The locus of the moving point whose coordinates are given by `(e^t+e^(-t),e^t-e^(-t))` where `t` is a parameter, is (a) `x y=1` (b) `x+y=2` (c) `x^2-y^2=4` (d) `x^2-y^2=2`

Options

A. `xy=1`
B. `x+y=2`
C. `x^2-y^2=4`
D. `x^2-y^2=2`

Explanation

To find the locus of the moving point whose coordinates are given by \( (e^t + e^{-t}, e^t - e^{-t}) \), we can follow these steps:

### Step 1: Define the coordinates
Let:
- \( x = e^t + e^{-t} \)
- \( y = e^t - e^{-t} \)

### Step 2: Express \( e^t \) and \( e^{-t} \) in terms of \( x \) and \( y \)
From the equations defined:
1. Adding the two equations:
\[ x + y = (e^t + e^{-t}) + (e^t - e^{-t}) = 2e^t \]
Thus, we have:
\[ e^t = \frac{x + y}{2} \]

2. Subtracting the second equation from the first:
\[ x - y = (e^t + e^{-t}) - (e^t - e^{-t}) = 2e^{-t} \]
Thus, we have:
\[ e^{-t} = \frac{x - y}{2} \]

### Step 3: Multiply the expressions for \( e^t \) and \( e^{-t} \)
Now, we can multiply \( e^t \) and \( e^{-t} \):
\[ e^t \cdot e^{-t} = 1 \]
Substituting the expressions we found:
\[ \left(\frac{x + y}{2}\right) \cdot \left(\frac{x - y}{2}\right) = 1 \]
This simplifies to:
\[ \frac{(x + y)(x - y)}{4} = 1 \]

### Step 4: Simplify the equation
Multiplying both sides by 4 gives:
\[ (x + y)(x - y) = 4 \]
Using the difference of squares, this can be rewritten as:
\[ x^2 - y^2 = 4 \]

### Conclusion
Thus, the locus of the moving point is given by the equation:
\[ x^2 - y^2 = 4 \]

### Final Answer
The correct option is (c) \( x^2 - y^2 = 4 \).

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