= Solution
{c}
For $u=1$, the inverse <parabolic polar coordinates> give $y=v$ and $x=(1-v^2)/2$. For $v=1$, they give $y=u$ and $x=(u^2-1)/2$. Thus the requested <curves> are the upper halves of the <parabolas>
$$
\boxed{u=1:\ x=\frac{1-y^2}{2},\ y\geq0},\qquad
\boxed{v=1:\ x=\frac{y^2-1}{2},\ y\geq0}.
$$
The first <parabola> has vertex $(1/2,0)$ and opens to the left; the second <parabola> has vertex $(-1/2,0)$ and opens to the right. Both pass through $(0,1)$ and are reflections of one another across the $y$-axis.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-v-resit-parabolic-regions.png]
{title=Parabolic coordinate curves and the two regions used in Question 1}
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