Ellipses were introduced in Introduction to Conic Sections, as one of several different curves (‘conic sections’) that are formed by intersecting a plane with an infinite double cone.
The prior section explores the standard
definition of an ellipse. The essential ideas are reviewed below for your convenience.
The purpose of this section is to explore the reflecting property of ellipses—that light emitted from one focus
is reflected through the other focus.
In an ellipse:
Light emitted from one focus
You can play with this reflecting property at right:
is reflected through the other focus.

(refresh the page to start over) 
To see why light from one focus is always reflected through the other focus, we need a few ideas.
We start by looking at a simple childhood game:
This game involves a wall (or fence) and two special spots marked on the ground. Two or more kids must run from one designated spot ($\,P_1\,$), touch the wall ($W$), and then run to a second designated spot ($P_2$). First arriver at $\,P_2\,$ wins the game! Of course, your running speed matters. But, the distance that you run also matters. To maximize your chances of winning, where should you touch the wall? Does it matter? 

You can ‘play with’ this game above:

Let's summarize and prove the results of the prior exploration.
Suppose you have a line, and two different points on the same side of the line. The shortest path
Indeed, the following two statements are equivalent:
If one statement is false, so is the other. For ease of reference in what follows, this equivalence is referred to as the ‘Point/Line/Point Equivalence’. 

Proof of the Point/Line/Point EquivalenceThe Point/Line/Point Equivalence is easy to prove, with a bit of cleverness.As shown at right, reflect $\,\overline{P_2W}\,$ about the line; call the end of the new (reflected) segment $\,P_2'\,$. The path from $\,P_1\,$ to $\,W\,$ to $\,P_2\,$ has the same length as the path from $\,P_1\,$ to $\,W\,$ to $\,P_2'\,$. 

Now it's clear how to minimize the distance $\,d(P_1,W) + d(P_2,W)\,$:

Mother Nature is extremely efficient! When light (or sound) travels between two points, it always takes the fastest route. (For light, this is called Fermat's Principle.) In particular, this is true of reflected light. If the light/sound is always traveling through the same medium (like air), then the fastest route is also the shortest route. The wellknown physics Law of Reflection states that when light/sound is reflected, the angles made with the reflecting surface must be equal. By the Point/Line/Point Equivalence, equal angles minimizes the distance $\,d(P_1,W) + d(P_2,W)\,$ over all points $\,W\,$ on the line. Make sure you understand what is being said here! You can certainly draw line segments that go from $\,P_1\,$ to $\,W\,$ to $\,P_2\,$, where $\,m_1 \ne m_2\,$. However, this wouldn't be a picture of reflected light. If it's reflected light, then the angles must be the same! 
$\,m_1\ne m_2\,$ definitely NOT reflected light 
reflected light $\,m_1 = m_2\,$ (minimizing the length of the red path over all points $\,W\,$ on the line) 
Here's the final idea that's needed, before putting everything together.
Consider an ellipse with ellipse constant $\,k\,$ and foci $\,F_1\,$ and $\,F_2\,$. Let $\,P\,$ be a typical point in the plane containing the ellipse.


These facts are fairly intuitive:
small
changes in point locations should produce small changes in the distances involved.
Also, you can see these facts ‘in action’ in the exploration above:
drag the point $\,P\,$ outside/on/inside the ellipse, observing the displayed sum of the distances.
If desired, move the foci and/or change the ellipse constant via the slider, and explore again!
The facts are also easy to prove, as follows.
Let $\,P\,$ be outside an ellipse with ellipse constant $\,k\,$, as shown at right. Then: traveling from $\,F_1\,$ to $\,P\,$ to $\,F_2\,$ is longer than traveling from $\,F_1\,$ to $\,I\,$ to $\,F_2\,$ (which equals $\,k\,$). Here are the details:
NOTE:
The formatting $$ \begin{align} A &= B\cr &= C\cr &> D\cr &= E \end{align} $$ is a shorthand for $$ A = B\ \ \text{and}\ \ B = C\ \ \text{and}\ \ C > D\ \ \text{and}\ \ D = E $$ from which one concludes that $\,A > E\,$. 

Next, let $\,P\,$ be inside an ellipse with ellipse constant $\,k\,$, as shown at right. Then: traveling from $\,F_1\,$ to $\,P\,$ to $\,F_2\,$ is shorter than traveling from $\,F_1\,$ to $\,I\,$ to $\,F_2\,$ (which equals $\,k\,$). Here are the details:
NOTE:
The formatting $$ \begin{align} A &< B\cr &= C\cr &= D\cr &= E \end{align} $$ is a shorthand for $$ A < B\ \ \text{and}\ \ B = C\ \ \text{and}\ \ C = D\ \ \text{and}\ \ D = E $$ from which one concludes that $\,A < E\,$. 
With all the prior facts in place, it's now easy to understand the ellipse reflecting property.
It must be shown that
light emitted from one focus is reflected through the other focus.
Refer to the sketch at right as you read.

On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
IN PROGRESS 