Reflective Property of a Parabola

We’ve been going the reflective property for a parabola in class and I wanted to make my own diagram to illustrate a few nice points (and all without calculus).

First without loss of generality I’m going to use the simplest quadratic equation $y=x^2$ but remember “There is only true parabola”

That said to set the context, for this parabola the directrix is at $y=-\frac{1}{4}$ and the focus F is at $(0,\frac{1}{4})$

1. By definition every point P on the parabola is equidistant from the focus F and a perpendicular line down to the directrix $\overline{\rm PN}$

2. That means by definition P is also on the perpendicular bisector of FN (every point on the perpendicular bisector is also equidistant from the two endpoints).

3. What’s super nice its easy to prove that no other point on the parabola also lies on the $\perp$ bisector.

   

   P’ is equidistant from F and N’ while P’N is the hypotenuse of a triangle with P’N’ and therefore P’ is not equidistant from P and N, not on the $\perp$ bisector.

4. That means the perpendicular bisector by definition is also the tangent to the parabola at point P since no other point beside P on the parabola lies on it.

5. Via the congruent triangles one can see the tangent line intersects the horizontal segment through vertex to the $\overline{\rm PN}$ exactly at its midpoint.

6. All three grey angles are congruent (two are from the perpendicular bisector and the last is from the transversal.

7. $\therefore$ that means a vertical wave or ray heading down will bounce off the parabola and hit the focus or put another way every vertical ray will converge at the focus.