The locus of the midpoints of all chords of the parabola $y^{2} = 4 a x$ through its vertex is another parabola with directrix

x = - a

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x = a

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x = 0

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x = - \frac{a}{2}

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Solution

The correct option is D $x = - \frac{a}{2}$
A point $A$ on the parabola $y^{2} = 4 a x$ can be written as $(a t^{2}, 2 a t)$ and $M$ is midpoint of the chord $O A$ .

The coordiante of point M are
$(\frac{a t^{2}}{2}, a t)$
Let $h = \frac{a t^{2}}{2}$ and $k = a t$
$t = \frac{k}{a}$
$\Rightarrow 2 h = a \frac{k^{2}}{a^{2}}$
$\Rightarrow k^{2} = 2 a h$
Replace $h$ by $x$ and $k$ by $y$

The locus will be
$y^{2} = 2 a x$
$y^{2} = 4 (\frac{a}{2}) x$

Hence, the directrix is
$x = - \frac{a}{2}$

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