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The locus of the midpoints of chords of the parabola $y^{2} = 4 a x$ which subtends a right angle at the vertex is

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$y^{2} = 2 a (x - 4 a)$

$y^{2} = 2 a (x + 4 a)$

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

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

answer is B.

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Detailed Solution

$y^{2} = 4 a x \to$ mid pt chord $s_{1} = s_{11}$

$y y_{1} - 2 a (x + x_{1}) = y_{1}^{2} - 4 a x_{1}$

hom $\to y^{2} = 4 a x (\frac{y y_{1} - 2 a x}{y_{1}^{2} - 2 a x_{1}})$

$a + b = 0 \to y^{2} = 2 a (x - 4 a)$

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