I have been trodding through the Course in Differential and Integral Calculus offered by R. Courant (Vol. I, Second Edition, 1954), specifically through the section on determining the maxima/minima of differentiable functions. In the problem set situated at the end of the subchapter there is a problem formulated as follows:
Given the parabola $y^2 = 2px$, $p>0$, and a point $P(x=\xi, y=\eta)$ within it ($\eta^2 < 2p\xi$), find the shortest path (consisting of two line segments) leading from $P$ to a point $Q$ on the parabola and then to the focus $F(x = \frac{1}{2}p, y = 0)$ of the parabola. Show that the angle $FQP$ is bisected by the normal to the parabola, and that $QP$ is parallel to the axis of the parabola. (Principle of the parabolic mirror.)
It has been several days since first embarked upon this problem, and I should say that I have not yet been able come up with any method other than that of brute force, i.e. of expressing the coordinates and distances between points in the Cartesian Plane directly, without taking to the introduction of any angles or transformations. The problem is that this „algebraic” way of dealing with the difficulties of this task is (how surprising!) being completely engulfed by that very algebra.
For instance, I assumed that point $Q$ possesses the ordinate of $y$, whence the abscissa of that point shall be given by $\frac{y^2}{2p}$. Then we have ourselves the following expressions:
$$FQ = \sqrt{\left(\frac{p}{2} - \frac{y^2}{2p}\right)^2 + y^2} = \sqrt{\left(\frac{p}{2} + \frac{y^2}{2p}\right)^2} = \lvert \frac{y^2}{2p} + \frac{p}{2} \rvert = \frac{y^2}{2p} + \frac{p}{2},$$
$$PQ = \sqrt{\left(\xi - \frac{y^2}{2p}\right)^2 + \left(\eta - y\right)^2}$$
The sum of the mentioned distances shall thus be represented by
$$L(y) = \frac{y^2}{2p} + \frac{p}{2} + \sqrt{\left(\xi - \frac{y^2}{2p}\right)^2 + \left(\eta - y\right)^2}$$
Therefore,
$$L’(y) = \frac{y}{p} + \frac{-\frac{y}{p}\left(\xi - \frac{y^2}{2p}\right) - (\eta - y)}{\sqrt{\left(\xi - \frac{y^2}{2p}\right)^2 + \left(\eta - y\right)^2}} = \frac{y\left[\sqrt{\left(\xi - \frac{y^2}{2p}\right)^2 + \left(\eta - y\right)^2} - \left(\xi - \frac{y^2}{2p}\right)\right] - p(\eta - y)}{p\sqrt{\left(\xi - \frac{y^2}{2p}\right)^2 + \left(\eta - y\right)^2}}$$
It is quite easy to observe that $y = \eta$ gives a root of $L’(y) = 0$. As $L’(y)$ is defined over the whole domain of $L(y)$, all the extrema of the latter function shall simultaneously correspond to roots of $L’(y) = 0$. If one takes the second the derivative of $L’’(y)$ and substitutes $y = \eta$, it is plain that $L’’(\eta) > 0$, making the mentioned point a minimum. The problem now lies within the demonstration of the fact that this is not merely a local minimum, but the global as well. Doing this by showing that $L’’(y) > 0$ is increasingly arduous due to the algebraic expression of $L’’(y)$ itself (I believe you can imagine the look of the formula judging from the expression for $L’(y)$), so I abandoned that idea (but it makes sense, if one observes the graph of $L’’(y)$). Thus I took another path, i.e. demonstrating that $L’(y) > 0 \, \forall y > \eta$ and $L’(y) < 0 \, \forall y < \eta$. Should I have succeeded, and I would have a perfect reason for omitting the cumbersome process of differentiating $L’(y)$, for then I would have myself the necessary and sufficent conditions for a minimum to exist at $y = \eta$, and for that minimum to be the global one, as it would be the only point where the derivative $L’(y)$ vanishes. For that matter, I have to show that
$$ y\left[\sqrt{\left(\xi - \frac{y^2}{2p}\right)^2 + \left(\eta - y\right)^2} - \left(\xi - \frac{y^2}{2p}\right)\right] - p(\eta - y) > 0 (<0) \, \forall y > \eta (< \eta),$$
but that turned out to be troublesome for $|y| < \eta$, once again, due to algebraic subtleties, although the graph shows that the statement should be correct. This is, chiefly, the reason I decided not to resort to any form of transformation or substitution, for I understood too well that if I were to do so, the algebra, if not becoming several orders of magnitude more unwieldy, at least emerging at the latter stages of my manipulations, ruining the whole effort.
What I wish to ask you of is to give me a hint. This is important, not a solution, or even a half-solution, but a hint. Something of the form „try making this transformation”, or „try to use this quantity as the variable” etc., without clarifying the immediate implications of such steps. Please, do not redirect me to any sources where I can find solutions as well. I should be incredibly grateful to those of you who give valuable advice. Thank you!
Edit: I should add that I do not require assistance on the part of the problem concerning line $PQ$ being parallel to the $y$-axis, or concerning the angle bisection. I shall work them out on my own, I just added them to the formulation for the sake of completeness.