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…
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
$d^2+(2 b+3 c)^2=0$
$d^2+(3 b+2 c)^2=0$
$d^2+(2 b-3 c)^2=0$
$d^2+(3 b-2 c)^2=0$
Solution
Points of intersection of given parabolas are $(0,0)$ and $(4 a, 4 a)$ $\Rightarrow$ equation of line passing through these points is $y=x$ On comparing this line with the given line $2 b x+3 c y+4 d=0$, we get $d=0$ and $2 b+3 c=0 \Rightarrow(2 b+3 c)^2+d^2=0$