Question

If $G$ be the geometric mean of $x$ and $y$, then $\frac{1}{G^{2}-x^{2}}+\frac{1}{G^{2}-y^{2}}=$

• A. $G^{2}$
• B. $\frac{1}{G^{2}}$
• C. $\frac{2}{G^{2}}$
• D. $3 G^{2}$

Answer 1

Answer: (B)

As given $G=\sqrt{x y}$
$\therefore \frac{1}{G^{2}-x^{2}}+\frac{1}{G^{2}-y^{2}}=\frac{1}{x y-x^{2}}+\frac{1}{x y-y^{2}}$
$=\frac{1}{x-y}\left\{-\frac{1}{x}+\frac{1}{y}\right\}=\frac{1}{x y}=\frac{1}{G^{2}}$