If `x=7+4sqrt3` and xy=1 then the value of `(1/x^2+1/y^2)` is

Options

A. 194
B. 57
C. `85sqrt3`
D. `(7+4sqrt3)/(7-4sqrt3)`

Answer

To solve the problem, we need to find the value of \( \frac{1}{x^2} + \frac{1}{y^2} \) given that \( x = 7 + 4\sqrt{3} \) and \( xy = 1 \).

### Step-by-Step Solution:

1. **Given Values**:
- \( x = 7 + 4\sqrt{3} \)
- \( xy = 1 \)

2. **Finding \( y \)**:
- Since \( xy = 1 \), we can express \( y \) in terms of \( x \):
\[ y = \frac{1}{x} \]

3. **Finding \( \frac{1}{y} \)**:
- From \( y = \frac{1}{x} \), we can also find \( \frac{1}{y} \):
\[ \frac{1}{y} = x \]

4. **Finding \( \frac{1}{x^2} + \frac{1}{y^2} \)**:
- We can rewrite \( \frac{1}{y^2} \) as \( \left(\frac{1}{y}\right)^2 \):
\[ \frac{1}{y^2} = \left(\frac{1}{\frac{1}{x}}\right)^2 = x^2 \]
- Therefore, we have:
\[ \frac{1}{x^2} + \frac{1}{y^2} = \frac{1}{x^2} + x^2 \]

5. **Finding \( \frac{1}{x} \)**:
- To find \( \frac{1}{x} \), we rationalize:
\[ \frac{1}{x} = \frac{1}{7 + 4\sqrt{3}} \]
- Multiply the numerator and denominator by the conjugate \( 7 - 4\sqrt{3} \):
\[ \frac{1}{x} = \frac{7 - 4\sqrt{3}}{(7 + 4\sqrt{3})(7 - 4\sqrt{3})} \]

6. **Calculating the Denominator**:
- Use the difference of squares:
\[ (7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 7^2 - (4\sqrt{3})^2 = 49 - 48 = 1 \]
- Thus:
\[ \frac{1}{x} = 7 - 4\sqrt{3} \]

7. **Finding \( x^2 + \frac{1}{x^2} \)**:
- We know:
\[ x + \frac{1}{x} = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 14 \]
- Now, square both sides:
\[ (x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2} \]
- Therefore:
\[ 14^2 = x^2 + 2 + \frac{1}{x^2} \implies 196 = x^2 + 2 + \frac{1}{x^2} \]

8. **Solving for \( x^2 + \frac{1}{x^2} \)**:
- Rearranging gives:
\[ x^2 + \frac{1}{x^2} = 196 - 2 = 194 \]

9. **Final Result**:
- Thus:
\[ \frac{1}{x^2} + \frac{1}{y^2} = 194 \]

### Conclusion:
The value of \( \frac{1}{x^2} + \frac{1}{y^2} \) is \( \boxed{194} \).