If a neq 0 and the line 2bx + 3cy + 4d = 0 pa | vedclass

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(A) The parabolas $y^2 = 4ax$ and $x^2 = 4ay$ are symmetric with respect to the line $y = x$. Therefore,the line $y = x$ passes through their points o

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$d^2 + (2b + 3c)^2 = 0$

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(A) The parabolas $y^2 = 4ax$ and $x^2 = 4ay$ are symmetric with respect to the line $y = x$. Therefore,the line $y = x$ passes through their points of intersection. Solving $y^2 = 4ax$ and $y = x$: $x^2 = 4ax \Rightarrow x(x - 4a) = 0 \Rightarrow x = 0$ or $x = 4a$. When $x = 0$,$y = 0$. When $x = 4a$,$y = 4a$. Thus,the points of intersection are $(0, 0)$ and $(4a, 4a)$. Since the line $2bx + 3cy + 4d = 0$ passes through $(0, 0)$: $2b(0) + 3c(0) + 4d = 0 \Rightarrow 4d = 0 \Rightarrow d = 0$. Since it also passes through $(4a, 4a)$: $2b(4a) + 3c(4a) + 4d = 0 \Rightarrow 8ab + 12ac + 4d = 0$. Since $d = 0$ and $a 
eq 0$,we have $8ab + 12ac = 0 \Rightarrow 4a(2b + 3c) = 0 \Rightarrow 2b + 3c = 0$. Therefore,$d^2 + (2b + 3c)^2 = 0^2 + 0^2 = 0$.

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

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