$\text{ The curve described parametrically by }x=t^2+t+1,\text{ and }y=t^2-t+1\text{ represents }$

$\frac{x+y}{2}=t^2+1,\frac{x-y}{2}=t$

Eliminating t , we get

$2(x+y)=(x-y{)}^2+4$

Since the second-degree terms form a perfect square, it
represents a parabola $(\text{ also },\mathrm{\Delta }\ne 0)$