If the line 2bx + 3cy + 4d = 0 passes through | vedclass

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(A) The given parabolas are $y^2 = 4ax$ and $x^2 = 4ay$. Solving these,we substitute $y = x^2 / (4a)$ into $y^2 = 4ax$,giving $(x^2 / (4a))^2 = 4ax$,w

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

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(A) The given parabolas are $y^2 = 4ax$ and $x^2 = 4ay$. Solving these,we substitute $y = x^2 / (4a)$ into $y^2 = 4ax$,giving $(x^2 / (4a))^2 = 4ax$,which simplifies to $x^4 = 64a^3x$. This yields $x(x^3 - 64a^3) = 0$,so $x = 0$ or $x = 4a$. The intersection points are $(0, 0)$ and $(4a, 4a)$. The line $2bx + 3cy + 4d = 0$ passes through both points. For $(0, 0)$: $2b(0) + 3c(0) + 4d = 0 \Rightarrow d = 0$. For $(4a, 4a)$: $2b(4a) + 3c(4a) + 4d = 0$. Since $d = 0$,this simplifies to $8ab + 12ac = 0$,which implies $4a(2b + 3c) = 0$. Assuming $a 
eq 0$,we get $2b + 3c = 0$. Combining $d = 0$ and $2b + 3c = 0$,we get $d^2 + (2b + 3c)^2 = 0$.

If the line $2bx + 3cy + 4d = 0$ passes through the points of intersection of $y^2 = 4ax$ and $x^2 = 4ay$,then

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