If $\mathrm{a}\ne 0$ and the line 2bx + 3cy + 4d = 0 passes through the point of intersection of the parabola's ${\mathrm{y}}^2=4\mathrm{ax}$ and ${\mathrm{x}}^2=4\mathrm{ay},$ then

 

$$\begin{gathered} \mathrm{Given} \mathrm{parabolas} {\mathrm{y}}^2=4\mathrm{ax} and {\mathrm{x}}^2=4\mathrm{ay} \\ By solving these two equations we get \\ \Rightarrow {\mathrm{x}}^4=16{\mathrm{a}}^2\left(4\mathrm{ax}\right) \\ \Rightarrow {\mathrm{x}}^3=64{\mathrm{a}}^3 \\ \Rightarrow x=0 or \mathrm{x}=4\mathrm{a} \\ \Rightarrow y=0 or \mathrm{y}=4\mathrm{a} \end{gathered}$$

point of intersection  are (0,0) and  $(4\mathrm{a}, 4\mathrm{a})$ 

it passes through (0,0) then d=0

 $$\begin{gathered} 2\mathrm{bx}+3\mathrm{cy}=0 \\ Again it passes through the point (4a,4a) then \\ 4a(2b+3c)=0 \\ \Rightarrow 2b+3c=0 \end{gathered}$$ 

then

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