Analytic Geometry (high school): Why is the sum of the distances from any point of the ellipse to the two foci the major axis? I don't understand where that formula came from. Could someone explain? For example any point $(x,y)$ on the ellipse from the two foci $(-c,0)$ and $(c,0)$ is equal to $2a$ where $2a$ is the distance of the major axis. Where did this idea come from? 

The formula for an ellipse centred at the origin is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \tag{1} $$ where a>b.
Now let us define a constant (note at this point I'm just defining a constant - nothing about foci)
$$f = \sqrt{a^2 - b^2}\tag{2}$$
Let us now pick 2 points on the x axis (f,0) and (-f,0).
The distance from point (f,0) to a point on the ellipse is (by Pythagoras)
$$r_1 = \sqrt{(x-f)^2 + y^2}$$
and the distance from point (-f,0) to the same point on the ellipse is
$$r_2 = \sqrt{(x+f)^2 + y^2}$$
Looking at $r_1$ first
$$r_1 = \sqrt{(x-f)^2 + y^2}\tag{3}$$
From (1)
$$y^2 = (1-\frac{x^2}{a^2})b^2\tag{4}$$
so substituting (4) into (3) we have
$$r_1 = \sqrt{(x-f)^2 + (1-\frac{x^2}{a^2})b^2}$$
Expanding this out we get
$$r_1 = \sqrt{x^2 -2fx + f^2 + b^2 - \frac{x^2b^2}{a^2}}$$
Rearranging
$$r_1 = \sqrt{x^2(1-\frac{b^2}{a^2}) -2fx + f^2 + b^2}$$
$$ r_1 = \sqrt{x^2(\frac{a^2-b^2}{a^2}) -2fx + f^2 + b^2}$$
but from (2) $f^2 = a^2 - b^2$ so 
$$ r_1 = \sqrt{x^2(\frac{f^2}{a^2}) -2fx + f^2 + b^2}$$
also from (2) $f^2 + b^2 = a^2 $ so 
$$ r_1 = \sqrt{x^2(\frac{f^2}{a^2}) -2fx + a^2}$$
and rearrange under the sq root
$$r_1 = \sqrt{(a - \frac{xf}{a})^2}$$
$$r_1 = a - \frac{xf}{a}$$
Similarly you can show that 
$$r_2 = a + \frac{xf}{a}$$
so
$$r_1+r_2 = 2a$$
