Analytic Geometry (high school): Why is the sum of the distances from any point of the ellipse to the two foci the major axis?

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 $=a(1+e\cos\theta)$

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 $$r_1+r_2 = 2a$$