Question

If a normal to a parabola $y^2=4ax$ make an angle $\varphi$ with its axis,then it will curve again at an angle

• A. $\tan {}^{-1}\left(2\tan \varphi \right)$
• B. $\tan {}^{-1}\left(\frac{1}{2}\tan \varphi \right)$
• C. $\cot {}^{-1}\left(\frac{1}{2}\tan \varphi \right)$
• D. $none$

Answer 1

The correct option is B $\tan {}^{-1}\left(\frac{1}{2}\tan \varphi \right)$

Normal to the parabola at $P$

$y=t_{1}x+2a{t}_1+a{t}_1^3$

$\underset{?}{slope}\tan \theta =-t_1-----\left(1\right)$

It reacts again at $Q(a{t}_2^2,2at,)$

$\therefore t_2=-t_1-\frac{2}{t_1}-----\left(2\right)$

Again between manual and parabola $=$ angle between normal and tangent.

$\therefore t_{2}y=x+a{t}_2^2{m}_2=\frac{1}{t_2}-----slope$

Let angle between manual and parabola be $Q$.

$\therefore \tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}=\frac{-t_1-\frac{1}{t_2}}{1-t_1\frac{1}{t_2}}$

Or, $\tan \theta =\frac{-(t_1{t}_2+1)}{(t_2-t_1)}[\because t_2=-t_1-\frac{2}{t_1}{t}_1{t}_2=-t_1^2-2]$

$\tan \theta =\frac{-(t_1^2-2+1)}{{-t}_1-\frac{2}{t_1}-t_1}$

Or, $\tan \theta =\frac{-(t_1^2-1)}{{-2t}_1-\frac{2}{t_1}}$

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

Or, $\tan \theta =-\frac{\varphi }{2}\frac{1}{2}\tan \varphi \left[from\left(1\right)\right]$

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