The locus of a point represent by
`x=(a)/(2)((t+1)/(t)),y=(a)/(2)((t-1)/(t))`, where `t=in R-{0}`, is

Options

A. `x^2+y^2=a^2`
B. `x^2-y^2=a^2`
C. `x+y=a`
D. `x-y=a`

Explanation

To find the locus of the point 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 can follow these steps:

### Step 1: Rewrite the equations
We start with the given equations:
\[ x = \frac{a}{2} \left( t + \frac{1}{t} \right) \]
\[ y = \frac{a}{2} \left( t - \frac{1}{t} \right) \]

### Step 2: Add the equations
Now, let's add the two equations:
\[ x + y = \frac{a}{2} \left( t + \frac{1}{t} \right) + \frac{a}{2} \left( t - \frac{1}{t} \right) \]
This simplifies to:
\[ x + y = \frac{a}{2} \left( (t + \frac{1}{t}) + (t - \frac{1}{t}) \right) \]
\[ x + y = \frac{a}{2} \left( 2t \right) \]
\[ x + y = a t \]

### Step 3: Express \( t \) in terms of \( x \) and \( y \)
From the equation \( x + y = a t \), we can express \( t \) as:
\[ t = \frac{x + y}{a} \]

### Step 4: Substitute \( t \) back into one of the original equations
Now, we can substitute \( t \) back into one of the original equations to find a relationship between \( x \) and \( y \). Let's use the equation for \( x \):
\[ x = \frac{a}{2} \left( \frac{x + y}{a} + \frac{1}{\frac{x + y}{a}} \right) \]
This simplifies to:
\[ x = \frac{a}{2} \left( \frac{x + y}{a} + \frac{a}{x + y} \right) \]
\[ x = \frac{1}{2} \left( x + y + \frac{a^2}{x + y} \right) \]

### Step 5: Rearranging the equation
Multiplying through by \( 2(x + y) \) to eliminate the fraction:
\[ 2x(x + y) = (x + y)^2 + a^2 \]
\[ 2x^2 + 2xy = x^2 + 2xy + y^2 + a^2 \]
This simplifies to:
\[ x^2 - y^2 - a^2 = 0 \]

### Step 6: Final equation
Thus, we can express the locus of the point as:
\[ x^2 - y^2 = a^2 \]

### Conclusion
The locus of the point represented by the given equations is a hyperbola defined by the equation:
\[ x^2 - y^2 = a^2 \]