Given the parameters $a,b>0$ we set $c:=\sqrt{a^2+b^2}$ and $e:=\large\frac{c}{a}$ (eccentricity),
the focal points are $F=(c,0)$ and $F'=(-c,0)$,
the directrix $L$ with the equation $x=\large\frac{a^2}{c}$
and the hyperbola $H=\{P\in\mathbb{R}^2:\large\frac{d(P,F)}{d(P,L)}\normalsize =e\}$.
How to prove this statement?
$P=(x,y)\in H\Rightarrow \large\frac{x^2}{a^2}-\frac{y^2}{b^2}\normalsize=1\Rightarrow|{d(P,F)-d(P,F')}|=2a\Rightarrow P=(x,y)\in H$
I've come so far:
Since we have $P=(x,y)\in H\iff \large\frac{d(P,F)}{d(P,L)}\normalsize =e$
The following must hold:
$\large\frac{\sqrt{(p_1-c)^2+(p_2-0)^2}}{d(P,L)}\normalsize =e\iff\large\frac{x^2}{a^2}\normalsize -\large\frac{y^2}{b^2}\normalsize =1$
$\iff \sqrt{(p_1-c)^2+(p_2-0)^2}-\sqrt{(p_1+c)^2+(p_2-0)^2}=2a$
But now I don't know how to rewrite $d(P,L)$ in terms of the euclidean metric, since $L$ is a line.
How do I go from here?