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

Answer: (D)

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

Answer 2

ax2+2bx+c=0
⇒ax2+2acx​+c=0(∵b2=ac)
⇒(xa​+c​)2=0
x2−a​c​​……(1)
Now, dx2+2ex+f=0
⇒d(ac​)+2e[−a​c​​]+f=0
⇒adc​+f=2eac​​
⇒ad​+cf​=2eac1​​
⇒ad​+cf​=b2e​[ as b=ae​]
∴ad​,be​,cf​ are in A.P.
So, the correct option is (D) : $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.