The locus represented by `x=a/2(t+1/t), y=a/2(t-1/t)` is

Options

A. an ellipse
B. a circle
C. a pair of lines
D. none of these

Answer

To find the locus represented by the equations \( x = \frac{a}{2}\left(t + \frac{1}{t}\right) \) and \( y = \frac{a}{2}\left(t - \frac{1}{t}\right) \), we will follow these steps:

### Step 1: Express \( x \) and \( y \) in terms of \( t \)

We have:
\[ x = \frac{a}{2}\left(t + \frac{1}{t}\right) \]
\[ y = \frac{a}{2}\left(t - \frac{1}{t}\right) \]

### Step 2: Square both equations

Squaring both equations, we get:
\[ x^2 = \left(\frac{a}{2}\left(t + \frac{1}{t}\right)\right)^2 = \frac{a^2}{4}\left(t^2 + 2 + \frac{1}{t^2}\right) \]
\[ y^2 = \left(\frac{a}{2}\left(t - \frac{1}{t}\right)\right)^2 = \frac{a^2}{4}\left(t^2 - 2 + \frac{1}{t^2}\right) \]

### Step 3: Simplify the squared equations

From the squared equations:
\[ x^2 = \frac{a^2}{4}\left(t^2 + 2 + \frac{1}{t^2}\right) \quad \text{(Equation 1)} \]
\[ y^2 = \frac{a^2}{4}\left(t^2 - 2 + \frac{1}{t^2}\right) \quad \text{(Equation 2)} \]

### Step 4: Subtract Equation 2 from Equation 1

Now, we will subtract Equation 2 from Equation 1:
\[ x^2 - y^2 = \frac{a^2}{4}\left(t^2 + 2 + \frac{1}{t^2}\right) - \frac{a^2}{4}\left(t^2 - 2 + \frac{1}{t^2}\right) \]

This simplifies to:
\[ x^2 - y^2 = \frac{a^2}{4}\left(4\right) \]

### Step 5: Final simplification

Thus, we have:
\[ x^2 - y^2 = a^2 \]

### Conclusion

The locus represented by the given equations is:
\[ x^2 - y^2 = a^2 \]

This is the equation of a hyperbola.

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