If the tangent to the curve given by x=t^{2}- | vedclass

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(D) Given the parametric equations of the curve are $x=t^{2}-1$ and $y=t^{2}-t$. First,we find the derivatives with respect to $t$: $\frac{dx}{dt} = 2

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$\frac{1}{2}$

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(D) Given the parametric equations of the curve are $x=t^{2}-1$ and $y=t^{2}-t$. First,we find the derivatives with respect to $t$: $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2t-1$. The slope of the tangent to the curve is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t-1}{2t}$. Since the tangent is parallel to the $X$-axis,its slope must be zero: $\frac{2t-1}{2t} = 0$. This implies $2t-1 = 0$,which gives $t = \frac{1}{2}$.

If the tangent to the curve given by $x=t^{2}-1$ and $y=t^{2}-t$ is parallel to the $X$-axis,then the value of $t$ is

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