Locus of mid-points of all chords of the parabola $y^{2} = 4 x$ which are drawn through the vertex is

y^{2} = 8 x

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y^{2} = 2 x

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

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x^{2} = 2 y

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Solution

The correct option is B $y^{2} = 2 x$
R.E.F image

$Q (a t^{2}, 2 a t)$

$P (x, y)$ be the mid pt.

$∴ P (\frac{a t^{2}}{2}, a t)$

Thus, $x = \frac{a t^{2}}{2}$ $y = a t$

$y^{2} = a^{2} t^{2}$

$\Rightarrow y^{2} = a^{2} \frac{2 x}{a}$

$\Rightarrow y^{2} = 2 a x$

$a = 1$ $y^{2} = 2 x$

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Similar questions

The locus of mid points of chords of the parabola $y^{2} = 8 x$ drawn through the vertex is a parabola whose

y^{2} = 4 a x

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

90 °

y^{2} = 4 a x

y^{2} = 2 a x .

x^{2} - y^{2} = 4,

y^{2} = 8 x,

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