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

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$\pm 1$

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

$\pm \frac{1}{\sqrt{3}}$

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

answer is D.

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Detailed Solution

$(x_{1}, y_{1}) = (2 t^{2}, 4 t) \Rightarrow y^{2} = 8 x$

$4 a = 8 = 2$

$|x_{1} + a| = 3 \Rightarrow |2 t^{2} + 2|$

$2 t^{2} = \pm 1 \Rightarrow t^{2} = \pm \frac{1}{2} \Rightarrow t = \pm \frac{1}{\sqrt{2}}$

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