Question

The parabolas : \(ax^2 + 2bx + cy = 0\) and \(dx^2 + 2ex + fy = 0\) 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 :

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

Hint:

Put \(y=1\) in both parabolas: the quadratics \(ax^2+2bx+c=0\) and \(dx^2+2ex+f=0\) share a root. G.P. gives \(b^2 = ac\), so the first quadratic has a double root \(x = -b/a\).

Answer 1

The Correct Option is A

On \(y = 1\) the parabolas become \(ax^2 + 2bx + c = 0\) and \(dx^2 + 2ex + f = 0\). A common point means a common value of \(x\).
Since \(a,b,c\) are in G.P., \(b^2 = ac\). The discriminant of the first equation is \(4b^2 - 4ac = 0\), so it has the repeated root \(x = -\frac{b}{a}\). This must also satisfy the second equation:
\(d\frac{b^2}{a^2} - \frac{2eb}{a} + f = 0\).
Using \(b^2 = ac\): \(d\frac{c}{a} - \frac{2eb}{a} + f = 0\), i.e. \(dc + fa = 2eb\).
Divide by \(ac = b^2\): \(\frac{d}{a} + \frac{f}{c} = \frac{2e}{b}\).
So \(\frac{d}{a}, \frac{e}{b}, \frac{f}{c}\) are in A.P.
Answer: (1) \(\frac{d}{a}, \frac{e}{b}, \frac{f}{c}\) are in A.P.