The locus of the center of the circle described on any focal chord of a parabola $y^{2} = 4 a x$ as diameter is

x^{2} = 2 a (y - a)

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x^{2} = - 2 a (y - a)

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y^{2} = 2 a (x - a)

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y^{2} = - 2 a (x - a)

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Solution

The correct option is C $y^{2} = 2 a (x - a)$
If $A (a {t_{1}}^{2}, 2 a t_{1}), B (a {t_{2}}^{2}, 2 a t_{2})$ be the extremities of a focal chord for the parabola
$y^{2} = 4 a x$ , then $t_{1} t_{2} = - 1$
We want to find locus of point $C (α, β)$
where $C$ is the center of the circle having $A B$ as the diameter.
$\Rightarrow α = \frac{a}{2} ({t_{1}}^{2} + {t_{2}}^{2}); β = a (t_{1} + t_{2})$
To eliminate $t_{1} t_{2}$
$β^{2} = a^{2} ({t_{1}}^{2} + {t_{2}}^{2} + 2 t_{1} t_{2}) = a^{2} (\frac{2 α}{a} - 2)$
$\Rightarrow β^{2} = 2 a (α - a)$
$\Rightarrow y^{2} = 2 a (x - a)$

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y^{2} = 4 a x

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