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? 
 A: As the center is the midpoint of the foci, so the center $O(0,0)$
As the foci lie on the major axis , so the equation of the major axis $y=0\implies$ the equation of the minor axis $x=0$
Now,  if the length of the major, minor axes be $2a,2b$ respectively with eccentricity $=e$
So, the equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Any point $P$ on the ellipse can be $P(a\cos\theta,b\sin\theta)$
So, the distance between $(a\cos\theta,b\sin\theta), (ae,0)$  is
$\sqrt{(ae-a\cos\theta)^2+(b\sin\theta-0)^2}$
$=\sqrt{a^2e^2+a^2\cos^2\theta-2a^2e\cos\theta+a^2(1-e^2)(1-\cos^2\theta)}$
$=a(1-e\cos\theta)$ as $0\le e<1,-1\le \cos\theta\le 1$ and $b^2=a^2(1-e^2)$
Similarly,  the distance between $(a\cos\theta,b\sin\theta), (-ae,0)$  is $=a(1+e\cos\theta)$
A: 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
$$r_1+r_2 = 2a$$
A: Suppose that the foci of the ellipse are $(c,0)$ and $(-c,0)$, and that the major axis runs from $(-x,0)$ to $(x,0)$. Then the length of the major axis is $2x$. At the same time, the distance from $(x,0)$ to $(c,0)$ is $(x-c)$, and the distance from $(x,0)$ to $(-c,0)$ is $x-(-c) = x+c$. Then the sum of these distances is
$$ (x-c) + (x+c) = 2x$$
So the sum of the distances is precisely equal to the length of the major axis.
A: OK, imagine you have tacks at the points $(-c,0$ and $(c,0)$, which hold each end of a string of length $2 a$.  We draw an ellipse by holding a pen taut against the string.  The sum of the distances to a point on the ellipse from each of the tack points is
$$\sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2 a$$
The trick is to manage the algebra so that the derivation is readable.  First, square both sides to get
$$(x-c)^2 + (x+c)^2 + 2 y^2 + 2 \sqrt{x^2+y^2+c^2+2 c x} \sqrt{x^2+y^2+c^2-2 c x} = 4 a^2$$
This simplifies a little to
$$x^2+y^2+c^2+\sqrt{(x^2+y^2+c^2)^2-4 c^2 x^2} = 2 a^2$$
Now we need to rid ourselves of this remaining square root by isolating it:
$$\begin{align}(x^2+y^2+c^2)^2-4 c^2 x^2 &= [2 a^2 - (x^2+y^2+c^2)]^2\\ &= 4 a^4 - 4 a^2 (x^2+y^2+c^2) + (x^2+y^2+c^2)^2 \end{align}$$
We have some fortuitous cancellation which leaves us with a quadratic.  Rearrange to get
$$(a^2-c^2) x^2 + a^2 y^2 = a^2 (a^2-c^2)$$
or, in standard form:
$$\frac{x^2}{a^2} + \frac{y^2}{a^2-c^2} = 1$$
Note that, for an ellipse, $a>c$.  We interpret $a$ to be the semimajor axis, $c$ to be the focal length, and $b=\sqrt{a^2-c^2}$ is the semiminor axis.
A: From the comments we know that the sum of the distances from a point $(x,y)$ on the ellipse to the two foci is a constant. Now, that's true for any point on the ellipse, so look at a point on the ellipse on the line connecting the two foci. In this case, we're looking at a point $(a,0)$ defining the semi-major axis, and the distance is $(a-c) + (a-(-c)) = 2a$.
