If the equations x=t^2+t+1 and y=t^2-t+1 repr | vedclass

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

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

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(A) Given equations are $x=t^2+t+1$ 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 $t = \frac{x-y}{2}$. Substitute $t$ into the expression for $x+y$: $x+y = 2\left(\left(\frac{x-y}{2}\right)^2 + 1\right)$ $x+y = 2\left(\frac{x^2+y^2-2xy}{4} + 1\right)$ $x+y = \frac{x^2+y^2-2xy}{2} + 2$ Multiplying by $2$: $2x+2y = x^2+y^2-2xy+4$ Rearranging the terms: $x^2-2xy+y^2-2x-2y+4=0$.

If the equations $x=t^2+t+1$ and $y=t^2-t+1$ represent a curve $C$ with parameter $t$,then the Cartesian equation of $C$ is

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