The locus of the middle points of focal chords of the parabola $y^{2} = 4 a x$ is

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

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

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

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

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Solution

The correct option is B $y^{2} = 2 a (x - a)$
Let the equation of the parabola be $y^{2} = 4 a x .$

Let $t 1, t 2$ be the extremities of the focal chord. Then $t 1 . t 2 = - 1.$

The equation of the circle on $t 1, t 2$ as diameter is

$(x - a t 2^{2}) (x - a t 2^{2}) + (y - 2 a t 1) (y - 2 a t 2) = 0$

or $x^{2} + y^{2} - a x (t 1^{2} + t 2^{2}) - 2 a y (t 1 + t 2) + a^{2} t 1^{2} t 1^{2} + 4 a^{2} t 1 t 2 = 0$

$\Rightarrow x^{2} + y^{2} - a x (t 1^{2} + t 2^{2}) - 2 a y (t 1 + t 2) - 3 a^{2} = 0.$

If $(α, β)$ be the centre of the circle, then $α = \frac{a}{2} (t 1^{2} + t 2^{2})$

$β = a (t 1 + t 2) \Rightarrow (t 1 + t 2)^{2} = \frac{β^{2}}{α^{2}} \Rightarrow t 1^{2} + t 2^{2} + 2 t 1 t 2 = \frac{β^{2}}{α^{2}} \Rightarrow \frac{2 α}{a} - 2 = \frac{β^{2}}{α^{2}}$

$\Rightarrow 2 a α - 2 a^{2} = β^{2} \Rightarrow β^{2} = 2 a (α - a) .$

Hence locus of $(α, β)$ is $y^{2} = 2 a (x - a) .$

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