Locus of trisection point of any arbitrary double ordinate of the parabola $x^{2} = 4$ by, is

9 x^{2} = b y

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3 x^{2} = 2 b y

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9 x^{2} = 4 b y

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9 x^{2} = 2 b y

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Solution

The correct option is C $9 x^{2} = 4 b y$
$x^{2} = 4 b y$
Parametric point $(2 b t, 4 b t^{2})$
$P = (\frac{2 b t}{3}, b t^{2}) \Rightarrow x = \frac{2 b t}{3}, y = b t^{2}$ Locus of $P$ is $9 x^{2} = 4 b y$

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