Q. If AM and GM of $x$ and $y$ are in the ratio $p:q$, then $x:y$ is
A) $p - \sqrt{p^{2} + q^{2}}:p + \sqrt{p^{2} + q^{2}}$
B) $p + \sqrt{p^{2} - q^{2}}:p - \sqrt{p^{2} - q^{2}}$
C) $p:q$
D) $p + \sqrt{p^{2} + q^{2}}:p - \sqrt{p^{2} + q^{2}}$
Explanation
(b)
Since, $\frac{x + y}{2\sqrt{xy}} = \frac{p}{q} \Rightarrow \frac{(x + y)^{2}}{4xy} = \frac{p^{2}}{q^{2}}$ …(i)
On subtracting both sides by1, we get
$\frac{(x - y)^{2}}{4xy} = \frac{p^{2} - q^{2}}{q^{2}}\ldots(ii)$
From Eqs. (i) and (ii), we get
$\left( \frac{x + y}{x - y} \right)^{2} = \frac{p^{2}}{p^{2} - q^{2}} \Longrightarrow \frac{x + y}{x - y} = \frac{p}{\sqrt{p^{2} - q^{2}}}$
$\Rightarrow \frac{2x}{2y} = \frac{p + \sqrt{p^{2} - q^{2}}}{p - \sqrt{p^{2} - q^{2}}}$
[by componendo-dividendo rule]
$\therefore x:y = (p + \sqrt{p^{2} - q^{2})}:(p - \sqrt{p^{2} - q^{2})}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.