Equidistance of Hyperbola equals Transverse Axis

Theorem
Let $K$ be an hyperbola whose foci are $F_1$ and $F_2$.

Let $P$ be an arbitrary point on $K$.

Let $d$ be the constant distance such that:
 * $\left\lvert{d_1 - d_2}\right\rvert = d$

where:
 * $d_1 = P F_1$
 * $d_2 = P F_2$

Then $d$ is equal to the transverse axis of $K$.

Proof

 * HyperbolaEquidistanceTransverseAxis.png

By the equidistance property of $K$:
 * $\left\lvert{d_1 - d_2}\right\rvert = d$

applies to all points $P$ on $K$.

Thus it also applies to the two vertices $V_1$ and $V_2$.

Observing the signs of $\left\lvert{d_1 - d_2}\right\rvert$ as appropriate:


 * $V_1 F_2 - V_1 F_1 = d$
 * $V_2 F_1 - V_2 F_2 = d$

Adding:


 * $\left({V_1 F_2 - V_2 F_2}\right) + \left({V_2 F_1 - V_1 F_1}\right) = 2 d$

But:
 * $V_1 F_2 - V_2 F_2 = V_1 V_2$
 * $V_2 F_1 - V_1 F_1 = V_1 V_2$

and so:
 * $2 V_1 V_2 = 2 d$

By definition, the transverse axis of $K$ is $V_1 V_2$.

Hence the result.