If PQ is doubled ordinate of the parabola $y^{2} = 4 a x$ then locus of its points of trisection is

{9 y}^{2} = 4 a x

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{9 y}^{2} = 16 a x

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{3 y}^{2} = 8 a x

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{9 y}^{2} = 8 a x

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Solution

The correct option is B ${9 y}^{2} = 4 a x$

We have,
$\begin{matrix} y^{2} = 4 a x A (a t^{2}, 2 a t) ∴ h = a t^{2} & 3 k = 2 a t ∴ {(3 k)}^{2} = 4 a (h) ∴ 9 k^{2} = 4 a h 9 y^{2} = 4 a x ∴ l o c u s i s 9 y^{2} = 4 a x \end{matrix}$

Hence, this is the answer.

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