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Q.
The curve described parametrically by $x=t^{2}+t+1$, and $y=t^{2}-t+1$ represents
Conic Sections
Solution:
$\frac{x+y}{2}=t^{2}+1, \frac{x-y}{2}=t$
Eliminating $t$, we get
$2(x+y)=(x-y)^{2}+4$
Since the second-degree terms form a perfect square, it represents a parabola (also, $\Delta \neq 0$ ).