The curve described parametrically by x = t^2 | vedclass

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(C) Given $x = t^2 + t + 1$ and $y = t^2 - t + 1$. Adding the two equations: $x + y = 2t^2 + 2 = 2(t^2 + 1)$. Subtracting the two equations: $x - y =

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$A$ parabola

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(C) Given $x = t^2 + t + 1$ and $y = t^2 - t + 1$. Adding the two equations: $x + y = 2t^2 + 2 = 2(t^2 + 1)$. Subtracting the two equations: $x - y = 2t$,which implies $t = \frac{x - y}{2}$. Substituting $t$ into the expression for $x + y$: $x + y = 2\left(\left(\frac{x - y}{2}\right)^2 + 1\right) = 2\left(\frac{(x - y)^2}{4} + 1\right) = \frac{(x - y)^2}{2} + 2$. Multiplying by $2$: $2(x + y) = (x - y)^2 + 4$. Expanding: $2x + 2y = x^2 - 2xy + y^2 + 4$. Rearranging: $x^2 - 2xy + y^2 - 2x - 2y + 4 = 0$. Comparing with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we have $a = 1, h = -1, b = 1$. Since $h^2 - ab = (-1)^2 - (1)(1) = 1 - 1 = 0$,the curve represents a parabola.

List-$I$	List-$II$
$I$. Vertex	$A$. $8$
$II$. Length of latus rectum	$B$. $(\frac{29}{10}, \frac{-38}{10})$
$III$. Directrix	$C$. $3x+4y-1=0$
$IV$. One end of the latus rectum	$D$. $(\frac{-2}{5}, \frac{-16}{5})$
	$E$. $6$

The curve described parametrically by $x = t^2 + t + 1$ and $y = t^2 - t + 1$ represents:

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