Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 15

The Euclidean distance is $D(x,y)=\sqrt{x^2+y^2}$. Along the parametric curve,This is at least $a^2$, with equality exactly when $t=0$. Therefore

The hyperbolic functions identity eliminates $t$ from the parametric curve:The restriction is essential: $\cosh t\ge 1$, so the parametrization traces only the right branch of the full hyperbola. Conversely, every point on that right branch is reached by taking $t=\mathrm{arsinh}(y/b)$.

Solving for $y$ gives $y=\pm (b/a)\sqrt{x^2-a^2}$ with $x\ge a$. For large $x$, the square root is $x-a^2/(2x)+O(x^{-3})$, so the two asymptotes areThe curve lies between these lines on the right side and has its nearest point, with vertical tangent, at $(a,0)$. As $t$ increases it travels from the lower unbounded end through that vertex to the upper unbounded end. The left hyperbola branch is not part of the given curve.

Put $c=\sqrt{a^2+b^2}$ and denote the distances to $(-c,0)$ and $(c,0)$ by $d_{-}$ and $d_{+}$. Direct expansion using $c^2=a^2+b^2$ givesSince $c>a>0$ and $\cosh t\ge 1$, both expressions inside the final square roots are positive. Hencefor 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 $2a$.