= Solution
Inverting $u+iv=\operatorname{arcosh}z$ gives $z=\cosh(u+iv)$, hence the <elliptic coordinates>
$$
x=\cosh u\cos v,\qquad y=\sinh u\sin v.
$$
Eliminating $v$ at fixed $u>0$ gives the <ellipse>
$$
\boxed{\frac{x^2}{\cosh^2u}+\frac{y^2}{\sinh^2u}=1.}
$$
Its focal distance is $\sqrt{\cosh^2u-\sinh^2u}=1$. At fixed $v$, eliminating $u$ instead gives the <hyperbola>
$$
\boxed{\frac{x^2}{\cos^2v}-\frac{y^2}{\sin^2v}=1.}
$$
Its focal distance is also one, so both families have foci $(\pm1,0)$. A fixed $v$ traces an appropriate half of the hyperbola; complementary coordinate values complete it. Degenerate values give the focal segment and portions of the coordinate axes. Away from these degeneracies the curves are orthogonal by part (i), or by $\mathbf r_u\cdot\mathbf r_v=0$ in the displayed parametrization. A branch of the inverse is used locally; its real part with $u\geq0$ is single-valued outside the focal segment.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1-q7-elliptic-coordinates.png]
{title=Confocal ellipses and hyperbolae generated by the complex inverse hyperbolic cosine}
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