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

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

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

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

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

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Solution

The correct option is A $d^{2} + (2 b + 3 c)^{2} = 0$
The point of intersection of $y^{2} = 4 a x$ and $x^{2} = 4 a y$ is $(0, 0)$ and $(4 a, 4 a)$
Therefore, given line is same as $y = x$
Slope $= \frac{- 2 b}{3 c} = 1$
$\Rightarrow 2 b + 3 c = 0$ ...(1)
Also line passes through $(0, 0)$
Therefore, $d = 0$ ...(2)
From (1) and (2): $d^{2} + (2 b + 3 c)^{2} = 0$
Ans: A

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