Question

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

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

Answer 1

The correct option is B $\frac{1}{G^2}$

$G=\sqrt{xy}$

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

$\Rightarrow \frac{1}{xy-x^2}+\frac{1}{xy-y^2}$

$\Rightarrow \frac{xy-y^2+xy-x^2}{(xy-x^2)(xy-y^2)}$

$\Rightarrow \frac{-(x^2+y^2-2xy)}{\left(x\right(y-x\left)\right)\left(y\right(x-y\left)\right)}$

$\Rightarrow \frac{-(x-y{)}^2}{-xy(x-y)}$

$\Rightarrow \frac{1}{xy}=\frac{1}{G^2}$