The normal at $(a p^{2}, 2 a p) on y^{2} = 4 a x$ meets the curve again at $(a q^{2}, 2 a q)$ then

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$p^{2} + p q + 2 = 0$

$p^{2} - p q + 2 = 0$

$q^{2} + p q + 2 = 0$

$p^{2} + p q + 1 = 0$

answer is A.

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Detailed Solution

Since the normal at $(a p^{2}, 2 a p) on y^{2} = 4 a x$ meet the curve again at $(a q^{2}, 2 a q)$ .Therefore,

$p x + y = 2 a p + a p^{3}$ passes through $(a q^{2}, 2 a q)$

$\begin{array}{ll} \Rightarrow p a q^{2} + 2 a q = 2 a p + a p^{3} \\ \Rightarrow p (q^{2} - p^{2}) = 2 (p - q) \\ \Rightarrow p (q + p) = - 2 \\ \Rightarrow p^{2} + p q + 2 = 0 . \end{array}$

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