P(a{t}_1^2,2{at}_1)$ and it intersect the curve again at $Q(a{t}_2^2,2{at}_2)$

Equation of normal at $P$ is

$y=-t_{1}x+2a{t}_1+a{t}_1^3......\left(i\right)$

Slope $=m=-t_1$

$\tan \varphi =-t_1$

Now we have to find the angle of intersection with the curve at $Q$. So, we have to find the angle between the normal at $P$ and tangent at $Q$

Equation of tangent at $Q$ is

$t_{2}y=x+{{at}_2}^{2}y=\frac{x}{t_2}+{at}_2$

Slope of tangent $=\frac{1}{t_2}$

$\tan \theta =\frac{|-t_1-\frac{1}{t_2}|}{1+(-t_1)\frac{1}{t_2}}\tan \theta =\frac{t_1{t}_2+1}{t_1-t_2}$

When the normal intersect the curve again $t_2=-t_1-\frac{2}{t_1}$

$\tan \theta =\frac{t_1(-t_1-\frac{2}{t_1})+1}{t_1-(-t_1-\frac{2}{t_1})}\tan \theta =\frac{-t_1}{2}$

substituting $t_1$ from slope $\left(i\right)$

$\tan \theta =\frac{\tan \varphi }{2}\theta ={\tan }^{-1}\left(\frac{\tan \varphi }{2}\right)$

Hence proved