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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Solution

The correct option is D

Given parabolas are $y^{2} = 4 a x$ .....(i)
$x^{2} = 4 a y$ .....(ii)
Putting the value of y from (ii) in (i), we get
$\frac{x^{4}}{16 a^{2}} = 4 a x \Rightarrow x (x^{3} - 64 a^{3}) = 0 \Rightarrow x = 0, 4 a$

from(ii), y=0, 4a. Let A $\equiv$ (0,0); B $\equiv$ (4a,4a)
Since, given line 2bx+3cy+4d=0 passes through A and B, $∴$ d=0 and 8ab+12ac=0 $\Rightarrow$ 2b+3c=0, $(∴ a \neq 0)$
Obviously, $a^{2} + (2 b + 3 c)^{2} = 0$

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