Show that the actual path is not necessarily one of minimum time. Hint: In the diagram, A is a source of light; CD is a cross section of a reflecting surface, and B is a point to which a light ray is to be reflected. APB is to be the actual path and AP'B, AP"B represent varied paths. Then show that the varied paths:

(a) Are the same length as the actual path if CD is an ellipse with A and B as foci.

(b) Are longer than the actual path if CD is a line tangent at P to the ellipse in (a).

(c) Are shorter than the actual path if CD is an arc of a curve tangent to the ellipse at P and lying inside it. Note that in this case the time is a maximum!

(d) Are longer on one side and shorter on the other if CD crosses the ellipse at P but is tangent to it (that is, CD has a point of inflection at P).

In the given diagram, A is a source of light; CD is a cross section of a reflecting surface, and B is a point to which a light ray is to be reflected. APB is to be the actual path and AP'B, AP"B represent varied paths.

When a ray of light strikes a smooth polished surface and bounces back, this is referred to as a light reflection. The incident light ray is said to be reflected off the surface when it lands on it. The reflected ray is the ray that rebounds back.

One of the features of an ellipse is that the total of distances to the foci A and B is constant for any point P on the ellipse, (i.e.$\left|PA\right|+\left|PB\right|=2a$). Therefore,$$\begin{gathered} \left|PA\right|+\left|PB\right|=\left|P\text{'}A\right|+\left|P\text{'}B\right| \\ \left|P\text{'}A\right|+\left|P\text{'}B\right|=\left|P"A\right|+\left|P"B\right| \end{gathered}$$......(1)

A) By geometric construction in the figure below, it is obvious that:$\left|AP\right|+\left|PB\right|\le \left|AP\text{'}\right|+\left|P\text{'}B\right|$.

B) Once again, proof is obtained using geometric construction. Because the curve CD is tangent to the ellipse and the inside of the ellipse's perimeter, the time situation is inverted. The following is taken from the diagram: $\left|AP\right|+\left|PB\right|\ge \left|AP\text{'}\right|+\left|P\text{'}B\right|$.

C) The proof is shown in the diagram below. $\left|AP\right|+\left|PB\right|$can be less than or greater than $\left|AP\right|+\left|PB\right|$, depending on where P' is$\left|AP\right|+\left|PB\right|$.

Hence, proved.