1. Find the intersection points of the parabolas

The parabolas are $y^2=4ax \quad \text{and} \quad x^2=4ay,$ with $a\neq 0$.

We need the common points satisfying both equations.

From $y^2=4ax \implies x=\frac{y^2}{4a}.$ Substitute into $x^2=4ay$: $\left(\frac{y^2}{4a}\right)^2=4ay.$ So, $\frac{y^4}{16a^2}=4ay$ $y^4=64a^3y$ $y(y^3-64a^3)=0.$ Thus, $y=0 \quad \text{or} \quad y=4a.$

• If $y=0$, then from $y^2=4ax$, we get $x=0$.
• If $y=4a$, then from $y^2=4ax$, $16a^2=4ax \implies x=4a.$

Hence the intersection points are $(0,0) \quad \text{and} \quad (4a,4a).$

2. Use the condition that the line passes through both points

The line is $2bx+3cy+4d=0.$

Since it passes through $(0,0)$, $2b(0)+3c(0)+4d=0 \implies 4d=0 \implies d=0.$

Since it also passes through $(4a,4a)$, $2b(4a)+3c(4a)+4d=0.$ Using $d=0$, $8ab+12ac=0$ $4a(2b+3c)=0.$ Given $a\neq 0$, $2b+3c=0.$

So we have $d=0 \quad \text{and} \quad 2b+3c=0.$

3. Match with the options

From the above, $d^2=0 \quad \text{and} \quad (2b+3c)^2=0.$ Therefore, $d^2+(2b+3c)^2=0.$

This matches Option D.

4. Check other options

• A: $d^2+(3b-2c)^2=0$ would require $d=0$ and $3b-2c=0$, not implied.
• B: $d^2+(3b+2c)^2=0$ would require $d=0$ and $3b+2c=0$, not implied.
• C: $d^2+(2b-3c)^2=0$ would require $d=0$ and $2b-3c=0$, not implied.
• D: $d^2+(2b+3c)^2=0$ is exactly what we obtained.

So the correct answer is D.