For proving the reflective property of a parabola, only the shape of the parabola would matter, the position would not matter. The vertex equation of a parabola is \(f(x) = \frac{(x - h)^2}{4p} + k\), but for proving the reflective property, \(f(x) = \frac{x^2}{4p}\) would suffice.
Let \((0, 0)\) be the vertex, \((0, p)\) be the focus and \((a, b)\) be a random point in the parabola. We will call these points \(V\), \(F\) and \(R\):
The tangent to \(f(x)\) at \(x\) has the gradient \(f'(x) = \frac{2x}{4p}\) or \(f'(x) = \frac{x}{2p}\), we will call this \(t(x)\). For example, \(t(a)\) is the gradient of the tangent at \(a\), which is \(\frac{a}{2p}\):
The directrix would be at \(y = -p\). Let \(D\) be a point at \((a, -p)\), and let \(E\) be a point at \((a, 0)\):
Let point \(M\) be the intersection between \(\overline{FD}\) and the \(x\)-axis.:
Since \(\overline{FV}\) and \(\overline{ED}\) are of the same length, then \(M\) is the midpoint of \(\overline{VE}\) and \(\overline{FD}\), which means \(M\) is at \((\frac{a}{2}, 0)\). Now lets try to find the gradient of line \(\overline{MR}\):
The gradient of \(\overline{MR}\) is the length of \(\overline{RE}\) divided by the length of \(\overline{ME}\), which is \(\frac{b}{a/2}\), or \(\frac{2b}{a}\). The value of \(b\) is \(f(a)\), or \(\frac{2a^2}{p}\), which means the gradient of \(\overline{MR}\) is \(\frac{2(a^2/4p)}{a}\):
Does the above seem familiar? This is equal to \(t(a)\), so the line segment \(\overline{MR}\) is the tangent to the parabola.
Let the angle θ be the angle between the \(t(a)\) and the line \(x = a\):
This means angle \(MRE\) is θ:
Lets call the angle \(FRM\) α:
The reflection property states that any ray perpendicular to the directrix will bounce off the parabola and pass through the focus:
In order to prove the reflective property, we need to show that the angles α and θ are equal:
By definition of a parabola, the lines \(\overline{FR}\) and \(\overline{RD}\) are equal, which makes the triangle \(FRD\) an isosceles triangle. Since \(M\) is the midpoint of line \(\overline{FD}\), then the line \(\overline{MR}\) cuts the isosceles triangle \(FRD\) by half, meaning that \(\overline{MR}\) is an angle bisector of \(FRD\), which would mean α and θ are equal.