= Solution
Solving for $y$ gives $y=\pm(b/a)\sqrt{x^2-a^2}$ with $x\geq a$. For large $x$, the <square root> is $x-a^2/(2x)+O(x^{-3})$, so the two <asymptotes> are
$$
\boxed{y=\frac ba x,\qquad y=-\frac ba x}.
$$
The 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.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2-hyperbola.png]
{title=The parametrized right branch of a hyperbola, its foci, nearest point, and asymptotes}
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