= Solution
For $ab\ne0$, the hyperbolic identity gives $x^2/a^2-y^2/b^2=1$. The <hyperbolic parametrization of a hyperbola> also imposes $x/a=\cosh\theta\geq1$, so the exact traced locus is
$$
\boxed{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\qquad\frac xa\geq1.}
$$
It is only one branch: the right branch for $a>0$ and the left branch for $a<0$. Its vertex is $(a,0)$ and its asymptotes are $y=\pm(b/a)x$. If $a=0$, $b\ne0$, the entire vertical axis is traced. If $b=0$, $a\ne0$, the locus is the ray $y=0$, $x/a\geq1$. Both zero give the origin.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-1-ellipse-hyperbola.png]
{title=A trigonometric ellipse and the single hyperbolic branch traced by real parameters}
{description=Original example with $a=2,b=1$. The ellipse is closed, whereas $(a\cosh\theta,b\sinh\theta)$ traces only the right branch of the hyperbola. Dashed lines show its two asymptotes.}
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