Reflective property of the tangent

The tangent of the parabola described by equation y=ax² has slope

${dy \over dx} = 2 a x = {2 y \over x}$

This line intersects the y-axis at the point (0,-y) = (0, - a x²), and the x-axis at the point (x/2,0). Let this point be called G. Point G is also the midpoint of line segment FQ:

$F = (0,f), \quad$

$Q = (x,-f), \quad$

${F + Q \over 2} = {(0,f) + (x,-f) \over 2} = {(x,0) \over 2} = \left({x \over 2}, 0\right).$

Since G is the midpoint of line FQ, this means that

$\| FG \| \cong \| GQ \|,$

and it is already known that P is equidistant from both F and Q:

$\| PF \| \cong \| PQ \|,$

and, thirdly, line GP is equal to itself, therefore:

$\Delta FGP \cong \Delta QGP$

It follows that $\angle FPG \cong \angle GPQ$ .

Line QP can be extended beyond P to some point T, and line GP can be extended beyond P to some point R. Then $\angle RPT$ and $\angle GPQ$ are vertical, so they are equal (congruent). But $\angle GPQ$ is equal to $\angle FPG$ . Therefore $\angle RPT$ is equal to $\angle FPG$ .

The line RG is tangent to the parabola at P, so any light beam bouncing off point P will behave as if line RG were a mirror and it were bouncing off that mirror.
Let a light beam travel down the vertical line TP and bounce off from P. The beam's angle of inclination from the mirror is $\angle RPT$ , so when it bounces off, its angle of inclination must be equal to $\angle RPT$ . But $\angle FPG$ has been shown to be equal to $\angle RPT$ . Therefore the beam bounces off along the line FP: directly towards the focus.

Conclusion: Any light beam moving vertically downwards in the concavity of the parabola (parallel to the axis of symmetry) will bounce off the parabola moving directly towards the focus. (See parabolic reflector.)

The same reasoning can be applied to a parabola whose axis is vertical, so that it can be specified by the equation

$y = ax^2 + bx + c \,$ .

The tangent has then a generic slope of

$m_{tan} = 2ax + b \,$ .

Reflection derivation, together with trigonometric angle addition rules, leads to the result that the reflected ray has a slope of

$m_{ref} = {m_{tan}^2 - 1 \over 2m_{tan}}$ .

It is generally easier to remember the reflective property of a parabola than its mathematical formula. Fortunately, the latter can be easily derived from the former.
Consider a point, P, such that light that is initially travelling parallel to the axis of symmetry is reflected from P along a line that is perpendicular to the axis of symmetry. Since the light is turned through 90 degrees by the reflection, the slope of the parabola at P must be 1.

$\frac {dy} {dx} = 2ax = 1$