A conic section with eccentricity $e$ is defined as :
Given a line L, a point $F$ not on $L$, and a positive number $e$. Let $d(X,L)$ denote the distance from a point $X$ to $L$. The set of all $X$ satisfying the relation $$\|X-F\| = e \cdot d(X,L)$$ is called a conic section with eccentricity $e$. The conic is called an ellipse if $e < 1$.
Now, taking a vector $N$ normal to $L$ with $\|N\| = 1$, we have $d(X,L) = |(X-P) \cdot N|$ with $P$ being any point on $L$, so the equation becomes $$\|X-F\| = e |(X-P) \cdot N|$$
Now, for an ellipse $(X-P) \cdot N$ is always negative or positive according to the choice of $N$.
Setting $P$ to be that point on $L$ nearest to $F$,and positioning $F$ in the negative half plane (in the sense that $(F-P) \cdot N < 0$, we get that $P-F = dN$ with $d = \|P-F\|$. So we can rewrite the basic relation as
$$\|X-F\| = e |(X-F) \cdot N - d|$$
This was the background, I wrote it to make it clear to you and to me :D
From here, I want to show that, if a conic is an ellipse with center at the origin then
$$\|X-F\| + \|X+F\|= 2a$$
with $a = ed+eF \cdot N$. Namely that the sum of distances from any point on an ellipse to its foci is constant.
Attempt :
if $F$ is on the negative half plane then $(X-F) \cdot N - d < 0$, so the basic relation can be written as
$$\|X-F\| = e (d - (X-F) \cdot N ) = ed+eF\cdot N -eX\cdot N = a -eX\cdot N$$
Now we can see that $$eX\cdot N + a = e((X+F)\cdot N + d) \stackrel{?}{=} \|X+F\|$$
If I was able to establish the last equality then it would be done, but I'm not able to do it.
Now, I saw that the problem involved a simmetric relation about the origin so if it works for $X$ it works also for $-X$, then substituting in the basic relation we get
$$\|X+F\| = e |(X+F) \cdot N + d| = e ((X+F) \cdot N + d) $$
because $(-X-F) \cdot N - d < 0$ . So , we have proven the theorem. Is this right?
There are two parallel line directrices which are separated by constant distance $D_1 D_2 = u_1+ u_2$ symmetrically placed about origin on both sides.
Multiply the relation by $e$ so that for a new constant $a$
$$ 2 a = e u_1+ e u_2 $$
which proves the given proposition.