Here is probably a very stupid question, but I couldn't find the answer myself. Also, I was unable of finding anywhere.
I bet all of you knows this obvious stuff that you are about to read, but I'll write anyway: Let $F_1$ and $F_2$ and $F$ to be arbitrarily non-coincident points. Let $L$ and $D$ be the directrix lines such that $F\notin L$ and $F\notin D$. The point $P$ belongs to the conic section. Notation $d(A, B)$ means the distance between geometrical objects $A$ and $B$ (points or lines).
$$ \mbox{Conics: } \begin{cases} \mbox{Ellipse: } && d(P, F_1) + d(P, F_2) = 2a \\ \mbox{Hyperbola: } && d(P, F_1) - d(P, F_2) = 2a \\ \mbox{Parabola: } && d(P, F) = d(P, L) \\ \end{cases} $$
For the appropriate values of $\epsilon$, we have each of those conics, with a single equivalent definition: $$d(P, F) = \epsilon\cdot d(P, D)$$.
Well, one can prove it by selecting point $P = (x, y)$, making the proper equations, with a few simplifications, on the first definition, and on the second, and show they are the same.
Here is my question: How can one prove equivalence between those two definitions very quickly? Efortlessly?
Also, if you are feeling generous, perhaps you can tell me: is there a way of proving this, using the whole time the $d(A, B)$ notation? How? =).
The little of what I got:
The proof for parabola is obvious. Then, basically, we have: $$ \mbox{Conics: } \begin{cases} \mbox{Ellipse: } && d(P, F_1) = 2a - d(P, F_2) = \epsilon\cdot d(P, D), && \epsilon < 1 \\ \mbox{Hyperbola: } && d(P, F_1) = 2a + d(P, F_2) = \epsilon\cdot d(P, D), && \epsilon > 1 \\ \end{cases} $$
Thus, we are basically proving that, for fixed point $F$, fixed line $D$, fixed constant $a>0$, there exists $\epsilon$ such that: $$ 2a \pm d(P, F) = \epsilon\cdot d(P, D), \quad\forall P\in\mathbb{R}^2 $$
Well.. that's it. I couldn't progress further. I'll edit if I make progresses....
