Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 5

The absolute value is nonnegative, so a real point must satisfy $1-x^2\ge 0$, or $-1\le x\le 1$. On that interval the two branches areThey form a closed lens symmetric about both coordinate axes, meeting at $(\pm 1,0)$ and reaching $(0,\pm 1)$. The parabolas outside the stated interval do not belong to the original locus.

Substituting $y=1/2$ into the absolute value relation gives $1/2=1-x^2$, so $x^2=1/2$. All real intersections are