Prove that the normal to the parabola at $Q$ bisects the angle $FQP$

Question

QUESTION: Consider the parabola $y^2=4x$. Let $P=(a,b)$ be any point inside the parabola, i.e. $b^2<4a$ and let $F$ be the focus of the parabola. Find the point $Q$ on the parabola such that $FQ+QP$ is minimum. Also show that the normal to the parabola at $Q$ bisects the angle $FQP$.

MY ANSWER: The first part of the problem is easy. I could do that. The answer to that came out to be $|a+1|$. But I am stuck at the second part. I cannot prove that the normal bisects that angle. Can anyone please help me out?

Thank you.

Answer 1

We take parabola equation as $$ y^2= 4 f x \tag1 $$

Bringing in $f$ instead of unity has the advantage of checking dimensions of equations and also assigning physical meaning to focal length of parabola $ f=OF $

Differentiate (1)

$$ 2 y y'= 4 f\rightarrow y'= \tan \phi = \dfrac{2f}{\sqrt{4fx}}=\sqrt{f/x} \tag2 $$ From the diagram we need to prove that

$$ \beta + \phi = \pi/2 -\phi $$ or $$ \beta = \pi/2 -2 \phi \tag3 $$

or $$ \tan \beta = \tan ( \pi/2-2 \phi) \tag4$$ LHS

$$\tan \beta =\dfrac{x-f}{y}=\dfrac{x-f}{2 \sqrt{fx}}=\dfrac{1-f/x}{2\sqrt{f/x}} \tag5 $$ RHS $$ \cot 2\phi = \dfrac{1- \tan^2\phi}{2 \tan \phi} =\dfrac{1-f/x}{2\sqrt{f/x}} \tag6$$

which tally, the normal at $Q$ reflects incident rays to focus $F$ and vice-versa.