= Solution
Put $c=\sqrt{a^2+b^2}$ and denote the distances to $(-c,0)$ and $(c,0)$ by $d_-$ and $d_+$. Direct expansion using $c^2=a^2+b^2$ gives
$$
\begin{aligned}
d_-^2&=(a\cosh t+c)^2+b^2\sinh^2t=(c\cosh t+a)^2,\\
d_+^2&=(a\cosh t-c)^2+b^2\sinh^2t=(c\cosh t-a)^2.
\end{aligned}
$$
Since $c>a>0$ and $\cosh t\geq1$, both expressions inside the final square roots are positive. Hence
$$
\boxed{d_--d_+=2a}
$$
for every point of this right branch. This is the <focal-distance identity for a hyperbola>; on the left branch the signed difference would reverse, while its absolute value would remain $2a$.
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