If G be the GM between x and y, then the value of `(1)/(G^(2) - x^(2)) + (1)/(G^(2) - y^(2))` is equal to

Options

A. `G^(2)`
B. `(2)/(G^(2))`
C. `(1)/(G^(2))`
D. `3G^(2)`

Answer

To solve the problem, we need to find the value of

\[ \frac{1}{G^2 - x^2} + \frac{1}{G^2 - y^2} \]

where \( G \) is the geometric mean (GM) of \( x \) and \( y \). The geometric mean \( G \) is given by:

\[ G = \sqrt{xy} \]

Thus, we can express \( G^2 \) as:

\[ G^2 = xy \]

Now, substituting \( G^2 \) into the expression we want to evaluate:

\[ \frac{1}{G^2 - x^2} + \frac{1}{G^2 - y^2} = \frac{1}{xy - x^2} + \frac{1}{xy - y^2} \]

Next, we can simplify each term in the expression:

1. **For the first term**:
\[ xy - x^2 = x(y - x) \]
Therefore,
\[ \frac{1}{xy - x^2} = \frac{1}{x(y - x)} \]

2. **For the second term**:
\[ xy - y^2 = y(x - y) \]
Therefore,
\[ \frac{1}{xy - y^2} = \frac{1}{y(x - y)} \]

Now we can rewrite the entire expression:

\[ \frac{1}{x(y - x)} + \frac{1}{y(x - y)} \]

To combine these fractions, we need a common denominator. The common denominator will be \( xy(y - x)(x - y) \).

Now, rewriting each term with the common denominator:

\[ \frac{y(x - y)}{xy(y - x)(x - y)} + \frac{x(y - x)}{xy(y - x)(x - y)} \]

Now, we can combine the numerators:

\[ \frac{y(x - y) + x(y - x)}{xy(y - x)(x - y)} \]

Expanding the numerator:

\[ y(x - y) + x(y - x) = yx - y^2 + xy - x^2 = 2xy - (x^2 + y^2) \]

Thus, we have:

\[ \frac{2xy - (x^2 + y^2)}{xy(y - x)(x - y)} \]

Now, we know that:

\[ x^2 + y^2 = (x + y)^2 - 2xy \]

So, substituting this back into our expression gives:

\[ 2xy - ((x + y)^2 - 2xy) = 4xy - (x + y)^2 \]

Finally, the expression simplifies to:

\[ \frac{4xy - (x + y)^2}{xy(y - x)(x - y)} \]

However, we can also notice that \( (y - x)(x - y) = -(x - y)^2 \), thus we can rewrite the denominator and simplify further.

The final value of the expression is:

\[ \frac{2}{xy} \]