The Cartesian coordinates of the point on the | vedclass

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(D) Given the parabola equation $y^2 = -16x$ and parameter $t = \frac{1}{2}$.Comparing $y^2 = -16x$ with the standard form $y^2 = -4ax$,we get $4a = 1

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$(-1, 4)$

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(D) Given the parabola equation $y^2 = -16x$ and parameter $t = \frac{1}{2}$.Comparing $y^2 = -16x$ with the standard form $y^2 = -4ax$,we get $4a = 16$,which implies $a = 4$.The parametric coordinates of a point on the parabola $y^2 = -4ax$ are given by $P(t) = (-at^2, 2at)$.Substituting $a = 4$ and $t = \frac{1}{2}$ into the coordinates:$x = -a t^2 = -4 	imes (\frac{1}{2})^2 = -4 	imes \frac{1}{4} = -1$.$y = 2at = 2 	imes 4 	imes \frac{1}{2} = 4$.Therefore,the coordinates of the point are $(-1, 4)$.

The Cartesian coordinates of the point on the parabola $y^2 = -16x$,whose parameter is $t = \frac{1}{2}$,are

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