If the line $2 b x+3 c y+4 d=0$ passes through the points of intersection of $y^2=4 a x$ and $x^2=4 a y$, then
If the line $2 b x+3 c y+4 d=0$ passes through the points of intersection of $y^2=4 a x$ 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
Given parabola, we have
$
x^2=4 a y \text { and } y^2=4 a x
$
Intersection point of these two parabolas are $(0,0)$ and $(4 a, 4 a)$.
Given that, $2 b x+3 c y+4 d=0$
passes through point of intersection.
Case I If it passes through $(0,0)$, we obtain
$
\begin{aligned}
2 b(0)+3 c(0)+4 d & =0 \\
\Rightarrow \quad d & =0
\end{aligned}
$
Case II If it passes through $(4 a, 4 a)$, we obtain
$
\begin{aligned}
2 b(4 a)+3 c(4 a)+4(0) & =0 \Rightarrow(2 b+3 c)(4 a)=0 \\
\Rightarrow \quad 2 b+3 c & =0
\end{aligned}
$
From Eqs. (i) and (ii), we get
$
d^2+(2 b+3 c)^2=0
$