Question

The parabolas : $a x^2+2 b x+c y=0$ and $d x^2+2 e x+f y=0$ intersect on the line $y=1$ If $a, b, c, d, e, f$ are positive real numbers and $a, b, c$ are in GP, then

• A. $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in G.P.
• B. $d, e, f$ are in A.P.
• C. $d, e, f$ are in G.P.
• D. $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.

Answer 1

The Correct Option is D

$a{x}^2+2bx+c=0$
$\Rightarrow a{x}^2+2\sqrt{acx}+c=0\left(\because b^2=ac\right)$
$\Rightarrow (x\sqrt{a}+\sqrt{c}{)}^2=0$
$x^2-\frac{\sqrt{c}}{\sqrt{a}}\dots \dots (1)$
$\text{Now,}d{x}^2+2ex+f=0$
$\Rightarrow d\left(\frac{c}{a}\right)+2e\left[-\frac{\sqrt{c}}{\sqrt{a}}\right]+f=0$
$\Rightarrow \frac{dc}{a}+f=2e\sqrt{\frac{c}{a}}$
$\Rightarrow \frac{d}{a}+\frac{f}{c}=2e\sqrt{\frac{1}{ac}}$
$\Rightarrow \frac{d}{a}+\frac{f}{c}=\frac{2e}{b}[\text{as}b=\sqrt{ae}]$
$\therefore \frac{d}{a},\frac{e}{b},\frac{f}{c}\text{are in A.P.}$
So, the correct option is (D) : $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.