The curve repersented by `x=5(t+(1)/(t)),y=(t-(1)/(t)),t ne 0` is

Options

A. a point of straight lines
B. an ellipse
C. a hyperbola
D. a rectangular hyperbola

Answer

To determine the nature of the curve represented by the parametric equations \( x = 5\left(t + \frac{1}{t}\right) \) and \( y = t - \frac{1}{t} \), we will eliminate the parameter \( t \) and find a relationship between \( x \) and \( y \).

### Step-by-step Solution:

1. **Start with the given parametric equations**:
\[ x = 5\left(t + \frac{1}{t}\right) \]
\[ y = t - \frac{1}{t} \]

2. **Express \( t + \frac{1}{t} \) in terms of \( x \)**:
From the equation for \( x \):
\[ \frac{x}{5} = t + \frac{1}{t} \]

3. **Square both sides**:
\[ \left(\frac{x}{5}\right)^2 = \left(t + \frac{1}{t}\right)^2 \]
Expanding the right-hand side:
\[ \left(t + \frac{1}{t}\right)^2 = t^2 + 2 + \frac{1}{t^2} \]
Thus, we have:
\[ \frac{x^2}{25} = t^2 + 2 + \frac{1}{t^2} \]

4. **Rearranging gives us**:
\[ t^2 + \frac{1}{t^2} = \frac{x^2}{25} - 2 \]

5. **Now, express \( t - \frac{1}{t} \) in terms of \( y \)**:
From the equation for \( y \):
\[ y = t - \frac{1}{t} \]

6. **Square both sides**:
\[ y^2 = \left(t - \frac{1}{t}\right)^2 \]
Expanding the right-hand side:
\[ \left(t - \frac{1}{t}\right)^2 = t^2 - 2 + \frac{1}{t^2} \]
Thus, we have:
\[ y^2 = t^2 - 2 + \frac{1}{t^2} \]

7. **Rearranging gives us**:
\[ t^2 + \frac{1}{t^2} = y^2 + 2 \]

8. **Set the two expressions for \( t^2 + \frac{1}{t^2} \) equal**:
\[ \frac{x^2}{25} - 2 = y^2 + 2 \]

9. **Rearranging this equation**:
\[ \frac{x^2}{25} - y^2 = 4 \]

10. **Multiply through by 25**:
\[ x^2 - 25y^2 = 100 \]

11. **Rearranging gives the standard form of the hyperbola**:
\[ \frac{x^2}{100} - \frac{y^2}{4} = 1 \]

### Conclusion:
The curve represented by the given parametric equations is a hyperbola.