$x - 3 = t^{2}, y = 4 t$ are the parametric equations of the parabola

y^{2} = 16 x

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y^{2} - 16 x + 48 = 0

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y^{2} - 16 x - 48 = 0

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y^{2} - 4 x + 16 = 0

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Solution

The correct option is B $y^{2} - 16 x + 48 = 0$
$y = 4 t \Rightarrow \frac{y}{4} = t$
$x - 3 = t^{2}$
$\Rightarrow x - 3 = {(\frac{y}{4})}^{2}$
$\Rightarrow 16 (x - 3) = y^{2}$

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