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 \( x = \frac{a}{2} \left( \frac{t+1}{t} \right) \) and \( y = \frac{a}{2} \left( \frac{t-1}{t} \right) \), we can follow these steps:

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

### Step 2: Simplify the equations
We can express \( x \) and \( y \) in terms of \( t \):
\[ x = \frac{a}{2} \left( 1 + \frac{1}{t} \right) \quad \text{and} \quad y = \frac{a}{2} \left( 1 - \frac{1}{t} \right) \]

### Step 3: Isolate \( \frac{1}{t} \)
From the equation for \( x \):
\[ \frac{1}{t} = \frac{2x}{a} - 1 \]
From the equation for \( y \):
\[ \frac{1}{t} = 1 - \frac{2y}{a} \]

### Step 4: Set the two expressions for \( \frac{1}{t} \) equal
Now we can set the two expressions for \( \frac{1}{t} \) equal to each other:
\[ \frac{2x}{a} - 1 = 1 - \frac{2y}{a} \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ \frac{2x}{a} + \frac{2y}{a} = 2 \]
Multiplying through by \( a \) results in:
\[ 2x + 2y = 2a \]
Dividing by 2:
\[ x + y = a \]

### Step 6: Find the equation of the locus
Now, we square both \( x \) and \( y \):
\[ x^2 + y^2 = a^2 \]

### Final Result
Thus, the locus of the point is given by:
\[ x^2 + y^2 = a^2 \]