The parabolas : a x ^ 2 + 2bx + cy = 0 and d ^ 2 + 2ex + fy = 0...

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,

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

$⇒ {ax}^{2} + 2\sqrt{acx} + c = 0\left(∵ {b}^{2} = ac\right)$

$⇒ ( x\sqrt{a} + \sqrt{c}{)}^{2} = 0$

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

Now, putting the value of $x \text{ and } {x}^{2} in$d{x}^{2} + 2ex + fy = 0$and taking$y = 1$we get,</p><p>$⇒ d\left(\dfrac{c}{a}\right) + 2e\left[- \dfrac{\sqrt{c}}{\sqrt{a}}\right] + f = 0$</p><p>$⇒ \dfrac{dc}{a} + f = 2e\sqrt{\dfrac{c}{a}}$</p><p>$⇒ \dfrac{d}{a} + \dfrac{f}{c} = 2e\sqrt{\dfrac{1}{ac}}$</p><p>$⇒ \dfrac{d}{a} + \dfrac{f}{c} = \dfrac{2e}{b}$as \left[b = \sqrt{ac}\right]$

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