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(D) The equation of the parabola is $y^{2} = x$. Comparing this with the standard form $y^{2} = 4ax$,we get $4a = 1$,so $a = \frac{1}{4}$.Any point on

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$\left(\frac{4}{9}, -\frac{2}{3}\right)$

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(D) The equation of the parabola is $y^{2} = x$. Comparing this with the standard form $y^{2} = 4ax$,we get $4a = 1$,so $a = \frac{1}{4}$.Any point on the parabola $y^{2} = 4ax$ in terms of parameter $t$ is given by $(at^{2}, 2at)$.Given the parameter $t = -\frac{4}{3}$,we substitute $a = \frac{1}{4}$ and $t = -\frac{4}{3}$ into the coordinates:$x = at^{2} = \frac{1}{4} 	imes \left(-\frac{4}{3}ight)^{2} = \frac{1}{4} 	imes \frac{16}{9} = \frac{4}{9}$.$y = 2at = 2 	imes \frac{1}{4} 	imes \left(-\frac{4}{3}ight) = \frac{1}{2} 	imes \left(-\frac{4}{3}ight) = -\frac{2}{3}$.Thus,the coordinates are $\left(\frac{4}{9}, -\frac{2}{3}ight)$.

The Cartesian coordinates of the point on the parabola $y^{2}=x$ whose parameter is $t = -\frac{4}{3}$ are

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