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 $\mathrm{y}=1$. If $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}, \mathrm{f}$ are positive real numbers and $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in G.P., then

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

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

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Sequences and Series

The parabolas :

a x^{2} + 2 b x + cy = 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 G.P., then

Answer: (C)

y = 1

the first parabola gives

a x^{2} + 2 b x + c = 0

; with

a, b, c

in G.P. so that

b^{2} = a c

, this is

(a x + c)^{2} = 0

, hence

x = - \frac{c}{a}

. The same point lies on the second parabola on

y = 1

d \frac{c}{a} - 2 e \frac{c}{a} + f = 0

. Dividing throughout by

a c = b

gives

\frac{d}{a} + \frac{f}{c} = \frac{2 e}{b}

, which is exactly the condition that

\frac{d}{a}, \frac{e}{b}, \frac{f}{c}

are in A.P.

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option