The exercise $13.25.12$ asks us to prove that on each branch of a hyperbola $\|X - F\| - \|X+F\| = constant$.
From previous discussions, we know that $\|X-F\| = |eXN - a|$ (equation $13.29$ in the book), $a = \frac{ed}{1 - e^2} < 0$ (section $13.22$), $e > 1$ (because it's hyperbola).
If we take $N = (1, 0)$ and $X = (x, y)$, then $XN = x$, so we get $\|X-F\| = |ex-a|$ and by having $-X$ in that equation, we get $\|X+F\| = |ex+a|$. Because it's hyperbola, I suppose we can assume that $|ex| < |a|$.
If that's all correct, then have two cases, one for each hyperbola branch:
Case 1. $ex > 0$ (right branch). $a < 0 < ex \implies 0 < ex - a \implies \|X-F\| = ex-a$. Similarly $-a < ex \implies 0 < ex + a \implies \|X+F\| = ex+a$. Taking the difference, we get $\|X - F\| - \|X+F\| = ex - a - ex - a = -2a > 0$.
Case 2. $ex < 0$ (left branch). $ex < a \implies ex - a < 0 \implies \|X-F\| = a-ex$. Similarly $ex < -a \implies ex + a < 0 \implies \|X+F\| = -ex-a$. Taking the difference, we get $\|X - F\| - \|X+F\| = a - ex + ex + a = 2a < 0$.
Did I make a mistake in the proof? Is the first difference positive, because the reference focal point is in the left branch of the hyperbola, as we started with $a < 0$?
Thanks!