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The locus of the point represented by $x = t^{2} + t + 1$ , $y = t^{2} - t + 1$ is

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$x^{2} + 2 xy + y^{2} - 2 x - 2 y + 4 = 0$

$x^{2} - 2 xy - y^{2} - 2 x + 2 y + 4 = 0$

$x^{2} - 2 xy + y^{2} + 2 x - 2 y + 4 = 0$

$x^{2} - 2 xy + y^{2} - 2 x - 2 y + 4 = 0$

answer is D.

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Detailed Solution

$x + y = 2 (t^{2} + 1), x - y = 2 t$
Eliminating
$\begin{array}{l} x + y = 2 [{(\frac{x - y}{2})}^{2} + 1] \\ \to 2 x + 2 y = x^{2} - 2 xy + y^{2} + 4 \\ \to x^{2} - 2 xy + y^{2} - 2 x - 2 y + 4 = 0 \end{array}$

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