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Q.
If $a \neq 0$ and the line $2bx + 3cy + 4d = 0$ passes through the points of intersection of the parabolas $y^2 = 4ax$ and $ x^2 = 4ay$, then
Conic Sections
Solution:
The given parabolas are $y^2 = 4ax$ and $x^2 = 4ay$.
Solving these, we get $A (0, 0), B (4a, 4a)$ as their points of intersection
Also the line $2bx + 3cy + 4d = 0 $ passes through $A$ and $B$.
$\therefore d=0$ and $2b\cdot4a +3c\cdot4a = 0 $
$ \Rightarrow a\left(2b+3c\right) = 0$
$ \Rightarrow 2b+3c = 0 \quad\left[\because a\ne0\right] $
$\therefore d^{2} +\left(2b+3c\right)^{2} = 0 $
$= 0+0 = 0$