The parabolas ax^2 + 2bx + cy = 0 and dx^2 + | vedclass

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(C) Given the parabolas $ax^2 + 2bx + cy = 0$ and $dx^2 + 2ex + fy = 0$ intersect on the line $y = 1$.At $y = 1$,the equations become $ax^2 + 2bx + c

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$\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in $A.P.$

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(C) Given the parabolas $ax^2 + 2bx + cy = 0$ and $dx^2 + 2ex + fy = 0$ intersect on the line $y = 1$.At $y = 1$,the equations become $ax^2 + 2bx + c = 0$ and $dx^2 + 2ex + f = 0$.Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$,so $b = \sqrt{ac}$.The first equation becomes $ax^2 + 2\sqrt{ac}x + c = 0$,which is $(\sqrt{a}x + \sqrt{c})^2 = 0$.Thus,$x = -\sqrt{\frac{c}{a}}$.Substituting this value of $x$ into the second equation $dx^2 + 2ex + f = 0$:$d(\frac{c}{a}) + 2e(-\sqrt{\frac{c}{a}}) + f = 0$.Dividing by $c$,we get $\frac{d}{a} + \frac{f}{c} = 2e\frac{1}{\sqrt{ac}}$.Since $b = \sqrt{ac}$,this simplifies to $\frac{d}{a} + \frac{f}{c} = \frac{2e}{b}$.This condition implies that $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in $A.P.$

The parabolas $ax^2 + 2bx + cy = 0$ and $dx^2 + 2ex + fy = 0$ intersect on the line $y = 1$. If $a, b, c, d, e, f$ are positive real numbers and $a, b, c$ are in $G.P.$,then

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