Question

If $G$ is 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. $3G^2$

Answer 1

The correct option is B

$\frac{1}{G^2}$

Explanation for the correct option:

Finding the value of $\frac{1}{G^2-x^2}+\frac{1}{G^2-y^2}$:

Since, $G$ is the geometric mean of $x$ and $y$, then

$G^2=x\times y$

Now, Substitute the value $G^2$ in the finding expression,

$\begin{array}{rcl}\frac{1}{G^2-x^2}+\frac{1}{G^2-y^2}& =& \frac{1}{xy-x^2}+\frac{1}{xy-y^2}\\ & =& \frac{1}{x\left(y-x\right)}+\frac{1}{y\left(x-y\right)}\\ & =& \frac{1}{x-y}\left[-\frac{1}{x}+\frac{1}{y}\right]\\ & =& \frac{1}{x-y}\left[\frac{-y+x}{xy}\right]\\ & =& \frac{1}{xy}\\ & =& \frac{1}{G^2}\\ & & \end{array}$

Hence, the correct option is B.