### Sequences And Series Question 12

#### Question 12 - 30 January - Shift 2

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

(1) d, e, f are in A.P.

(2) $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in G.P.

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

(4) d, e, f are in G.P.

## Show Answer

#### Answer: (3)

#### Solution:

#### Formula: Condition for three no. in AP (iii)

$ax^{4}+2 bx+c=0$

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

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

$x^{2}-\frac{\sqrt{c}}{\sqrt{a}}$

Now, $dx^{2}+2 ex+f=0$

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

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

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

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

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