Question

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

(A) $G^2$

(B) $\frac{1}{G^2}$

(C) $\frac{2}{G^2}$

(D) $3 G^2$

Answer 1

$\because G=\sqrt{x y}$

Now,

$\frac{1}{G^2-x^2}+\frac{1}{G^2-y^2}=\frac{1}{x y-x^2}+\frac{1}{x y-y^2}$

$=\frac{1}{x-y}\left(-\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{x y}=\frac{1}{G^2}$

So, The correct option will be (B).