For the parabola represented in the parametri | vedclass

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(A) Given parametric 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

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$2$

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(A) Given parametric 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 $t = \frac{x-y}{2}$.Substituting $t$ into the sum equation: $x+y = 2\left(\left(\frac{x-y}{2}\right)^2+1\right)$.$x+y = 2\left(\frac{(x-y)^2}{4}+1\right) = \frac{(x-y)^2}{2}+2$.Multiplying by $2$: $2(x+y) = (x-y)^2+4$.Rearranging: $(x-y)^2 = 2(x+y-2)$.This is in the form $Y^2 = 4aX$,where $Y = x-y$,$X = x+y-2$,and $4a = 2$.The length of the latus rectum is $4a = 2$.

For the parabola represented in the parametric form by $x=t^2+t+1$ and $y=t^2-t+1$,the length of the latus rectum is

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