If (2 t^2, 4 t) is a point on the parabola y^ | vedclass

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(D) The given equation of the parabola is $y^2 = 8x$,which is of the form $y^2 = 4ax$. Comparing,we get $a = 2$.The focus of the parabola is $(a, 0) =

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$\pm \frac{1}{\sqrt{2}}$

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(D) The given equation of the parabola is $y^2 = 8x$,which is of the form $y^2 = 4ax$. Comparing,we get $a = 2$.The focus of the parabola is $(a, 0) = (2, 0)$.The focal distance of a point $(x_1, y_1)$ on the parabola $y^2 = 4ax$ is given by $|x_1 + a|$.Given the point is $(2t^2, 4t)$ and focal distance is $3$,we have:$|2t^2 + 2| = 3$Since $2t^2 + 2$ is always positive for real $t$,we have:$2t^2 + 2 = 3$$2t^2 = 1$$t^2 = \frac{1}{2}$$t = \pm \frac{1}{\sqrt{2}}$Thus,option $(D)$ is correct.

If $(2 t^2, 4 t)$ is a point on the parabola $y^2 = 8x$ such that its focal distance is $3$,then $t =$

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