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

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(A) The given parabolas are $y^2 = 4ax$ $(i)$ and $x^2 = 4ay$ $(ii)$.Substituting $y = \frac{x^2}{4a}$ from $(ii)$ into $(i)$,we get $(\frac{x^2}{4a})

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

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(A) The given parabolas are $y^2 = 4ax$ $(i)$ and $x^2 = 4ay$ $(ii)$.Substituting $y = \frac{x^2}{4a}$ from $(ii)$ into $(i)$,we get $(\frac{x^2}{4a})^2 = 4ax$,which simplifies to $x^4 = 64a^3x$.This gives $x(x^3 - 64a^3) = 0$,so $x = 0$ or $x = 4a$.For $x = 0$,$y = 0$. For $x = 4a$,$y = 4a$. The intersection points are $A(0, 0)$ and $B(4a, 4a)$.The line $2bx + 3cy + 4d = 0$ passes through $A(0, 0)$,so $2b(0) + 3c(0) + 4d = 0$,which implies $d = 0$.The line also passes through $B(4a, 4a)$,so $2b(4a) + 3c(4a) + 4d = 0$.Since $d = 0$,this becomes $8ab + 12ac = 0$. Dividing by $4a$ (given $a 
eq 0$),we get $2b + 3c = 0$.Thus,$d = 0$ and $2b + 3c = 0$,which implies $d^2 + (2b + 3c)^2 = 0$.

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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