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

Given,

$a{x}^2+2bx+cy=0$ and $d^2+2ex+fy=0$ intersect on the line $y=1$,

And $a,b,c$ are in G.P.

So, $b^2=ac$

Now putting the value of $b$ in $a{x}^2+2bx+cy=0$and taking $y=1$ we get,

$a{x}^2+2bx+c=0$

$\Rightarrow {\mathrm{ax}}^2+2\sqrt{\mathrm{ac}x}+\mathrm{c}=0 \left(\because {\mathrm{b}}^2=\mathrm{ac}\right)$

$\Rightarrow (x\sqrt{a}+\sqrt{c}{)}^2=0$

$\Rightarrow x=-\frac{\sqrt{c}}{\sqrt{a}}$ and $x^2=\frac{c}{a} .......\left(1\right)$

Now, putting the value of $x \& x^2$ in $d{x}^2+2ex+fy=0$ and taking $y=1$ we get,

$\Rightarrow d\left(\frac{c}{a}\right)+2e\left[-\frac{\sqrt{c}}{\sqrt{a}}\right]+f=0$

$\Rightarrow \frac{dc}{a}+f=2e\sqrt{\frac{c}{a}}$

$\Rightarrow \frac{d}{a}+\frac{f}{c}=2e\sqrt{\frac{1}{ac}}$

$\Rightarrow \frac{d}{a}+\frac{f}{c}=\frac{2e}{b}$  as $\left[b=\sqrt{ac}\right]$

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