Given `P=(1,0)` and `Q=(-1,0)` and R is a variable point on one side of the line `PQ` such that `/_RPQ - /_RQP = pi /4`. The locus of the point `R` is

Options

A. `y^(2)-x^(2)+2xy-1=0`
B. `x^(2)-y^(2)+2xy+1=0`
C. `y^(2)+x^(2)-2xy=1`
D. `y^(2)-x^(2)-2xy+1=0`

Answer

To find the locus of the point \( R \) given the conditions in the problem, we will follow these steps:

### Step 1: Understand the points and angles
We have two fixed points \( P(1, 0) \) and \( Q(-1, 0) \). The point \( R(h, k) \) is a variable point on one side of the line \( PQ \). We know that the difference between the angles \( \angle RPQ \) and \( \angle RQP \) is \( \frac{\pi}{4} \).

### Step 2: Define the angles
Let:
- \( \alpha = \angle RPQ \)
- \( \beta = \angle RQP \)

According to the problem, we have:
\[ \alpha - \beta = \frac{\pi}{4} \]

### Step 3: Apply the tangent difference formula
Using the tangent of the difference of angles, we have:
\[ \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \]
Since \( \tan\left(\frac{\pi}{4}\right) = 1 \), we can set up the equation:
\[ \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} = 1 \]

### Step 4: Express \( \tan \alpha \) and \( \tan \beta \)
The slopes (tangents) can be expressed as:
- For \( RP \):
\[ \tan \alpha = \frac{k - 0}{h - 1} = \frac{k}{h - 1} \]
- For \( RQ \):
\[ \tan \beta = \frac{k - 0}{h + 1} = \frac{k}{h + 1} \]

### Step 5: Substitute into the equation
Substituting these into the equation gives:
\[ \frac{\frac{k}{h - 1} - \frac{k}{h + 1}}{1 + \frac{k}{h - 1} \cdot \frac{k}{h + 1}} = 1 \]

### Step 6: Simplify the left side
The left side becomes:
\[ \frac{k\left(\frac{1}{h - 1} - \frac{1}{h + 1}\right)}{1 + \frac{k^2}{(h - 1)(h + 1)}} \]
Calculating the numerator:
\[ \frac{k\left(\frac{(h + 1) - (h - 1)}{(h - 1)(h + 1)}\right)}{1 + \frac{k^2}{(h - 1)(h + 1)}} = \frac{k\left(\frac{2}{(h - 1)(h + 1)}\right)}{1 + \frac{k^2}{(h^2 - 1)}} \]

### Step 7: Set the equation to 1
Setting the fraction equal to 1 leads to:
\[ k \cdot 2 = (h^2 - 1) + k^2 \]

### Step 8: Rearranging the equation
Rearranging gives:
\[ k^2 - 2k + (h^2 - 1) = 0 \]

### Step 9: Recognizing the locus
This is a quadratic equation in \( k \). For \( R \) to exist, the discriminant must be non-negative:
\[ (-2)^2 - 4 \cdot 1 \cdot (h^2 - 1) \geq 0 \]
This simplifies to:
\[ 4 - 4(h^2 - 1) \geq 0 \implies 4 - 4h^2 + 4 \geq 0 \implies 8 - 4h^2 \geq 0 \implies 2 - h^2 \geq 0 \implies h^2 \leq 2 \]

### Step 10: Final locus equation
Thus, the locus of point \( R \) can be expressed as:
\[ k^2 - h^2 - 2kh + 1 = 0 \]
Replacing \( h \) with \( x \) and \( k \) with \( y \), we get:
\[ y^2 - x^2 - 2xy + 1 = 0 \]

### Final Answer
The locus of the point \( R \) is given by the equation:
\[ y^2 - x^2 - 2xy + 1 = 0 \]