$\begin{array}{1 1}(A)\;x=-a \\(B)\;x=\large\frac{-a}{2} \\(C)\;x=0 \\(D)\;x=\large\frac{a}{2} \end{array}$

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Take any point on the parabola $(at^2,2at)$

Therefore the mid point $(h,k)=\bigg(\large\frac{at^2+a}{2},\frac{2at+0}{2}\bigg)$

$2h=a(t^2+1),at=k$

$2h= a (\large\frac{k^2}{a^2}+1)$

So,locus is $2xa=y^2+a^2$

$y^2=2a(x-a/2)$

Direction : x $=\large\frac{-a}{2}+\frac{a}{2}$

$\qquad=0$ or y axis

Hence C is the correct answer.

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