Solution
= Solution
Set $X=x^2$ and $Y=y^2$ in the two <conic sections>. A real intersection requires $X,Y\ge0$. Solving the resulting <linear equations> gives
$$
X=\frac3{2a+1},\qquad Y=\frac{a-1}{2a+1}.
$$
For $a>0$ the denominator and $X$ are positive, so intersections exist exactly when $a\ge1$. At $a=1$ they are $(\pm1,0)$; for $a>1$ both choices of sign for each square root give intersections. Consequently
$$
\boxed{0<a<1\quad\text{gives no shared points}.}
$$