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If $a \neq 0$ and the line $2 b x + 3 c y + 4 d = 0$ passes through the points of intersection of the parabolas $y^{2} = 4 a x$ and $x^{2} = 4 a y$ then

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$d^{2} + {(2 b - 3 c)}^{2} = 0$

$d^{2} + {(3 b + 2 c)}^{2} = 0$

$d^{2} + {(2 b + 3 c)}^{2} = 0$

$d^{2} + {(3 b - 2 c)}^{2} = 0$

answer is C.

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Detailed Solution

Point of intersection of parabolas are $(0, 0), (4 a, 4 a)$
Origin lies on the line $\Rightarrow d = 0$
$(4 a, 4 a)$ lies on the line $\Rightarrow 2 b + 3 c = 0$

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