Answer

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**Hint:**We have given two equations in the form of x and y, compare both equations and add and subtract it, after that compare both the equations, it will be similar to the parabola equation, we will get an answer.

**Complete step-by-step answer:**

We have, $$x = {t^2} + t + 1$$ .... (i)

and $$y = {t^2} - t + 1$$ .... (ii)

Now, $$x + y = 2(1 + {t^2})$$ .... (iii)

and $$x - y = 2t$$.... (iv)

Now, from Eqs. (iii) and (iv), we get

$$\eqalign{

& x + y = 2[1 + {\left( {\dfrac{{(x - y)}}{2}} \right)^2}] \cr

& \Rightarrow x + y = 2[\dfrac{{4 + {x^2} + {y^2} - 2xy}}{4}] \cr} $$

$$ \Rightarrow {x^2} + {y^2} - 2xy - 2x - 2y + 4 = 0\;{\text{ }}\;{\text{ }}\;$$ .... (v)

On comparing with, we get

$$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$$

We get, $$a = 1,b = 1,c = 4,h = - 1,g = - 1,f = - 1$$

$$\vartriangle = abc + 2fgh - a{f^2} - b{g^2} - c{h^2}$$

Now,

$$\eqalign{

& \vartriangle = 1 \cdot 1 \cdot 4 + 2\left( { - 1} \right)\left( { - 1} \right)\left( { - 1} \right) - 1 \times {\left( { - 1} \right)^2} - 1 \times {\left( { - 1} \right)^2} - 4{\left( { - 1} \right)^2} \cr

& = 4 - 2 - 1 - 1 - 4 \cr

& = - 4 \cr} $$

, therefore, $$\vartriangle \ne 0$$

and $$ab - {h^2} = 1 \cdot 1 - {\left( 1 \right)^2} = 1 - 1 = 0$$

**So, it is the equation of a parabola.**

**Note:**We knew the equation of parabola, i.e. $$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$$. After comparing both the given equations it gets similar to the parabola equation, so the answer is the equation is of parabola.

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