If the coordinates of a vartiable point P be `(t+(1)/(t), t-(1)/(t))`, where t is the variable quantity, then the locus of P is

Options

A. xy = 8
B. `2x^(2)-y^(2)=8`
C. `x^(2)-y^(2)=4`
D. `2x^(2)+3y^(2)=5`

Answer

To find the locus of the variable point \( P \) with coordinates \( (t + \frac{1}{t}, t - \frac{1}{t}) \), we will follow these steps:

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

### Step 2: Square both equations
To eliminate \( t \), we will square both equations:
1. \( x^2 = \left(t + \frac{1}{t}\right)^2 = t^2 + 2 + \frac{1}{t^2} \)
2. \( y^2 = \left(t - \frac{1}{t}\right)^2 = t^2 - 2 + \frac{1}{t^2} \)

### Step 3: Expand both squared equations
From the first equation:
\[ x^2 = t^2 + 2 + \frac{1}{t^2} \]
From the second equation:
\[ y^2 = t^2 - 2 + \frac{1}{t^2} \]

### Step 4: Rearrange both equations
Rearranging both equations gives:
1. \( x^2 - 2 = t^2 + \frac{1}{t^2} \)
2. \( y^2 + 2 = t^2 + \frac{1}{t^2} \)

### Step 5: Set the equations equal to each other
Since both expressions equal \( t^2 + \frac{1}{t^2} \), we can set them equal:
\[ x^2 - 2 = y^2 + 2 \]

### Step 6: Simplify the equation
Rearranging gives:
\[ x^2 - y^2 - 4 = 0 \]
or
\[ x^2 - y^2 = 4 \]

### Step 7: Final equation of the locus
The final equation of the locus is:
\[ x^2 - y^2 = 4 \]

### Conclusion
The locus of the point \( P \) is represented by the equation \( x^2 - y^2 = 4 \).

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